1
Way to Success
A COMPLETE GUIDE
FOR
+2 CHEMISTRY
ALONG WITH
PRACTICE BOOK
UNITWISE GOVT. EXAMINATION Q & A ( 2006 – 2017 )
& Additional questions
( Based on Govt. Question paper answer Key }
Q -70 Compulsory problems (Solved)
Prepared by
Dr.S.Denis Arockiaraj M.Sc., M.Phil., M.Ed., Ph.D Associate Professor of Chemistry
St.Joseph’s College, Trichy
Published by
Way to Success Publications
------For subject related clarifications------
Mail us : [email protected] &
[email protected] Call us : 9843113114
Visit us : www.waytosuccess.org You can download free study materials from our website
2
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Name of the Book
Way to Success, A Complete Guide for +2 Chemistry
With Practice book
New Edition
Published by
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Old Karur Road, Melachinthamani,
Trichy – 620 002 Tamil Nadu.
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3
CONTENT
1 Atomic Structure – II 07
2 Periodic Classification – II 20
3 p-Block Elements 29
4 d- Block Elements 48
5 f- Block Elements 59
6 Coordination Compounds and Bio coordination Compounds 65
7 Nuclear Chemistry 79
8 Solid State – II 89
9 Thermodynamics – II 98
10 Chemical Equilibrium – II 107
11 Chemical Kinetics – II 119
12 Surface Chemistry 129
13 Electro Chemistry - I 143
14 Electro Chemistry - II 157
15 Isomerism in Organic Chemistry 169
16 Hydroxy Derivatives 179
17 Ethers 199
18 Carbonyl Compounds 207
19 Carboxylic Acids 227
20 Organic Nitrogen Compounds 246
21 Biomolecules 263
22 Chemistry in Action 271
Q – 70. a Hydroxy Derivatives 277
Q – 70. b d- Block Elements 292
Q – 70. c Carbonyl Compounds 301
Q – 70. d Electro Chemistry - I 314
Dear Students,
We wish you to get 150/150 in 12th
Standard Public Examination by
using our Way to Success – 12th
Standard Chemistry Guide.
Let your success start with our Way To Success.
Note : At most care is taken to fulfill your requirements without mistakes. If you find any
mistakes, kindly inform us by log in our e-mail id [email protected]. Corrections will be
updated then and there in our website www.waytosuccess.org and that will be carried out in
our further edition.
4
BLUE PRINT
Unit
No Topic
No. of
1 Mark
questions
No. of
3 Mark
questions
No. of
5 Mark
questions
Part-III
No. of
5 Mark
questions
Part-IV
Total
Marks
1 Atomic Structure II 2 1 1 - 10
2 Periodic Classification II 1 1 - 1 09
3 p-Block Elements II 1 2 - 1 12
4 d-Block Elements 2 2 1 1* 18
5 f- Block Elements 2 - 1 - 07
6 Coordination Compounds 1 - 1 1 11
7 Nuclear Chemistry 1 1 - 1 09
8 Solid State II 1 1 - 1 09
9 Thermodynamics II 2 1 1 - 10
10 Chemical Equilibrium II 2 1 1 - 10
11 Chemical Kinetics II 1 2 1 - 12
12 Surface Chemistry 3 1 - 1 11
13 Electro Chemistry I 1 1 - 1+1* 14
14 Electro Chemistry II - - 1 1 10
15 Isomerism in Organic
Chemistry - 1 - 1 08
16 Hydroxy Derivatives 1 2 - 1* 12
17 Ethers 2 - 1 - 07
18 Carbonyl Compounds 1 1 1 1* 14
19 Carboxylic acids 1 1 1 1 14
20 Organic Nitrogen
Compounds 3 1 - 1 11
21 Bio molecules 2 - - 1 07
22 Chemistry in Action - 1 1 - 08
* Compulsory Problem (Q-70)
5
How to get 150
150
1. Study one mark questions from all the units according to the blue print.
2. Study three mark questions from the units given below. Give more importance to the units given
in bold.
PART – II 3 Mark questions
Answer any 15 questions
31. Atomic Structure – II 38. Solid State – II 45. Iso. in Org. Chemistry
32. Periodic Class.– II 39. Thermodynamics – II 46. Hydroxy Derivatives
33. p-Block Elements 40. Chem. Equilibrium – II 47. Hydroxy Derivatives
34. p-Block Elements 41. Chemical Kinetics – II 48. Carbonyl Compounds
35. d- Block Elements 42. Chemical Kinetics – II 49. Carboxylic Acids
36. d- Block Elements 43. Surface Chemistry 50. Org. Nitrogen Compounds
37. Nuclear Chemistry 44. Electro Chemistry - I 51. Chemistry in Action
3. Study five mark questions ( Part – III & IV ) from the units given below. Give more importance
to the units given in bold.
PART – III 5 Mark questions
Answer any 7 questions choosing at least two questions from each section
Section – A Section – B Section – C
52. Atomic structure
53. d-Block elements
54. f-Block elements
55. Coordination chem
56. Thermodynamics
57. Che. Equilibrium
58. Che.Kinetics
59. Electrochem II
60. Ethers
61. Carbonyl comp.
62. Carboxylic acids
63. Che. in action
6
PART – IV 5 Mark questions
Question number 70 is compulsory. Answer any 3 from remaining questions
64 a) Periodic classification b) p-Block elements
65 a) Coordination chem b) Nuclear chem
66 a) Solid state b) Surface chem
67 a) Electro chem I b) Electro chem II
68 a) Isomerism in org. b) Carboxylic acids
69 a) Org. Nitrogen comp b) Bio molecules
70 a) Hydroxy derivatives b) d-Block elements
or c) Carbonyl comp. d) Electro chem I
4. Slow learners can study the following units and score more than 100 marks.
Unit No. Topic Marks
1. Atomic Structure – II 10
2. Periodic Classification – II 09
3. p-Block Elements II 12
5. f- Block Elements 07
6. Coordination Compounds 11
7. Nuclear Chemistry 09
8. Solid State – II 09
9. Thermodynamics – II 10
10. Chemical Equilibrium – II 10
12. Surface Chemistry 11
17. Ethers 07
22. Chemistry in Action 08
For Copies:
Contact following phone Numbers
9787609090, 9787201010
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1. ATOMIC STRUCTURE – II
Blue print 1 Mark = 2 3 Mark = 1 5 Mark = 1 Total marks = 10
Govt. Examination questions
1 MARK
1. En = – 2n
313.6, If the value of En = –34.84 to which value „n‟ corresponds [Jun-08, Mar-16, Sep-16]
a) 4 b) 3 c) 2 d) 1
2. Dual character of an electron was explained by
[Mar-2006, Sep-2007, Sep-2008, June-2009, June-2010, Mar-2013, Sep-2015]
a) Bohr b) Heisenberg c) de-Broglie d) Pauli
3. de-Broglie équation is [June-2006, Sep-2009, Mar-2012, Sep-2012, Mar-2014, June-2014]
a) λ = h
mv b) λ = hmv c) λ =
m
hv d) λ =
mv
h
4. Which of the following particle having same kinetic energy, would have the maximum de-Broglie wave
length? [Mar-2007, Sep-2014]
a) α-particle b) proton c) β-particle d) neutron
Note : λ =mv
h ( Mass of β-particle is very less and hence it has maximum λ )
5. If the energy of an electron in the second Bohr orbit of H-atom is –E, what is the energy of the electron
in the Bohr‟s first orbit? [June-2011]
a) 2E b) – 4E c) –2E d) 4E
Note: En = 2
1
n
E
E2 =
2
1
2
E
E1 = 4 E2 E1 = 4 (– E ) = – 4 E
6. The bond order of oxygen molecule is [Sep-2010, Sep-2011]
a) 2.5 b) 1 c) 3 d) 2
7. The hybridisation in SF6 molecule is [Mar-2008, Mar-2010, Sep-2013]
a) sp3
b) sp3d
2 c) sp
3d d) sp
3d
3
8. The intramolecular hydrogen bonding is present in
[June-2007, Mar-2009, Mar-2011, June-2010, June-2013, June-2015, Mar-17]
a) o-nitrophenol b) m-nitrophenol c) p-nitrophenol d) None
9. The momentum of a particle which has de-Broglie wave length of 1Ǻ is [Mar-2006, Mar-2011]
a) 6.6 × 10–23
kg ms–1
b) 6.6 × 10–24
kg ms–1
c) 6.6 × 10–34
kg ms–1
d) 6.6 × 1034
kg ms–1
Note: λ =p
h p =
h
=
10
-34
10
106.6
= 6.6 × 10
–24 (λ = 1Ǻ = 10
–10 m )
10. Which one of the following experiments confirmed the wave nature of electron? [Mar-2010]
a) G.P. Thomson’s gold foil experiment b) Black body radiation
c) Photoelectric effect d) Milliken‟s oil drop experiment
Wave character of electron Particle character of electron
1. Davisson and Germer‟s experiment
2. G.P. Thomson‟s experiment
[Diffraction of electrons]
1. Scintillation
2. J.J Thomson experiment
3. Milliken‟s oil drop experiment
4. Black body radiation
5. Photoelectric effect
11. The circumference of the circular orbit of the electron must be an integral multiple of -------- [Sep-06]
a) Frequency b) Momentum c) Mass d) Wavelength
Note: 2πa = n λ
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12. Number of spherical nodes in 2s- orbital is [June-2009]
a) 1 b) 2 c) 3 d) 4
Note: Number of nodes in s-orbital = n – 1 ( n = principal quantum number )
13. The bond order of nitrogen molecule is [June-2006, June-2012, June-2013]
a) 0 b) 1 c) 2 d) 3
14. In a molecule eight electrons are present in bonding molecular orbitals and four electrons are present in
antibonding molecular orbitals. Its bond order is [Mar-2008]
a) 3 b) 4 c) 2.5 d) 2
15. Energy levels of molecular orbitals have been determined experimentally by [Sep-2007]
a) spectroscopic studies b) x-ray diffraction c) crystallographic studies d) none
16. Molecular orbital with the least energy is ------------- [Sep-2011]
a) 1s b) *1s c) 2py d) *2py
17. The type of hybridization of I in ICl4ˉ is [Sep-2012]
a) sp3
b) sp3d
2 c)dsp
2 d) sp
3d
18. The nature of hybridization in IF7 molecule is [Sep-2006, Sep-2010, Mar-2014]
a) sp3d
2 b) sp
3d
4 c) sp
3d
3 d) sp
3d
4
19. The hybridization involved in XeF6 is [June-2007, June-2011]
a) sp3d
3 b) sp
3d
2 c) sp
3d d) sp
3
20. The type of hybridization of S in SO42ˉ is [Mar-2009]
a) sp3 b) sp
3d c) sp d) sp
2
21. The hybridization in CO32ˉ ion is [Sep-2009]
a) sp2
b) sp3
c) sp d) sp3d
22. The hybridization in PCl5 molecule is [June-2012, Mar-2013]
a) sp3d
b) sp
3d
c) sp
3 d) sp
2
23. The type of hybridization of N in NH4+
ion is [Mar-2012]
a) sp3
b) sp3d c) sp d) sp
2
24. Inter-molecular hydrogen bonding is present in [Mar-2007]
a) HF b) H2O c) ethanol d) all the above
25. Water exists in liquid state. This is due to [June-2008]
a) high boiling point b) low boiling point c) freezing point is zero d) hydrogen bond
26. Intermolecular hydrogen bonding is present in [Sep-2008]
a) orthonitrophenol b)salicylic acid c) orthohydroxybenzaldehyde d) HF
27. The molecular orbitals are fi1led according to [Sep-2013]
a) Pauli's exclusion principle b) Hund‟s rule c) Aufbau principle d) all the above
28. In a molecule when Nb = 8 and Na = 2, then the bond order is [June-2014]
a) 3 b) 4 c) 2.5 d) 2
29. Shape of p-orbital is [Sep-2014]
a) Spherical b) Clover leaf c) dumb-bell d) doughnut
30. Which molecule is relatively more stable? [Mar-2015]
a) O2 b) H2 c) Li2 d) N2 Note: Stability of molecule Bondorder; Bond order of H2 = 1, Li2 = 1, O2 = 2, N2 = 3
31. The value of Bohr radius for hydrogen atom is [Mar-2015]
a) 0.529 × 10–8
cm b) 0.529 × 10–10
cm c) 0.529 × 10–6
cm d) 0.529 × 10–12
cm
32. The bond order of hydrogen molecule is [June-2015]
a) 0 b) 1 c) 3 d) 2
32. The bond order of lithium molecule is [Sep-2015]
a) 0 b) 1 c) 3 d) 2
33. The total valence electrons in BeCl2 [Mar-2016]
a) 18 b) 12 c) 16 d) 14 Note: EC of Be (Z = 4) : 2, 2 EC of Cl (Z= 17) : 2, 8, 7 The total valence electrons = 2 + 7 + 7 = 16
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34. The energy of electron in an atom is given by En = [Jun-2016]
a) – 22
42
hn
me4π b) –
22
22
hn
me2π c) –
22
42
hn
me2π d) –
22
4
hn
me2
35. The hybridisation in XeOF4 molecule is [Jun-2016]
a) sp3
b) sp3d
2 c) sp
3d d) sp
3d
3 36. The charge of an electron is determined by [Sep-2016]
a) Thomson
b) de-Broglie
c) Milliken d) Einstein 37. sp
2 hybridisation is not present in ------------- ion [Mar-2017]
a) CO32ˉ b) SO4
2ˉ c) NO3ˉ d) NO2ˉ
3 MARK
1. State Heisenberg’s uncertainty principle
[March-06, Sep-07, March-08, Sep-09, Mar-11, Sep-11, Jun-12, Mar-13, Jun-15, Mar-17]
“It is impossible to measure simultaneously both the position and velocity (or momentum) of a
microscopic particle with absolute accuracy or certainty.”
∆x . ∆p ≥ 4π
h
where, Δx = uncertainty in the position of the particle and
Δp = uncertainty in the momentum of the particle.
h = Planck‟s constant
2. Give the differences between particle and wave [June-2006, June-2014, Sep-2014]
No Particle Wave
1 Particle is localized in space.
i.e. A particle occupies a well-defined
position in space
Wave is delocalized in space.
i.e. A wave is spread out in space
2 Particles do not interfere.
i.e. When a particular space is occupied
by one particle, the same space cannot
be occupied simultaneously by any
other particle.
Waves interfere.
i.e. Two or more waves can coexist in the
same region of space.
3 When a number of particles are present
in a given region of space, their total
value is equal to their sum
When a number of waves are present in a
given region of space, due to interference,
the resultant wave can be larger or
smaller than the individual waves.
3. What is the significance of negative electronic energy?
[Jun-2008, June-2009, Jun-2010, Mar-2012, Sep-2012, Mar-2014, Jun-2016]
The energy of an electron at infinity is arbitrarily assumed to be zero. This state is called zero-energy
state. When an electron moves and comes under the influence of nucleus, it does some work and spends
its energy in this process. Thus, the energy of the electron decreases and it becomes less than zero ie. it
acquires a negative value.
4. Define hybridisation [Mar-2009, June-2013, Mar-2015]
Hybridization is the concept of intermixing of the orbitals of an atom having nearly the same energy to
give exactly equivalent orbitals with same energy, identical shapes and symmetrical orientations in
space.
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5. What is bond order? [June-2007, Mar-2010, Mar-2016]
Bond order is defined as half the difference between the number of electrons in bonding molecular
orbitals (Nb) and the number of electrons in antibonding molecular orbitals (Na).
Bond order = 2
NN ab
6. Give the conditions for effective hydrogen bonding [Sep-2006, June-2011]
1. High electronegativity of the atom bonded to hydrogen atom so that bond is sufficiently polar.
2. Small size of the atom bonded to hydrogen so that it is able to attract the bonding electron pair
effectively.
3. Only F, O and N form strong hydrogen bonds because they have high value of electronegativity and
small atomic size.
7. Why He2 is not formed? [March-2007, Sep-2008, Sep-2010, Sep-2013, Sep-2016]
The electronic configuration of helium (Z = 2) is 1s2.
As each helium atom contains two electrons, there will be four electrons in He2 molecule.
He2 : (σ1s)2 (σ*1s)
2
He HeHe2
1s 1s
1s
*1s
En
erg
y
Molecular orbitalsAtomic orbital Atomic orbital
Bond order = 2
NN ab =
2
22 = 0
As the bond order for He2 is zero, this molecule does not exist.
8. Define orbital [Sep-2015]
An orbital is the region of space around the nucleus within which the probability of finding an electron
of given energy is maximum
5 MARK
1. Discuss the Davisson and Germer experiment [March-2007, Sep-2015, Mar-17]
1. A beam of electrons obtained from a heated tungsten filament is accelerated by using a high,
positive potential.
2. When this fine beam of accelerated electrons is allowed to fall on a large single crystal of nickel, the
electrons are scattered from the crystal in different directions.
3. The diffraction pattern so obtained is similar to the diffraction pattern of X-rays by Bragg‟s
experiment. Since X-rays have wave character, therefore, the electrons must also have wave
character.
4. The wave length of the electrons as determined by the diffraction experiments were found to be in
agreement with the values calculated from de-Broglie equation.
5. From the above discussion, it is clear that an electron behaves as a wave.
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Nickel crystal
diffracted
electrons
photographic
plate
Incident beam of
electrons
diffraction pattern
2. Derive de-Broglie’s equation. What is its significance?
[Sep-2006, June-2007, Mar-2011, Sep-2013, June-2015, Mar-2016]
The wavelength of the wave associated with any material particle can be calculated by analogy with
photons as follows.
If the photon has wave character, its energy is given by
E = h (according to plank‟s quantum theory)
Where, h = Plank‟s constant = frequency of the wave
If the photon has particle character, its energy is given by
E = mc2 ( according to Einstein equation)
Where m = mass of photon c = velocity of light
From the above equations we get,
h = mc2
But ν=λ
c
Therefore, λ
hc= mc
2
Or λ= mc
h
de Broglie pointed out that the above equation is applicable to any material particle. The mass of the photon
is replaced by the mass of the material particle and the Velocity „c‟ of photon is replaced by the velocity „v‟
of the material particle.
Thus for any material particle like electron, we may write.
λ= mv
h or λ=
p
h
Where, mv = p (the momentum of the particle)
The above equation is called de Broglie equation and is called de Broglie wavelength.
Significance of de Broglie equation
de Broglie equation relates the particle character with the wave character of matter.
3. Briefly explain Molecular Orbital Theory. [or] What are the assumptions (postulates) of
molecular orbital theory? [March-2008, June-2012, Sep-2012, Sep-2014, Mar-2015]
1. In a molecule, electrons are present in new orbitals called molecular orbitals.
2. Molecular orbitals are formed by combination of atomic orbitals of equal energies
3. The number of molecular orbitals formed is equal to the number of atomic orbitals undergoing
combination.
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4. Two atomic orbitals can combine to form two molecular orbitals. One of these two molecular
orbitals one has a lower energy and the other has a higher energy. The molecular orbital with lower
energy is called bonding molecular orbital and the other with higher energy is called anti
bonding molecular orbital.
5. The shapes of molecular orbitals depend upon the shapes of combining atomic orbitals.
6. The bonding molecular orbitals are represented by by σ (sigma), π (pi), δ (delta) and the
antibonding molecular orbitals are represented by σ*, π*, δ*.
7. The molecular orbitals are filled in the increasing order of their energies, starting with orbital of
least energy. (Aufbau principle)
8. A molecular orbital can accommodate only two electrons and these two electrons must have
opposite spins. (Pauli’s exclusion principle)
9. While filling molecular orbitals of equal energy, pairing of electrons does not take place until all
such molecular orbitals are singly filled with electrons having parallel spins. (Hund’s rule)
4. Explain the formation of N2 molecule by molecular orbital theory
[June-2008, Sep-2008, June-2009, June-2010, Sep-2010, June-2011, Mar-2014]
N ( Z = 7 ) : 1s2 2s
2 2p
3
Each nitrogen atom has 7 electrons, hence, in N2 molecule there are 14 electrons.
Electronic configuration of N2 :KK (2s)2 < (*2s)
2 < (2px)
2 = (2py)
2 < (2pz)
2
Here (1s)2 < (
*1s)
2 = KK { K shell of the two atoms}
Molecular orbital energy level diagram of N2
Atomic
OrbitalsAtomic
Orbitals
Molecular
Orbitals
En
ergy
N NN2
2s 2s
2p2p
2s
*2s
2pz
2px
* *2py
2pz*
2px 2py
1. Bond Order = 2
NN ab =
2
28 = 3
2. Nature of bond : Since bond order is 3, triple bond is present in N2 molecule. N N
3. Diamagnetic nature: Since all the electrons in N2 are paired, it is a diamagnetic molecule.
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5. Explain the formation of O2 molecule by molecular orbital theory
[March-2006, June-2006, Sep-2007, Mar-2010, Sep-2011, Mar-2013, June-2014]
O ( Z = 8 ) : 1s2 2s
2 2p
4
Each oxygen atom has 8 electrons, hence, in O2 molecule there are 16 electrons.
Electronic configuration of O2:KK (2s)2<(*2s)
2<(2pz)
2 < (2px)
2 = (2py)
2 < (
*2px)
1 = (
*2py)
1
Here (1s)2 < (
*1s)
2 = KK { K shell of the two atoms}
Molecular orbital energy level diagram of O2
Atomic
OrbitalsAtomic
Orbitals
Molecular
Orbitals
En
erg
y
O OO2
2s 2s
2p2p
2s
*2s
2pz
2px
* *2py
2pz*
2px 2py
1. Bond Order = 2
NN ab =
2
48 = 2
2. Nature of bond : Since bond order is 2, double bond is present in O2 molecule O = O
3. Paramagnetic Nature : O2 molecule contains two unpaired electrons (*2px1 and *2py
1).
So it is paramagnetic.
6. Explain the shape of p-orbitals [Sep-2016]
For p-orbital , l = 1 m = – 1, 0 , + 1
p- orbital can have three orientations. (px, py, pz)
p- orbitals have directional character.
shape : dumb – bell
2px, 2py and 2pz orbitals are equal in energy (degenerate orbitals) but differ in orientations.
2px orbital lies along x – axis, 2py orbital lies along y - axis and 2pz orbital lies along z - axis
Each p – orbital consists of two lobes
Two lobes of each p-orbital are separated by a nodal plane (a plane having zero electron density)
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X X X X
Y Y Y Y
Z Z ZZ
2px Orbital 2py Orbital 2p
z Orbital Nodal plane for
2px Orbital
For 2px orbital yz plane is the nodal plane.
all the p–orbitals (2p, 3p …) have similar shapes.
size 2p < 3p < 4p ….
7. Explain the shape of d-orbitals [Jun-2016]
For d – orbital l = 2, and m = –2, – 1, 0, +1, +2
So, d- orbitals, can have 5 orientations zxyzxy ddd ,, , 22 yxd
and 2z
d
zxyzxy ddd ,, and 22 yxd
orbitals have clover leaf shape
2zd orbital has a dumb- bell shape with a doughnut shaped electron cloud in the centre.
dxy lies in between x and y axes and lie in xy plane
dyz lies in between y and z axes and lie in yz plane
dzx lies in between z and x axes and lie in zx plane
2 2x -yd
lies along the x and y axes
2zd lies along z – axis
node : each d–orbitals have 4 lobes and two nodes.
X Y X X X
Y YZ Z Z
dxy d
yzd
zx dx2
- y2 d
z2
8. The uncertainty in the position of a moving bullet of mass 10 g is 10
–5 m. Calculate the uncertainty
in its velocity [Mar-2009]
Δx = 10–5
m m = 10 g = 10 × 10–3
kg = 10–2
kg h = 6.626 ×–
kg m2 s
–1
Substituting these values in the equation for uncertainty principle
∆x . ∆p = 4π
h
∆x . m∆v = 4π
h
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∆v = Δxm4
h
=
52
-34
10107
224
106.626
= 5.27 × 10
–28 m sec
–1
9. The wavelength of a moving body of mass 0.1 mg is 3.310 × 10-29
m. Calculate its kinetic energy.
(h = 6.626 × 10-34
J.s) [Sep-2009]
m = 0.1 mg = 0.1 × 10–3
g = 0.1 × 10– 6
kg = 10–7
kg
K.E = ½ mv2 = ½ ×10
–7 v
2
= mv
h or v =
m
h =
29
-34
1031.310
106.6267
= 200 ms–1
∴ K.E = ½ ×10–7
× (200)2 = 2×10
–3 J
10. A moving electron has 4.55 × 10 – 25
joules of kinetic energy. Calculate its wavelength.
[ mass = 9.1× 10 – 31
kg and h = 6.626 ×10 – 34
kg m2 s
– 1] [Mar-2012]
K.E = 4.55 × 10 – 25
½ mv2 = 4.55 × 10
– 25
½ × 9.1×10 – 31
v2 = 4.55 × 10
– 25
v2 =
31101.9
2104.55 -25
= 10
6 v = 10
3 m s
– 1
= mv
h =
310101.9
106.62631
-34
= 7.25 ×10 – 7
m
11. Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is
5.7 ×105 m/sec (h = 6.626 × 10
−34 kg m
2 s
−1, mass of the electron = 9.1 × 10
−31 kg) [June-2013]
∆x . m∆v = 4π
h
∆x = Δvm4
h
=
531
34
107.5101.97
224
106.626
= 1 × 10–10
m
Additional questions
1 MARK
1. Which explains the wave nature of an electron?
a) scintillation b) black body radiation c) interference d) photoelectric effect
2. de-Broglie relationship has no significance for
a) an electron b) a proton c) a neutron d) an iron ball
3. Mathematical expression of Heisenberg uncertainty principle is
a) ∆x . ∆p ≥ 4π
h b) ∆x . ∆m ≥
4π
h c) ∆x . ∆v ≥
4π
h d) ∆x . ∆y ≥
4π
h
4. Energy of an electron of a hydrogen atom is (in KJ mol–1
)
a) En = + 2n
1312 b) En = –
2n
1312 c) En = –
n
1312 d) En = +
n
1312
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5. Shape of s-orbital is
a) spherical b) clover c) dumb-bell d) dough nut
6. Shape of dxy-orbital is
a) dough nut b) clover leaf c) dumb-bell with doughnut d) spherical
[ 22 yxzxyzxy d,d,d,d
= clover leaf shape, 2zd = dumb-bell with doughnut shape]
7. The probability of finding an electron is same in all directions at a given distance from the nucleus for
a) s-orbital b) p-orbital c) d-orbital d) f-orbital
8. 1s and 2s orbitals do not differ in
a) size b) shape c) energy d) number of nodes
9. 2px , 2py and 2pz orbitals differ in
a) size b) shape c) energy d) orientation
10 The total number of spherical nodes for any s-orbital is equal to
a) l –1 b) n –1 c) m –1 d) 0
11. A node refers to the region of
a) maximum probability of finding an electron b) more electron density
c) zero probability of finding an electron d) none of these
12. Number of spherical nodes in 3s- orbital is
a) 1 b) 2 c) 3 d) 4
13 For 2px orbital ------------- plane is the nodal plane
a) yz b) xz c) xy d) x
14. The bond order of H2 molecule is
a) 1 b) 0 c) 2 d) 1.5 Molecule H2 He2 Li2 N2 O2
Bond order 1 0 1 3 2
15 Which of the following is paramagnetic?
a) H2 b) N2 c) O2 d) He2
16. Stability of a molecule is directly proportional to
a) electron density b) number of neutrons c) bond order d) electro negativity
17. If Nb > Na the molecule is
a) stable b) unstable c) explosive d) highly reactive
18. If Nb = Na the molecule is
a) stable b) unstable c) explosive d) highly reactive
19. Which of the following is more stable?
a) N2 b) O2 c) Li2 d) all the above
20. The correct order of stability or bond dissociation energy is
a) N2 < O2 < Li2 b) N2 > O2 > Li2 c) N2 = O2 < Li2 d) N2 < O2 = Li2
21. The correct order of bond length is
a) N2 < O2 < Li2 b) N2 > O2 > Li2 c) N2 = O2 < Li2 d) N2 < O2 = Li2 22. ------------ concept is used in the construction of electron microscope and in the study of surface
structure of solids by electron diffraction
a) de-Broglie b) photoelectric effect c) blackbody radiation d) all the above
23. Hydrogen bonding exists in
a) skin b) blood c) bones d) all the above
24. The strength of hydrogen bond is in the order
a) H-F….H > H-O….H > H-N….H b) H-F….H = H-O….H > H-N….H
c) H-O….H < H-N….H < H-F….H d) H-O….H = H-N….H = H-F….H
25. Intramolecular hydrogen bonding is present in
a) orthonitrophenol b)salicylic acid c) Salicylaldehyde d) All the above
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3 MARK.
1. Explain the formation of H2 molecule by molecular orbital theory
It is formed by the combination of two hydrogen atoms. Each hydrogen atom in the ground state has one
electron in 1s orbital. Therefore, in all there are two electrons in hydrogen molecule which are present in
lower most σ1s molecular orbital.
The molecular orbital electronic configuration of hydrogen molecule is (σ1s)2
En
ergy
H HH2
1s 1s
1s
*1s
Molecular orbitalsAtomic orbital Atomic orbital
i) Bond order = 2
NN ab =
2
02 = 1
ii) Nature of bond : The two hydrogen atoms in a molecule of hydrogen are bonded by a single
covalent bond. H-H
iii) Diamagnetic character : Since no unpaired electron is present in hydrogen molecule, it is
diamagnetic in nature.
2. Give the Bohr’s quantum condition for stationary states
The circumference of the circular orbit of the electron should be an integral multiple of the wavelength
of de Broglie wave, otherwise the wave cannot be smoothly continuous.
2πa = nλ, (n = 1, 2, 3, ....)
This orbit is not allowed
3. What is meant by hydrogen bonding? Hydrogen bonding comes into existence as a result of dipole-dipole interactions between the molecules
in which hydrogen atom is covalently bonded to a highly electronegative atom.
It is a weak bond because it is merely an electrostatic force and not a chemical bond.
H
F
H
F
H
F
4. Give the consequences of intra molecular hydrogen bonding
1. Intramolecular hydrogen bonding decreases the boiling point of the compound.
2. Intramolecular hydrogen bonding decreases the solubility of the compound in water by
restricting the possibility of intermolecular hydrogen bonding.
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5 MARK
1. Describe G.P. Thomson’s experiment to prove wave character of electrons A beam of electrons is allowed to pass through thin foil of gold. A diffraction pattern is observed on the
photographic plate placed perpendicular to the direction of the beam. This confirms the wave nature of
electrons.
beamof electrons
incident
thin foil
of gold photographic
plate
diffractedelectrons
diffraction pattern
2. Explain two types of hydrogen bonding with suitable examples
1. Inter molecular hydrogen bonding
Inter molecular hydrogen bond is formed between the two molecules of the same or different
compounds.
a) Hydrogen fluoride, H - F
In the solid state, hydrogen fluoride consists of long zig-zag chains of molecules associated by hydrogen
bonds as shown below:
H
F
H
F
H
F
Therefore, hydrogen fluoride is represented as (HF)n
b) Water
In water molecule, the electronegative oxygen atom forms two polar covalent bonds with two hydrogen
atoms. The oxygen atom due to its higher electronegativity acquires partial negative charge and the two
hydrogen atoms acquire partial positive charge.
δ+
δ+
H – O – H
δ2-
The negatively charged oxygen forms two hydrogen bonds with two positively charged hydrogen atoms
of two neighbouring molecules.
O
HH
O
HH
O
HH
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2. Intra molecular hydrogen bonding
This type of bond is formed between hydrogen atom and N, O or F atom of the same molecule. This
type of hydrogen bonding is commonly called chelation and is more frequently found in organic
compounds. Intra molecular hydrogen bonding is possible when a six or five membered rings can be
formed.
O
H
N
O
O
O
H
C
O
OH
O
H
C
O
H
O-Nitrophenol Salicylic Acid Salicylaldehyde
Intramolecular hydrogen bonding (chelation) decreases the boiling point of the compound and also its
solubility in water by restricting the possibility of intermolecular hydrogen bonding.
3. Give the importance of hydrogen bonding
1. Life would have been impossible without liquid water which is the result of intermolecular
H – bonding in it.
2. Hydrogen bonding increases the rigidity and strength of wood fibres and thus makes it an article of
great utility to meet requirements of housing, furniture etc.
3. The rigidity and tensile strength of cotton and silk is due to the presence of hydrogen bonding.
4. Most of our food materials such as carbohydrates and proteins also consist of hydrogen bonding.
5. H – bonding is also present in various tissues, organs, skin, blood and bones
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2. PERIODIC CLASSIFICATION – II
Blue print 1 Mark = 1 3 Mark = 1 5 Mark = 1 Total marks = 9
Govt. Examination questions 1 MARK
1. On moving down the group, the radius of an ion [Sep-2006, June-2009, Mar-2013, Sep-2015]
a) Decreases b) Increases c) No change d) None of these
2. Effective nuclear charge (Z*) can be calculated by using the formula
[June-2006, June-2008, Mar-2011, Mar-2014, June-2014, Mar-17]
a) Z* = Z – S b) Z* = Z + S c) Z* = S – Z d) Z = Z* – S
3. Among the following which has the maximum ionisation energy
[June-2011, Sep-2011, June-2012, June-2015]
a) Alkali elements b) Alkaline elements c) Halogens d) Noble gases
4. The electron affinity of an atom is [Mar-2009, Jun-2010, Jun-2016]
a) directly proportional to its size b) inversely proportional to its size
c) is independent of its size d) none of these
5. Among the following which has higher electron affinity value [Sep-2008, Sep-2009, Mar-2012]
a) Fluorine b) Chlorine c) Bromine d) Iodine
6. The scale which is based on an empirical relation between the energy of a bond and the
electronegativities of bonded atoms is [Mar-2007]
a) Pauling scale b) Mulliken’s scale
c) Sanderson’s scale d) Alfred and Rochow’s scale
7. Electron affinity is expressed in [June-2013, Mar-2016]
a) kJ b) J c) kJ mol d) kJ mol-1
8. The bond length of Cl2 molecule is (in Å) [Sep-2010]
a) 0.74 b) 1.44 c) 1.98 d) 2.28
9. The order of ionization energy [Mar-2006, Sep-2016]
a) s < p < d < f b) s > p > d > f c) s > d > p > f d) s < d < p < f
10. Noble gases have --------- electron affinity [Sep-2012, Sep-2014, Mar-2015]
a) High b) Low c) Zero d) Very low
11. When XA>>XB, A–B bond is [June-2007, Sep-2007, Mar-2010]
a) polar covalent b) non-polar covalent c) Ionic d) metallic
12. Among the following which has higher electronegativitv value? [Sep-2013]
a) Fluorine b) Chlorine c) Bromine d) Iodine
13. The metal having maximum electron affinity is [Mar-2008]
a) sodium b) calcium c) gold d) silver
3 MARK
1. Define electron affinity [March-2008, Sep-2014]
Electron affinity or electron gain enthalpy is the amount of energy released when an isolated gaseous
atom accepts an electron to form a monovalent gaseous anion.
Atom (g) + Electron Anion (g) + Energy
2. The value of Cl-Cl bond distance is 1.98Ǻ. What is the atomic radius of chlorine? [Sep-2013]
r ( Cl ) = 2
Cl)d(Cl =
2
98.1 = 0.99 Ǻ
3. The experimental value of d(C – Cl) is 1.76 Ǻ. What is the atomic radius of carbon?(rCl = 0.99 Ǻ)
[June-2007]
d(C – Cl) = r(C) + r(Cl)
r(C) = d(C – Cl) – r(Cl)
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= 1.76 – 0.99
= 0.77 Å
4. The experimental value of d(Si – C) is 1.93 Ǻ. If the radius of carbon is 0.77 Ǻ , calculate the
radius of silicon. [Mar-2015]
d(Si – C) = r(Si) + r(C)
r(Si) = d(Si – C) – r(C)
= 1.93 – 0.77
= 1.16 Å
5. Calculate the effective nuclear charge experienced by the 4s electron in potassium atom.
[Sep-2007, Sep-2012]
19 K = (1s2) (2s
2 2p
6) (3s
2 3p
6) (4s
1)
Inner shells (n –1)th shell nth shell
Effective nuclear charge, Z* = Z – S
Z* = 19 – [ ( 0.85 × No. of electrons in (n –1)th
shell ) +
( 1 × total number of electrons in the inner shells ) ]
= 19 – [ ( 0.85 × 8 ) + ( 1 × 10 ) ]
= 2.2
6. Screening constant of potassium ion (K+) is 11.25. Calculate its effective nuclear charge [Jun-2016]
Effective nuclear charge, Z* = Z – S
Z* = 19 –11.25
= 7.75
7. Calculate the effective nuclear charge of the last electron in an atom whose configuration is 1s2 2s
2
2p6 3s
2 3p
5 [Mar-2016]
(1s2) (2s
2 2p
6)
(3s
2 3p
5)
Inner shell (n –1)th shell nth shell
Effective nuclear charge (Z*) = Z – S
= 17 – [(0.35 × No. of other electrons in nth
shell) +
(0.85 × No. of electrons in (n –1)th
shell) +
(1 × total number of electrons in the inner shells)]
= 17 – [(0.35 × 6) + (0.85 × 8) + (1 × 2)]
= 17 – 10.9 = 6.1
8. Why the first ionisation energy of Be is greater than that of Li? [June-2006, June-2011]
Li Be
Nuclearcharge (Z) = 3 Nuclearcharge (Z) = 4
EC: 1s2 2s
1 EC: 1s
2 2s
2
Nuclear charge of Be is greater than Li.
Higher the nuclear charge, greater would be the force of
attraction between nucleus and outermost electron.
Hence, the first I.E. of Be is greater than that of Li.
9. Ionisation energy of Be is greater than that of B. Why? [Sep-2006, June-2008, Mar-2014]
Be B
Nuclearcharge (Z) = 4 Nuclearcharge (Z) = 5
EC: 1s2 2s
2 EC: 1s
2 2s
22p
1
Be atom has paired electrons in the 2s-orbital
Fully filled 2s-orbital is more stable due to symmetry.
More energy is needed to remove an electron from Be-
atom.
Hence, Be has high I. E.
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10. Compare ionisation energy of C and B. [Jun-2012, Mar-17]
B C
Nuclearcharge (Z) = 5 Nuclearcharge (Z) = 6
EC: 1s2 2s
22p
1 EC: 1s
2 2s
22p
2
Nuclear charge of C is greater than B.
Higher the nuclear charge, greater would be the force of
attraction between nucleus and outermost electron.
Hence, the first I.E. of C is greater than that of B.
11. Ionisation energy of F is greater than that of O. Why? [Sep-2009, Sep-2016]
O F
Nuclearcharge (Z) = 8 Nuclearcharge (Z) = 9
EC: 1s2 2s
22p
4 EC: 1s
2 2s
22p
5
Nuclear charge of F is greater than O.
Higher the nuclear charge, greater would be the force of
attraction between nucleus and outermost electron.
Hence, the first I.E. of F is greater than that of O.
12. Compare the ionisation energy of Ne with F [Mar-2009, June-2010]
F Ne
Nuclearcharge (Z) = 9 Nuclearcharge (Z) = 10
EC: 1s2 2s
2 2p
5 EC: 1s
2 2s
2 2p
6
Nuclear charge of Ne is greater than F.
Higher the nuclear charge, greater would be the force of
attraction between nucleus and outermost electron.
Hence, the first I.E. of Ne is greater than that of F.
13. Compare the first ionisation energy of Al with Mg [Mar-2012]
Mg Al
Nuclearcharge (Z) = 12 Nuclearcharge (Z) = 13
EC: 1s2 2s
2 2p
6 3s
2 EC: 1s
2 2s
2 2p
6 3s
2 3p
1
One has to remove 3p-electron in case of aluminium and 3s
electron in the case of magnesium.
s-electrons remain closer to the nucleus than p-electrons of the
same valence shell.
So it is easier to remove the p- electron than the s-electron.
Thus, the first ionization energy of aluminium is lower than
that of magnesium.
14. Compare the ionisation energy of N and O [March-2007]
N O
Nuclearcharge (Z) = 7 Nuclearcharge (Z) = 8
EC: 1s2 2s
2 2p
3 EC: 1s
2 2s
22p
4
N atom has half-filled p-orbital
Half- filled p-orbital is more stable due to symmetry.
More energy is needed to remove an electron from N-
atom.
Hence, N has high I. E.
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15. Larger the size of atom lesser is the ionization energy. Explain [Sep-2011]
The larger the size of atom, lesser is the ionization energy. This is due to the fact that electrons are
tightly held in smaller atoms whereas in large atoms, electrons are held quite loose, i.e., lesser energy is
required for removal of electrons from larger atoms than the smaller one. Hence ionization energy is
lower for larger atoms and higher for smaller atoms.
16. Electron affinity of Be, Mg, Ca and N are almost zero. Why? [Mar-2010, June-2013]
4Be : 1s2 2s
2 12Mg : 1s
2 2s
2 2p
6 3s
2 20Ca : 1s
2 2s
2 2p
6 3s
2 3p
6 4s
2
Fully filled valence s-orbitals are most stable due to symmetry. So, these elements ( Be, Mg, Ca ) do not
have tendency to accept electrons and therefore, the electron affinities are zero.
7N : 1s2 2s
2 2px
1 2py
1 2pz
1
Half filled p-orbital is more stable due to symmetry. So, nitrogen does not have tendency to accept
electrons and therefore, the electron affinity is zero.
17. Electron affinity of fluorine is less than that of chlorine. Why?
[March-2006, June-2009, Sep-2010, Mar-2013, Sep-2015]
1. The size of F-atom is very small and hence 2p-subshell becomes compact.
( 9F = 1s2 2s
2 2p
5 ).
So, addition of an extra electron produces high electron density which increases strong electron-
electron repulsion. This repulsion is responsible for less tendency of F-atom to accept electron.
2. Because of small size of fluorine there occurs large crowding of electrons around the nucleus. This
crowding is able to screen the nucleus. Because of this, effective nuclear charge gets decreased.
Thus, the electron is having less attraction during addition. Hence electron affinity of F-atom gets
decreased.
18. Mention the disadvantage of Pauling and Mulliken scale. [Sep-2008, June-2014, June-2015]
The disadvantage of Pauling’s scale:
Bond energies are not known with any degree of accuracy for many solid elements.
The disadvantage of Mulliken scale:
Electron affinities with the exception of a few elements are not reliably known.
19. Calculate the electronegativity value of fluorine on Mullikan’s scale from the following data:
Ionisation potential of F = 17.4 eV / atom. Electron affinity of F = 3.62 eV / atom
[Mar-2011]
Electronegativity = 2.82
E.AI.E
= 2.82
62.34.71
= 3.75
5 MARK
1. Explain Pauling’s Method to determine ionic radii
[Mar-06, Sep-09, Mar-10, Jun-10, Sep-2010, Sep-11, Mar-12, Mar-2014, Jun-14, Mar-17]
1. Pauling has calculated the radii of the ions on the basis of the observed internuclear distances in four
crystals: NaF, KCl, RbBr and CsI.
2. In each ionic crystal the cations and anions are isoelectronic with inert gas configuration.
NaF crystal : Na+ - 2, 8
F–
- 2, 8 Ne type configuration
KCl crystal : K+ - 2, 8, 8
Cl–
- 2, 8, 8 Ar type configuration
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3. The cations and anions of an ionic crystal are assumed to be in contact with each other and hence
the sum of their radii will be equal to the inter nuclear distance between them.
r(C+) + r(A
–) = d (C
+–A
–) ….. (1)
where,
r(C+) = radius of cation, C
+
r(A–) = radius of anion, A
–
d(C+–A
–) = internuclear distance between C
+ and A
– ions in C
+A
– ionic crystal
4. For a given noble gas configuration, the radius of an ion is inversely proportional to its effective
nuclear charge. i.e.
r(C+)
)(CZ
1*
…….. (2)
r(A–)
)(AZ
1*
…….. (3)
where,
Z*(C+) & Z*(A
–) are the effective nuclear charges of cation (C
+) and anion (A
–) respectively.
On combining (2) & (3)
)(CZ
)(AZ
)r(A
)r(C*
*
…….. (4)
5. Hence the above two equations (1) & (4) can be used to evaluate the values of r(C+) and r(A
–)
provided that the values of d(C+–A
–), Z*(C
+) and Z*(A
–) are known.
2. Explain the factors which affect ionization energy. [Jun-2008, Jun-2013, Sep-16]
1. Size of atom
The ionization energy decreases with the increasing size of atom. This is due to the fact that
electrons are tightly held in smaller atoms whereas in large atoms, electrons are held quite loose,
i.e., lesser energy is required for removal of electrons from larger atoms than the smaller one.
Ionization energy
atomofSize
1
2. Magnitude of nuclear charge
The higher the nuclear charge of protons in the nucleus, the higher is the ionization energy. Because
of the higher nuclear charge, the electrons are bound with more force and hence higher energy will
be required for their removal.
Ionization energy
rgeNuclearcha
3. Effect of number of electrons in the inner shells (Screening or shielding effect)
The electron to be removed is shielded from the nucleus by the electrons in the inner shell. Thus, the
electron in the valence shell experiences less attraction from the nucleus. Hence the ionization
energy will be low.
Ionization energy
effect Screening
1
4. Effect of shape of orbital
As s-electrons remain closer to the nucleus than p-,d-, and f-electrons of the same valence shell, the
ionization energy decreases in the order : s > p > d > f
5. Effect of arrangement of electrons
The more stable the electronic arrangement, the greater is the ionization energy. As the noble gases
have completely filled, stable electronic arrangements, they show maximum ionization energy.
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3. Explain the factors which affect electron affinity
[June-2006, Sep-2006, Mar-2007, June-2007, Mar-2009, June-2011, June-12, Sep-13, Sep-14, June-15]
1. Atomic size
Electron affinityatomofSize
1
Smaller the size of an atom, greater is its electron affinity. As the size of atom increases, the
nuclear attraction for adding electron decreases. Consequently, atom will have less tendency to
attract additional electron towards itself.
2. Effective nuclear charge
Electron affinity Effective nuclear charge.
As the effective nuclear charge of atom increases, nuclear attraction for adding electron increases.
3. Shielding or Screening Effect
Electron affinityeffect Screening
1
The incoming electron is shielded from the nuclear attraction by the electrons in the inner shell.
Therefore, greater the number of inner shells, less will be the electron affinity.
4. Electronic Configuration
Electron affinities of inert gases are zero. This is because their atoms have stable ns2 np
6
configuration in their valence shell and there is no possibility for addition of an extra electron.
4. Explain the Pauling’s scale for the determination of electronegativity
[Mar-2008, Sep-2012, Mar-2013, Mar-2015]
Pauling’s scale is based on an empirical relation between the energy of a bond and the
electronegativities of bonded atoms.
Consider a bond A-B between two dissimilar atoms A and B of a molecule AB.
Let the bond energies of A-A, B-B and A-B bonds be represented as EA-A, EB-B and EA-B respectively.
It may be seen that the bond dissociation energy of A-B is almost higher than the geometric mean of the
bond dissociation energies of A-A and B-B bonds i.e.,
EA-B > BBAA EE
=EA-B – BBAA EE
The difference () is related to the difference in the electronegativities of A and B according to the
following equation
= (XA – XB)2
Δ = (XA – XB)
0.208 Δ = (XA – XB)
Here, XA and XB are the electronegativities of A and B respectively. The factor 0.208 arises from the
conversion of Kcals to electron volt.
Considering arbitrarily the electronegativity of hydrogen to be 2.1, Pauling calculated electronegativities
of other elements with the help of this equation.
Disadvantage of Pauling scale
The bond energies are not known with any degree of accuracy for many solid elements.
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5. Explain Mulliken’s scale of electro negativity. Mention its advantage and disvantage. [Jun-2016]
According to Mulliken’s method, electronegativity could be regarded as the average of the ionization
energy and electron affinity of an atom
Electronegativity = 2
E.AI.E
Mulliken used ionisation energy and electron affinity values measured in electron volts and values were
found to be 2.8 times higher than Pauling values. Therefore the commonly accepted Pauling values are
more nearly obtained by
Electronegativity = 2.82
E.AI.E
If the values of ionisation energy and electron affinity are measured in kJ mol-1
, then
(96.48 kJ mol-1
= 1eV )
Electronegativity = 96.482.82
E.AI.E
= 540
E.AI.E
The advantage of Mulliken scale:
Different electronegativity values can be obtained for different oxidation states of the same element.
The disadvantage of Mulliken scale:
Electron affinities with the exception of a few elements are not reliably known.
6. How do electronegativity values help to find out the nature of bonding between atoms? [Sep-2007, June-2009, Mar-2011, Sep-2015]
Electronegativity Nature of A-B bond Representation Example
XA = XB,
i.e. XA- XB = 0 Non polar covalent bond A–B H –H
XA > XB
i.e. XA - XB is small Polar covalent bond A
δ––B
δ+ O
δ –-H
δ+ in H2O
XA >> XB
i.e. XA - XB is very large Ionic bond A
– B
+ Na
+Cl
–
7. Explain how electronegativity values help to find out the percentage of ionic character in a polar
covalent bond. [Sep-2008]
Electronegativity
difference Bond character Bond type Representation
(XA–XB) = 1.7 ionic character = 50%
covalent character =50% 50% ionic and 50% covalent A
δ ––B
δ+
(XA–XB) < 1.7 ionic character < 50%
covalent character >50% Predominantly covalent A–B
(XA–XB) > 1.7 ionic character > 50%
covalent character <50% Predominantly ionic A
– B
+
8. Explain the variation of ionizsation energy along the group and period [Mar-2016]
Variation of ionization energy down a Group:
In a group, the ionisation potential decreases from top to bottom. On moving down the group size of
atom is increased and as a result the outermost electron is less powerfully attracted by the nucleus which
results in the decrease in I.E Variation of ionization energy along a period:
In a period, the value of ionisation potential increases from left to right. On moving from left to right in
a period nuclear charge is increased and as a result the outermost electron is strongly attracted by the
nucleus which results in the increase in I.E
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Additional questions
1 MARK
1. The value of C-C distance found experimentally in a saturated hydrocarbon is
a) 1.34 Å b) 1.36 Å c) 1.54 Å d) 1.56 Å
2. Pick the correct statement
a) Carbon has more nuclear charge than boron
b) The size of carbon atoms is larger than boron
c) Carbon forms electron deficient compounds
d) Carbon forms ionic compounds
3. Comparing the ionisation energy of fluorine with carbon, fluorine has
a) higher ionisation energy b) lower ionisation energy
c) same ionisation energy d) none of these
4. Across the period, electron affinity
a) decreases b) increases
c) decrease and the increases d) increase and then decreases
5. Which one of the following has the largest ionization energy?
a) K b) Na c) Be d) Ne
6. Which one of the following has the lowest ionization energy?
a) Li b) Be c) B d) C
7. The atomic radius increases as we move down a group because
a) effective nuclear charge increases b) atomic number decreases
c) atomic mass increases d) additive electrons are accommodated in new electronic level
8. Slater rules are used to calculate
a) atomic number b) screening constant c) atomic mass d) atomic radius
9. Which of the following electrons has the minimum screening effect?
a) d-electron b) f-electron c) s-electron d) p-electron
10. As we go down the group, ionization potential
a) increases b) does not change c) decreases d) varies regularly
11. Which one of the following elements has the lowest ionization energy?
a) Be b) B c) C d) N
12. The element with the highest ionization potential is
a) B b) C c) N d) O
13. Which of the following has the highest electron affinity?
a) inert gas b) lanthanides c) alkali metals d) halogens
14. For a molecule with XA–XB = 1.7 the percentage ionic character is
a) 25% b) 50% c) 75% d) zero
15. Na+ and F
–are isoelectronic with
a) Ne b) K c) Ca d) all the above
16. K+ and Cl
– are isoelectronic with
a) He b) Ar c) Ne d) Xe
17. Second electron affinity of sulphur and oxygen
a) positive b) zero c) exothermic d) negative
18. Electron affinity is negative for
a) Zn b) Cd c) both a and b d) none of these
19. Which of the following has non-polar covalent bond?
a) H2 b) H2O c) NaCl d) KCl
20. In the same valence shell -------- electrons are closer to the nucleus
a) s-electrons b) p-electrons c) d-electrons d) f-electrons
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3 MARK 1. Define ionization energy
The amount of energy required to remove the most loosely bound electron from an isolated atom in the
gaseous state in known as Ionisation Energy.
2. Define electro negativity
Electronegativity is defined as the relative tendency of an atom in a molecule to attract the shared pair of
electrons towards itself.
H F (fluorine is more electronegative than hydrogen)
3. Compare the IE of Na with Mg
Na Mg
Nuclearcharge (Z) = 11 Nuclearcharge (Z) = 12
EC: 1s2 2s
2 2p
6 3s
1 EC: 1s2 2s
2 2p
6 3s
2
Nuclear charge of Mg is greater than Na.
Higher the nuclear charge, greater would be the force of attraction
between nucleus and outermost electron.
Hence, the first I.E. of Mg is greater than that of Na.
4. Why inert gases show maximum IE?
Inert gases or noble gases have completely filled (ns2np
6), more stable electronic configuration.
Removal of electron from more stable electronic configuration becomes very difficult. So, inert gases
show maximum IE.
5. Why do elements of group 17 possess high electron affinity?
Elements of group 17 (halogens) possess high electron affinity. The reason for this is that by picking up
an electron halogens attain the stable noble gas electronic configuration.
Eg: F 1s22s
22p
5 + e
- 1s
22s
22p
6 (completely filled, more stable, noble gas configuration)
6. Electron affinity of Noble gases is zero. Why?
Electron affinities of noble gases are zero. These atoms possess ns2np
6 configuration in their valence
shells. This fully filled electronic configuration is more stable due to symmetry. So, there are no chances
for the addition of an extra electron. Thus, the electron affinities of noble gases are zero.
7. Among all the metals electron affinity of gold is high. Why?
Of all the metals, the E.A. of gold is comparatively high. This is due to the higher effective nuclear
charge and poor shielding of the nucleus by d-electrons.
8. Calculate the electronegativity value of chlorine on Mullikan’s scale from the following data:
Ionisation potential of Cl = 13 eV / atom. Electron affinity of Cl = 4 eV / atom
Electronegativity = 2.82
E.AI.E
= 2.82
413
= 3.03
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3. p -BLOCK ELEMENTS
Blue print 1 Mark = 1 3 Mark = 2 5 Mark = 1 Total marks = 12
Govt. Examination questions
1 MARK
1. An element which was burnt in limited supply of air to give oxide A which on treatment with water
gives an acid. B. Acid B on heating gives acid C which gives yellow precipitate with AgNO3 solution. A
is [Mar-2007, Mar-2009]
a) SO2 b) NO2 c) P2 O3 d) SO3
2. The compound with garlic odour is [June-2010, Mar-2012]
a) P2 O3 b) P2O5 c) H3PO3 d) H3PO4
3. The compound with garlic taste is [Jun-2013, Sep-2015, Sep-2016]
a) H3PO4 b) H3PO3 c) P2O3 d) P2O5
PCl3 Pungent odour
P2O3 Garlic odour
H3PO3 Garlic taste
PH3 Rotten fish odour
4. The shape of PCl5 is [Sep-2011, June-2014]
a) pyramidal b) trigonal bipyramidal c) linear d) tetrahedral
5. The compound used as smoke screen [Sep-2008, Sep-2010, Mar-2014]
a) PCl3 b) PCl5 c) PH3 d) H3PO3
6. Among the halogen acid, the weakest acid is [Mar-2010, Sep-2013]
a) HF b) HCl c) HBr d) HI
7. The noble gases are unreactive because they [Sep-2007, Mar-2013]
a) have same number of electrons b) have an atomicity of one
c) are gases with low densities d) have stable electronic configuration.
8. The shape of XeF4 is [June-2012, Mar-2015]
a) tetrahedral b) octahedral c) square planar d) pyramidal
9. The lightest gas which is non-inflammable is [June-2011, June-2015]
a) He b) H2 c) N2 d) Ar
10. The toxic element of boron family is [Sep-2006]
a) Boron b) Indium c) Thallium d) Gallium
11. The compound used to arrest the bleeding is [June-2008, Mar-17]
a) K2SO4 b) Potash alum K2SO4 . Al2(SO4)3 . 24H2O c) Al2(SO4)3 d) KI
12. The general electronic configuration of carbon family is [Mar-2006]
a) ns2 np
2 b) ns
2 np
3 c) ns
2 np
1 d) ns
2 np
4
Group General electronic
configuration
13 Boron family ns2 np
1
14 Carbon family ns2 np
2
15 Nitrogen family ns2 np
3
16 Oxygen family ns2 np
4
17 Halogens ns2 np
5
18 Inert gases ns2 np
6
13. The metalloid among the following is [June-2006]
a) Pb b) P c) Ge d) Sn
14. Which of the following does not belong to group 14? [June-2007]
a) C b) Si c) Ga d) Pb
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15. Which of the following is second most abundant in earth‟s crust? [Sep-2012]
a) C b) Si c) Ge d) Sn
16. An element which belongs to Group 14 is soft in nature, does not react with pure water, but dissolves in
water containing dissolved air. Then the element is [Sep-2009]
a) C b) Ge c) Pb d) Ti
17. Which of the following has the property of etching on glass? [Mar-2008]
a) HI b) HF c) HBr d) HCl
18. Which of the following shows negative oxidation state only? [June-2009]
a) Br b) F c) Cl d) I
19. Inert gas used in beacon lights for safety of air navigation is [Mar-2011]
a) Helium b) Argon c) Neon d) Xenon
20. One can draw the map of building on glass plate by [Sep-2014]
a) HI b) HF c) HBr d) HCl
21. Without lone pair of electrons in AX3 type of interhalogen compounds, its shape is [Mar-2016]
a) trigonal bipyramidal b) tetrahedral c) T-shape d) square planar
22. The metal which exists as a liquid at room temperature is [Jun-2016]
a) Germanium b) Indium c) Thallium d) Gallium
3 MARK 1. What is potash alum and how is it prepared? [March-2008, Sep-2016]
Potash Alum is K2 SO4. Al2(SO4)3. 24 H2O
It is manufactured from alunite or alum stone.
Alunite or alum stone is K2SO4. Al2(SO4)3. 4Al(OH)3. It is finely powdered and boiled with dilute
sulphuric acid, the aluminium hydroxide part changes into aluminium sulphate. When a little more
potassium sulphate in calculated amount is added, the alum is crystallised.
2. Write the uses of potash alum [Jue-2012]
1. It is used in purification of water
2. It is used in water proofing of textiles, in dyeing and in paper industry.
3. It is used to arrest bleeding
3. How is burnt alum obtained? [June-2014, Jun-2016]
When heated, potash alum ( K2 SO4. Al2(SO4)3. 24 H2O ) melts at 365 K and on further heating loses
the whole of its water of crystallisation and swells up. The swollen mass so produced is called burnt
alum.
4. What is inert pair effect? [Mar-2009, Mar-2016]
The tendency of being less availability of ns electrons in bonding is known as inert pair effect. The inert
pair effect increases down the group with the increase in atomic number.
5. Write a note on plumbo solvency
[Jun-06, Sep-06, Jun-07, Jun-09, Mar-10,Sep-10, Jun-11, Mar-13, Sep-14, Jun-15, Mar-17]
Lead is not attacked by pure water in the absence of air, but water containing dissolved air has a solvent
action on it due to the formation of lead hydroxide (a poisonous substance). This phenomenon is called
Plumbo solvency.
2Pb + O2 + 2H2O 2Pb(OH)2
6. H3PO3 is diprotic or dibasic acid. Why? [Sep-2009, Mar-2010, June-2013, June-2014, Sep-2014]
H3PO3 is a dibasic acid. It combines with NaOH to form two types of salts.
H3PO3 + NaOH NaH2PO3 + H2O
Sodium dihydrogen Phosphite
H3PO3 + 2NaOH Na2HPO3 + 2H2O
Disodium hydrogen Phosphite
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7. Prove that H3PO4 is a tribasic acid (triprotic acid) [March-2006, March-2013, Mar-2015]
H3PO4 is a tribasic acid. It combines with NaOH to form three types of salts.
H3PO4 + NaOH NaH2PO4 + H2O
Sodium dihydrogen phosphate
H3PO4 + 2NaOH Na2HPO4 + 2H2O
Disodium hydrogen phosphate
H3PO4 + 3NaOH Na3PO4 + 3H2O
Sodium phosphate
8. Prove that P2O5 ( or P4O10 ) is a powerful dehydrating agent.
[Sep-2006, March-2007, Mar-2009, June-2010, Mar-2012, Sep-2012, Mar-2014]
Phosphorus pentoxide extracts water from many inorganic compound including sulphuric acid, nitric
acid and several organic compounds. It is therefore, used as a powerful dehydrating agent.
H2SO4 4 10P O
SO3 + H2O
2HNO3 4 10
P ON2O5 + H2O
RCONH2 4 10
P O RCN + H2O
Amide Nitrile
9. How will you prove that H3PO3 is a powerful reducing agent? [June-2006]
H3PO3 is a powerful reducing agent because it has P-H bond. It reduces silver nitrate solution into
silver.
2AgNO3 + H2O + H3PO3 2Ag + H3PO4 + 2HNO3
10. Show that PH3 is a powerful reducing agent. [Sep-2016]
PH3 is a powerful reducing agent. When it is passed through the salt solutions, corresponding metal is
formed.
PH3 + 6AgNO3 + 3H2O 6Αg + 6HNO3 + H3PO3
11. How is orthophosphoric acid prepared in laboratory? [June-2008]
In the laboratory orthophosphoric acid is be prepared by boiling a mixture of red phosphorus with 50%
nitric acid. Iodine acts as a catalyst.
P + 5HNO3 H3PO4 + 5NO2 + H2O
12. What is the action of heat on phosphorus acid? [June-2009]
When H3PO3 is heated it undergoes auto-oxidation and reduction to form ortho phosphoric acid and
phosphine.
4H3 PO3 3H3PO4 + PH3
13. What is the action of heat on ortho phosphoric acid? [Mar-2011]
On heating it gives pyrophosphoric acid at 523 K and at 589 K gives metaphosphoric acid
2H3PO4 K523
H4P2O7 + H2O
H4P2O7 K589
2HPO3 + H2O
14. How silver nitrate reacts with ortho phosphoric acid? [Mar-2016]
On reaction with silver nitrate, it gives yellow precipitate of silver phosphate.
H3PO4 + 3AgNO3 Ag3PO4 + 3HNO3
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15. Write note on Holme’s signal [Sep-2007, Sep-2009]
Containers which have a perforated bottom and a hole at the top are filled with calcium phosphide and
calcium carbide. These are thrown into the sea. Water enters the container through the bottom and reacts
with calcium carbide and calcium phosphide to give acetylene and phosphine. Phosphine gets ignited
spontaneously as it comes in contact with air and also ignites acetylene. Thus a bright red flame is
produced which is accompanied by huge smoke due to the burning of phosphine. This serves as a signal
to the approaching ships.
Ca3P2 + 6H2O 2 PH3↑ + 3Ca(OH)2
CaC2 + 2H2O C2H2 ↑ + Ca(OH)2
16. Draw the electronic structure [electron dot formula] of a) PCl3 b) PCl5 c) H3PO3
d) H3PO4 e) H4P2O7 (Any 2) [Sep-08, Mar-08, Jun-11, Jun-15, Sep-15, Mar-17]
a) PCl3 b) PCl5
PCl
Cl
Cl x
xx
x
x
x- Electron of P
- Electron of Cl P
Cl
Cl
Cl
x
x
x
Cl x
Clx
c) H3PO3 d) H3PO4 e) H4P2O7
( Phosphorus acid ) ( orthophosphoric acid ) ( pyrophosphoric acid )
P
H
O
OO HH
x x
x
x
x P
H
O
OO HH
x x
x
x
x
O
P
H
O
OH
x x
xx
x
O
O x P Ox H
Oxx
H
x
O
- Electron of P
- Electron of O
- Electron of H
x
17. Discuss the oxidising power of fluorine [June-2007, Sep-2011, June-2013]
Oxidizing property of the halogen is due to high electron affinity of halogen atoms. The oxidizing
power decreases from fluorine to iodine. Fluorine is the strongest oxidising agent. It oxidises other
halide ions to halogens. In solution (or) when dry
F2 + 2X– 2F
– + X2 ( X
– = Cl
–, Br
–, I
– )
Halogen of low atomic number oxidises the halide ion of higher atomic number.
18. Why is HF not stored in glass bottles? [March-2007, Mar-2011, Sep-2013]
HF cannot be stored in glass or silica bottles as it attacks silicates and silica.
Na2SiO3 + 6HF Na2SiF6 + 3H2O
SiO2 + 4HF SiF4 + 2H2O
So, HF is stored in wax bottles.
19. Write the uses of fluorine [Sep-2012]
Refer-5 mark – Q-7
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20. What are interhalogen compounds? How are they formed? [Sep-2008, June-2012, Mar-2015]
Each halogen combines with another halogen to form several compounds known as interhalogen
compounds.
They are prepared by direct combination or by the action of a halogen on a lower interhalogen.
Cl2 + F2 K473
2ClF
Cl2 + 3F2 K573
2ClF3
IF5 + F2 K573
IF7
21. How are xenon fluorides prepared? [Sep-2011]
Xenon forms three binary fluorides XeF2 , XeF4 and XeF6 by the direct union of elements under
appropriate experimental conditions.
Xe + F2 K673
XeF2
Xe + 2F2 K673
XeF4
Xe + 3F2 K573
XeF6
22. Give any three uses of helium. [Sep-2007, Mar-2014]
1. Because of its lightness and non-inflammability helium is used to filling balloons for meteorological
observations
2. Because of its lightness it is used in inflating aeroplane tyres.
3. A mixture of oxygen and helium is used in the treatment of asthma.
23. Write any three uses of neon? [June-2008, June-2010, Sep-2010, Mar-2012, Sep-2013, Sep-2015]
1. Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
2. Neon is used in beacon lights for safety of air navigation as the light possesses fog and storm-
penetrating power.
3. Neon light is used in botanical gardens as it stimulates growth and helps the formation of
chlorophyll.
24. Write any three uses of lead? [June-2016]
1. Lead is used for making lead pipes
2. It is used for making telegraph and telephone wires
3. It is used for making bullets and lead accumulators
5 MARK
1. What are the uses of silicones? [Mar-2007, Mar-2012, Sep-2012, Mar-2014]
No. Types of silicones Uses
1 Silicones They withstand high temperature and hence used as
insulators for electric motors.
2 Silicone fluids
(Straight chain polymers
of 20 to 500 units)
They are water repellent because of the organic side group
and hence used in waterproofing textiles, as lubricants and
as polish.
3 Silicone rubber They retain their elasticity even at low temperatures and resist
chemical attack and hence they are mixed with paints to
make them damp-resistant.
4 Silicone resins
(Cross-linked polymer)
They are used as non-stick coating for pans and are used in
paints and varnish.
5 Silicone oils They are highly stable and non-volatile even on heating and
hence used for high temperature oil bath and high vacuum
pump.
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2. How is lead extracted from its ore? [June-2008, Sep-2011]
1. Ore: Galena PbS
2. Concentration: Froth floatation process.
3. Smelting in a Reverberatory furnace:
The concentrated ore is roasted in a reverberatory furnace. During roasting, galena is partly oxidized
to lead monoxide and partly to lead sulphate.
Reverberatory furnace
2PbS + 3O2 2 PbO + 2SO2
PbS + 2O2 PbSO4
More of galena is then added. Lead sulphide reacts with the two oxidised products giving lead.
PbS + 2PbO 3Pb + SO2
PbS + PbSO4 2Pb + 2SO2
4. Purification of Lead
Lead extracted by the above method contains impurities such as silver, copper, tin, bismuth, gold
and iron. It is refined by the following processes.
a. Liquation
The impure metal is heated on a sloping hearth. Lead melts and flows down the slope. The
infusible impurities remain on the hearth.
b. Desilverisation
Silver is removed by either Pattinson‟s process or Park‟s process.
c. Electrolytic refining
Very pure lead is obtained by this process.
Anode : Impure lead
Cathode : Very pure lead
Electrolyte : Lead fluosilicate + Hydrofluosilicic Acid (PbSiF6) (H2SiF6)
The metallic impurities which are more electropositive than lead, such as iron and tin, go into the
solution while the rest of the impurities are thrown down as anode mud.
3. Write any five anomalous nature of Fluorine or How does fluorine differ from other halogens?
[Mar-2008, Mar-2009, Jun-2010, Sep-2010, Mar-2013, Sep-16]
1. Fluorine is the most reactive element among halogen. This is due to the minimum value of F–F
bond dissociation energy.
2. Hydrofluoric acid is a weak acid whereas the other hydrohalic acids are strong acids. This is
because HF is associated due to hydrogen bonding. .
...... H– F...... H– F..... H–F.....
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3. Being strongly electronegative it can have only a negative oxidation state while the other halogens
can have negative as well as positive oxidation state.
4. HF attacks glass while others do not.
5. AgF is soluble in water while the other AgX are insoluble.
6. It forms two types of salts with metals. NaF and NaHF2
7. Fluorine does not form polyhalides because of the absence of d-orbitals in its valence shell. Thus
we have Cl3–, Br3
–, I3
– ions but no F3
– ion.
4. How is fluorine isolated from fluorides? [Mar-2006, Sep-2009, June-2011, June-2014, Mar-2015]
Dennis’ Method:
In this fluorine is prepared by the electrolysis of fused sodium or potassium hydrogen fluoride (perfectly
dry) Electrolysis is carried out between graphite electrodes in a V-shaped electrically heated copper
tube. The ends of the tube are covered with copper caps into which the graphite electrodes are fixed
with bakelite cement. The copper tube is thickly lagged to prevent loss of heat.
KHF2 KF + HF
HF H+ + F¯
2H+ + 2e
– H2 (At cathode)
2F – – 2e
– F2 (At anode)
Fluorine liberated at the anode is passed through the U-tube containing sodium fluoride. This removes
the hydrogen fluoride vapours coming with fluorine.
NaF + HF NaHF2
5. i) Prove that H3PO4 is a tribasic acid
ii) Prove that H3PO3 is a powerful reducing agent [June-2007]
Ans: Refer – 3 Mark – Q -7 & 9
6. i) Prove that P2O5 ( or P4O10 ) is a powerful dehydrating agent.
ii) Discuss the oxidising power of fluorine [Sep-2008]
Ans: Refer – 3 Mark – Q -8 & 15
7. Write any five uses of fluorine [Mar-2016]
1. Fluorine is used in the manufacture of freons ( dichlorodifluoro methane CF2 Cl2 ). These are used
as refrigerants in refrigerators, deep freezers and air conditioners.
2. CaF2 is used as flux in metallurgy.
3. NaF is used as a preservative to prevent fermentation and also for preventing dental cavities 4. SF6 is used as an insulating material in high voltage equipment 5. Teflon is used as container to store hydrofluoric acid 6. UF6 is used in the separation of U
235 from U
238
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8. Write a note on etching on glass. [Jun-2016]
The action of hydrofluoric acid on silica and silicates is used for etching glass.
Na2SiO3 + 6HF Na2SiF6 + 3H2O
SiO2 + 4HF SiF4 + 2H2O
The glass article is first covered with a film on wax. The design to be etched is now drawn on the waxed
surface and is then exposed to the action of hydrofluoric acid. Now the glass can be very soon etched.
The wax is finally washed off with turpentine.
9. Explain the structure of AX type interhalogen compounds [Sep-2007]
Type AX (ClF) Electronic structure of Chlorine atom, in the ground state and hybridized state is represented as in Fig.
The dotted arrow indicates electron contributed by the fluorine atom.
3s 3p 3d
sp3 hybridisation
A X
AX
Linear Molecule
Atom in Ground state
Hybridisation sp
3
Number of bond pair of electrons 1
Number of lone pair of electrons 3
Shape (without lone pair) Linear
10. Explain the structure of AX5 type interhalogen compounds [Sep-2007]
Type AX5 (F5)
5s 5p 5d
sp3d2 Hybridisation
Iodine Atom in Ground state
Only one unpaired
electron permits one
covalent bond
Five unpaired
electrons permit five
covalent bonds
Formation of IF5
molecule by sp3d2
hybridisation
Iodine Atom in Excited state
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I
F
F
FF
F
Hybridisation sp3d
2
Number of bond pair of electrons 5
Number of lone pair of electrons 1
Shape with lone pair Octahedral
Shape without lone pair Square pyramidal
11. How are noble gases isolated from air by Ramsay - Raleigh’s method?
[June-2006, Sep-2006, Mar-2010, Sep-2013, June-2015, Mar-17]
A mixture of air and oxygen is constantly admitted into a glass globe of about 50 litres capacity. Two
platinum electrodes are introduced and a discharge from a transformer of about 6000 - 8000 volts is
passed by the action of which nitrogen and oxygen rapidly combine to form oxides of nitrogen. The
oxides are dissolved out in a solution of sodium hydroxide continuously circulated through the flask.
N2 + O2 2 NO
2 NO + O2 2NO2
2NO2 + 2NaOH NaNO3 + NaNO2 + H2O
Oxygen if any is removed by introducing alkaline pyrogallol in the globe. The supply of air and electric
discharge is shut after some time and the remaining mixture of noble gases is pumped out.
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12. How are noble gases separated by Dewar’s method?
[June-2009, Mar-2011, June-2012, Sep-2014, Sep-2015]
Principle:
The mixture of noble gases is separated into individual constituents by the use of coconut charcoal
which adsorbs different gases at different temperatures.
He, Ne, Ar, Kr, Xe
Brought in contact with charcoal at 173 K
Adsorbed Unadsorbed
Ar, Kr, Xe He, Ne
Brought in contact with
charcoal at 93 K
Adsorbed Unadsorbed
HeNe
Brought in contact with fresh
charcoal at the temperature
of the liquid air
Adsorbed on fresh charcoal
Not adsorbed
Ar Kr, Xe
(remain in the first charcoal)
Temperature of the
first charcoal is raised
to 183 K
Kr is set free Xe remain adsorbed in the charcoal.
It is recovered by heating.
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Additional questions 1 MARK
1. Which of the following does not belong to group 13?
a) B b) Al c) Ge d) In
2. Which of the following is most abundant in earth‟s crust?
a) C b) Si c) Ge d) Sn
Most abundant element in earth’s crust = oxygen
Second most abundant element in earth’s crust = silicon
3. Which shows only -1 oxidation state?
a) fluorine b) bromine c) chlorine d) iodine
4. Halogens belong to the group number
a) 14 b) 15 c) 17 d) 18
5. Which is not known?
a) XeF6 b) XeF4 c) XeO3 d) ArF6
6. Which of the following has highest first ionisation energy?
a) He b) Ne c) Ar d) Kr
7. Which of the following is used in aircraft?
a) Boron b) Aluminium c) Thallium d) Gallium
8. Which of the following is used in making transistors and thermistors?
a) Boron b) Aluminium c) Thallium d) Indium
9. Potash alum is
a) KOH . Al(OH)3 b) KMnO4.Al2(SO4)3 c) K2SO4.Al2(SO4)3.24H2O d) none
10. The aqueous solution of potash alum is acidic due to
a) the hydrolysis of Al2(SO4)3 b) the hydrolysis of K2SO4
c) the oxidation of potash alum d) none of these
11. The compound used in the purification of water is
a) PbSO4 b) Potash alum c) PH3 d) KI
12. The general electronic configuration of p-block elements is
a) ns2 np
2 b) ns
2 np
1-6 c) ns
2 np
0,1 d) ns
1 np
1-5
13. Modern solid-state electronic devices are made by using
a) Si & Ge b) Al & C c) As & Bi d) Sb & P
14. Silicones are water repellent because of the presence of
a) Si-O bond b) organic side group c) OH-group d) none of these
15. The sulphide ore of lead is
a) Galena b) Cerrusite c) Anglesite d) Lead ochre
16. Litharge is
a) PbCl2 b) Pb2O3 c) Pb3O4 d) PbO
17. Red lead is
a) Pb3O4 b) PbO c) PbS d) PbSO4
18. Lead is used for making
a) telegraph wires b) telephone wires c) bullets d) all the above
19. Which of the following alloys are made by using lead?
a) solder b) pewter c) type metal d) all of these
20. Which of the following compounds of lead is used as an additive to petrol to prevent knocking?
a) Lead fluosilicate b) chloroplumbic acid c) tetraethyl lead Pb(C2H5)4 d) none
21. Which of the following elements is used to make pesticides?
a) Nitrogen b) Arsenic c) Indium d) Carbon
22. The pharmaceutical that contains bismuth is
a) Pepto-bismol b) bismuth arsenide c) BHC d) none of these
23. The hybridisation and shape of PCl3 molecule
a) sp3, pyramidal b) sp
2, pyramidal c) sp
3d, trigonal planar d) sp, planar
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24. The hybridisation and shape of PCl5 molecule
a) sp3, pyramidal b) sp
2, pyramidal c) sp
3d, trigonal bipyramidal d) sp, planar
25. Which of the following compounds undergo sublimation?
a) PCl5 b) PCl3 c) PH3 d) H3PO4
26. Which is a powerful dehydrating agent?
a) P2O3 b) P2O5 c) PH3 d) H3PO3
27. Which is a dibasic acid?
a) H3PO3 b) H3PO4 c) H4P2O7 d) HPO3
28. Which is a tribasic acid?
a) H3PO3 b) H3PO4 c) H4P2O7 d) HPO3
29. Which of the following compounds are used as reducing agents?
a) H3PO3 b) PH3 c) both a and b d) H3PO4
30. H3PO3 is a powerful reducing agent because of the presence of
a) O-H bond b) P-H bond c) P-O bond d) none
31. Which of the following compounds undergoes auto-oxidation and reduction on heating?
a) H3PO3 b) H3PO4 c) PCl5 d) PH3
32. In the laboratory preparation of orthophosphoric acid from red phosphorus and nitric acid, the catalyst
used is
a) Pt b) iodine c) MnO2 d) iron
33. Deliquescent crystalline solid is
a) H3PO4 b) PH3 c) PCl3 d) none
34. Which is used as souring agent in the preparation of soft drinks?
a) H3PO4 b) H3PO3 c) HPO3 d) PCl3
35. Orthophosphoric acid reacts with silver nitrate to give
a) yellow precipitate of silver phosphate Ag3PO4 b) white precipitate of silver nitrite
c) red precipitate of silver phosphate d) none of these
36. The substances used in Holme‟s signal are
a) Ca3P2 & CaC2 b) CaCO3& CaC2 c) Ca3(PO4)2 & CaCO 3 d) none of these
37. In Holme‟s signal the gases liberated due to the hydrolysis of calcium phosphide and calcium carbide are
a) Phosphine and acetylene b) Phosphine and ethylene
c) Acetylene and O2 d) CH4 and PH3
38. In Holme‟s signal the gas which gets ignited spontaneously as it comes in contact with air is
a) Phosphine b) acetylene c) O2 d) CH4
39. The first four elements in group 16 [O, S, Se, Te] are collectively called as
a) halogens b) Chalcogens or ore forming elements c) noble gases d) none
40. The most abundant element which constitutes 46.6% of earth‟s crust is
a) O b) S c) B d) Al
41. Which is known as Brim stone?
a) Ca b) S c) Si d) P
42. Which is a radioactive element?
a) As b) Ne c) Polonium d) Selenium
43. The group 17 elements are known as
a) halogens or salt producer b) chalcogens c) rare gases d) none of these
44. The strongest oxidising agent is
a) F b) Cl c) Br d) I
45. Among the hydracids, HF is a liquid due to
a) intermolecular hydrogen bonding b) ionic bonding c) covalent bonding d) none
46. The correct order of acidic strength is
a) HF < HCl < HBr < HI b) HF >HCl >HBr >HI c) HCl< HF< HBr< HI d) none
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47. Fluorine is the most reactive element among the halogens. This is due to the minimum value of
a) electronegativity b) F-F bond dissociation energy c) I.E d) electron affinity
48. Which is soluble in water?
a) AgF b) AgCl c) AgBr d) AgI
49. Fluorine can have only negative oxidation states because it is
a) strongly electronegative b) strongly electropositive c) a non-metal d) none
50. Fluorine does not form polyhalides (F3–) because of the absence of
d) unpaired electron b) d-orbitals in its valence shell c) p-orbital d) none
51. Aqueous HF can be stored in
a) glass bottles b) wax bottles c) silica bottles d) none of these
52. Two types of salts formed by fluorine are
a) NaF and NaHF2 b) NaF and Na2F c) NaF and NaF2 d) NaF and NaH2F
53. Freons [CF2Cl2] are used as
a) prservatives b) refrigerants c) antiseptics d) anaesthetics
54. Which is used as a flux in metallurgy?
a) SF6 b) CaF2 c) NaF d) UF6
CaF2 used as a flux in metallurgy
NaF used as a preservative to prevent fermentation and for
preventing dental cavities
SF6 Used as an insulating material in high voltage equipment
UF6 Used in the separation of U235
from U238
Teflon Used as a container to store hydrofluoric acid
55. The shape of ClF is
a) trigonal bipyramidal b) tetrahedral c) linear d) Trigonal planar
[ClF, BrF, BrCl, IBr, ICl = Linear]
56. The shape of IF7 is
a) pentagonal bipyramidal b) linear c) square planar d) terahedral
Molecule Geometry or Shape
With lone pair Without lone pair
ClF3 Trigonal bipyramidal T-shaped
IF5, BrF5 Octahedral Square pyramidal
57. Group 18 elements are known as
a) rare gases b) inert gases c) noble gase d) all the above
58. The shape of XeF2 is
a) linear b) bent „v‟ shaped c) angular d) none
59. The shape of XeF4 is
a) linear b) tetrahedral c) squareplanar d) none
60. The shape of XeF6 is
a) planar b) distorted octahedral c) square pyramidal d) none
61. Number of bond pair and lone pair in XeF6
a) 6 Bp and 1 Lp b) 6 Bp and 2 Lp c) 6 Bp and 3 Lp d) 5 Bp and 1 Lp
62. The inert gas used for filling balloons and for inflating aeroplane tyres is
a) Helium b) Neon c) Argon d) Krypton
63. Which of the following is used by deep-sea divers and in the treatment of asthma?
a) He-O2 mixture b) Ar-O2 mixture c) He-N2 mixture d) Xe-O2 mixture
64. Inert gas used to produce superconducting magnets is
a) Helium b) Argon c) Neon d) Xenon
65. Which of the following is used as cryogenic agent to carry out low temperature-experiments?
a) liquid helium b) liquid hydrogen c) liquid neon d) liquid oxygen
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66. Which of the following is used in fluorescent bulbs for advertisement display?
a) helium b) hydrogen c) neon d) radon
67. Which of the following is used to protect electrical instruments from high voltage?
a) He-O2 mixture b) Ar-O2 mixture c) He-Ne mixture d) Xe-Ne mixture
68. ---------- light is used in botanical garden as it stimulates growth and helps the formation of
chlorophyll
a) Helium b) Neon c) Argon d) Krypton
69. Which of the following is used in gas filled electric lamps?
a) He + 26% N2 b) Ar + 26% N2 c) Ar + 26% O2 d) Ne + 26% N2
70. ------------ is used in radioactive research and in radiotherapy for the treatment of cancer
a) Neon b) Argon c) Krypton d) Radon
3 MARK
1. How will you prepare pure phosphine in the laboratory?
It is obtained by boiling white phosphorus with 30-40% solution of caustic soda in an inert atmosphere
of CO2.
4P + 3NaOH + 3H2O PH3 + 3NaH2PO2
Sodium hypophosphite
Phosphine so obtained is impure. It is passed into an aqueous solution of hydrogen iodide, PH4I is
formed. PH4I is heated with KOH or NaOH, pure phosphine is obtained.
PH3 + HI PH4I
PH4I + NaOH PH3 + NaI + H2O
2. Give the uses of PH3
1. Smoke screens
When PH3 burns it produces smoke which is dense enough to serve as smoke screens.
2. Holme’s signal
Containers which have a perforated bottom and a hole at the top are filled with calcium phosphide and
calcium carbide. These are thrown into the sea. Water enters the container through the bottom and reacts
with calcium carbide and calcium phosphide to give acetylene and phosphine. Phosphine gets ignited
spontaneously as it comes in contact with air and also ignites acetylene. Thus a bright red flame is
produced which is accompanied by huge smoke due to the burning of phosphine. This serves as a signal
to the approaching ships.
Ca3P2 + 6H2O 2 PH3↑ + 3Ca(OH)2
CaC2 + 2H2O C2H2 ↑ + Ca(OH)2
3. What is the action of water on a) PCl3 and b) PCl5?
a) PCl3
It is violently hydrolysed by water giving phosphorus acid
PCl3 + 3H2O 3HCl + H3PO3
b) PCl5
It is violently hydrolysed by water giving phosphorus oxychloride or phosphoric acid depending upon
the quantity of water.
PCl5 + H2O waterntinsufficie
POCl3 + 2HCl
PCl5 + 4H2O waterofxcessE
H3PO4 + 5HCl
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4. What is the action of water on a) P2O3 and b) P2O5?
a) P2O3
It reacts with cold water to give phosphorus acid.
P2O3 + 3H2O 2H3PO3
It reacts with hot water vigorously to form inflammable phosphine.
2P2O3 + 6H2O PH3 ↑ + 3H3 PO4
b) P2O5
It reacts with moisture to form metaphosphoric acid.
P4O10 + 2H2O 4HPO3
When the solution is boiled, the metaphosphoric acid is changed to orthophosphoric acid.
HPO3 + H2O H3PO4
5. Write the uses of ortho phosphoric acid
1. H3PO4 is used in the preparation of HBr and HI as a substitute for sulphuric acid.
2. It is used as souring agent in the preparation of soft drinks.
3. It is used in the preparation of phosphate salts of sodium, potassium and ammonium.
4. It is used in the manufacture of phosphatic fertilizers.
6. Why H2O is a liquid while H2S is a gas?
In water hydrogen is connected to more electronegative oxygen atom. So, water molecules form inter
molecular hydrogen bonding and hence water is a liquid.
In H2S hydrogen is connected to less electronegative sulphur atom. So, H2S molecules do not form inter
molecular hydrogen bonding and hence they exist as discrete molecules. Thus H2S is a gas.
7. Fluorine atom is more electronegative than iodine atom yet, HF has lower acid strength than HI.
Explain? HF is associated due to the inter molecular hydrogen bonding. So, it cannot give H
+ readily and
therefore hydrofluoric acid is a weak acid whereas the other hydrohalic acids are strong acids.
...... H– F...... H– F..... H–F.
The acidic character of HX are in the following order:
HF < HCl < HBr < H
8. Why is HF a liquid while other hydrogen halides are gases?
Except HF, all hydrogen halides are gases. HF is a liquid because of inter molecular hydrogen bonding.
H – F ....... H –F ....... H–F ....... H–F
9. What are freons? Write their uses.
Chloro fluoro carbon compounds are known as freons. These non-toxic, non-combustible and volatile
liquids are used as refrigerants in refrigerators, deep freezers and air conditioners. The most common,
freon is known as dichlorodifluoro methane CF2 Cl2
10. What are Chalcogens? Why are they called so?
The first four elements of the group 16 [oxygen, sulphur, selenium, tellurium] are collectively called
Chalcogens or ore forming elements, because many metal ores occur as oxides and sulphides.
11. Why is He-O2 mixture used by deep sea divers?
Helium oxygen mixture is used by deep-sea divers in preference to nitrogen oxygen mixtures. It is much
less soluble in blood than N2. This prevents “bends” which is the pain caused by formation of nitrogen
bubbles in blood veins when a diver comes to the surface.
12. Why is helium used in balloons?
Because of its lightness and non-inflammability helium is used to filling balloons for meteorological
observations.
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5 MARK
1. Write a note on xenon fluoride compounds
Xenon forms three binary Fluorides XeF2, XeF4, and XeF6 by the direct union of elements under
appropriate experimental conditions.
Xe + F2 K673
XeF2
Xe + 2F2 K673
XeF4
Xe + 3F2 K573
XeF6
Properties:
XeF2, XeF4 and XeF6 are colourless crystalline solids subliming readily at 298K. They are powerful
fluorinating agents. They are readily hydrolysed by even traces of water. For example.
2Xe F2 + 2 H2O 2Xe + 4HF + O2
Structure:
The structure of the three xenon fluorides can be deduced from VSEPR theory. XeF2 and XeF4 have the
linear and square planar structure respectively. XeF6, has 7 electron pairs (6 bonding pairs and one lone
pair) and thus have a distorted octahedral structure in the gas phase.
2. Write a note on interhalogen compounds or interhalogens
Each halogen combines with another halogen to form several compounds known as interhalogen
compounds. The less electronegative element is written first. In naming also, the less electronegative
element is mentioned first. They are divided into four types.
AX AX3 AX5 AX7
ClF
BrF
BrCl
ICl
IBr
ClF3
BrF3
ICl3
BrF5
IF5
IF7
They are prepared by direct combination or by the action of a halogen on a lower interhalogen, the
product formed depends on the conditions.
Cl2+ F2 (equal volume) K473
2ClF (AX type)
2 + Cl2 liquid (equi molar) Cl (AX type)
Cl2 + 3F2 (excess) K573
2 ClF3 (AX3 type)
Br2 + 3F2 (diluted with nitrogen) 2Br F3(AX3 type)
Br2 + 5F2 (excess) BrF5 (AX5 Type)
2 + 5F2 (Excess) 2IF5 (AX5 Type)
F5 + F2 (Excess) K573
F7 (AX7 Type)
The bonds are essentially covalent because of the small electronegativity difference.
The interhalogens are generally more reactive than the halogens (except F) because the A-X bond is
weaker than the X–X bond in the halogens.
3. Explain the structure of AX3 type interhalogen compounds
Type AX3 (ClF3):
Compounds of the type AX3 have trigonal bipyramidal structure,
Trigonal Bipyramidal structure arises out of sp3d hybridisation involved in the formation of this
compound. The three dotted arrows indicate electrons contributed by the three fluorine atoms (without
lone pair it is T-shaped).
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3s 3p 3d
sp3d Hybridisation
Atom in Ground state
Atom in Excited state
Only one unpaired
electron permits one
covalent bond
Three unpaired
electrons permit
three covalent bonds
Formation of ClF3
molecule by sp3d
hybridisation
Cl
F
F
F
Trigonal Bipyramidal
Hybridisation sp3d
Number of bond pair of electrons 3
Number of lone pair of electrons 2
Shape with lone pair Trigonal bipyramidal
Shape without lone pair T-shaped
4. Explain the structure of AX7 type interhalogen compounds
This compound has a pentagonalbipyramidal structure since this is formed by sp3d
3 hybridisation.
5s 5p 5d
sp3d3 Hybridisation
Iodine Atom in Ground state
Only one unpaired
electron permits one
covalent bond
Seven unpaired
electrons permit
seven covalent bonds
Formation of IF7
molecule by sp3d3
hybridisation
Iodine Atom in Excited state
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I
F
F
F
F
F
F
F
Pentagonalbipyramidal
Hybridisation sp3d
3
Number of bond pair of electrons 7
Number of lone pair of electrons 0
Shape Pentagonal bipyramidal
5. What are silicones? How are they prepared?
1. Silicones are a group of organosilicon polymers.
2. The hydrolysis of alkyl-substituted chlorosilanes produce long-chain polymers called silicones.
SiCl Cl
R
R
H2O
SiOH OH
R
R
SiOH OH
R
R
SiOH OH
R
R
+ SiOH O
R
R
Si OH
R
R
3. The dialkyldichlorosilane R2SiCl2 on hydrolysis gives rise to straight chain polymers and, since an
active OH group is left at each end of the chain, polymerisation continues and the chain increases in
length.
Si O
R
R
Si OH
R
R
SiOH O
R
R
Si O
R
R
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4. The hydrolysis of trialkylmonochlorosilane R3SiCl yields hexaalkylsiloxane.
Si O
R
R
Si
R
R
RR
5. The hydrolysis of alkyl tricholorosilane RSiCl3 gives a very complex crosslinked polymer.
Si O
O
Si RR
O
O
O
Si O
O
Si RR
O
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4. d-BLOCK ELEMENTS
Blue print 1 Mark = 2 3 Mark = 2 5 Mark = 1+1* Total marks = 18
Govt. Examination questions 1 MARK
1. The general electronic configuration of d-block elements is [Sep-2006, Sep-2011]
a) (n-1)d1-10
ns0-2
b) (n-1) d1-5
ns2
c) (n-1)d0 ns
1 d) None of these
correct E.C is (n-1)d1-10
ns1-2
2. Formation of coloured ions is possible when compounds contains [Sep-2010, Sep-2016]
a) paired electrons b) unpaired electrons c) lone pairs of electrons d) none of the above
3. The electronic configuration of chromium is [Sep-2008, June-2009, June-2010, Mar-2014]
a) [Ar] 3d6 4s
0 b) [Ar] 3d
5 4s
1 c) [Ar] 3d
4 4s
2 d) [Ar] 3d
3 4s
2 4p
1
4. Para magnetism is the property of [Mar-2008, Mar-2012, Mar-2013, June-2013, Sep-2015]
a) paired electrons b) completely filled electronic subshells
c) unpaired electrons d) completely vacant electronic subshells
5. The correct electronic configuration of copper atom is [Sep-2009]
a) [Ar] 3d10
4s1
b) [Ar] 3d10
4s2
c) [Ar] 3d9 4s
2 d) [Ar] 3d
5 4s
2 4p
4
6. Silver salt used in photography is [June-2011, Sep-2013, Mar-2016]
a) AgCl b) AgNO3 c) AgF d) AgBr
7. Which of the following compounds will not give positive chromyl chloride test? [Mar-06, June-13, Mar-15]
a) CuCl2 b) HgCl2 c) ZnCl2 d) C6H5Cl
8. Which of the ions will give colourless aqueous solution? [Mar-2009, Mar-2012, Mar-2014]
a) Ni2+
b) Fe2+
c) Cu2+
d) Cu+
9. Which of the following compounds is not coloured? [Jun-2012, Jun-2016]
a) Na2CuCl4 b) Na2CdI4 c) K4 [Fe(CN)6] d) K3 [Fe(CN)6]
10. Choose the wrong statement regarding K2Cr2O7 [June-2008]
a) It is a powerful oxidizing agent b) It is used in tanning industry
c) It is soluble in water d) It reduces ferric sulphate to ferrous sulphate
11. Which compound is formed when excess of KCN is added to an aqueous solution of copper sulphate?
[Mar-2011, Sep-2012, Mar-17]
a) Cu2(CN)2 b) K2[Cu(CN)6] c) K[Cu(CN)2] d) Cu2(CN)2 + (CN)2
12. Which of the following has the maximum number of unpaired electrons? [Mar-2007, June-2010]
a) Mn2+
b) Ti3+
c) V3+
d) Fe2+
13. The chemical composition of slag formed during the smelting process in the extraction of copper is
a) Cu2O + FeS b) FeSiO3 c) CuFeS2 d) Cu2S + FeO [June-2006]
14. The transition element with the lowest atomic number is [Mar-2006, June-2007]
a) Scandium b) Titanium c) Zinc d) Lanthanum
15. Which transition element show highest oxidation state? [June-06, Mar-10, June-12, June-14, Sep-15]
a) Sc b) Ti c) Os d) Zn
16. The metal used in galvanizing iron sheets is [Sep-2006]
a) chromium b) zinc c) copper d) silver
17. Bordeaux mixture contains -------- [Mar-2007, June-2011, June-2014]
a) AgNO3 + HNO3 b) ZnSO4 + H2SO4 c) CuSO4 + Ca(OH)2 d) KMnO4 + HCl
18. K2Cr2O7 reacts with KI and dil.H2SO4 and liberates [June-2007]
a) O2 b) I2 c) H2 d) SO2
19. The colour of purple of Cassius is ------- [Sep-2007]
a) purple b) blue c) bluish green d) apple green
20. Silver obtained from silver coin is purified by fusion with [Mar-2008]
a) AgNO3 b) HNO3 c) H2SO4 d) borax
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21. The number of unpaired electrons in Ti3+
is 1.Its magnetic moment in BM is [June-2008]
a) 1.414 b) 2 c) 1.732 d) 3
22. The catalyst used in the manufacture of polythene is [Sep-2008]
a) V2O5 b) Fe c) Mo d) TiCl4
Process Catalyst
Synthesis of NH3 by Haber‟s process Fe = catalyst Mo = promoter
Oxydation of SO2 to SO3 V2O5
Manufacture of polythene TiCl4
23. A metal which precipitates gold from its aurocyanide complex is [Mar-2009]
a) Cr b) Ag c) Pt d) Zn
24. The reagent which is added first in the separation of silver from silver coin is [June-2009]
a) Conc.H2SO4 b) Conc.HCl c) Conc.HNO3 d) Aquaregia
25. The substance used in making ruby red glass and high class pottery is [Sep-2009]
a) colloidal silver b) purple of Cassius c) ruby silver d) ruby copper
26. Spitting of silver can be prevented by covering the molten metal with a layer of [Mar-2010]
a) charcoal b) sand c) silver bromide d) borax
27. If the magnetic moment value is 5.92 M, the number of unpaired electron is [Sep-2010]
a) 5
b) 3
c) 4
d) 6
28. Which one of the following has the maximum magnetic moment? [Mar-2011]
a) 3d2
b) 3d6
c) 3d7
d) 3d9
29. Which of the following pairs have almost equal radii? [Sep-2011]
a) Mo, W
b) Y, La
c) Zr, Hf
d) Nb, Ta
30. The most malleable and ductile of all metal is [Sep-2012]
a) silver
b) gold
c) copper
d) zinc
31. Ferrochrome is an alloy of ------ [Sep-2007]
a) Cr,C,Fe,Ni b) Cr,Co,Ni,C c) Fe,Cr d) Cr,Ni,Fe
32. The metals present in Nichrome alloy [Mar-2013]
a) Cr, Ni, Fe
b) Cr, Co, Ni
c) Cr, Fe
d) Cr, Fe, Cu
Alloys Composition
Brass Cu + Zn
Bronze Cu + Sn
Gun metal Cu + Sn + Zn
Ferrochrome Cr + Fe
Nichrome Cr + Ni + Fe
Stainless steel Cr + Fe + Ni + C
Stellite Cr + Co + Ni + C
33. A „d‟ block metal ion has a magnetic moment of 1.732 BM. The number of unpaired electrons is
a) 1 b) 2 c) 3 d) 4 [Sep-2013]
34. Purity of blister copper is [Sep-2014]
a) 100 % b) 98 % c) 90 % d) 12 %
35. Excess of sodium hydroxide reacts with zinc to form [Sep-2014, June-2015]
a) Zn - H2 b) Na2ZnO2 c) ZnO d) Zn(OH)2 36. The alloy used in the manufacture of resistance wires is [Mar-2015]
a) Ferrochrome b) Bronze c) Nichrome d) Stellite 37. In metal Carbonyls the oxidation state oI the metal is [June-2015]
a) 0 b) +2 c) + 4 d) +6
38. Pick out the colourless ion of transition metal from the following [Mar-2016]
a) Zn2+
b) Cu2+
c) Fe3+
d) Mn2+
39. The incomplete series is [Jun-2016]
a) 3d-series
b) 4d-series
c) 6d-series
d) 5d-series
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40. Which silver salt is used in hair dye? [Sep2016]
a) AgCl
b) AgBr
c) AgNO3
d) AgI 41. The transition element used for making calorimeters is [Mar-2017]
a) Cr b) Ni c) Zn d) Cu
3 MARK
1. Why are transition metal ions coloured? [Sep-2009, Sep-2011]
1. The colour of transition metal ions is due to the presence of unpaired electrons in it and the energy
gap between two energy levels in the same d-subshell being small.
2. Hence very small amount of energy is required for excitation of electrons from one energy level to
the other. The energy can be easily provided by the visible light.
3. The electron absorbs visible light of a particular wave length and transmits the rest of the light. The
transmitted light has a colour complementary to the absorbed colour.
2. Why Zn2+
salts are white while Ni2+
salts are coloured? [June-2009, Sep-2012, June-2013]
E.C of Zn [Ar] 3d10
4s2
E.C of Zn2+
[Ar] 3d10
(completely filled d-orbital)
Zn2+
ion is colourless because of the absence of vacant d-orbitals to which electrons can be excited.
E.C of Ni [Ar] 3d8 4s
2 E.C of Ni
2+ [Ar] 3d
8
Ni2+
has 3d8 configuration and therefore excitation of electron from lower energy d-orbital to higher
energy d-orbital is possible. The energy can be easily provided by the visible light. So, Ni2+
salts are
coloured.
3. Explain why d-block elements exhibit variable oxidation states?
[Sep-2006, Mar-2007, June-2007, Mar-2009, Mar-2011, Mar-2013, June-2014, June-2015]
Transition elements exhibit variable oxidation states (or) variable valencies in their compounds. This
property is due to the following reasons.
1. These elements have several ( n – 1 ) d and ns electrons.
2. The energies of ( n – 1 ) d and ns orbitals are fairly close to each other.
4. Why do transition elements form complexes? [June-2006, Sep-2007, June-2008, June-2009, Mar-
2010, June-2010, Sep-2010, Mar-2012, Mar-2014, Mar-2015]
Transition elements form complexes due to,
1. Small size and high positive charge density.
2. Presence of vacant ( n – 1 ) d orbitals which are of appropriate energy to accept lone pair and
unshared pair of electrons from the ligands for bonding with them.
5. Why do transition metals form alloys? [Sep-2013, Mar-17]
Transition metals form alloys with each other. This is because they have almost similar size and the
atoms of one metal can easily take up positions in the crystal lattice of the other.
Eg. Alloys of Cr-Ni, Cr-Ni-Fe
6. Why most of the transition metals and their compounds are used as catalyst? [Sep-2014, Sep-2015]
The catalytic activity of transition metals is due to the following reasons: 1. They show a variety of oxidation states and thereby can form intermediate products with various reactants. 2. They are capable of forming interstitial compounds which can adsorb and activate the reacting species.
7. A substance is found to have a magnetic moment of 3.9 BM. How many unpaired electrons does it
contain? [Mar-2006]
µ = )2n(n BM
3.9 = )2n(n
( 3.9 )2 = n ( n + 2 )
n ( n + 2 ) = 15
n = 3
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8. Explain why Mn2+
is more stable than Mn3+
? [Mar-2006, June-2012]
E.C of Mn2+
[Ar] 3d5
E.C of Mn3+
[Ar] 3d4
Mn2+
has half-filled d-orbital. Half-filled d-orbital is more stable due to symmetry. So, Mn2+
is more
stable than Mn3+
.
9. What happens when KI solution is added to an aqueous solution of copper sulphate? [Sep-2009]
When KI is added to a solution of CuSO4, a white precipitate of cuprous iodide is produced.
CuSO4 + 2KI Cu I2 + K2SO4
( unstable )
2CuI2 Cu2I2 + I2
( white ppt )
10. What is the action of heat on copper sulphate crystal?
[Mar-2008, Sep-2010, June-2012, Sep-2014, June-2015]
On heating CuSO4.5H2O loses its water of crystallization and decomposes at 720°C to give cupric oxide
and sulphur trioxide.
CuSO4 . 5H
2O CuSO
4 . H
2O CuSO
4 CuO + SO
3
100oC 230oC 720oC
4 H2O H
2O
(Blue) (White)
11. What is philosopher’s wool? How is it formed? [Sep-2013, Jun-2016]
When heated in air at 773 K, Zinc burns to form a white cloud of Zinc oxide which settles to form a
wooly flock called philosopher‟s wool.
2Zn + O2 2ZnO
12. Explain chrome plating. [Sep-2006 , Mar-2008, Mar-2010, Mar-17]
Anode = plate of lead
Cathode = article to be plated
Electrolyte = chromic acid + sulphuric acid
During electrolysis chromium deposits on the article.
Generally the articles are first plated with nickel and then subjected to chromium plating.
13. What is spitting of silver? How is it prevented?
[Sep-2007, Sep-2008, June-2010, June-2011, Mar-2013, Mar-2015]
Molten silver absorbs about twenty times its volume of oxygen which it again expels on cooling.
Globules of molten silver are thrown off. This is called “spitting of silver”.
This can be prevented by covering the molten metal with a layer of charcoal.
14. What is aqua regia? Give the reaction of gold with aqua regia [Mar-07, Mar-11, June-11, Mar-14]
Aqua regia is the mixture of 3 parts of con.HCl + 1 part of con.HNO3
Gold dissolves in aquaregia to form auric chloride.
2Au + 9 HCl + 3HNO3 2AuCl3 + 6H2O + 3NOCl
Auric chloride Nitrosyl chloride
15. Show that K2Cr2O7 is a powerful oxidizing agent [June-2006]
1. K2Cr2O7 oxidises ferrous sulphate to ferric sulphate
K2Cr2O7 + 7H2SO4 + 6 FeSO4 K2SO4 + Cr2 (SO4)3 + 3 Fe2 (SO4)3 + 2H2O
2. K2Cr2O7 oxidises H2S to sulphur
K2Cr2O7 + 4H2SO4 + 3H2S K2SO4 + Cr2 (SO4)3 + 7H2O + 3S
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16. Write a note about chromyl chloride test [Mar-2009, Mar-2012, June-2014]
When salt containing chloride is treated with K2Cr2O7 and con. H2SO4 orange red vapours of chromyl
chloride are obtained.
K2Cr2O7 + 4KCl + 6H2SO4 2CrO2Cl2 + 6 KHSO4 + 3H2O
Chromylchloride
This reaction is used in the detection of chloride ions in qualitative analysis.
17. What is lunar caustic? How is it prepared? [Sep-2012]
Silver nitrate is known as Lunar caustic. It is prepared by dissolving silver in dil.HNO3
3Ag + 4HNO3 3AgNO3 + 2H2O + NO↑
18. What is the action of heat on silver nitrate? [June-2013]
On heating AgNO3 decomposes in two stages
2AgNO3 K723
2AgNO2 + O2
Silvernitrite
AgNO2 K980
Ag + NO2
19. What is the action of Zn on hot NaOH solution? [Sep-2008]
Zinc dissolves in hot NaOH solution forming sodium zincate.
Zn + 2NaOH Na2ZnO2 + H2
Sodium zincate
20. What is the reaction of CuSO4 with KCN? [June-2008]
A yellow precipitate of cupric cyanide is first formed with KCN and it decomposes to give cyanogen gas.
CuSO4 + 2KCN Cu(CN)2 + K2SO4
2Cu(CN)2 Cu2(CN)2 + (CN)2
cyanogen
21. Explain electrolytic refining of copper. [June-2007]
Anode = impure copper
Cathode = pure copper
Electrolyte = CuSO4 solution + dil. H2SO4
When electric current is passed through the electrolytic solution pure copper get deposited on the
cathode, impurities settle near the anode in the form of sludge called anode mud.
22. What is purple of cassius? How is it prepared? Mention its use [Sep-2011]
Purple of cassius is a form of colloidal gold with stannic hydroxide.
Preparation:
It is prepared by mixing very dilute solution of gold chloride with stannous chloride solution.
2AuCl3 + 3 SnCl2 2Au↓+ 3SnCl4
The gold thus precipitated is adsorbed by stannic hydroxide formed by the hydrolysis of SnCl4.
SnCl4 + 4H2O Sn(OH)4 + 4HCl
Use: It is used in making ruby-red glass and high class pottery.
23. Write two alloys of copper and their uses [Sep-2015] Alloy Composition Uses
Brass Cu + Zn For making utensils, condenser tubes and wires
Bronze Cu + Sn For making cooking utensils, statues and coins
24. Give the percentage composition and use of Nichrome [Mar-2016, Sep-2016] Alloy Composition Use
Nichrome Ni = 60%
Cr = 15%
Fe = 25%
It is used in resistance wires for electrical
heating
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25. What is the action of heat on K2Cr2O7 ? [Mar-2016]
On heating, K2Cr2O7 decomposes to give potassium chromate with evolution of O2 gas.
4K2Cr2O7 4K2CrO4 + 2Cr2O3 + 3 O2
26. What is Bordeaux mixture? Mention its use [Jun-2016]
A mixture of copper sulphate and lime is known as Bordeaux mixture. It is used as fungicide.
27. Give the reaction of ammoniacal silver nitrate with formic acid. [Sep-2016]
Ammoniacal silver nitrate is Tollen‟s reagent. It is reduced to silver mirror by formic acid.
2AgNO3 + 2NH4OH Ag2O + 2NH4NO3 + H2O
Ag2O + HCOOH 2Ag↓+ H2O + CO2
5 MARK
1. Write note on aluminothermic process [Sep-2008, Mar-2009, Jun-2016, Sep-2016]
Cr2O3 is reduced to chromium metal by Aluminothermic process.
Cr2O3+Al
BaO2+Mg powder
Fireclay crucible
Sand
Magnesium ribbon
Chromic oxide is mixed with powdered Aluminium in the ratio 3:1 and is placed in a large fire clay
crucible.
A mixture of barium peroxide and Mg powder is placed over this.
The crucible is surrounded by sand which prevents loss of heat by radiation.
The mixture is ignited by a piece of Mg ribbon.
During this process a large amount of heat is liberated, in which Cr2O3 is reduced to chromium.
The molten chromium is collected in the crucible and aluminium oxide is removed as slag.
Cr2O3 + 2Al 2Cr + Al2O3 + 468.6 kJ
2. How is zinc extracted? [Mar-06, Jun-09, Sep-10, Mar-11, June-11, June-12, Mar-14, Sep-15, Mar-16]
1. Ore: Zinc blende, ZnS
2. Concentration
The ore is crushed and then concentrated by froth-floatation process.
3. Roasting
The concentrated ore is then roasted in the presence of excess of air at about 1200 K.
2 ZnS + 3O2
2ZnO + 2SO2
4. Reduction
Zinc oxide is mixed with powdered coke and heated to 1673 K in a fire clay retort, in which ZnO is
reduced to zinc metal
ZnO + C K1673
Zn + CO
5. Electrolytic refining
Anode : Impure zinc
Cathode : Thin sheet of pure zinc
Electrolyte : Zinc sulphate acidified with dil.H2SO4
On passing electric current, pure zinc get deposited at the cathode.
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3. How is silver extracted from its chief ore?
[June-2006, Sep-2007, June-2008, Mar-2012, Mar-2013, Sep-2014, June-2015, Mar-17]
1. Ore: Argentite Ag2S
Silver is extracted from the argentite ore by the Mac-Arthur and Forrest’s cyanide process.
2. Concentration: The crushed ore is concentrated by froth-floatation process
3. Treatment of the ore with NaCN
The concentrated ore is treated with 0.4-0.6% solution of sodium cyanide for several hours. The
mixture is continuously agitated by a current of air, so that Ag present in the ore is converted into
soluble sodium argento complex.
Ag2S + 4NaCN ⇌ 2Na [Ag(CN)2] + Na2S
Sodium argento cyanide (soluble)
4. Precipitation of silver
The solution containing sodium argento cyanide is filtered to remove insoluble impurities and
filtrate is treated with zinc dust, silver gets precipitated.
2Na [Ag(CN)2] + Zn Na2[Zn(CN)4] + 2Ag ↓
5. Electrolytic refining
Anode : Impure silver
Cathode : Thin sheet of pure silver
Electrolyte : Silver nitrate acidified with 1% nitric acid
On passing electricity pure silver gets deposited at the cathode.
4. How is gold extracted?[Sep-06, Jun-07, Mar-08, Sep-09, Jun-10, Sep-12, Jun-13, Sep-13, Jun-14, Mar-15]
1. Ore: Alluvial sand or gravel
Gold is extracted from sulphide ore by the Mac-Arthur Forrest cyanide process.
2. Crushing, grinding and concentration
The gold ore is crushed and powdered, and then concentrated by froth floatation process.
3. Roasting
The concentrated ore is roasted in excess of air or oxygen when impurities of sulphur, arsenic and
tellurium are oxidised and escape as their volatile oxides.
4. Treatment with KCN
The finely powdered roasted ore is then treated with a dilute (0.5%) solution of KCN in presence of
excess of air for 24 hours. As a result, gold dissolves in KCN to form a soluble complex called
potassium aurocyanide.
4Au + 8KCN + 2H2O + O2 4K[Au(CN)2] + 4KOH
Pot. aurocyanide
5. Precipitation of gold
The above solution containing potassium aurocyanide is filtered to remove insoluble impurities and
then treated with zinc dust. Gold being less electropositive than zinc, it gets precipitated.
2K[Au(CN)2] + Zn K2[Zn(CN)4] + 2 Au ↓
The precipitated gold is recovered by filtration .It is further purified by electrorefining.
6. Electrolytic refining
Anode : Impure gold
Cathode : Thin sheet of pure gold
Electrolyte : Auric chloride (AuCl3) acidified with 10-12% HCl
On passing current pure gold gets deposited on the cathode.
5. How is potassium dichromate prepared from chrome iron ore? [Mar-2007, Mar-2010, Sep-2011]
1. Conversion of chrome iron ore to sodium chromate
The powdered ore is mixed with Na2CO3 and quick lime, and then roasted in a reverberatory
furnace with free expose to air.
4FeO.Cr2O3 + 8Na2CO3 + 7O2 8Na2CrO4 + 2Fe2O3 + 8CO2 ↑
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2. Conversion of Na2CrO4 to Na2Cr2O7
Sodium chromate solution so obtained is filtered and treated with con.H2SO4, when sodium
chromate is converted to sodium dichromate.
2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O
3. Conversion of sodium dichromate into potassium dichromate
Hot concentrated solution of sodium dichromate is treated with KCl, when potassium dichromate,
being much less soluble than sodium salt, crystallizes out on cooling.
Na2Cr2O7 + 2KCl K2Cr2O7 + 2NaCl
Additional questions
1 MARK 1. Paramagnetism is common in
a) p-block elements b) d-block elements c) s-block elements d) f-block elements
2. The colour of [Ti(H2O)6]3+
ion is due to
a) d-d transition b) Presence of water molecules
c) Inter atomic transfer of electrons d) None of the above
3. d-block elements form coloured ions because
a) they absorb some energy for d-s transition b) they absorb some energy for p-d transition
c) they absorb some energy for d-d transition d) they do not absorb any energy
4. Copper is extracted from
a) cuprite b) copper glance c) malachite d) copper pyrites
5. Sodiumthiosulphate is used in photography because of its
a) Oxidizing behaviour b) Reducing behaviour
c) Complexing behaviour d) Photochemical behaviour
6. In the extraction of Cu, the reaction which does not take place in Bessemer converter is
a) 2CuFeS2 + O2 Cu2S + FeS + SO2 b) 2Cu2S + 3O2 2Cu2O + 2SO2
c) 2Cu2O + Cu2S 6Cu + SO2 d) 2FeS + 3O2 2FeO + 2SO2
7. Select the wrong statement
a) All cuprous salts are blue in colour b) Transition metals are highly reactive
c) All cuprous salts are white in colour d) Mercury is a liquid metal
8. For a transition metal ion, the effective magnetic moment in BM is given by the formula
a) 1)(nn b) 1)(nn c) 2)(nn d) 2)(n1)(nn
9. The correct statement in respect of d-block elements is
a) They are all metals b) They show variable valency.
c) They form coloured ions and complex salts d) All above statement are correct
10. Among the following statement, the incorrect one is
a) Calamine and siderite are carbonates b) Argentite and cuprite are oxides
c) Zinc blende and pyrites are sulphides d) Malachite and azurite are ores of copper
11. In d-block elements the last electron enters the d-orbital of -------- shell
a) penultimate b) outermost c) valence d) (n-2)
12. Zirconium and Hafnium have almost equal atomic radii due to --------
a) lanthanide contraction b) actinide contraction c) unpaired electrons d) all the above
13. Sc3+
and Ti4+
are colourless because they have
a) completely filled d-orbital b) empty d-orbital c) d5
d ) none of these
14. Zn2+
and Cu+ are colourless because they have
a) completely filled d-orbital b) empty d-orbital c) d5
d) none of these
15. In the first transition series ------- exhibits maximum number of oxidation states (+2 to +7)
a) Mn b) Fe c) Zn d) Cr
16. The transition element in lower oxidation states form
a) ionic compounds b) covalent compounds c) explosive compounds d) none of these
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17. The transition element in higher oxidation states form
a) ionic compounds b) covalent compounds c) explosive compounds d) none of these
18. The highest oxidation state (+8) is shown by
a) ruthenium Ru b) osmium Os c) both a and b d) Cr
19. Oxidation number of Ni in Ni(CO)4 is
a) +4 b) –4 c) 0 d) +2
20. Oxidation number of Fe in Fe(CO)5 is
a) +5 b) –5 c) 0 d) +2
21. Froth floatation is employed for the concentration of
a) sulphide ores b) oxide ores c) both d) none
22. Gravity separation method is employed for the concentration of
a) sulphide ores b) oxide ores c) both d) none
23. In the smelting of copper ore, the Matte formed in the lower layer is
a) Cu2S + FeS b) Cu2O + FeO c) CuS + FeO d) Cu2S + FeO
24. The impure copper (98% pure or 2% impure) is called
a) blister copper b) matte c) Bordeaux d) none
25. When Cu is heated with O2 below 1370 K ------ is formed
a) cuprous oxide b) cupric oxide CuO c) CuCO3 d) none
26. When Cu is heated with O2 above 1370 K ------ is formed
a) cuprous oxide Cu2O b) cupric oxide c) CuCO3 d) none
27. The metal that does not react with dil. HNO3 is
a) Cr b) Zn c) Ag d) Cu
28. Which acid renders chromium passive (inactive) ?
a) HCl b) con. HNO3 c) con. H2SO4 d) dil.H2SO4
29. The metal which is not attacked by alkali (NaOH) is
a) Cu b) Ag c) Au d) all the above
30. When Zn reacts with O2, white cloud of ZnO settles to form a wooly flock called ----------
a) calamine b) Philosopher’s wool c) lithopone d) none
31. Gold dissolves in a) H2SO4 b) HCl c) aqua regia d) none
Aqua regia: 3 parts of con.HCl + 1 part of con.HNO3
32. Cr2O3 is reduced to Cr by
a) Aluminothermic process b) Mond‟s process c) Contact process d) none
33. In aluminothermic process, the fire clay crucible is surrounded by sand to
a) reduce alumina b) reduce Cr2O3 c) prevent loss of heat by radiation d) none
34. Which is true about aluminothermic process?
a) Al is reducing agent b) Al2O3 is slag c) exothermic d) all the above
35. Ignition mixture used in aluminothermic process is
a) BaO2 + Mg b) Al2O3 c) Cr2O3 d) none
36. Mac-Arthur & Forrest‟s cyanide process is employed in the extraction of
a) Ag b) Au c) both a and b d) none
37. Silver coins are alloys of
a) Ag + Au b) Ag + Cu c) Ag + Cr d) Ag + Fe
38. Silver obtained from coins is purified by fusion with
a) Na2CO3 b) borax c) C d) NaHCO3
39. Molten silver absorbs 20 times its volume of
a) CO2 b) O2 c) N2 d) H2
40. Which is used in photography?
a) AgBr b) K2Cr2O7 c) both a and b d) none
41. Gold which is mixed with quartz in ancient rocks is known as
a) Vein gold b) placer gold c) blister gold d) none
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42. Gold which is scattered in sand and gravel is known as
a) Vein gold b) placer gold c) blister gold d) none
43. Pure gold is a) 24 carats b) 12 carats c) 18 carats d) 100 carats
44. Ornaments are made of 18 carats gold. It contains
a) 18 parts by weight of Cu alloyed with 6 parts by weight of Au
b) 18 parts by weight of Au alloyed with 6 parts by weight of Cu
c) 24 parts by weight of Cu alloyed with 18 parts by weight of Au
d) 24 parts by weight of Au alloyed with 18 parts by weight of Cu
45. Tollen‟s reagent is
a) basic copper sulphate b) ammoniacal silver nitrate c) ZnCO3 d) none
46. Ammoniacal silver nitrate is reduced to silver mirror by
a) formic acid b) formaldehyde c) glucose d) all the above
47. Purple of cassius is a form of
a) colloidal silver b) colloidal gold c) colloidal starch d) colloidal sulphur
48. Which of the following is used for making surgical instruments?
a) nichrome b) ferrochrome c) stainless steel d) Stellite
Alloys Use
Bronze Statues
Ferrochrome Burglar proof safe
Nichrome Resistance wires
Stainless steel Cutlery
Chrome nickel steel armour plates
49. Chromium salts are used in
a) mordants b) colourd glass c) pottery d) all the above
50. Which is used for dental filling?
a) Silver amalgam b) zinc amalgam c) Blue vitriol d) Lunar caustic
51. Lunar caustic is
a) CuSO4.5H2O b) K2Cr2O7 c) ZnCO3 d) AgNO3
52. Which of the following is used in making ointment for curing skin diseases?
a) zinc carbonate b) silver nitrate c) copper sulphate d) purple of Cassius
K2Cr2O7 Calico printing, dyeing, photography
CuSO4.5H2O
Blue vitriol
Germicide and insecticide in agriculture, Calico printing
Bordeaux mixture (CuSO4+lime): fungicide
AgNO3 Lunar caustic Marking inks and hair dyes, Silvering of mirrors
ZnCO3 Calamine Ointment for curing skin diseases, Cosmetics, Pigment for rubber
Purple of Cassius Ruby-red glass and high class pottery
3 MARK
1. Why d-block elements are named as transition elements?
The d-block elements are called transition elements because these represent a transition from highly
electropositive elements (metals) of s-block to least electropositive elements (non-metals) of p-block.
2. Zn, Cd, Hg do not form coloured compounds. Why?
E.C of Zn, Cd, Hg : d10
s2 (completely filled d-orbitals)
Zinc, Cadmium and Mercury do not form coloured compounds because of the absence of vacant d-
orbitals to which electrons can be excited.
3. Sc3+
and Zn2+
ions are colourless. Why?
E.C of Sc [Ar] 3d1 4s
2 E.C of Sc
3+ [Ar] 3d
0
Sc3+
ion is colourless because of the absence of d-electrons.
E.C of Zn [Ar] 3d10
4s2
E.C of Zn2+
[Ar] 3d10
(completely filled d-orbital)
Zn2+
ion is colourless because of the absence of vacant d-orbitals to which electrons can be excited.
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4. [Ti (H2O)6]3+
is coloured while [Sc (H2O)6]3+
is colourless. Explain.
E.C of Ti 3d2 4s
2 E.C of Ti
3+ 3d
1
Ti3+
has 3d1 configuration and therefore excitation of electron from lower energy d-orbital to higher
energy d-orbital is possible. The energy can be easily provided by the visible light. So, Ti3+
is coloured.
E.C of Sc 3d1 4s
2 E.C of Sc
3+ 3d
0 (empty d-orbital)
Sc3+
has 3d0 configuration. So, it is colourless because of the absence of d-electrons.
5. Why does Mn(II) show maximum paramagnetic character among the bivalent ions of the first
transition series?
E.C of Mn [Ar] 3d5 4s
2 E.C of Mn
2+ [Ar] 3d
5
Mn2+
has maximum number of unpaired electrons (5). So, Mn(II) shows maximum paramagnetic
character.
6. What is the action of heat on copper in the presence of oxygen?
2Cu + O2 Below1370K
2CuO
Black cupric oxide
4Cu + O2 Above1370K
2Cu2O
Red cuprous oxide
7. Write a note about fineness of gold
Gold content of an alloy of gold is termed its fineness and is generally expressed in carats. Pure gold is
24 carats. Ornaments are made of 18 carat gold, which is an alloy containing 18 parts by weight of gold
alloyed with 6 parts by weight of copper.
8. What is blue vitriol? How is it prepared?
Copper Sulphate penta hydrate (CuSO4.5H2O) is called as Blue vitriol. In laboratory it is prepared by
dissolving cupric oxide (or) cupric hydroxide (or) cupric carbonate in dilute H2SO4
CuO + H2SO4 CuSO4 + H2O
Cu(OH)2 + H2SO4 CuSO4 + 2H2O
CuCO3 + H2SO4 CuSO4 + H2O + CO2↑
9. How is silver extracted from silver coins?
Silver coins are Ag-Cu alloys. Ag can be obtained from such an alloy by dissolving the alloy in
concentrated HNO3, a solution containing mixture of AgNO3 and Cu(NO3)2 is obtained. The solution is
boiled to expel excess of nitric acid, then the solution is treated with con.HCl, silver is precipitated as
AgCl. AgCl is separated and converted to silver by fusing with excess Na2CO3.
4AgCl + 2Na2CO3 4Ag↓+ 4NaCl + 2CO2 + O2
The silver thus obtained, is purified by fusion with borax and followed by electrolytic purification.
5 MARK 1. How is zinc carbonate prepared? What is the action of heat on it? Write its uses
Zinc carbonate occurs in nature as calamine.
Preparation: When NaHCO3 is added to Zinc sulphate solution, ZnCO3 is obtained.
ZnSO4 + 2NaHCO3 ZnCO3 + Na2SO4 + H2O + CO2 ↑Action of Heat: On heating at 300°C , Zinc carbonate decomposes into Zinc oxide and carbon dioxide
ZnCO3 Co300
ZnO + CO2
Uses
i) It is used in making ointment for curing skin diseases.
ii) It is used in the preparation of cosmetics.
iii) It is used a pigment for rubber.
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5. f - BLOCK ELEMENTS
Blue print 1 Mark = 2 3 Mark = 0 5 Mark = 1 Total marks = 7
Govt. Examination questions 1 MARK
1. The most common oxidation state of lanthanides is Jun-2008, Sep-2011, Sep-2016]
a) +2 b) +1 c) +3 d) +4
2. The elements in which the extra electron enters (n-2) f orbitals are called [Mar-2006]
a) s- block elements b) p- block elements c) d- block elements d) f-block elements
3. The Lanthanide contraction is due to [Mar-2010, Mar-2011, Sep-2014]
a) Perfect shielding of 4f electron b) Imperfect shielding of 4f electron
c) Perfect shielding of 3d electron d) Imperfect shielding of 3d electron
4. Ceria is used in [June-2011, June-2014]
a) toys b) tracer bullets c) gas lamp materials d) none of the above
5. --------- is used in gas lamp material [Mar-07, June-07, Sep-10, Sep-12, June-13, Mar-15]
a) MnO2 b) CeO2 c) N2O5 d) Fe2O3
6. Alloys of Lanthanides are called as [Sep-2007, Mar-2008, June-2008, Sep-2009, June-2010, Mar-2011]
a) Mish-metals b) Metalloids c) Plate metals d) Actinides
7. ---------- form oxocations
[June-06, Sep-07, June-09, Sep-10, Mar-12, June-12, Mar-13, Sep-13, Sep-15, Mar-17]
a) Lanthanides b) Actinides c) Noble gases d) Alkali metals
8. Maximum oxidation state exhibited by Lanthanides is [Sep-2008, Mar-2009, Sep-2012, Mar-2013]
a) +1 b) +2 c) +3 d) +4 9. The isotope used as a power source in long mission space probes is [Mar-06, Sep-06, Mar-10, June-11]
a) U-235 b) Pu-235 c) Pu-238 d) U-238
10. The radioactive lanthanide is -------- [June-06, Mar-07, June-07, Mar-09, June-12, June-13, Sep-13]
a) terbium b) lutetium c) promethium (Pr) d) gadolinium
11. Among the lanthanide elements, with the increase in atomic number the tendency to act as reducing
agent [Sep-2006]
a) increases b) decreases c) no change d) none of these
12. The common oxidation state of actinide is --------- [Mar-2008, Mar-2014, Jun-2016]
a) +2 b) +3 c) +4 d) +6
13. ThO2 is used in --------- [Sep-2008]
a) toys b) tracer bullets c) gas lamp materials d) dyeing cotton
14. Oxidation state of uranium in UF6 and in UO2Cl2 is --------- [June-2009, June-2014, Mar-2016]
a) +4 b) +3 c) +6 d)+2
15. Which is fissionable? Or Which is used in nuclear fission reaction? [June-2010]
a) U-235 b) U-232 c) both a and b d) none of these
16. Which is used as fuel in nuclear power plants and as a component in nuclear weapons? [Sep-11, Mar-12]
a) U-235 b) U-232 c) Pr d) Lu
17. According to Fajan’s rule decrease in size of Ln3+
ion in Ln(OH)3 [Sep-2009]
a) increases the covalent character b) decreases the covalent character
c) increases the basic character d) increases the ionic character
18. Which of the following lanthanides have no partly filled 4f sub shell but have electrons in 5d sub shell?
a) Ce b) La & Lu c) Pm d) Nd [Mar-2014]
19. Important ore of Lanthanides is [Sep-2014]
a) Pitch blende b) Monozite c) Gypsum d) Chromite 20. The actinide contraction is due to [Mar-2015]
a) Perfect shielding of 5f electron b) Imperfect shielding of 4f electron
c) Imperfect shielding of 5f electron d) Perfect shielding of 4f electron
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21. The electronic configuration of Achnides is [June-2015]
a) [Rn] 5f 0 – 14
6d0 7s
0 b)
[Rn] 5f
0 – 14 6d
0– 2 7s
0 c)
[Rn] 5f
0 – 14 6d
0– 2 7s
1 d)
[Rn] 5f
0 – 14 6d
0– 2 7s
2
22. The elements in which the extra elechons enter (n –2) f -orbitals are called
a) s-block elements b) p-block elements c) d-block elements d) f-block elements
23. The major constituents of mish metals are [Sep-2015]
a) Ce, Nd, Th b) Th, Ac, Ce c) Ce, La, Nd d) Ce, Lr, U 24. Lanthanides are extracted from [Mar-2016, Jun-2016]
a) Limonite b) Monazite c) Magnetite d) Cassiterite
25. -------------- is used as fuel in nuclear power plants [Sep-2016]
a) U-236 b) Pu-238 c) U-235 d) Pu-235 26. Which of the following is used in making parts of jet engines? [Mar-2017]
a) Mg-alloys containing 30% mishmetal and 1% Zr b) Cu alloys containing1% mishmetal
c) Ni-alloy containing 12% mishmetal and 30% Zr d) none of the above
5 MARK 1. Discuss the position of lanthanides in the periodic table [March-2008, Sep-2013]
1. The 15 elements from La to Lu (lanthanum to lutetium) present in the 6th
period of the periodic table
are called lanthanides.
2. In lanthanides the last electron enters in the 4f orbital. So, they are called 4f block elements.
3. The electronic configuration of lanthanides is [Xe] 4f 1-14
5d1
6s2
4. These elements (La to Lu) closely resemble one another in properties as their outer two shells are
similarly occupied and therefore they are considered together.
5. Lanthanides are placed below the main body of the periodic table to give greater importance to
periodicity.
2. What is lanthanide contraction? Discuss its causes and consequences [Jun-06, Mar-07, Sep-07,
Mar-09, Sep-11, Mar-12, Jun-12,Sep-12, Mar-13, Jun-13,Jun-15, Sep-15,Mar-17]
The size of M3+
ions decreases as we move through the lanthanides from lanthanum to lutetium. This
steady decrease in ionic radii of M3+
cations in the lanthanide series is called Lanthanide contraction.
Cause of lanthanide contraction
The lanthanide contraction is due to the imperfect shielding of one 4f electron by another in the same
sub shell. As we move along the lanthanide series, the nuclear charge and the number of 4f electrons
increase by one unit at each step. However, due to imperfect shielding, the effective nuclear charge
increases causing a contraction in electron cloud of 4f-subshell.
Consequences of lanthanide contraction
1. Basicity of ions
Due to lanthanide contraction, the size of Ln3+
ions decreases regularly with increase in atomic number.
La3+Lu3+
> >.......................
According to Fajan’s rule, decrease in size of Ln
3+ ions increase the covalent character and
decreases the basic character between Ln3+
and OH– ion in Ln(OH)3.
La(OH)3 > ……………… > Lu(OH)3
More basic Less basic There is regular decrease in their ionic radii.
2. Regular decrease in their tendency to act as reducing agent, with increase in atomic number.
3. Due to lanthanide contraction, second and third rows of d-block transition elements are quite close
in properties.
4. Due to lanthanide contraction, these elements occur together in natural minerals and are difficult to
separate.
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3. Write any five differences between lanthanides and actinides
[March-2006, June-2008, June-2010, Mar-2011, Sep-2014, Mar-2016]
No Lanthanides Actinides
1 Binding energies of 4f electrons are higher. Binding energies of 5f electrons are lower.
2 4f electrons have greater shielding effect. 5f electrons have poor shielding effect.
3 They do not form complexes easily. They have much greater tendency to form
complexes.
4 Except promethium, they are
non-radioactive.
All of them are radioactive.
5 Their compounds are less basic. Their compounds are more basic.
6 They do not form oxocations They form oxocations UO+
4. Mention the oxidation states and three uses of lanthanides. [Sep-2008]
Oxidation states:
The common oxidation state exhibited by all the lanthanides is +3. Some elements exhibit +2 and +4
states as uncommon oxidation states.
La - +3
Ce - +3, +4, +2
Pr - +3, +4
Nd - +3, +4, +2
Uses of lanthanides: Refer – Q - 6
5. Write the uses of lanthanides and actinides [June-2009, Sep-2009, Mar-2015]
Use of lanthanides:
No. Lanthanides Uses
1 Pyrophoric alloy
( La+Ce+Nd+Fe +Al+Ca+C+Si )
Cigarette lighters, toys, flame throwing tanks
and tracer bullets.
2 Ceria (CeO2) and Thoria (ThO2) Gas lamp materials
3 Cerium salts Dyeing cotton, lead storage batteries and as
catalyst.
4 Lanthanides Reducing agents in Lanthanido-thermic
processes to yield pure metals
5 Mish-metals (Alloys of Lanthanides)
( La+Ce+Nd+Fe +Ca )
Production heat resistant steel, stainless steel
and instrumental steel.
6 Mg-alloys containing 30% mishmetal
and 1% Zr
Making parts of jet engines
Uses of Actinides:
Actinides Uses
U-235 Fuel in nuclear power plants,
Component in nuclear weapons.
Plutonium - 238 Power source in long mission space probes.
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6. Describe the extraction of Lanthanides from monazite sand
[Sep-2006, Jun-2007, Mar-2010, Sep-2010, Jun-2011, Mar-2014, Jun-2014, Jun-2016, Sep-2016]
Monazite sand
Heat with H2SO
4
for several hours210 oC
Grey Mud
Cold water
Unreacted monazite sand Filtrate containing
SiO2, TiO
2, ZrSiO
4
(recycle sand)
Ln3+, Th4+, H3O+
HSO4-, SO
42-, H
2PO
4-
Neutralise to proper acidity
or add HF
Filtrate containingLanthanide andPhosphate ions
NaOH or
Oxalic acid
Lanthanide hydroxidesor Oxalates
Precipitate of Th3(PO
4)
4
or Precipitate of Th
Individual lanthanides are separated by a suitable physical method.
1. Anhydrous fluorides
and chlorides
Ca 1270 K
( argon atmosphere )Individual metal
2. Trifluorides of
lanthanides Ca, LiPure metal
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Additional questions 1 MARK
1. The electronic configuration of Lanthanides is
a) [Xe]4f 0 5d
0 6s
0 b) [Xe] 4f
1-7 5d
1 6s
1 c) [Xe]4f
1-14 5d
1 6s
2 d) [Xe]4f
1-14 5d
1-10 6s
2
2. The lanthanide contraction is responsible for the fact that
a) Zn and Y have about the same radius b) Zr and Nb have similar oxidation state
c) Zr and Hf have about the same radius d) Zr and Zn have the same oxidation state
3. Metallothermic processes involving Lanthanides are called as
a) Aluminothermic process b) Lanthanido-thermic process
c) Reduction process d) Oxidation process
4. Lanthanides are separated by
a) Fractional distillation b) Steam distillation c) Fractional Crystallisation d) Sublimation 5. f-block elements are also known as
a) inner transition elements b) rare earth elements c) both a and b d) none
6. Lanthanides belong to
a) 4f-block b) 5f-block c) 6f-block d) 4d-block
7. Actinides belong to
a) 4f-block b) 5f-block c) 6f-block d) 4d-block
8. Which is more covalent?
a) La(OH)3 b) Ce(OH)3 c) Lu(OH)3 d) Gd(OH)3
9. Which is more basic?
a) La(OH)3 b) Ce(OH)3 c) Lu(OH)3 d) Gd(OH)3
10. Second and third rows of d-block elements are quite close in properties. This is due to
a) Actinide contraction b) lanthanide contraction c) both d) none
11. Maximum oxidation state exhibited by Actinides is
a) +3 b) +4 c) +5 d) +6
12. Which of the following elements have almost equal atomic radii due to lanthanide contraction and
difficult to separate?
a) Zirconium (Zr) & Hafnium (Hf) b) La & Sc c) Cr & Co d) none
13. Pyrophoric alloy is used in
a) cigarette lighters & toys b) flame throwing tanks c) tracer bullets d) all the above
14. Cerium salts are used in
a) dyeing cotton b) lead storage batteries c) as catalyst d) all the above
15. Lanthanides are used in metallothermic reactions due to their extraordinary
a) oxidizing property b) reducing property c) catalytic property d) none of the above
16. Mish-metals are used for the production of
a) heat resistant steel b) stainless steel c) instrumental steels d) all the above
5 MARK 1. Discuss the general characteristics of lanthanides
The Lanthanide series include fifteen elements i.e. lanthanum (La) to lutetium (Lu). Lanthanum and
Lutetium have no partly filled 4f-subshell.
1. Electronic configuration
The fourteen electrons are filled in Ce to Lu with configuration [Xe] 4f 1-14
5d1
6s2
2. Oxidation states
The common oxidation state exhibited by all the lanthanides is +3 (Ln3+
) in aqueous solutions and
in their solid compounds. Some elements exhibit +2 and +4 states as uncommon oxidation states.
La - +3
Ce - +3, +4, +2
Pr - +3, +4
Nd - +3, +4, +2
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3. Radii of tripositive lanthanide ions
The size of M3+
ions decreases as we move through the lanthanides from lanthanum to lutetium.
This steady decrease in ionic radii of M3+
cations in the lanthanide series is called Lanthanide
contraction.
Cause of lanthanide contraction: Refer – Govt. Exam-5 mark –Q-2
2. Discuss the general characteristics of actinides The fifteen elements from actinium to lawrencium constitute the actinide series of the periodic table.
1. Electronic configuration
The general electronic configuration of actinides is [Rn] 5f 0,1-14
6d 0,1-2
7s2
where Rn stands for radon core.
2. Oxidation states
These elements shows the oxidation states of +2, +3, +4, +5 and +6. Out of these, +4 oxidation state
is most common state.
3. Radii of M3+
and M4+
ions
The ionic radii of actinide elements decrease gradually as we move along the actinide series. The
steady decrease in the ionic radii with increase in nuclear charge is called actinide contraction and is
analogous to lanthanide contraction.
Cause of actinide contraction
Cause of actinide contraction is the imperfect shielding by 5f-electrons. As we proceed from one
element to the next one in actinide series, the nuclear charge increases by +1 at each next element
which is not compensated due to poor shielding effect of 5f orbitals due to their more diffuse
shape. Hence as the atomic number increases, the inward pull experienced by 5f-electrons increase.
Consequently steady decrease in size occurs in the actinide series.
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6. COORDINATION COMPOUNDS AND BIO-COORDINATION COMPOUNDS
Blue print 1 Mark = 1 3 Mark = 0 5 Mark = 2 Total marks = 11
Govt. Examination questions 1 MARK
1. An example of a chelating ligand is [Mar-2007, Mar-2011, June-2015, Mar-2016]
a) NO2 –
b) Chloro c) Bromo d) en
2. The geometry of complex ion [Fe(CN)6]4–
is [Sep-2010]
a) tetrahedral b) square planar c) Octahedral d) triangular
3. The geometry of [Ni(CN)4]2–
is [Sep-2009, Mar-2012, Sep-2016]
a) Tetrahedral b) Square planar c) Triangular d) Octahedral
4. An example of an ambidentate ligand is [June-2006, Sep-2013]
a) CN– b) Cl
– c) NO2
– d) I
–
5. [FeF6]4–
is paramagnetic because [Mar-2008, Mar-2010, Mar-2013]
a) F– is a weaker ligand b) F
–is a strong ligand
c) F– is a flexidentate ligand d) F
– is a chelating ligand
6. In [FeII(CN)6]
4–, the central metal ion is [June-2007, Sep-2007]
a) Fe b) Fe+2
c) Fe+3
d) CN–
7. The coordination number of Ni(II)
in [Ni(CN)4]2–
is [June-2011]
a) 2 b) 4 c) 5 d) 6
8. Paramagnetic moment is expressed in [June-2012, June-2013]
a) Debye unit b) K Joules c) BM d) ergs
9. The type of isomerism found in the complexes [Co(NO2)(NH3)5]SO4 and [Co(SO4)(NH3)5]NO2 [Sep-12]
a) Hydrate isomerism b) Coordination isomerism c) Linkage isomerism d) Ionisation
10. The type of isomerism found in the complexes [Pt(NH3)4] [CuCl4] and [Cu(NH3)4] [Pt Cl4] is [Mar-06]
a) Ionisation isomerism b) Coordination isomerism c) Linkage isomerism d) ligand isomerism
11. Which of the following is cationic complex? [Sep-2006, Mar-2009]
a) K4[Fe(CN)6] b) [Cu(NH3)4]Cl2 c) K3[Cr(C2O4)3] d) K3[Fe(CN)6]
12. The coordination number of Cr (III) in [Cr(H2O)4Cl2] Cl.2H2O is [June-2008]
a) 3 b) 4 c) 6 d) 2
13. An example of bidentate chelating ligand is [Sep-2008, June-2010]
a) NO2– b) NO3
– c) en d) SO4
2–
14. The oxidation number of Nickel in the complex ion, [NiCl4]2–
is [June-2009]
a) +1 b) 4 c) +2 d) 6
15. The coordination number of Ni(II)
in [NiCl4]2–
is [Sep-2011]
a) 2 b) 4 c) 5 d) 6
16. Chlorophyll is a -------------------- complex [Mar-2014]
a) Magnesium-Porphyrin b) iron-Porphyrin c) Copper-Porphyrin d) Nickel-Porphyrin
17. The type of isomerism found in the complexes [Co(NO2) (NH3)5]SO4 and [Co(SO4) (NH3)5] NO2 is
a) Hydrate isomerism b) Coordination isomerism
c) Linkage isomerism d) Ionisation isomerism [Jun-2014]
18. A metal ion from the first transition series forms an octahedral complex with magnetic moment of 4.9
BM and another octahedral complex which is diamagnetic The metal ion is [Sep-2014]
a) Fe2+
b) Co2+
c) Mn2+
d) Ni2+
19. The function of ferridoxin is [Mar-2015]
a) photosynthesis b) storage and transport of oxygen c) electron transfer d) sensitiser
20. Valence bond theory does not explain the property of complex compound [Sep-2015]
a) geometry b) magnetic c) nature of ligand d) colour
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21. cis-[Pt(NH3)2Cl2] is used as [Jun-2016]
a) Antidotes for heavy metal poisoning b) Synthetic detergent c) Anti tumour drug d) Masking agent
22. An example of a complex compound having coordination number 4. [Mar-2017]
a) K4[Fe(CN)6] b) [Co(en)3]Cl3 c) [Fe(H2O)6]Cl3 d) [Cu(NH3)4]Cl2
5 MARK 1. For the following complex mention a) Name b) Central metal ion c) Ligand d) Coordination
number e) Oxidation number f) Geometry g) Charge on the complex ion h) Nature of the
complex ( any five )
K4[Fe(CN)6] [Sep-2006, Mar-2007, Sep-2007, Mar-2011, Sep-2013]
IUPAC name Potassiumhexacyanoferrate(II)
Central metal ion Fe2+
Ligand CN– cyano
Coordination Number 6
Oxidation Number +2
Geometry or Shape Octahedral
Charge on the complex ion – 4
Nature of the complex Anionic complex
2. [Cu(NH3)4] SO4 [Sep-2009]
IUPAC name Tetraamminecopper(II)sulphate
Central metal ion Cu2+
Ligand NH3 ammine
Coordination Number 4
Oxidation Number +2
Geometry or Shape Square planar
Charge on the complex ion +2
Nature of the complex Cationic complex
3. K3[Cr(C2O4)3].3H2O [June-2013]
IUPAC name Potassium tris (oxalato) chromate(III) trihydrate
Central metal ion Cr3+
Ligand C2O42–
Oxalato
Coordination Number 6
Oxidation Number III or +3
Geometry or Shape Octahedral
Charge on the complex ion –3
Nature of the complex Anionic complex
Electronic configuration d3
Isomers optical isomers are possible
4. [Co(NH3)6]Cl3 [June-2009]
IUPAC name hexaamminecobalt(III)chloride
Central metal ion Co3+
Ligand NH3 ammine
Coordination Number 6
Oxidation Number +3
Geometry or Shape Octahedral
Charge on the complex ion +3
Nature of the complex Cationic complex
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5. [Cr(en)3]Cl3 [Mar-2014]
IUPAC name tris(ethylenediamine) chromium (III) chloride
Central metal ion Cr3+
Ligand en (ethylenediamine)
Coordination Number 6
Oxidation Number +3
Geometry or Shape Octahedral
Charge on the complex ion +3
Nature of the complex Cationic complex
6. [Co(NH3)3 (NO2)3] [June-2014]
IUPAC name triamminetrinitrocobalt(III)
Central metal ion Co3+
Ligand NH3 ammine & NO2– nitro
Coordination Number 6
Oxidation Number +3
Geometry or Shape Octahedral
Charge on the complex ion 0
Nature of the complex Neutral complex
7. [Co(NH3)4Cl2] NO2 [Mar-2015]
IUPAC name tetraamminedichlorocobalt(III)nitrite
Central metal ion Co3+
Ligand NH3 ammine & Cl– chloro
Coordination Number 6
Oxidation Number +3
Geometry or Shape Octahedral
Charge on the coordination sphere +1
Nature of the complex Cationic complex
8. [Co(NH3)5Cl]2+
[Jun-2016]
IUPAC name pentaamminechlorocobalt(III)ion
Central metal ion Co3+
Ligand NH3 ammine & Cl– chloro
Coordination Number 6
Oxidation Number +3
Geometry or Shape Octahedral
Charge on the coordination sphere +2
Nature of the complex Cationic complex
9. Discuss the postulates of Werner’s theory of coordination compounds
[Mar-2006, June-2006, June-2007, Sep-2007, June-2008, Sep-2009, June-2010, Sep-2010, June-2011,
Sep-2011, Sep-2012, Sep-2013, Sep-2014, Sep-2015, Mar-2016]
1. Every metal atom has two types of valencies
i) Primary valency or ionisable valency
ii) Secondary valency or non ionisable valency
2. The primary valency corresponds to the oxidation state of the metal ion. The primary valency of the
metal ion is always satisfied by negative ions
3. Secondary valency corresponds to the coordination number of the metal ion or atom. The secondary
valencies may be satisfied by either negative ions or neutral molecules.
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4. The molecules or ion that satisfy secondary valencies are called ligands.
5. The ligands which satisfy secondary valencies must project in definite directions in space. So the
secondary valencies are directional in nature whereas the primary valencies are non-directional
in nature.
6. The ligands have unshared pair of electrons. These unshared pair of electrons are donated to central
metal ion or atom in a compound. Such compounds are called coordination compounds.
7. Werner‟s representation of [Co(NH3)6]Cl3 : In this representation, the primary valencies (dotted lines) are satisfied by the three chloride ions.
The secondary valencies (solid lines) are satisfied by the six ammonia molecules.
Co
NH3
NH3
NH3
NH3
H3N
H3N
Cl
Cl
Cl
Defects of Werner’s theory
Werner‟s theory does not explain the magnetic and spectral properties coordination compounds
10. Explain Coordination isomerism and Ionisation isomerism with examples
[Jun-07, Mar-08, Sep-08, Jun-09, Sep-10, Mar-12, Jun-12, Mar-13, Mar-15, Mar-17]
Coordination isomerism
In a bimetallic complex, both complex cation and complex anion may be present. In such a case the
distribution of ligands between the two coordination spheres can vary, giving rise to isomers called the
coordination isomers. This phenomenon is called coordination isomerism.
Example:
[CoIII
(NH3)6] [CrIII
(CN)6] and [CrIII
(NH3)6] [CoIII
(CN)6] Hexamminecobalt(III)hexacyanochromate(III) Hexamine chromium (III) hexacyano cobaltate (III)
Ionisation isomerism
Coordination compounds having the same molecular formula but forming different ions in solution are
called ionisation isomers. This property is known as ionisation isomerism.
Example:
[Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br
red-violet red
pentaamminebromocobalt(III) sulphate pentaamminesulphatocobalt (III) bromide
The red-violet isomer yields sulphate ion and the red isomer furnishes bromide ion in solution.
11. Explain Hydrate, Linkage and Ligand isomerism in coordination compounds with examples
[Mar-2010 ]
Hydrate isomerism or Solvate isomerism
“CrCl3.6H2O” may contain 4, 5, (or) 6 coordinated water molecules.
1. [Cr(H2O)4Cl2]Cl.2H2O
Tetraaquadichlorochromium(III) chloride dihydrate
2. [Cr(H2O)5Cl]Cl2.H2O
Pentaaquachlorochromium(III) chloride monohydrate
3. [Cr(H2O)6]Cl3
Hexaaquachromium(III) chloride
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These isomers on reaction with AgNO3 to test for Cl–ions, would find 1,2, and 3 Cl
– ions in solution
respectively.
Linkage isomerism
Linkage isomerism occurs with ambidentate ligands such as SCN – / NCS
– and NO2
– / ONO
–
These ligands are capable of coordinating in more than one way.
Example:
[Co(NH3)5ONO]Cl2
pentaamminenitritocobalt(III) chloride - O attached
[Co(NH3)5 NO2]Cl2
pentaamminenitrocobalt(III) chloride - N attached
Ligand isomerism
Ligand isomerism arises from the presence of ligands which can adopt different isomeric forms.
Example: The ligand diaminopropane may have the amine groups in the terminal (1,3-) positions or in
the 1,2-positions.
H2N-CH
2-CH
2-CH
2-NH
2 H2N-CH
2-CH-CH
3
NH2
12. Discuss the postulates of Valence Bond theory of coordination compounds
[Mar-2009, Mar-2012, Mar-2014, Jun-2014, Jun-2015, Jun-2016]
1. The central metal atom/ion makes available a number of vacant orbitals equal to its coordination
number.
2. These vacant orbitals form covalent bonds with the ligand orbitals.
3. A covalent bond is formed by the overlap of a vacant metal orbital and filled ligand orbitals. This
complete overlap leads to the formation of a metal ligand,σ(sigma) bond.
4. A strong covalent bond is formed only when the orbitals overlap to the maximum extent. This
maximum overlapping is possible only when the metal vacant orbitals undergo a process called
„hybridisation‟. A hybridised orbital has a better directional characteristics than an unhybridised
one.
5.
Coordination
Number
Types of
hybridization
Geometry or
Shape
2 sp linear
4 sp3 tetrahedral
4 dsp2 square planar
6 d2sp
3 octahedral
6 sp3d
2 octahedral
6. Magnetic moment:
Complexes which have one or more unpaired electrons are paramagnetic. They are attracted by an
external field. The paramagnetic moment is given by the following spin-only formula.
μs = )2n(n BM
μs = spin-only magnetic moment n = number of unpaired electrons
BM = Bohr magneton, the unit which expresses the magnetic moment.
When the complex does not contain any unpaired electron, it is diamagnetic.
Defects of Valence bond theory
Valence bond theory fails to account for various magnetic, electronic and spectroscopic properties of
coordination compounds.
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13. Discuss hybridisation and magnetic property of Ni(NH3)4]2+
using VB Theory.
[June-2006, Sep-2006, Mar-2007, June-2010, Mar-2013]
In Ni(NH3)4]2+
nickel is in +2 oxidation state.
Nickel atom : Outer electronic configuration 3d8
4s2
Number of unpaired electrons = 2. It is para magnetic.
μs = )22(2 BM = 8 = 2.83 BM
Since the hybridisation is sp3, the geometry of the molecule is tetrahedral
Hybridisation sp3
Geometry or shape tetrahedral
No. of unpaired electrons 2
Magnetic moment 2.83 BM
Magnetic character paramagnetic
14. Discuss the hybridisation and magnetic property of [Ni(CN)4]2–
using VB Theory.
[June-2006, Mar-2007, June-2010, Sep-2011, Mar-2013]
In [Ni(CN)4]2–
nickel is in +2 oxidation state.
Nickel atom : Outer electronic configuration 3d8
4s2
The ligand CN
– is a powerful ligand. Hence it forces the unpaired electrons to pair up in d-orbitals.
Hence this complex ion does not contain unpaired electrons. It is diamagnetic.
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The geometry of the molecule is square planar.
Hybridisation dsp2
Geometry or shape square planar
No. of unpaired electrons 0
Magnetic moment 0
Magnetic character diamagnetic
15. Discuss the hybridisation and magnetic property of [FeF6]4–
using VB Theory.
[Mar-2006, Mar-2011, June-2011, Sep-2011, Sep-2012, June-2013]
In [FeF6]4–
iron is in +2 oxidation state.
Iron atom: Outer electronic configuration 3d6 4s
2
Number of unpaired electrons = 4
μs = )24(4 = 24 = 4.9 BM
The complex contains 4 unpaired electrons and it is paramagnetic.
The molecular geometry is octahedral.
Hybridisation sp3d
2
Geometry or shape octahedral
No. of unpaired electrons 4
Magnetic moment 4.9 BM
Magnetic character paramagnetic
16. Discuss hybridisation and magnetic property of [Fe(CN)6]4–
using VB Theory
[Mar-2006, Mar-2011, June-2011, Sep-2012, June-2013]
In [Fe(CN)6]4–
iron is in +2 oxidation state.
Iron atom: Outer electronic configuration 3d6 4s
2
Fe atom : Outer electronic configuration 3d6 4s
2
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In [Fe(CN)6]4 –
complex the CN –
ligand is a powerful ligand, it forces the unpaired electrons in 3d level
to pair up inside. Hence the complex has no unpaired electron. The molecule is diamagnetic.
The molecular geometry is octahedral
Number of unpaired electron = 0 µs = 0
Hybridisation d2 sp
3
Geometry or shape octahedral
No. of unpaired electrons 0
Magnetic moment 0
Magnetic character diamagnetic
17. In what way [FeF6]4–
differs from [Fe(CN)6]4–
[Mar-2009]
No. [FeF6]4–
[Fe(CN)6]4–
1 Fe 2+
is connected to weak ligand F –
Fe2+
is connected to strong ligand CN –
2 Contains 4 unpaired electrons Has no unpaired electron
3
Spin-only magnetic moment
s = 4(4+2) = 24 = 4.9 BM
Spin-only magnetic moment is zero
4 Paramagnetic Diamagnetic
5 Sp3d
2 hybridisation d
2 Sp
3 hybridisation
18. [Ni(CN)4]2–
diamagnetic, whereas [NiCl4]2–
is paramagnetic. Explain. [ June-2008, June-2012]
[Ni(CN)4]2–
is diamagnetic. [ Refer – Q-13]
In [NiCl4]2 –
nickel is in +2 oxidation state.
Nickel atom : Outer electronic configuration 3d8 4s
2
[NiCl4]
2 – has two unpaired electrons, so it is paramagnetic.
19. [Ni(CN)4]2–
diamagnetic, whereas [Ni (NH3)4]2+
is paramagnetic. Explain. [ Jun-15, Mar-17]
Refer – Q -13 & 14
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20. Discuss the structure and function of haemoglobin [Mar-2010, Sep-2014, Sep-2015, Mar-2016]
1. Haemoglobin is iron-porphyrin complex. It is a tetramer.
2. The working part of haemoglobin is a heme group containing an Fe2+
cation coordinated to four
nitrogen atoms of porphyrin group and one nitrogen atom of histidine group. The sixth octahedral
site is available to bind oxygen molecule.
3. Haemoglobin in the red blood cells carries oxygen from the lungs to the tissues.
4. It delivers the oxygen molecule to myoglobin in the tissues.
5. When the oxygen has been released for cell respiration, haemoglobin loses its bright red colour and
becomes purple.
6. It then combines with the waste carbon dioxide produced by the cells and deposits in the lungs so
that the gas can be exhaled.
21. How chlorophyll is important in environmental chemistry? Mention its function [Mar-08, Sep-16]
1. Chlorophyll is a magnesium - porphyrin complex.
2. The magnesium is at the centre of the modified porphyrin ring
3. The oxidation state of magnesium is +2 (Mg2+
). The modified porphyrin acts as the ligand.
4. There are several kinds of chlorophyll that vary slightly in their molecular structure. In plants,
chlorophyll „a‟ is the pigment directly responsible for the transformation of light energy to chemical
energy. Hence in plants, the green pigment chlorophyll helps photosynthesis.
5. The conversion of atmospheric carbondioxide and atmospheric moisture into carbohydrate and
molecular oxygen in the presence of sunlight, by the plant is called as photosynthesis. Chlorophyll
acts as a light sensitiser in this important process.
x CO2 + y H2O Cx(H2O)y + O2
Chlorophyll
Sunlight 6. Photosynthesis requires, in addition to chlorophyll, the help of four other metal complexes, a
manganese complexes, two iron complexes and a copper complex.
7. All oxygenated animals take molecular oxygen through haemoglobin and release CO2. But
chlorophyll helps in the conversion of atmosphere CO2 into molecular oxygen which acts as a fuel
for human cell.
22. Write a note on a) Coordination sphere and b) Chelates c) Neutral ligand [Sep-2008]
a) Coordination sphere In a coordination compound, central metal ion and the ligands are enclosed within square bracket is
called as coordination sphere. This represents a single constituent unit.
Example: [Fe(CN)6]4–
, [Cu(NH3)4]2+
These ions do not ionise to give the test for constituent ions.
b) Chelates
Polydentate ligands (eg-ethylene diamine) are capable of forming more than one bond with the central metal
atom. This results in the formation of complexes with the ring structures which are known as metal chelates.
Hence the ring forming groups are described as chelating agents (or) polydentate ligands.
Cu
N
NCH2
CH2H2
H2
NH2
NH2CH2
CH2
2+
2 NH2CH2CH2NH2 + Cu2+
c) Neutral ligand
Neutral molecules which donate a pair of electron to the central metal are called neutral ligands. They are
named as such without any special name. But water is written as „aqua‟ and ammonia is written as ammine.
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23. Explain the following: a) ligand b) coordination number [Sep-2016]
a) Ligand
An ion (or) a molecule that donates a pair of electrons to the central metal is called as ligand. Hence, in
a complex compound ligands act as Lewis bases. Eg- NH3, CN–
b) Coordination number
Coordination number is the total number of the chemical bonds formed between the central metal ion
and the ligands. For example in K4[Fe(CN)6] the coordination number of Fe(II) is 6.
Additional questions
1 MARK 1. Which is a double salt?
a) K2SO4.Al2(SO4)3.24H2O b) NaCl c) K4[Fe(CN)6] d) KCl
2. The geometry of [Cu(NH3)4]2+
complex ion
a) Linear b) Tetrahedral c) Square planar d) Angular
3. The oxidation number of Nickel in the complex ion, [NiCl4]2–
is
a) +1 b) –1 c) +2 d) –2
4. Which is not an anionic complex?
a) [Cu(NH3)4]Cl2 b) K4[Fe(CN)6] c) K3[Fe(CN)6] d) [NiCl4]2–
5. The name of [PtIV
(NH3)2Cl2]2+
is
a) Diamminedichloroplatinum(IV) ion b) Diamminedichloroplanitate(IV)
c) Diamminedichloroplatinum d) Dichlorodiammineplatinum(IV) ion
6. For a compound K4[Fe(CN)6] 4K+ + [Fe(CN)6]
4–, the complex ion is
a) K+ b) CN
– c) Fe
II d) [Fe(CN)6]
4–
7. A metal ion from the first transition series forms an octahedral complex with magnetic moment of
4.9BM and another octahedral complex which is diamagnetic. The metal ion is
a) Fe2+
b) Co2+
c) Mn2+
d) Ni2+
8. Which of the following is double salt?
a) Potash alum K2SO4.Al2(SO4)3.24H2O b) Mohr‟s salt FeSO4.(NH4)2SO4.6H2O
c) both a and b d) K4[Fe(CN)6]
9. A coordination compound will not give
a) simple cation & complex anion b) simple anion & complex cation
c) complex cation & complex anion d) simple cation & simple anion
10. Which is bidentate ligand?
a) H2O b) Ethylene diamine c) NH3 d) Cl–
bidentate ligands Ethylene diamine H2N-CH2-CH2-NH2 , Oxalato C2O42–
11. Which is ambidentate ligand?
a) SCN– b) NO2
– c) both a and b d) none
12. Which is a positive ligand?
a) hydrazinium NH2-NH3+ b) NH2-NH2 c) NH4
+ d) H2O
13. The name of H2O ligand is
a) aqua b) hydroxide c) water d) none
14. The oxidation number of Cr in K3[Cr(C2O4)3].3H2O is
a) +2 b) +3 c) 0 d) -3
15. Coordination number of Cr in K3[Cr(C2O4)3].3H2O is
a) 3 b) 4 c) 5 d) 6
16. The geometry of [Cr(C2O4)3]3–
ion is
a) Tetrahedral b) Square planar c) Triangular d) Octahedral
17. The name of SO42–
ligand is
a) Sulphito b) sulphato c) sulphate d) sulphido
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18. Coordination number of Cr in [Cr(en)3]Cl3 is
a) 3 b) 4 c) 5 d) 6
19. Coordination number of Pt in [PtCl2(NH3)2] is
a) 3 b) 4 c) 5 d) 6
20. Number of Cl–
ions furnished by [Cr(H2O)5Cl]Cl2. H2O is
a) 1 b) 2 c) 3 d) 0
21. Number of Cl– ions furnished by [Cr(H2O)4Cl2] Cl. 2H2O is
a) 1 b) 2 c) 3 d) 0
22. Number of Cl– ions furnished by [Cr(H2O)6] Cl3 is
a) 1 b) 2 c) 3 d) 0
23. 1 mole of which of the following will give 2 moles of AgCl with AgNO3 solution?
a) [Cr(H2O)4Cl2] Cl. 2H2O b) [Cr(H2O)5Cl]Cl2. H2O c) [Cr(H2O)6] Cl3 d) none
24. Geometrical isomerism or cis-trans isomerism is not possible in ------- complexes
a) octahedral b) tetrahedral c) square planar d) all of these
25. Which of the following hybridisation will give square planar geometry to a complex?
a) sp3 b) dsp
2 c) d
2sp
3 d) sp
3d
2
26. Spin-only magnetic moment of a complex is 3.87 BM. Then the number of unpaired electrons present in
the complex is
a) 1 b) 2 c) 3 d) 4
27. Formation of red 2,2‟-bipyridyl and 1,10-phenanthroline complexes are used as a test for
a) Fe2+
b) Cu2+
c) Zn2+
d) Ca2+
28. Which of the following chelating ligands is used to estimate Ni2+
in gravimetric analysis?
a) DMG b) en c) oxine d) oxalato
29. Which of the following chelating ligands is used to estimate Al3+
in gravimetric analysis?
a) DMG b) en c) oxine d) oxalato
30. Which of the following is a chelating complex?
a) Ni(DMG)2 b) Al(oxine)3 c) [Cu(en)2]2+
d) all of these
31. The chelating ligand which is used in the complexometric titration and as masking agent is
a) EDTA b) en c) oxalato d) none
32. Which of the following metals can be extracted from their ores by formation of stable complexes of
cyanide ion?
a) Ag b) Au c) both a and b d) none
33. Which is used for the elimination of radioactive and other toxic elements(Pb2+
) from the body?
a) EDTA b) en c) oxalato d) none
34. Which is used as an anti-tumour drug?
a) cis-[Pt(NH3)2Cl2] b) trans-[Pt(NH3)2Cl2] c) Ni(DMG)2 d) Al(oxine)3
35. The chelating agent which is present in synthetic detergents and sequesters hard-water cations is
a) EDTA b) en c) tripolyphosphate d) oxine
36. The function of haemoglobin is
a) transport and storage of oxygen b) photosynthesis c) electron transfer d) none
37. The function of chlorophyll is
a) transport and storage of oxygen b) photosynthesis c) electron transfer d) none
38. The function of ferredoxin and rubredoxin is
a) transport and storage of oxygen b) photosynthesis c) electron transfer d) none
39. Both haemoglobin and myoglobin contain
a) Fe2+
b) Ca2+
c) Mg2+
d) Co3+
40. Chlorophyll contains
a) Fe2+
b) Ca2+
c) Mg2+
d) Co3+
41. Which is a magnesium-porphyrin complex?
a) haemoglobin b) myoglobin c) Chlorophyll d) none
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42. Which is a iron-porphyrin complex?
a) haemoglobin b) myoglobin c) Chlorophyll d) both a and b
43. In photosynthesis, chlorophyll acts as a
a) light sensitiser b) retardant c) oxygen carrier d) none
44. Haemoglobin is a ------------
a) monomer b) dimer c) trimer d) tetramer
45. Myoglobin is a ------------
a) monomer b) dimer c) trimer d) tetramer
5 MARK
1. Explain Geometrical (or) cis-trans isomerism in coordination compounds with examples
Geometric isomers are possible for both square planar and octahedral complexes, but not tetrahedral
complexes
In a cis-isomer two identical (or) similar groups are adjacent to each other whereas in a trans-isomer
they are diametrically opposite to each other.
i) Square planar complexes of the type [Ma2b2]n+
where a and b are monodentate ligands.
[Pt (NH3)2 Cl2]
Pt
Cl Cl
H3N NH3
Pt
Cl
H3N Cl
NH3
Cis Trans
cis-diamminedichloroplatinum (II) trans-diamminedichloroplatinum (II)
[Pd(NH3)2 (NO2)2]
H3N NH3 H3N
NH3
Pd Pd
O2N NO2
NO2
O2N
Cis Trans
cis-diamminedinitropalladium (II) trans-diamminedinitropalladium(II)
ii) An octahedral complex of the type [Ma4b2] where a and b are monodentate ligands, exists as
two geometrical Isomers:
[Co(NH3)4 Cl2]+
Co
Cl
Cl
NH3
NH3
H3N
H3N
+
Co
Cl
NH3 H3N
H3N
+
Cl
NH3
Cis Trans
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iii) The octahedral complex of the type, [Ma3b3]where a and b are monodentate ligands also exist
as geometrical isomers
[Rh(py)3Cl3] exist as cis-(1,2,3 trichlorocomplex) and trans-(1,2,6-trichloro complex) isomers as
represented below:
Rh
Cl
Cl
py
py
py
Rh
Cl
Cl
Cl
Cl
pypy
py
Cis Trans
2. Explain Optical Isomerism in coordination compounds with examples
This is a phenomenon in which certain organic or inorganic compounds have the property of rotating
plane polarised light. The compounds which exhibit this property are called optical isomers. The optical
isomers of a compound have identical physical and chemical properties. The only distinguishing
property is that the isomers rotate the plane of polarised light either to the left or right. In a coordination
compound of type [PtCl2(en)2]2+
, two geometrical isomers are possible. They are cis and trans. Among
these two isomers, cis isomer shows optical activity because the whole molecule is asymmetric and it
is non superimposable on its mirror image.
Pt
Cl
Cl
en
en
Pt
lC
lC
en
en
Optical isomers of cis [PtCl2 (en)2]
2+
3. Write the uses of coordination compounds
1. Dyes and Pigments
a) madder dye (red) – a complex of hydroxyanthraquinone
b) copper phthalocyanine coplex - blue pigment
2. Analytical Chemistry
a) Colour Tests:
Formation of red 2,2‟-bipyridyl and 1,10-phenanthroline complexes as a test for Fe2+
b) Gravimetric Analysis:
Here chelating ligands are often used to form insoluble complexes e.g. Ni(DMG)2 and Al(oxine)3.
c) Complexometric Titrations and Masking Agents :
An example of this is the use of EDTA in the volumetric determination of a wide variety of metal ions
in solution, e.g. Zn2+
, Pb2+
,Ca2+
,Co2+
,Ni2+
,Cu2+
, etc. By careful adjustment of the pH and using suitable
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indicators, mixtures of metals can be analysed, e.g. Bi3+
in the presence of Pb2+
. Alternatively, EDTA
may be used as a masking agent to remove a metal ion which would interfere with the analysis of a
second metal ion present.
3. Extraction of Metals
Sometimes certain metals can be leached from their ores by formation of stable complexes.
e.g. Ag and Au as complexes of cyanide ion.
4. Bio-Inorganic Chemistry
Naturally occurring complexes include haemoglobin, chlorophyll, vitamin B12 etc.
5. Removal of toxic elements EDTA is used to speed up the elimination of harmful radioactive and other toxic elements from the
body. (e.g. Pb2+
). In these cases a soluble metal chelate is formed.
6. Chemotherapy
An example here is the use of cis-Pt(NH3)2Cl2 as an anti-tumour drug.
7. Synthetic detergents
Synthetic detergents containing chelating agents such as tripolyphosphate. The chelating agent
sequesters hard-water cations, rendering them incapable of interfering with the surfactant.
4. Write a note on
a) Central metal ion b) Oxidation state
a) Central metal ion In the complex ion, the metal accepts a pair of electrons from the donor atoms (ligands). The acceptor
metal cation is referred to as central metal cation. Hence, central metal cation in a complex serves as a
lewis acid.
b) Oxidation state
Oxidation state or oxidation number is the number of electrons the central metal has lost to form the
cation.
5. Write a note on types of ligands
i) unidentate ligands
Ligands which bind to the central metal ion through a single donor atom is called as unidentate ligand
(one molecule of the ligand forms only one coordinate bond with the metal). Eg- Cl–, H2O
ii) polydentate ligands
Ligands which are capable of forming more than one bond with the central metal atom (or) ion are
called polydentate ligands.
When a single ligand has two coordinating positions, it is called bidentate ligand and when there are
three coordinating positions available, it is called a tridentate ligand and so on. For example,
ethylenediamine is a bidentate ligand because it has two amino groups each of which can donate a pair
of electrons.
NH2-CH2-CH2- NH2
iii) ambidentate ligands
Ligands which are capable of coordinating in more than one way to the central metal are called
ambidentate ligands. Eg- SCN–
/ NCS– and NO2
– / ONO
–
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7. NUCLEAR CHEMISTRY
Blue print 1 Mark = 1 3 Mark = 1 5 Mark = 1 Total marks = 9
Govt. Examination questions 1 MARK
1. The most penetrating radiations are [Sep-2009, June-2015]
a) α rays b) β rays c) γ rays d) all are equally penetrating
2. Which one of the following particles is used to bombard 13 Al 27
to give 15 P 30
and a neutron
[June-2006, Sep-2007, Mar-2009, Sep-2010]
a) α particle b) deuteron c) proton d) neutron
3. The reaction 5 B 8 4 Be
8 takes place due to [June-2008]
a) α decay b) β decay c) electron capture d) positron decay
4. Radioactivity is due to [Mar-2007, Jun-2010, Jun-2016]
a) Stable electronic configuration b) Stable nucleus
c) Unstable nucleus d) Unstable electronic configuration
5. In the following radioactive decay, 92 x 232
89 y 220
, how many α and β particles are ejected?
[Sep-2013, Sep-2014]
a) 3α and 3β b) 5α and 3β c) 3α and 5β d) 5α and 5β
6. 92U235
nucleus absorbs a neutron and disintegrates into 54 Xe 139
, 38 Sr94
and x. What will be the product
x? [Mar-2011, Mar-2013, Mar-17]
a) 3 neutrons b) 2 neutrons c) α particle d) β particle
7. Loss of a β-particle is equivalent to [Sep-2006, Jun-2009, Sep-2011, Sep-2016]
a) Increase of one proton only b) Decrease of one neutron only
c) Both (a) and (b) d) None of these 8. After 24 hours 0.125 g of the initial quantity of 1 g of a radioactive isotope is left out. The half-life
period is [Mar-2006, June-2012, Sep-2012]
a) 24 hours b) 12 hours c) 8 hours d) 16 hours
9. When 7N15
is bombarded with a proton it gives 6C12
and --------- [June-2007]
a) α-particle b) β-particle c) neutron d) proton
10. In a nuclear reaction -------- is/are balanced on both sides [Mar-2008]
a) mass b) number of atoms c) mass number d) atomic number and mass number
11. Half-life period of a radioactive element is 1500 years. The value of disintegration constant in terms of
seconds is [Sep-2008]
a) 0.1465 × 10 –10
s–1
b) 0.2465 × 10–10
s–1
c) 0.1465 × 10 –8
s–1
d) 0.3645 × 10–-10
s–1
12. Half-life period of a radioactive element is 100 seconds. Its average life period is [Mar-2010]
a) 100 seconds b) 50 seconds c) 200 seconds d) 144 seconds
13. Half-life of 79Au198
nucleus s 150 days. The average life is [Mar-2012]
a) 216 days b) 21.6 days c) 261 days d) 26.1 days
14. In a reaction 5B8 4Be
8 + ? [June-2011]
a) α-particle b) β-particle c) electron capture d) positron
15. β-particle is represented as [June-2013]
a) +1e0 b) –1e
0 c) 1H
1 d) 2He
4
16. In the nuclear reaction 90Th232
82Pb208
the number of αand βparticles emitted are [Mar-2014]
a) 1α and 4β b) 2α and 2β c) 6α and 4β d) 8α and 4β
17. In the nuclear reaction 92U238
82Pb206
the number of αand βparticles emitted are
[June-2014, Mar-2015, Sep-2015]
a) 7α and 5β b) 6α and 4β c) 4α and 3β d) 8α and 6β
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18. The average life of a radioactive element is [Mar-2016]
a) 2λ
1
b)
21t
0.693
c) 1.44 t ½ d) 14.4 t ½
Note: The average life of a radioactive element is = λ
1 =
0.693
t 21
= 1.44 t ½
3 MARK
1. What is Q value of a nuclear reaction? [June-2006, Mar-2016]
The amount of energy absorbed or released during nuclear reaction is called Q-value of nuclear reaction.
Q value = ( mp – mr ) 931 MeV
Where, mr = Sum of the masses of reactants
mp = Sum of the masses of products
In the case of energy absorbed then mp > mr, then Q value will be positive.
In the case of energy released, mr > mp, and hence Q value will be negative.
2. Explain the principle behind the Hydrogen bomb [Mar-2006, Mar-2010, Mar-2015]
Hydrogen bomb is based on the fusion reactions of hydrogen to form helium producing large amount
of energy.
Hydrogen bomb consists of an arrangement for nuclear fission in the centre surrounded by a mixture of
deuterium (1H2) and lithium isotope (3Li
6). Fission reaction provides the high temperature necessary to
start the fusion.
Fusion reactions take place in hydrogen bomb
i) Fission heat + neutrons
ii) 3Li6
+ 0n1 1H
3 + 2He
4 + 4.78 MeV
1H2 + 1H
3 2He4 + 0n
1 + 17.6 MeV
3. What is spallation reaction? Give an example [Sep-2014]
These are the reactions in which high speed projectiles may chip a heavy nucleus into several fragments.
29Cu63
+ 2He4 + 400 MeV 17Cl
37 + 14 1H
1 + 16 0n
1
4. What is binding energy of nucleus? [Sep-2012]
Whenever a nucleus is formed, certain mass is converted into energy. Hence for atom, the atomic mass
is lower than the sum of masses of protons, neutrons and electrons present. The difference in mass is
termed as “mass defect” ( Δm ).
Binding energy, ΔE = Δm C2.
So, binding energy of nucleus is the energy equivalent of mass defect.
5. Write three differences between chemical reactions from nuclear reactions [Sep-2006, Sep-2013]
Refer – 5 mark – Q - 4
6. Write the uses of radio carbon dating [Mar-2011]
1. Carbon dating has proved to be a great tool for correlating facts of historical importance.
2. It is very useful in understanding the evolution of life, and rise and fall of civilizations.
7. Calculate the decay constant for Ag108
if its half life is 2.31 minutes [Sep-10, Mar-14, Jun-15]
=
½t
0.693
= 2.31
0.693 = 0.3 min
–1
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8. Calculate the average life of 79Au198
having t ½ = 150 days [June-2008]
Average life, (Tau) = λ
1 =
0.693
t 21
= 1.44 t ½
= 1.44 × 150
= 216 days.
9. The decay constant for 6C14
is 2.31 ×10–4
year –1
calculate the half life period [Sep-2008]
t ½ = λ
0.693 =
4-102.31
0.693
= 3000 years
10. Calculate the disintegration constant if half life period is 100 seconds [June-2007, June-2009]
=
½t
0.693 =
001
0.693 = 0.00693 s
-1 = 6.93 × 10
–3 s
–1
11. The half-life period of U238
is 140 days. Calculate the average life time
[June-2010, June-2011, Mar-2013, Sep-2015]
Average life, (Tau) = λ
1 = 1.44 t ½ = 1.44 × 140 = 201.6 days
12. The half life of Th233
is 1.4 × 1010
years. Calculate its disintegration constant. [June-2014]
=
½t
0.693 =
10 1.4
0.69310
= 0.495 x 10–10
year –1
13. How many αand βparticles will be emitted by an element 84 A 218
is changing to a stable isotope
of 82 B 206
? [Mar-2009]
Let „a‟ and „b‟ be the number of α and β particles emitted during the change
84A218
82B206
+ a 2He4 + b–1e
0
Comparing the mass numbers,
218 = 206 + 4a + b × 0
4a = 218 – 206
4a = 12
a = 3
Comparing the atomic numbers
84 = 82 + 2 ×a + (–1)b
= 82 + 2a - b
2a – b = 84 –82 = 2
2 ( 3 ) – b = 2
b = 6 – 2
= 4
Number of α- particle emitted = 3 Number of β- particles emitted = 4
14. 92U238
undergoes a series of changes by emitting αand βparticles and finally 82Pb206
is formed.
Calculate the number of α and β particles emitted during the change. [Sep-2011]
Let „a‟ and „b‟ be the number of α and β particles emitted during the change
92U238 82Pb
206 + a 2He
4 + b –1e
0
Comparing the mass numbers,
238 = 206 + 4a + b × 0
4a = 238 – 206
4a = 32
a = 8
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Comparing the atomic numbers
92 = 82 + 2 ×a + (–1) b
= 82 + 2a – b
2a–b = 92 – 82 = 10
2(8) – b = 10
b = 16 –10
= 6
Number of α- particle emitted = 8 Number of β- particles emitted = 6
15. In the following radioactive decay, 92 x232
y220
, how many αand β particles are ejected?
[Sep-2007]
92x232
89y220
+ a 2He4 + b –1e
0
Comparing the mass numbers,
232 = 220 + 4a + b × 0
4a = 232 – 220
4a = 12
a = 3
Comparing the atomic numbers
92 = 89 + 2 ×a + (–1)b
= 89 + 2a – b
2a–b = 92 - 89 = 3
2(3) – b = 3
b = 6–3
= 3
Number of α- particle emitted = 3 Number of β- particles emitted = 3
16. The atomic masses of Li, He and proton are 7.01823 amu, 4.00387 amu and 1.00715 amu
respectively. Calculate the energy evolved in the reaction,
3Li7+1H
1 22He
4+energy (1 amu = 931 MeV) [Mar-2007]
Mass of reactants = mass of Li + mass of H
= 7.01823 + 1.00715
= 8.02538 amu
Mass of products = 2 ×mass of He
= 2 ×4.00 387
= 8.00 774 amu
Mass loss during change = (8.02538-8.00774) amu
= 0.01764 amu
Energy evolved during reaction
= 0.0176 × 931 MeV
= 16.423 MeV
17. Calculate the Q value of the following nuclear reaction: 13Al27
+ 2He4 14Si
30 + 1H
1 + Q
The exact mass of 13Al27
= 26.9815 amu, 14Si30
= 29.9738 amu, 2He4 = 4.0026 amu and 1H
1 = 1.0078
amu [Sep-2009, June-2013, Mar-17]
Q value = ( mp – mr ) 931 MeV
= [ ( 29.9738 + 1.0078 ) – ( 26.9815 + 4.0026 ) ] 931
= [– 0.0025] 931
= – 2.329 MeV
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18. Neutron bombardment fragmentation of U235
occurs according to the equation:
92U235
+ 0 n1 42Mo
98 + 54Xe
136 + x – 1e
0 + y 0 n
1
Calculate the values of x and y [Mar-2012]
92U235
+ 0 n1 42Mo
98 + 54Xe
136 + x –1e
0 + y 0 n
1
Comparing the atomic numbers
92 + 0 = 42 + 54 + (x ×y × 0)
92 = 96 – x
x = 4
Comparing the mass numbers
235 +1 = 98 + 136 + (x × 0) + (y × 1)
236 = 234 + y
y = 2
19. Calculate the number of αand βparticles emitted when 90Th232
nucleus is converted into 82Pb208
Let „a‟ and „b‟ be the number of α and β particles emitted during the change [Mar-2008]
90Th232
82pb208
+ a 2He4 + b –1e
0
Comparing the mass numbers,
232 = 208 + 4a + b 0)
4a = 232 – 208
= 24
a = 6
Comparing the atomic numbers
90 = 82 + 2 ×a + (–1)b
= 82 + 2a – b
a – b = 90 – 82 = 8
2(6) – b = 8
b = 12 – 8
= 4
Number of α - particle emitted = 6 Number of β- particles emitted = 4
20. Neutron bombardment fragmentation of U235
occurs according to the equation:
92U235
+ 0 n1 42Mo
95 + 57La
139 + x – 1e
0 + y 0 n
1
Calculate the values of x and y [June-2012]
92U235
+ 0 n1 42Mo
95 + 57La
139 + x –1e
0 + y 0 n
1
Comparing the atomic numbers
92 + 0 = 42 + 57 + (x ×y × 0)
92 = 99 – x
x = 7
Comparing the mass numbers
235 +1 = 95 + 139 + (x × 0) + (y × 1)
236 = 234 + y
y = 2
21. Complete the following nuclear reactions
i) 7N15
(p, α) ......... ii) 11Na23
(n, β) ......... iii)19K39
(p, d) ......... [Sep-2016]
i) 7N15
+ 1H1 6C
12 + 2He
4
ii) 11Na23
+ 0n1 12Mg
24 + –1e
0
iii) 19K39
+ 1H1 19K
38 + 1H
2
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5 MARK
1. Explain Nuclear fission reaction with examples [June-2008]
Nuclear fission is the process in which a heavy nucleus breaks up into two lighter nuclei of almost equal
size with the release of an enormous amount of energy.
The process is usually accompanied by emission of neutrons.
92U
+ 0 n 1
56Ba 141
+ 36 Kr 92
+ 3 0 n 1
+ 200 MeV
Mechanism of fission
In the fission process, the heavy nucleus absorbs a neutron and forms an unstable compound nucleus.
The compound nucleus then breaks up to give fission products.
Further, the neutrons released from the fission of first uranium atoms can hit three other uranium atoms.
In this way a chain reaction is set up resulting into the liberation of an enormous amount of energy.
92U235 + 0n
1
92U236
56Ba140 + 36Kr93 + 3 0n1
54Xe144 + 38Sr90 + 2 0n1
55Cs144 + 37Rb90 + 2 0n1
This fission process is self multiplying process and hence a tremendous amount of energy is released in
a very short interval of time. Therefore, explosion takes place. Atom bomb is based on nuclear fission
process.
2. Explain the nuclear reactions that take place in Sun (Stars)
[June-2007, Sep-2012, Sep-2013, Sep-2015, Jun-2016 (3 mark)]
The energy of the sun is due to the fusion of hydrogen nuclei into helium nuclei which in going on
inside it all the time.
The various fusion reactions taking place in the sun are as follows:
Proton - proton chain reaction:
1H1 + 1H
1
fusion 1H
2 + +1e
0 + energy
1H2 + 1H
1
fusion 2He
3 + energy
2He3 + 1H
1
fusion 2He
4 + +1e
0 + energy
The overall reaction may be written as,
41H1
fusion 2He
4 +2 +1e
0 + energy
3. Explain the principle behind the Hydrogen bomb [Mar-2007, Sep-2011]
Refer – 3 mark – Q - 2
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4. Differentiate chemical reactions from nuclear reactions
[ Mar-2009, Mar-2011, Mar-2012, June-2012, Mar-2014, Mar-2016]
No. Chemical reactions Nuclear reactions
1 These reaction involve some loss,
gain or overlap of outer orbital
electrons of the reactant atoms.
Nuclear reactions involve emission of
alpha, beta and gamma particles from
the nucleus.
2 A chemical reaction is balanced in
terms of mass only
Nuclear reaction is balanced in terms of
both mass and energy.
3 The energy changes in any chemical
reaction are very much less the
nuclear reaction.
The energy changes are very large in
nuclear reactions.
4 In chemical reactions, the energy is
expressed in terms of kilojoules per
mole.
In nuclear reactions, the energy
involved is expressed in MeV (Million
electron volts) per individual nucleus.
5 No new element is produced since
nucleus is unaffected.
New element / isotope may be produced
during the nuclear reaction.
5. How do nuclear fusion reactions differ from fission reactions? [Sep-2009]
No. Nuclear fission Nuclear fusion
1 A heavier nucleus breaks into
two lighter nuclei
Lighter nuclei fuse together to form a
heavier nucleus
2 It does not require high
temperature
Very high temperature is required for
fusion to take place (108 K)
3 A chain reaction sets in It is not a chain reaction
4 Atom bomb is based on nuclear
fission
Hydrogen bomb is based on nuclear
fusion
5 Fission process can be controlled
and the energy released can be
used for constructive purposes
Fusion process cannot be controlled
and the energy released can be used
for destructive purposes
6. Write a note on Radio carbon dating
[Jun-06, Sep-06, Sep-07, Mar-08, Mar-10, Jun-10, Sep-10, Mar-13,Mar-15, Mar-17]
Radio carbon dating is a method which is used to determine the age of wood or animal fossils. This
method is based on the fact that 6C14
, radioactive isotope of carbon is formed in the upper atmosphere
by reaction with neutrons (from cosmic rays).
7 N 14
+ 0 n 1
6 C 14
+ 1 H 1
The C14
atoms thus produced are rapidly oxidised to 14
CO2 which in turn is incorporated in plants as
result of photosynthesis. Animals too consume C14
by eating plants. On death, organisms cease to take
in fresh carbonations. Carbon-14 begins to decay.
6 C 14
7 N 14
+ –1 e 0
In 5700 years a fossil (plant or animal) will lose half the amount of carbon-14 present in its living state.
Therefore by knowing either the amount of C14
or the number of β-particles emitted per minute per
gram of carbon at the initial and final stages, the age of carbon material can be determined by the
following equation.
t =
wooddeadinCofAmount
woodfreshinCofAmountlog
0.693
t2.303
14
142
1
Uses
1. Carbon dating has proved to be a great tool for correlating facts of historical importance.
2. It is very useful in understanding the evolution of life, and rise and fall of civilizations.
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7. Explain the use of radioactive isotopes in the Study of reaction mechanism [Sep-2008, June-2015]
1. Mechanism of photosynthesis in plants
A small quantity of Radioactive CO2* containing radioactive oxygen O18
is mixed with ordinary
carbondioxide and the process is carried out. It has been found that oxygen gas evolved along with sugar
formation is non-radioactive. Therefore O2 produced comes from water and not from carbondioxide. So
the correct mechanism is as follows.
6C𝑂2* + 6H2O C6H12O6* + 6O2
2. Study of hydrolysis of ester
The mechanism of ester hydrolysis can be studied by using water labelled with O18
. The hydrolysis of
an ester by water enriched with radioactive oxygen is indicated as :
O
OR
R C + H - O*- H
O
R C
O*H
+ R-OH
Therefore it is the acid and not alcohol produced which is radioactive confirming the above mechanism.
8. Explain the use of radioactive isotopes in medicine ( any five )
[Mar-2006, June-2009, June-2011, June-2013, June-2014, Sep-2014, Jun-2016, Sep-16]
Isotope Use
Tritium Measure water content of the body
Carbon – 11 Brain scan
Carbon – 14 Radio immunology
Iron – 59 Diagnosis of anemia
Cobalt – 60 Treatment of cancer
Mercury – 197 Kidney scan
Gold – 198 Curing of cancer
Sodium – 24 Location of blood clots and circulatory disorders
Phosphorous – 32 Detection of eye tumours
Iodine –131 Diagnosis of damaged heart muscles and hyper thyroidism
Additional questions 1 MARK
1. The phenomenon of radioactivity was discovered by
a) Madam curie b) Pierre curie c) Henry Becquerrel d) Rutherford
2. Which of the following is used as neutron absorber in the nuclear reactor?
a) Water b) Deuterium c) Some compound of uranium d) Cadmium 3. All radioactive reactions follow
a) zero order b) first order c) second order d) third order
4. An α-particle is equal to the bundle of
a) two protons and two neutrons b) two protons and two electrons
c) two protons and four neutrons d) four protons and two neutrons
5. Which is fast moving electron?
a) α-particle b) β-particle c) γ-ray d) positron
6. Radio active heavy nuclei decay by a series of α-emission or β-emission, finally resulting in the
formation of a stable isotope of --------
a) lead (Pb208
) b) cobalt c) uranium d) none
7. Bombarding particle is called as
a) projectile b) ejectile c) recoil nucleus d) none
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8. 17Cl37
+ 1H2 18Ar
38 + X. X is --------
a) 1H1 b) 0n
1 c) –1e
0 d) 1H
2
9. Q value of a nuclear reaction will be positive in the case of reactions which
a) absorb energy b) release energy c) emit heat d) none
10. Reactions in which high speed projectiles chip a heavy nucleus into several fragments are known as
a) nuclear fusion b) nuclear fission c) spallation d) none
11. Which is used in nuclear reactor for producing power?
a) U235
b) U232
c) Mo95
d) none
12. Energy released from the fission of one U235
is
a) 200 MeV b) 100 MeV c) 2000 MeV d) 2 MeV
13. In nuclear reactor ---------- is used to take away heat from the core
a) oil b) inert gas c) heavy water d) none
14. Reactions in which lighter nuclei are fused together to form a heavy nucleus are known as
a) nuclear fusion b) nuclear fission c) spallation d) none
15. In fusion reactions, the mass of heavier nucleus formed is -------- the total mass of two lighter nuclei
a) less than b) more than c) equal to d) none
16. Nuclear fusion reaction takes place at very high temperature of about -------- K
a) 103 b) 10
2 c) 10
8 d) 10
17. Which of the following is known as thermonuclear reaction?
a) nuclear fusion b) nuclear fission c) spallation d) none
18 Hydrogen bomb is based on
a) nuclear fusion b) nuclear fission c) spallation d) none
19. Radiocarbon dating is based on
a) C12
b) C14
c) C11
d) C13
20. Half life of C14
is
a) 5700 years b) 500 years c) 3700 years d) 2700 years
21. Carbon-14 undergoes
a) α-decay b) β-decay c) γ-decay d) none
22. Sun is giving out energy at the rate of
a) 1.7 × 1023
ergs/sec. b) 1.7 × 1033
ergs/sec. c) 3.7 × 1023
ergs/sec. d) 3.7 × 1033
ergs/sec.
23. O2 produced during photosynthesis comes form
a) CO2 b) H2O c) both d) glucose
24. cobalt-60 is used in
a) brain scan b) kidney scan c) treatment of cancer d) none of these
25. Which of the following is used to find the thickness of coatings and level of liquids in tanks?
a) sodium-24 b) cobalt-60 c) strontium-90 d) iron-59
3 MARK
1. Define radio activity The phenomenon of spontaneous disintegration of certain atomic nuclei resulting in the emission of
radioactive rays is called radioactivity.
2. How is average life calculated?
Average life, (Tau) = λ
1 =
0.693
t 21
= 1.44 t ½
3. The half life of Th233
is 1.4 ×1010
years. Calculate its disintegration constant
=
½t
0.693
= 10 1.4
0.69310
= 0.495 x 10–10
year –1
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4 The atomic mass of Th is 232 and its atomic number is 90. In terms of its radioactivity six α and
four βparticles are emitted. What is the mass number and atomic number of the product?
90Th232
xAy + 6 2He
4 + 4 –1e
0
Comparing the mass numbers,
232 = y + 6(4) + 4(0)
y = 232–24 = 208
Mass number of the product = 208
Comparing the atomic numbers,
90 = x + 6(2) + 4(–1)
90 = x + 8
x = 82
Atomic number of the product = 82
5 MARK
1. Write a note on nuclear power generator or nuclear reactor
A nuclear reactor or nuclear power generator is a kind of furnace for carrying out the controlled fission
of a radioactive material like U235
for producing power. The core of the nuclear reactor produces heat
through nuclear fission. Heavy water at high pressure takes heat away from the core. In the heat
exchanger, the heavy water inside the reactor gives up its heat to water outside the reactor, which boils
to form steam. The steam is taken away to drive turbines that make electricity. In Tamilnadu atomic
power stations generating electricity are situated at Kalpakkam and another one is situated at
Koodamkulam.
2. Explain Nuclear fusion reaction with an example
When lighter nuclei moving at a high speed are fused together to form a heavy nucleus, the process is
called nuclear fusion.
In fusion reaction, the mass of heavier nucleus formed is less than the total mass of two lighter nuclei.
Thus, just like a fission reaction, the source of energy in a fusion reaction is also the disappearance of
mass, which gets converted into energy. Nuclear fusion reaction takes place at very high temperature of
about 108K. Therefore, this reaction is called thermonuclear reaction.
1H2 + 1H
3 2He
4 + 0n
1 + Energy
Deuterium Tritium Helium
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8. SOLID STATE - II
Blue print 1 Mark = 1 3 Mark = 1 5 Mark = 1 Total marks = 9
Govt. Examination questions 1 MARK
1. The Bragg’s equation is [Mar-2011, June-2014]
a) λ = 2d sinθ b) nd = 2λ sinθ c) 2λ = nd sinθ d) nλ = 2d sinθ
2. The crystal structure of CsCl is [ Sep-2006 ]
a) Simple cubic b) face-centred cubic c) Tetragonal d) Body centred cubic
3. An example for Frenkel defect is [June - 2012]
a) NaCl b) AgBr c) CsCl d) FeS
4. An example of metal deficiency defect [Mar-2008, Sep-2008]
a) NaCl b) AgCl c) CsCl d) FeS
Crystal
defect
Schottky
Defect
Frenkel
Defect
Metal excess
Defect
Metal deficiency
Defect
Example NaCl AgBr NaCl FeO, FeS
5. An ion leaves its regular site and occupies a position in the space between the lattice sites. This defect is
called as [March-2007]
a) Schottky defect b) Frenkel defect c) Impurity defect d) Vacancy defect
6. Semiconductors which exhibit conductivity due to the flow of excess negative electrons are called
[June- 2006, June-2013]
a) Superconductors b) n-type semiconductors c) p-type semiconductors d) Insulators
7. In the Bragg’s equation for diffraction of X-rays,’n’ represents [Jun-2008, Sep-2016]
a) number of moles b) Avogadro number c) quantum number d) Order of reflection
8. In a simple cubic cell, each point on a corner is shared by [Mar -2006, June-2007, Sep-2010]
a) One unit cell b) Two unit cell c) 8 unit cell d) 4 unit cell
Type of atom Corner
atom
Body centred
atom
Face centred
atom
Edge
atom
Shared by
(No.of unit cells) 8 1 2 4
9. The total number of atoms per unit cell in bcc is [Sep-2007, Sep-2009, June-2010, Sep-2011, Mar-2015]
a) 1 b) 2 c) 3 d) 4
10. The total number of atoms per unit cell in fcc is [Mar-2013]
a) 1 b) 2 c) 3 d) 4
Unit cell Simple Cubic BCC FCC Edge centred
No.of atoms per unit cell 1 2 4 4
11. Rutile is [June – 2011, Sep-2014, Mar-2016]
a) TiO2 b) Cu2O c) MoS2 d) Ru
12. The coordination number of ZnS is [Mar-2010]
a) 3 b) 4 c) 6 d) 8
13. The crystal lattice with coordination number four is [Sep-2012]
a) CsCl b) ZnO c) BN d) NaCl
Crystal CsCl NaCl FeS ZnO ZnS BN
Coordination
number 8 6 6 4 4 3
14. The number of close neighbours in a body centred cubic lattice of identical spheres is [Mar-09, Mar - 12]
a) 6 b) 4 c) 12 d) 8
15. Which one of the following crystal has 8 : 8 structure? [Sep-2013]
a) MgF2 b) CsCl c) KCl d) NaCl
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16. The size of the anion in Frenkel defect crystal is [June-2009]
a) larger than the cation b) smaller than the cation
c) equal in size with cation d) both are larger in size
17. The smallest repeating unit in space lattice which when repeated over and again results in the crystal of
the given substance is called [Mar-2014]
a) Space lattice b) Crystal lattice c) Unit cell d) Isomorphism
18. The number of chloride ions that surrounds the central Na+ ion in NaCl crystal is------[June-15, Mar-17]
a) 12 b) 8 c) 6 d) 4
19. The coordination number of BN is [Sep-2015]
a) 3 b) 4 c) 6 d) 8
20. The crystals which are good conductors of heat and electricity are [Jun-2016]
a) Ionic crystals b) Molecular crystals c) Metallic crystals d) Covalent crystals
3 MARK
1. Sketch the (a) simple cubic (b) face-centred cubic and (c) body centred cubic lattices
[Mar-2009, June-2012, Mar-2016]
2. Write a note on the assignment of atoms per unit cell in fcc [Sep – 2011]
A face atom is shared equally between two unit cells and therefore a face atom contributes only 2
N f to
the unit cell.
The total number of atoms per unit cell in fcc = 4312
6
8
8
2
N
8
N fc
Nf = Number of atoms at the faces.
3. What are superconductors?
[Mar-2006, June-2006, Sep-2008, June-2009, Mar-2010, Mar-2013, June-2015]
The ability of certain ultra cold substances to conduct electricity without resistance is called super
conductivity. This superconductivity state is a state in which a material has virtually zero electrical
resistance. Substances having this property are called super conductors.
4. What is meant by superconducting transition temperature? [June-2007, Sep-2014]
The super conducting transition temperature ‘Tc’ of a material is defined as a critical temperature at
which the resistivity of the material is suddenly changed to zero. Thus at that temperature a material is
changed from normal material to superconductor.
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5. Determine the number of CsCl units per unit cell. [Mar-2007]
CsCl is body centered cubic system. The chloride ions are at the corners of a cube where as Cs+ ion is at
the centre of the cube or vice versa. Each Cs+ ion is connected with eight Cl
– ions and each Cl
– is
connected with eight Cs+ ions.
Cs+
Cl-
Number of chloride ions per unit cell 18
8
8
Nc
Number of cesium ion per unit cell 11
1
1
Nb
Thus number of CsCl units per unit cell is one.
6. State Bragg’s law [June-2011, Sep-2013]
The fundamental equation which gives a simple relation between the wavelength of the X-rays, the
interplanar distance in the crystal and the angle of reflection, is known as Bragg’s equation.
where n = order of reflection
λ = wavelength of X-ray
d = interplanar distance in the crystal
θ = angle of reflection
7. Give three uses of super conductors [June-2010, Sep- 2010, June-2014, Mar-17]
Refer – 5 mark – Q - 4
8. What is a vitreous state? [Sep-2006, Sep-2007, Mar-2012, Mar-2014, Jun-2016]
Vitreous state is a condition in which certain substance can exist, lying between the solid and liquid
states. Eg: Glass
9. What is molecular crystal? Give an example [Mar-2008, Mar-2011]
1. The lattice points in molecular crystals consist of molecules which do not carry any charge.
2. The forces binding the molecules together are of two types
i) Dipole-dipole interaction and ii) Vanderwaal’s forces.
3. Dipole-dipole forces occur in solids which consists of polar molecules e.g., ice.
4. The Vanderwaal’s forces are more general and occur in all kinds of molecular solids.
10. Write note on Frenkel defect [ June-2008 ]
Refer – 5 mark – Q - 2
11. How are glasses formed? [Sep-2009, Sep-2012, June-2013, Mar-2015]
When certain liquids are cooled rapidly there is no formation of crystals at a definite temperature, such
as occurring on slow cooling. The viscosity of the liquid increases steadily and finally a glassy
substance is formed.
12. Write a note on Metal excess defect [ Sep-2015]
If a crystal of NaCl is heated in sodium vapour, it acquires a yellow colour. This yellow colour is due to
the formation of a non-stoichiometric compound of NaCl in which there is a slight excess of sodium
ions. This defect is called the metal excess defect.
n = 2d sinθ
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13. Write a note on metal deficiency defect [Sep-2016]
In certain cases, one of the positive ions is missing from its lattice site and the extra negative charge is
balanced by some nearby metal ion acquiring additional charges instead of original charge. This type of
defect is generally found in compounds of transition metals which can exhibit variable valency.
FeO and FeS show this type of defects.
5 MARK
1. Write the properties of ionic crystals [June-2007, Mar-2011, Sep-2011, June-2014, Mar-2015]
1. The melting and boiling points of ionic crystals are very high.
2. Ionic crystals are hard and brittle.
3. Ionic crystals are insulators in the solid state.
4. Ionic solids are good conductors when dissolved in water.
5. Ionic crystals are soluble in water and also in other polar solvents.
2. Write a note on Schottky defect and Frenkel defect (Or)
Write about the most common point defects [Mar-2006, Sep-2008, Mar-2009, June-2010, Sep-2010,
Mar-2012, Sep-2012, Mar-2013, Sep-2013, Sep-2014, June-2015, Sep-2015, Mar-2016]
Schottky defect
1. Schottky defect is caused if some of the lattice points are unoccupied. The points which are
unoccupied are called lattice vacancies.
2. The number of missing positive and negative ions is the same and thus, the crystal remains neutral.
3. Schottky defect appears generally in ionic crystals in which the positive and negative ions do not
differ much in size.
4. Example: NaCl
Cl is missingNa+ is missing
Frenkel defect 1. Frenkel defect arises when an ion occupies an interstitial position between the lattice points.
2. This defect occurs generally in ionic crystals in which the anion is much larger in size than the
cation.
3. The crystal remains neutral since the number of positive ions is the same as the number of negative
ions.
4. Example: AgBr
Ag+
Ag+
Ag+
Ag+
Ag+
Br -
Br -
Br -
Br -
Br -
Ag+
Ag+ Ag
+
Ag+
Ag+
Br -
Br - Br
-Br
-
Ag+
Ag+
Ag+
Ag+
Br -
Br - Br
-Br
-
Ag+
Ag+
Ag+
Ag+
Br -
Br - Br
-Br
- Br -
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3. Explain the nature of glass [March-2007, June-2013, Mar-17]
1. Glasses are optically isotropic
2. On heating without any sharp transition, glass passes into a mobile liquid.
3. They do not have definite geometrical forms.
4. Glasses are regarded as amorphous solids or super cooled liquids
5. Thus, glassy or vitreous state is a condition in which certain substance can exist, lying between the
solid and liquid states.
4. What is super conductivity? Give its uses [June-2008, June-2011, June-2012]
The ability of certain ultra cold substances to conduct electricity without resistance is called super
conductivity. This superconductivity state is a state in which a material has virtually zero electrical
resistance. Substances having this property are called super conductors.
Temperature
Super conductor
Tc
Normal
metal
Res
isti
vit
y
The super conducting transition temperature ‘Tc’ of a material is defined as a critical temperature at
which the resistivity of the material is suddenly changed to zero. Thus at that temperature a material is
changed from normal material to superconductor.
At the extremely low temperatures, vibration of the nuclei of certain atoms slows down so much and
they synchronise with the passing waves of electrons in a flow of electric current. When this happens,
resistance to electric current disappears.
Applications of superconductors
1. It is a basis of new generation of energy saving power systems. Super conducting generators are
smaller in size and weight when we compare with conventional generators.
These generators consume very low energy and so we can save more energy.
2. High efficiency ore separating machines may be built using superconducting magnets.
3. Superconducting solenoids are used in Nuclear Magnetic Resonance Imaging equipment which is a
whole body scan equipment.
5. Explain Bragg’s spectrometer method
[Jun-2006, Sep-2006, Sep-2007, March-2008, Jun-2009, Sep-2009, Mar-2010, Mar-2014, Sep-16]
This method is used for studying crystals using X-rays.
The apparatus consists of a X-ray tube from which a narrow beam of X-rays is allowed to fall on the
crystal mounted on a rotating table.
The rotating table is provided with scale and vernier, from which the angle of incidence, θcan be
measured.
An arm which is rotating about the same axis as the crystal table carries an ionisation chamber. The
rays reflected from the crystal enter into the ionization chamber and ionise the gas present inside.
Due to the ionisation, current is produced which is measured by electrometer.
The current of ionisation is a direct measure of intensity of reflected beam from the crystal.
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Turn table
ionisation
chamber
Path of
Recorder
Diffracted X-rays
Slit
X-ray tube
incident
X-rays
Crystal
Scale to measure
rotation of crystal
For different angles of incidence, the corresponding ionisation current is measured from the
electrometer. These values are plotted in the form of graph.
5.9o11.85o 18.15o
Ion
isat
ion
curr
ent
n=1
n=2 n=3
For sodium chloride, the maximum reflection or peaks for 100 plane occurred at,
θ 5.9° 11.85° 18.15°.
Sin θ 0.103 0.205 0.312
Ratio 1 2 3
These peaks represent the first, second and third order reflections.
The ratio confirms the correctness of Bragg’s equation.
6. What is Bragg’s equation? Give its significance [Jun-2016]
The fundamental equation which gives a simple relation between the wavelength of the X-rays, the
interplanar distance in the crystal and the angle of reflection, is known as Bragg’s equation.
where n = order of reflection
λ = wavelength of X-ray
d = interplanar distance in the crystal
θ = angle of reflection
n = 2d sinθ
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Significance of Bragg’s equation
1) If we use X-rays of known wavelength (λ), then the inter atomic distance (d) in an unknown crystal
can be calculated. On the other hand, if we use a crystal whose inter atomic distance‘d’ is known,
then the wavelength of X-rays can be calculated.
2) The Bragg’s equation gives the essential condition for diffraction to occur.
3) When the experiment is done, there will be a maximum reflection at a particular angle θ. That angle
is noted. It corresponds to first order reflection (n=1). If the angle ‘θ’ is increased, maximum
reflection occurs at some other higher angle. It corresponds to second order reflection (n=2).
Similarly, third, fourth and higher order of reflection occur at certain specific angles. The values of
angles obtained are in accordance with the Bragg’s equation. Hence Bragg’s equation is
experimentally verified.
Additional questions
1 MARK
1. A regular three dimensional arrangement of identical points in space is called
a) Unit cell b) Space lattice c) Primitive d) Crystallography
2. The ability of certain ultra cold substances to conduct electricity without resistance is called
a) Semiconductor b) Conductor c) Superconductor d) Insulator
3. Semiconductors are used as
a) rectifiers b) transistors c) solar cells d) all the above
4. Wave length of X-rays is in the order of
a) 10–8
m b) 10–8
cm c) 108 m d) 10
8 cm
5. Laue carried out X-ray diffraction experiment using ---------- crystal
a) zinc sulphide b) nickel c) sodium chloride d) copper
6. Molecular crystal is
a) diamond b) ice c) sodium chloride d) graphite
7. Covalent crystal is
a) ice b) diamond c) nickel d) NaCl
8. Number of CsCl units per unit cell is
a) 1 b) 2 c) 3 d) 4
9. Which is body centred cubic?
a) TiO2 b) CsCl c) NaCl d) CO2
10. Number of corners in a cubic unit cell is
a) 1 b) 2 c) 4 d) 8
No.of corners No.of faces No.of body centre No. of edges
8 6 1 12
3 MARK 1. Define the terms: space lattice and unit cell
A regular three dimensional arrangement of points (atoms, ions or molecules) in space is called space
lattice Unit cell is the smallest fundamental repeating portion of a crystal lattice from which the crystal
is built by repetition in three dimension.
2. Write a note on the assignment of atoms per unit cell in sc
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In a simple cubic unit cell, atoms are present only at the corners. Each atom at the corner is shared
equally by eight other unit cells. Hence the contribution of each atom to the unit cell is 8
1
The total number of atoms per unit cell 18
8
8
Nc
3. Write a note on the assignment of atoms per unit cell in bcc
In a bcc lattice, the body centred atom belongs exclusively to the unit cell.
The total number of atoms per unit cell in bcc 2111
1
8
8
1
N
8
N bc
4. Write a note on the assignment of atoms per unit cell in Edge centred cubic lattice An edge atom and edge centred is common to four unit cells and there are twelve edges of the unit cell.
The contribution from each edge atom is therefore 4
1
The total number of atoms per unit cell in edge centre 4314
12
8
8
4
N
8
N ec
5. What is rutile? Draw its structure
Rutile is TiO2
Ti
Oxygen
6. Write a note on semiconductors Elements like silicon and germanium are non conductors at ordinary temperature. However, they exhibit
appreciable conductivity upon addition of impurities as Arsenic and Boron. The resulting materials are
called semiconductors.
Semiconductors which exhibit conductivity due to the flow of excess negative electrons are called n-
type semiconductors.
Semiconductors which exhibit conductivity due to the positive holes are called p-type semiconductors.
Uses of semi conductors:
Semiconductors find application in modern devices as rectifiers, transistors and solar cells.
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5 MARK
1. Explain the different types of crystals
i) Molecular Crystals
The lattice points in molecular crystals consist of molecules which do not carry any charge.
The forces binding the molecules together are of two types
i) Dipole-dipole interaction and ii) Vanderwaal’s forces.
Dipole-dipole forces occur in solids which consists of polar molecules e.g., ice.
The Vanderwaal’s forces are more general and occur in all kinds of molecular solids.
ii) Covalent Crystals
The lattice in covalent crystals consists of atoms linked together by a continuous system of covalent
bonds. e.g Diamond.
iii) Metallic Crystals
Metallic crystal consists of an assemblage of positive ions immersed in a sea of mobile electrons. Thus,
each electron belongs to a number of positive ions and each positive ion belong to a number of
electrons. The force that binds a metal ion to a number of electrons within its sphere of influence is
known as metallic bond. This force of attraction is strong and is thus responsible for a compact solid
structure of metals.
iv) Ionic Crystals
In ionic crystals, the units occupying lattice points are positive and negative ions. Each ion of a given
sign is held by coulombic forces of attraction to all ions of opposite sign. The forces are very strong.
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9. THERMODYNAMICS - II
Blue print 1 Mark = 2 3 Mark = 1 5 Mark = 1 Total marks = 10
Govt. Examination questions 1 MARK
1. The amount of heat exchanged with the surrounding at constant temperature and pressure is called
[June-2006, Mar-2007]
a) ΔE b) ΔH c) ΔS d) ΔG
2. All the naturally occurring processes proceed spontaneously in a direction which leads to
[Mar-2009, Sep-2010, Sep-2012]
a) decrease of entropy b) increase in enthalpy
c) increase in free energy d) decrease of free energy
3. In an adiabatic process, which of the following is true? [Mar-2014, Mar-2016]
a) q = w b) q = 0 c) ΔE = q d) P ΔV = 0
4. When a liquid boils, there is [Mar-2007, June-2007, June-2009, June-2014, Sep-2015]
a) an increase in entropy b) a decrease in entropy
c) an increase in heat of vapourisation d) an increase in free energy
5. If ΔG for a reaction is negative, the change is [Jun-2006, Sep-2007, Mar-2008, Sep-2008, Sep-2016]
a) Spontaneous b) Non-spontaneous c) Reversible d) Equilibrium
6. Which of the following does not result in an increase in the entropy?
[Mar-2006, Mar-2010, Mar-2013, Mar-17]
a) crystallisation of sucrose from solution b) rusting of iron
c) conversion of ice to water d) vapourisation of camphor
7. In which of the following process, the process is always non-feasible?
[Sep-2009, Mar-2011, Mar-2012, June-2012, Mar-2015]
a) ΔH > 0, ΔS > 0 b) ΔH < 0, ΔS > 0 c) ΔH > 0, ΔS < 0 d) ΔH < 0, ΔS < 0
8. Change in Gibb‟s free energy is given by [Sep-2006, June-2008, June-2009, Sep-2013, Jun-2016]
a) ΔG = ΔH + TΔS b) ΔG = ΔH – TΔS c) ΔG = ΔH × TΔS d) None of the above
9. For the reaction 2Cl (g) Cl2 (g) , the signs of ΔH and ΔS respectively are
[Sep-2006, Sep-2007, June-2013, June-2015]
a) +, – b) +, + c) –, – d) –, +
10. The entropy change involved in the process H2O (s) H2O (l ) at 0°C and 1atm pressure involving
ΔH fusion = 6008 J mol–1
is --------- [June-2007, June-2011]
a) 22.007 J mol–1
k–1
b) 22.007 J mol k–1
c) 220.07 J mol–1
k–1
d) 2.2007 J mol–1
k–1
11. The change of entropy for the process, H2O (liq) H2O (vapor) involving ΔHvap= 40850 J mol–1
at
373K is -------- [Mar-2006, Sep-2008, Sep-2010]
a) 120 J mol–1
k–1
b) 9.1 × 10–3
J mol–1
k–1
c) 109.52 J mol–1
k–1
d) 9.1 × 10–4
J mol-1
k–1
12. Entropy (S) and the entropy change (ΔS) of a process [Mar-2008, Sep-2011, Mar-2014]
a) are path functions b) are state functions c) are constants d) have no values
13. H2O(l) H2O(g). In this process the entropy [June-2008]
a) remains constant b) decreases c) increases d) becomes zero
14. The percentage efficiency of a heat engine that operates between 127oC and 27
oC is
[Mar-2009, June-2015]
a) 20% b) 50% c) 100% d) 25%
15. Net work obtained from a system (– ΔG ) is equal to [Sep-2009, Sep-2011]
a) w + P ΔV b) w – P ΔV c) – w + P ΔV d) – w – P ΔV
16. Thermodynamic condition for irreversible spontaneous process at constant T and P is
[Mar-2010, Mar-2013]
a) ΔG < 0 b) ΔS < 0 c) ΔG > 0 d) ΔH > 0
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17. According to Trouton‟s rule, the value of change in entropy of vaporization is [Mar-2011]
a) 21 cal deg–1
mol–1
b) 12 cal deg–1
mol–1
c) 21 kcal deg mol–1
d) 12 kcal deg mol–1
18. A process accompanied by increase in entropy tends to be [Jun-2012, Sep-2016]
a) spontaneous b) non-spontaneous c) reversible d) non-feasible
19. Which of the following are State functions? [June-2010]
a) q b) Δq c) w d) ΔS
20. SI unit of entropy is [Sep-2013]
a) J deg–1
mol–1
or EU b) Cal–1
K c) Cal K d) J K 21. Standard free energies of formation of elements are taken as [Jun-2010, Sep-2014, Jun-2016]
a) positive b) negative c) zero d) infinitive 22. Free energy (G) and the free energy change (∆G) correspond to the [June-2011]
a) system only b) surrounding only c) system and surrounding d) all of these 23. In SI unit 1 eu is [Mar-2012]
a) 41.84 EU b) 4.184 EU c) 4184 EU d) 418.4 EU 24. The entropy change for the following process possessing ∆H(transition) as 2090 J mol
–1
1 mol Sn (α, 13oC) ⇌ 1 mol Sn (β, 13
oC) is [Sep-2012]
a) 22.007 J mol–1
K–1
b) 7.307 J mol–1
K–1
c) 0.314 J mol–1
K–1
d) 109.52 J mol–1
K–1
25. For an isothermal process, the entropy change of the universe during a reversible process is ------------
[June-2013]
a) Zero b) More c) Less d) None of the above
26. Which is the correct statement of IInd
1aw of Thermodynamics ? [June-2014]
a) A process accompanied by decrease in entropy tends to be spontaneous
b) Efficiency of a machine can be cent percent
c) It is impossible to transfer heat from a cold body to hot body by a machine without doing any
work
d) It is possible to convert the input energy completely to work by a machine
27. The percentage efficiency of a heat engine working between the temperature T1 K and T2 K (T1 > T2) is
given by [Sep-2014]
a) 100T
TT
2
12
b) 100
T
TT
1
21
c) 100
T
T
2
1 d) 100T
T
1
2
28. The change in entropy for a system and surrounding are – 0.228 JK – 1
and + 0.260 JK – 1
respectively.
Then entropy change of the universe is [Mar-2015]
a) – 0.0313 JK – 1
b) +0.0313 JK – 1
c) + 0.877 JK – 1
d) – 0.877 JK – 1
Note: ΔS universe = ΔS system + ΔS surrounding
= (– 0.228 JK – 1
) + ( + 0.260 JK – 1
)
= + 0.032 JK – 1
29. The expression G = H – TS is valid for [Sep-2015]
a) constant pressure processes b) constant temperature and pressure processes
c) constant temperature processes d) constant volume processes 30. In a chemical reaction, when number of molecules of products are more than the number of molecules
of reactants [Mar-2016]
a) entropy increases b) entropy decreases
c) heat of vapourisation increases d) free energy increases
31. The liquid that deviates from Trouton‟s rule is [Mar-2017]
a) Hydrchloric acid b) Sulphuric acid c) Phosphoric acid d) acetic acid
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3 MARK
1. What is entropy? What are the units of entropy? [March-2006, Mar-2015, Mar-17]
Entropy is a measure of randomness or disorder of the molecules of a system. It is a state function.
S = T
q
S = entropy, q = heat involved, T = temperature of the process
Units of entropy:
cgs unit of entropy is cal K–1
(or) eu
SI unit of entropy is J K–1
(or) EU (1 eu = 4.184 EU)
2. State Trouton’s rule [June-2009]
According to Trouton‟s rule, the heat of vaporisation ( ΔHvap ) in calories per mole divided by the
boiling point of the liquid in Kelvin is a constant equal to 21 cal deg–1
mole–1
and is known as the
entropy of vapourisaiton.
ΔSvap = b
vap
T
ΔH = 21 cal deg
–1 mol
–1
3. What type of liquids deviate from Trouton’s rule? [Sep-2006, June-2011]
Liquids Reason for deviation from Trouton’s rule
Hydrogen and Helium Low boiling points. They boil only a little above 0 K
Water and alcohol High boiling points due to hydrogen bond & high ΔHvap
Acetic acid Molecules are partially associated in the vapor phase and
possess very low entropy of vaporization which is very
much less than 21 cal deg–1
mol–1
4. What is Gibb’s free energy? [June-2006]
Gibb‟s free energy G is defined as,
G = H – TS
where H = enthalpy or heat content of the system,
T = Temperature in Kelvin
S = entropy
5. How ΔG is related to ΔH and ΔS? What is the meaning of ΔG = 0? [Sep-2010]
ΔG = ΔH – TΔS
Where, ΔG = free energy change
ΔH = changes in enthalpy
ΔS = changes in entropy
T = Temperature in Kelvin
If ΔG = 0 then the process is in equilibrium (reversible)
6. What is the nature of reaction when ΔG > 0, ΔG < 0, ΔG = 0 ? [Sep-2011]
ΔG > 0 : Non spontaneous ΔG < 0 : Spontaneous ΔG = 0 : Equilibrium
7. Give Kelvin – Planck statement of second law of thermodynamics
Refer – 5 mark – Q - 1 [Mar-2007, Sep-2012, Jun-2013, Jun-2016]
8. Give Clausius statement of second law of thermodynamics [June-2007, Sep-2014]
Refer – 5 mark – Q - 1
9. Give Entropy statement of second law of thermodynamics [Sep-2008]
Refer – 5 mark – Q - 1
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10. What is the entropy change of an engine that operates at 100°C when 453.6 k.cal of heat is
supplied to it? [Sep-2007, Sep-2009, June-2012]
ΔS rev = T
q rev =
373
453.6 =1.216 k Cal K
–1 = 1216 Cal K
–1
11. For a chemical reaction the values of ΔH and ΔS at 300 K are –10 k Cal mol–1
and
20 Cal deg–1
mol–1
respectively. What is the value of ΔG of the reaction? [June-2008, June-2010]
ΔH = –10 k Cal mol–1
= –10,000 Cal mol–1
ΔS = 20 Cal.deg–1
mol–1
ΔG = ΔH – TΔS
= –10,000 – (300 × 20)
= –16000 Cal mol–1
12. For a chemical reaction the values of ΔH and ΔS at 400 K are –10 k Cal mol–1
and
20 Cal deg–1
mol–1
respectively. What is the value of ΔG of the reaction? [Mar-2016]
ΔH = –10 k Cal mol–1
= –10,000 Cal mol–1
ΔS = 20 Cal.deg–1
mol–1
ΔG = ΔH – TΔS
= –10,000 – (400 × 20)
= –18000 Cal mol–1
13. Calculate the molar heat of vapourisation of the ideal liquid CCl4 (B.Pt 76.7oC and
ΔS =87.864 J K-1
mol–1
) [Mar-2008]
ΔSvap = b
vap
T
ΔH
ΔS = 87.864 J K–1
mol–1
& Tb = 76.7 + 273 = 349.7 K
Δ H vap = 87.864 × 349.7 = 30726 J mol–1
= 30.726 kJ mol–1
14. Calculate the molar heat of vapourisation of the ideal liquid CHCl3 (B.Pt 61.5oC) [Mar-12, Sep-16]
ΔSvap = b
vap
T
ΔH
Δ H vap = (21 cal mol–1
K–1
× 4.184) (273 + 61.5) K
= 29390 J mol–1
= 29.390 kJ mol–1
15. Calculate the change of entropy for the process, water (liq) to water vapour involving
ΔHvap = 40850 J mol –1
at 373 K [Mar-2010, Sep-2013, June-2015]
ΔSvap = b
vap
T
ΔH =
373
40850 = 109.517 J K
–1 mol
–1
16. Calculate the entropy change involved in the conversion of 1 mole of ice at 0oC and 1 atm to liquid
at 0oC and 1 atm. The enthalpy of fusion per mole of ice is 6008 J mol
–1 [Mar-2009]
H2O ( s ) C0
H2O ( l )
Ice water
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ΔSfusion = m
fusion
T
ΔH
ΔHfusion = 6008 J mol–1
& Tm = 0oC + 273 = 273 K
ΔSfusion = 273
6008 = 22.007 J K
–1 mol
–1
17. Calculate the maximum % efficiency possible from a thermal engine operating between 110°C
and 25°C [Mar-2011, June-2014]
% efficiency =
1
2
T
T1 × 100 T1 = 110°C + 273 = 383 K
T2 = 25°C + 273 = 298 K
=
383
2981 × 100
= 22.2 %
18. Calculate the entropy change for the following process possessing ΔH(transition) =2090 Jmol–1
1 mole Sn (α, 13ºC) ⇌ 1 mole Sn (β, 13ºC) [Mar-2013]
T(transition) = 13°C + 273 = 286 K
ΔS(transition) = transition
transition
T
ΔH =
286
2090 = 7.307 J K
–1 mol
–1
19. Calculate the entropy increase in the evaporation of 1 mole of a liquid when it boils at 100°C
having heat of vaporisation at 100°C as 540 cal / gm. [Mar-2014]
T = 100°C + 273 = 373 K
ΔH vap = 540 cal / g = 540 × 18 = 9720 cal / mol {1 mol of water = 18 g }
ΔSvap = b
vap
T
ΔH =
373
9720 = 26.06 cal K
–1 mol
–1
20. Calculate the standard entropy of formation ΔSo
f of CO2 (g). Given the standard entropies of
CO2(g), C (s), O2 (g) as 213.6, 5.740 and 205.0 JK– 1
respectively. [Sep-2015]
C (s) + O2 (g) CO2 (g)
ΔSo
f (compound) = ∑ S
ocompound - ∑ S
oelements
= So CO2 (g) - [ S
o C(s) + S
o O2 (g) ]
= 213.6 – [ 5.740 + 205.0]
= 213.6 – 210.74
ΔSof CO2 = 2.86 JK
– 1
5 MARK
1. State the various statements of second law of thermodynamics [Mar-2006, June-2006, Sep-2006,
Mar-2008, Sep-2009, Mar-2010, June-10, Sep-10, June-11, Sep-11, Mar-13, Sep-13, June-15, Mar-16]
1. Kelvin – Planck statement
“It is impossible to construct an engine operating in a complete cycle which will absorb heat from a
single body and convert it completely to work without leaving some changes in the working system”.
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2. Clausius statement
“It is impossible to transfer heat from a cold body to a hot body by a machine without doing some
work”.
3. Entropy statement
„A process accompanied by increase in entropy tends to be spontaneous”.
Entropy is a measure of randomness or disorder of the molecules of a system
A system always spontaneously changes from ordered to a disordered state. Therefore entropy of a
spontaneous process is constantly increasing.
4. Efficiency statement
“Efficiency of a machine can never be cent percent”.
The Efficiency of any machine is given by the value of ratio of output to input energies.
% efficiency = input
output× 100
Consider a heat engine which has an initial temperature T1 and final temperature as T2, then if T1>T2
then when some amount of heat is being converted into work, T2 is the lowered temperature. The
efficiency „η‟ is given by,
% efficiency = 1
21
T
TT × 100
According to II law of thermodynamics it is impossible to have a machine or heat engine which
converts the input energy completely into output energy or output work without any amount of heat
or energy being absorbed by the machine. Hence, % efficiency can never be achieved as cent
percent.
% efficiency =
1
2
T
T1 × 100
By II law, T2 < T1 and % efficiency less than 100.
2. What are the characteristics of entropy?
[Jun-2008, Sep-2008, Mar-2009, Mar-2012, Sep-2012, Jun-2013, Jun-2014, Sep-2014, Jun-2016, Sep-16]
1. The term „S‟ entropy is evolved from the formulation of II law of thermodynamics. It is a state
function.
2. Entropy change „ΔS‟ of a system is defined as the constant equal to the ratio of the heat change
accompanying a process at constant temperature to the temperature of the system.
ΔS rev = T
q rev
3. A spontaneous process is accompanied by increase in the „disorder‟ (or) entropy of the molecules
constituting the system.
4. The energy of the universe remains constant although the entropy of the universe tends to a
maximum.
5.
Spontaneous process ΔS = positive or ΔS > 0
Non spontaneous process ΔS = negative or ΔS < 0
Equilibrium process ΔS = 0
6. Units of entropy:
cgs unit of entropy is cal K–1
(or) eu
SI unit of entropy is J K–1 (or) EU (1 eu = 4.184 EU)
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3. Write the characteristics of Gibbs free energy G
[Mar-2007, June-2007, Sep-2007, June-2009, Mar-2011, Mar-2014, Mar-2015, Mar-17]
1. Gibbs free energy, G = H –TS
where H and S are the enthalpy and entropy of the system respectively. T = temperature.
G is a state function.
2. G is an extensive property
3. G has a single value for the thermodynamic state of the system.
4. G and ΔG values correspond to the system only. There are three cases of ΔG in predicting the nature
of the process.
Spontaneous process ΔG = negative or ΔG < 0
Non spontaneous process ΔG = positive or ΔG > 0
Equilibrium process ΔG = 0
5. ΔG = ΔH – TΔS
But according to I law of thermodynamics,
ΔH = ΔE + PΔV and ΔE = q – w
ΔG = ΔE + PΔV – TΔS
ΔG = q – w + PΔV – TΔS
But ΔS = T
q and TΔS = q
ΔG = q – w + PΔV – q = –w + PΔV
(or)
–ΔG = w – PΔV
Network = maximum obtainable work from the system – work of expansion
4. State Trouton’s rule. What type of liquids deviate from Trouton’s rule? [June-2012, Sep-2015]
Refer – 3 mark – Q – 2 & 3
Additional questions
1 MARK 1. Entropy can be calculated by the formula
a) S = T
q b) S =
q
T c) S = q + T d) S = q –T
2. Liquids which deviate from Trouton‟s rule
a) liquid H2 & liquid He b) alcohol & water c) acetic acid d) all the above
3. Liquids which obey Trouton‟s rule
a) CCl4 b) CHCl3 c) H2S d) all are above
4. Which of the following has low entropy of vapourisation?
a) CCl4 b) CHCl3 c) H2S d) CH3COOH
5. Water and alcohol deviate from Trouton‟s rule because they have
a) very high boiling point as a result of hydrogen bonding b) low boiling point
c) high density d) low viscosity
6. Hydrogen and helium deviate from Trouton‟s rule because they have
a) very high boiling point b) very low boiling point c) high density d) low viscosity
7. For spontaneous process
a) ΔH < T ΔS b) ΔH –T ΔS = –ve c) ΔH –T ΔS < 0 d) all are correct
8. For non-spontaneous process (non-feasible process)
a) ΔS < 0 or ΔS = –ve b) ΔH > 0 or ΔH = +ve
c) ΔG > 0 or ΔG = + ve d) all are correct
9. For equilibrium or reversible process
a) ΔS = 0 b) ΔG = 0 c) ΔH = 0 d) all are correct
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10. Which of the following are path functions?
a) q and w b) G and H c) S and T d) all the above
11. Which of the following are State functions?
a) S b) G c) H d) all the above
12. cgs unit of entropy is
a) Cal deg-1
mol-1
or eu b) cal-1
K c) Cal K d) J K
13. In which of the following process entropy is increased?
a) H2O (s) H2O (l) b) I2 (s) I2 (g)
c) H2O (l) H2O (g) d) all the above
14. Standard entropies of all substances at any temperature above 0 K or –273oC always have
a) +ve value b) –ve value c) zero d) none of these
15. Gibb‟s free energy is given by
a) G = H – TS b) G = H + TS c) G = T – HS d) G = HS – T
16. Entropy of vaporization of acetic acid in cal deg–1
mol–1
is
a) greater than 21 b) less than 21 c) equal to 21 d) none of these
17. Expansion of a gas against low external pressure is-------------- process
a) spontaneous b) non-spontaneous c) reversible d) cyclic
3 MARK 1. Define standard entropy
The entropy of a pure substance at 25°C (298 K) and 1 atm pressure is called the standard entropy, S°.
2. Define standard entropy change of formation Standard entropy change of formation, ΔS°f is defined as the entropy of formation of 1 mole of
compound from the elements present in the standard conditions. [25°C (298 K) and 1 atm]
ΔS°f can be calculated for chemical compounds using the S° values of elements from which the
compound is formed.
ΔS°f compound = o
elements
o
compoundSS
3. Predict the feasibility of a reaction when
a) both ΔH and ΔS increase b) both ΔH and ΔS decrease
c) ΔH decreases but ΔS increases d) ΔH increases but ΔS decreases
a) both ΔH and ΔS increase: The reaction is spontaneous only at high temperature
b) both ΔH and ΔS decrease: The reaction is spontaneous only at low temperature
c) ΔH decreases but ΔS increases: The reaction is always spontaneous
d) ΔH increases but ΔS decreases: The reaction is always non-spontaneous
4. Define standard free energy
The standard free energy value (G°) of all substances either elements or compounds may be calculated
from H° and S° values at standard conditions of temperature (298 K) and pressure (1 atm) and the
substance being present in the standard state.
G° = H° – TS°
5. Define standard free energy change
Standard free energy change of a reaction is the difference between the total sum of the standard free
energies of products and the total sum of the standard free energies of reactants.
oreactionΔG = o
reactantsoproducts GG
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6. Mention the essential thermodynamic condition for spontaneity in a chemical reaction
Spontaneous
(irreversible)
Equilibrium
(reversible)
Non spontaneous
(nonfeasible)
ΔG = – ve or ΔG < 0
ΔH = – ve or ΔH < 0
ΔS = +ve or ΔS > 0
ΔG = 0
ΔH = 0
ΔS = 0
ΔG = +ve or ΔG > 0
ΔH = +ve or ΔH > 0
ΔS = – ve or ΔS < 0
7. When does entropy increase in a process?
1. In a chemical reaction, when number of molecules of products is more than the number of
molecules of reactant entropy increases.
2. In physical process, when a solid changes to liquid, when a liquid changes to vapour and when a
solid changes to vapour, entropy increase in all these processes.
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10. CHEMICAL EQUILIBRIUM–II
Blue print 1 Mark = 2 3 Mark = 1 5 Mark = 1 Total marks = 10
Govt. Examination questions 1 MARK
1. State of chemical equilibrium is [Mar-08, Jun-08, Jun-10, Jun-13, Sep-14, Jun-15, Sep-15, Jun-16]
a) dynamic b) stationery c) none d) both
2. If the equilibrium constants of the following reactions are 2A ⇌ B is K1 and B ⇌ 2A is K2 then
[Sep-2006, Mar-2009, Sep-2010, Mar-2012, Mar-2014, Mar-2016]
a) K1 = 2K2 b) K1 =
2K
1 c) K2 = (K1)
2 d) K1 =
22K
1
3. In the reversible reaction 2HI ⇌H2 + I2 Kp is [Mar-2007, Sep-2008, June-2009, Sep-2013, Sep-2015]
a) greater than Kc b) less than Kc c) Equal to Kc d) Zero
4. In the equilibrium N2 + 3H2 ⇌2NH3 the maximum yield of ammonia will be obtained with the process
having [June-09, Sep-09, Mar-10, Mar-11,Mar-13, Mar-17]
a) low pressure and high temperature b) low pressure and low temperature
c) high temperature and high pressure d) high pressure and low temperature
5. For the homogeneous gas reaction at 600 K, 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) the equilibrium
constant Kc has the unit [Mar-2006, June-2006, Sep-2008, Mar-2009, June-2014, Mar-2015]
a) (mol dm–3
) –1
b) mol dm–3
c) (mol dm–3
)10
d) (mol dm–3
) –9
6. In the synthesis of NH3 between N2 and H2 reaction the unit of Kp is [Sep-2007]
a) lit2 mol
–1 b) atm
–2 c) lit atm
–1 d) atm
–1
Note: [ Unit of Kc = ( mol dm–3
) ∆ng
& Unit of Kp = ( atm ) ∆ng
]
7. Two moles of ammonia gas are introduced into a previously evacuated 1.0 dm3 vessel in which it
partially dissociates at high temperature. At equilibrium 1.0 mole of ammonia remains. The equilibrium
constant Kc for the dissociation is [Sep-2011, June-2012]
a) 16
27 (mole dm
–3)2
b) 8
27 (mole dm
–3)2
c) 4
27 (mole dm
–3)2
d) None of these
8. An equilibrium reaction is endothermic. If K1 and K2 are the equilibrium constants at T1 and T2
temperatures respectively and if T2 is greater than T1 then [Jun-2007, Sep-2016]
a) K1 is less than K2 b) K1 is greater than K2 c) K1 is equal to K2 d) None
9. 2H2O(g) + 2Cl2(g) ⇌ 4HCl(g) + O2(g) the value of Kp and Kc are related as [Mar-06, Mar-11, Mar-17]
a) Kp = Kc b) Kp > Kc c) Kp < Kc d) Kp = Kc =0
10. If the equilibrium constant for the formation of product is 25, the equilibrium constant for the
decomposition of the same product is [June-2006]
a) 25 b) 25
1 c) 5 d) 625
11. The ∆ng in a homogeneous reaction is positive, then [Sep-06, Jun-10, Sep-10, Jun-11, Sep-14, Mar-15]
a) Kp = Kc b) Kp < Kc c) Kp > Kc d) Kp = 2
K c
∆ng = + ve or ∆ng > 0 Kp > Kc
∆ng = – ve or ∆ng < 0 Kp < Kc
∆ng = 0 Kp = Kc
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12. Which of the following has negative value of ∆ng? [Mar-2008]
a) H2 + I2 ⇌ 2HI b) PCl5 ⇌ PCl3 + Cl2
c) 3H2 + N2 ⇌ 2NH3 d) 2H2O + 2Cl2 ⇌ 4HCl + O2
13. In which of the following gaseous reactions Kp < Kc? [Sep-2012]
a) PCl5(g) ⇌ PCl3(g) + Cl2 (g) b) H2(g) + I2(g) ⇌ 2HI (g)
c) N2(g) + 3H2(g) ⇌ 2NH3(g) d) CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
14. In the formation of HI fromH2 and I2, Kp = Kc because [June-2013]
a) ng = 2 b) ng = 1 c) ng = 0 d) ng = –1
15. Which of the following gaseous equilibria is favoured by increase in temperature? [Mar-2007]
a) N2O4 ⇌ 2NO2 ∆H = + 59 kJ mol–1
b) N2 + 3H2⇌ 2NH3 ∆H = – 22 kCal mol–1
c) 2SO2 + O2⇌ 2SO3 ∆H = –47 kCal mol–1
d) both b and c
16. The maximum yield of NH3 by Haber‟s process is [June-2007]
a) 78 % b) 97 % c) 37 % d) 89 %
17. In a reaction 2O3 ⇌ 3O2 the value of Kc is [Sep-2007]
a)
22
33
O
O b)
33
22
O
O c)
2
3
3
2
O
O d)
2
3
O
O
18. H2 + I2 ⇌ 2HI . The equilibrium constant Kc for this reaction is 16. Kp is [June-2008, June-2015]
a) 16
1 b) 4 c) 64 d) 16
19. The optimum range of temperature used in contact process for the synthesis SO3 is [Sep-2008]
a) 400 oC – 450
oC b) 1800
oC – 2700
oC c) 500
oC–550
oC d) 350
oC–450
oC
20. The rate constant of the forward and reverse reactions are 8 × 10–5
and 2 × 10–4
respectively. The value
of Kc is [Sep-2009, Sep-2013]
a) 0.04 b) 0.02 c) 0.2 d) 0.4
21. The relationship between Kp and Kc for the equilibrium 2H2O(g) + 2Cl2(g) ⇌ 4HCl(g) + O2(g)
[Mar-2011, Mar-2013]
a) Kp = Kc b) Kp = Kc (RT) 2
c) Kp = Kc (RT)1
d) Kp = Kc (RT) – 2
22. The equilibrium constant for the reaction 2A⇌B is 25 mol–1
dm3 at 900 K. What is the equilibrium
constant for the reaction B⇌2A in dm–3
mol at the same temperature? [Mar-2010]
a) 25 b) 625 c) 0.04 d) 0.4
23. Forward reaction takes place when [June-2011]
a) Q < Kc b) Q > Kc c) Q = Kc d) Kc = Q
1
24. The fraction of total moles of reactant dissociated is called [Sep-2011]
a) dissociation equilibrium b) degree of association
c) degree of dissociation d) dissociation constant
25. The equilibrium constant Kc for A⇌B is 2.5 × 10–2
. The rate constant of forward reaction is 0.05 sec– 1
.
Therefore, the rate constant of the reverse reaction is [Mar-2012]
a) 2 sec– 1
b) 0.2 sec– 1
c) 2 minute– 1
d) 0.2 minute– 1
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26. If the equilibrium constants of the reactions 2HI ⇌ H2 + I2 and H2 + I2 ⇌ 2HI are K1 and K2
respectively, then [June-2012]
a) K1 = 2K2 b) K1 =
2K
1 c) K2 = (K1)
2 d) K1 =
22K
1
27. In an equilibrium when Q < Kc then [Sep-2012]
a) Forward reaction is favoured b) Reverse reaction is favoured
c) Both forward and reverse reactions are favoured d) none of these
28. Presence of moisture in contact process [Mar-2014]
a) activates the catalyst b) deactivates the catalyst
c) increases the product d) makes the catalyst porous
29. The maximum yield of SO3 by in Contact process is [June-2014]
a) 97 % b) 37 % c) 50 % d) 47 %
30. The mole ratio of SO2 and O2 in contact process is [Mar-2016]
a) 1 : 2 b) 2 : 1 c) 3 : 1 d) 1 : 3
31. For the equilibrium PCl5 ⇌ PCl3 + Cl2 the factor that favours the reverse reaction is [Jun-2016]
a) increase in concentration of PCl5 b) increase in concentration of Cl2
c) no change in concentration of Cl2 d) decrease in concentration of Cl2
32. The pressure required to get maximum yield of ammonia by Haber‟s process: [Sep-2016]
a) 300 – 500 atm b) 500 – 700 atm c) 700 – 1200 atm d) 100 – 300 atm
3 MARK
1. Why do equilibrium reactions referred to as dynamic equilibrium? [Sep-2010]
At equilibrium, both forward and reverse reactions take place endlessly and simultaneously with equal
rates. So, equilibrium reactions are referred to as dynamic equilibrium
2. What is equilibrium constant? [Sep-2009]
Equilibrium constant is defined as the ratio of product of molar concentrations of products to the
product of molar concentrations of reactants under equilibrium conditions.
A + B ⇌ C + D
Kc = BA
DC
3. Define reaction quotient. Mention its use. How is it related to equilibrium constant?
[Sep-2007, June-2008, June-2010, Mar-2011, Mar-2014, June-2015, Mar-2016]
Reaction quotient „Q‟ is defined as the ratio of product of initial concentrations of products to the
product of initial concentrations of reactants under non-equilibrium conditions.
a A + b B ⇌ l L + m M
Let [A], [B], [L] and [M] be the actual concentrations present before the occurrence of equilibrium.
These concentrations are considered as the non-equilibrium concentration conditions and the reaction
quotient „Q‟ is given as
Q = ba
ml
BA
ML
Reaction quotient „Q‟ is used to find the direction in which an equilibrium reaction may proceed under
different sets of initial concentrations of reactants and products which are not equal to the respective
equilibrium concentrations.
If Q = Kc then the reaction is at equilibrium
If Q > Kc then the reverse reaction is favoured
If Q < Kc then the forward reaction is favoured
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4. Write the equilibrium constant for the following [Sep-2006, Jun-2014, Jun-2016]
i) H2O2 (g) ⇌ H2O (g) + ½ O2 (g) Kc =
22
21
22
OH
OOH Kp =
2O2H
21
2OO2H
p
pp
ii) CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g) Kc = OHCO
HCO
2
22 Kp =
O2HCO
2H2CO
pp
pp
iii) N2O4 (g) ⇌ 2NO2 (g) Kc = 42
22
ON
NO Kp =
4O2N
2
2NO
p
p
5. State Le Chatelier’s principle [March-07, Mar-10, Sep-11, Mar-12, Mar-13, Mar-15, Mar-17]
“If a system at equilibrium is subjected to a disturbance or stress, then the equilibrium shifts in the
direction that tends to nullify the effect of the disturbance or stress”
6. What is the relationship between formation equilibrium constant and dissociation constant?
Give an example. [Mar-2008, Sep-2016]
Consider the formation equilibrium reaction of SO3, from SO2 and O2 gases,
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
The equilibrium constant, Kc, is given by
Kc =
22
2
23
OSO
SO dm
3 mol
–1
In the dissociation equilibrium reaction of SO3, the reactants become products and vice versa.
2SO3 (g) ⇌ 2SO2 (g) + O2 (g)
The dissociation equilibrium constant Kc‟ is given by
K’c =
23
22
2
SO
OSO =
cK
1 mol dm
–3
Usually, the equilibrium constant of the dissociation equilibrium is the reciprocal of the equilibrium
constant of the formation equilibrium reaction.
7. Dissociation of PCl5 decreases in presence of increase in Cl2 why? [Mar-2009, June-2011]
PCl5 ⇌ PCl3 + Cl2
According to Le Chatelier principle, in a chemical equilibrium, increase in the concentration of the
products results in shifting the equilibrium in favour of the reactants. So, increase in the concentration of
Cl2 favours backward reaction and hence dissociation of PCl5 is decreased.
8. What happens when Δng = 0, Δng = –ve, Δng = +ve in a gaseous reaction? [June-2007, Sep-2015]
Kp = Kc ( RT ) Δng
When Δng = 0, then Kp = Kc ( RT ) 0
, Kp = Kc
When Δng = +ve, then Kp = Kc ( RT ) Δng
, Kp > Kc
When Δng = -ve, then Kp = Kc (RT )
–Δng , Kp =
Δngc
RT
K , Kp < Kc
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9. Give one gaseous equilibrium reaction as an example for the following:
i) Δng = 0 ii) Δng = + ve [Sep-2013]
i) H2(g) + I2(g) ⇌ 2HI(g) Δng = 2 – 2 = 0
ii) 2H2O(g) + 2Cl2 (g) ⇌ 4HCl (g) + O2 (g) Δng = 5 – 4 = 1 ( + ve )
10. Calculate Δng for the following reactions [Sep-2014]
a) N2 (g) + 3H2 (g) ⇌ 2NH3 (g) b) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) c) N2O4 (g) ⇌ 2NO2(g)
a) Δng = 2 – 4 = –2 b) Δng = 2 – 1 = 1 c) Δng = 2 – 1 = 1
11. Explain the effect of temperature on the following equilibrium by using Le Chatelier’s principle.
N2O4 (g) ⇌ 2NO2(g) ∆H = + 59 kJ mol–1
[June-2012]
If the forward reaction in a chemical equilibrium is endothermic then the reverse reaction is exothermic.
According to Le Chatlier‟s principle, increase of temperature would shift the equilibrium in the
direction in which heat is absorbed. This means that increase of temperature favours endothermic
reaction.
N2O4 (g) 2NO2(g) H = +59 kJ mol-1endothermic
exothermic
In the above equilibrium, the forward reaction is endothermic. So, increase of temperature favours
forward reaction.
12. The equilibrium constant Kc for A(g) ⇌ B(g) is 2.5 × 10–2
. The rate constant of the forward reaction
is 0.05 sec–1
. Calculate the rate constant of the reverse reaction. [March-2006]
Kc = r
f
k
k
2.5 × 10–2
= rk
0.05
kr = 10 2.5
0.052-
= 2 sec–1
13. In the equilibrium H2 + I2 ⇌ 2HI the number of moles of H2, I2 and HI are 1,2,3 moles respectively.
Total pressure of the reaction mixture is 60 atm. Calculate the partial pressures of H2, I2 and HI
in the mixture [June-2006]
Partial pressures of H2 = Total pressure × mole fraction of H2
= Total pressure × molesofno.Total
Hofmolesno.of 2
= 60 × 6
1 = 10 atm
Partial pressures of I2 = 60 × 6
2 = 20 atm
Partial pressures of HI = 60 × 6
3 = 30 atm
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14. For the reaction A + B ⇌ 3C at 25C, a 3 litre volume reaction vessel contains 1, 2 and 4 moles of
A, B and C respectively at equilibrium. Calculate the equilibrium constant Kc of the reaction at
25C [Sep-2008]
A + B ⇌ 3C
No. of moles at equilibrium 1 2 4
Equilibrium molar concentration.
V
n
3
1
3
2
3
4
Kc =
BA
C 3
=
3
2
3
1
3
43
= 10.66 mol dm–3
15. Degree of dissociation of PCl5 at 1 atm and at 25o C is 0.2. calculate Kp [Sep-2012]
For PCl5 dissociation equilibrium,
Kp = 2
2
x1
Px
P = 1 atm, x = 0.2
Kp =
2
2
2.01
12.0
Kp = 0.042 atm
16. Dissociation equilibrium constant of HI is 2.06 × 10–2
at 458C. At equilibrium, concentrations of
HI and I2 are 0.36M and 0.15M respectively. What is the equilibrium concentration of H2 at
458C. [ June-2009]
2HI ⇌ H2 + I2
Kc = 2
22
HI
IH
2.06 × 10– 2
= 2
2
0.36
0.15H
∴ [H2] = 1.779 × 10– 2
M
17. For the equilibrium 2NOCl(g) ⇌2NO(g) + Cl2(g) the value of the equilibrium constant Kc is
3.75×10–6
at 790°C. Calculate Kp for this equilibrium at the same temperature [June-2013]
Kp = Kc ( RT ) Δng
{For the equilibrium ng = 3 – 2 = 1}
Kp = Kc ( RT ) 1
Kp = 3.75 × 10–6
× 0.0821 × 1063
{R = 0.0821 L atm K –1
mol –1
, T = 790°C + 273 = 1063 K}
Kp = 3.27 × 10– 4
atm
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5 MARK
1. Derive the relation Kp = Kc (RT)Δng
for a general chemical equilibrium reaction
[Mar-2006, Jun-2007, Sep-2008, Mar-2010, Sep-2010, Mar-2012, Mar-2014, Jun-2014, Sep-16]
Consider a general chemical equilibrium reaction in which the reactants and products are in gaseous phase.
a A + b B + c C ⇌ l L + m M + n N
Kp = cba
nml
CBA
NML
ppp
ppp
where p is the partial pressure of the respective gases. In terms of molar concentrations of reactants and
products
Kc = cba
nml
CBA
NML
From gas equation, PV = n RT V
n=
RT
P
Molar concentration = litresinVolume
molesofNumber =
V
n=
RT
P
Ci = RT
Pi
Molar concentration of A ie., [A] = RT
pA and [A]
a =
a
RT
pA
Similarly, we can relate molar concentrations of B,C,L,M and N with their partial pressures.
Substituting, concentration terms by partial pressures,
Kc = cba
nml
RT
p
RT
p
RT
p
RT
p
RT
p
RT
p
CBA
NML
= cba
nml
CBA
NML
ppp
ppp
cba
nml
RT
1
RT
1
= Kp
cbanml
RT
1
= Kp
ng
RT
1
Kc = ngRT
K p
∴ Kp = Kc (RT)Δng
where Δng = total number of stoichiometric moles of gaseous products – total number of stoichiometric
moles of gaseous reactants.
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2. Using the law of chemical equilibrium derive the expressions for Kc and Kp for the formation of
HI [Sep-2009, Sep-2015]
Derivation of Kc
Let „a‟ moles of H2 and „b‟ moles of I2 gases are present in „V‟ litres of the reaction vessel.
H2 (g) + I2 (g) ⇌ 2HI (g)
Initial number of moles a b 0
Number of moles reacted x x -
Number of moles remaining at equilibrium a –x b–x 2x
Equilibrium molar concentration
V
n
V
xa
V
xb
V
2x
Kc =
22
2
IH
HI
=
V
b
V
a
V
22
xx
x
= xx
x
ba
V
V
4 2
2
2
Kc = xx ba
4x 2
Derivation of Kp
In this equilibrium reaction, the number of moles of the products is equal to the number of moles of the
reactants (Δng = 0)
H2 (g) + I2 (g) ⇌ 2HI (g)
Δng = 2 – (1+1) = 2 – 2 = 0
Kp = Kc (RT) Δng
Kp = Kc (RT) 0 ∴ Kp = Kc
∴ Kp = xx ba
4x 2
3. Derive the expressions for Kc and Kp for decomposition of PCl5
[Mar-2007, Mar-2011, Sep-2011, June-2012, Mar-2013, June-2013, Mar-2015, Mar-17]
Derivation of Kc
Let „a‟ moles of PCl5 vapour be present in „V‟ litres initially.
The dissociation equilibrium of PCl5 in gaseous state is written as
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
Initial number of moles a 0 0
Number of moles reacted x - -
Number of moles remaining at equilibrium a–x x x
Equilibrium molar concentration
V
n V
a x
V
x
V
x
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Kc =
5
23
PCl
ClPCl
=
V
aVVx
xx
=
x
x
a
V
V2
2
Kc = Va
2
x
x
If initially 1 mole of PCl5 is present then Kc = V1
2
x
x
x = degree of dissociation = initiallypresentmolesofnumberTotal
ddissociatemolesofNumber
Derivation of Kp
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
Kp =
5PCl
2Cl3PCl
p
pp atm
Kp = 2
2
1
p
x
x
atm
4. Apply Le Chatlier’s principle for the synthesis of NH3 by Haber’s process.
[June-2006, Mar-2008, June-2010, June-2011]
Synthesis of NH3 by Haber‟s process
N2(g) + 3H2(g) 2NH3(g) H0f = 22 kCal mol-1
Fe catalyst
1 Pressure Increase of pressure will decrease the volume.
Increase of pressure will shift the above equilibrium in the forward
direction in which there is a decrease in number of moles (or volume).
So, to get more NH3 300 - 500 atm pressure is applied on 3 : 1 mole
ratio of H2 : N2 gas mixture.
2 Temperature Formation of NH3 is exothermic. According to Le Chatlier‟s principle,
decrease of temperature favours exothermic reaction.
Low temperature favours the formation of NH3.
However, at low temperature the time to reach the equilibrium
becomes very long. Hence an optimum temperature close to
500°C-550°C is maintained.
3 Catalyst Iron catalyst is used to attain the equilibrium quickly.
4 Removal of
NH3
Steam is passed to remove away the ammonia as and when it is formed so
that the equilibrium remains shifted towards the product side.
5 Yield The maximum yield of NH3 is 37%
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5. Apply Le Chatlier’s principle for the synthesis of SO3 by Contact process. [Sep-2007, June-2008, Mar-2009, Sep-2013, Sep-2014, June-2015]
Synthesis of SO3 by Contact process
2SO2(g) + O2(g) 2SO3(g) H0f = 47 kCal mol-1
V2O5 catalyst
1 Pressure Increase of pressure will decrease the volume.
Increase of pressure will shift the above equilibrium in the forward
direction in which there is a decrease in number of moles (or volume).
So, to get more SO3 700 - 1200 atm pressure is applied on 2 : 1 mole
ratio of SO2 : O2 gas mixture.
2 Temperature Formation of SO3 is exothermic. According to Le Chatlier‟s principle,
decrease of temperature favours exothermic reaction.
Low temperature favours the formation of SO3.
However, at low temperature the time to reach the equilibrium
becomes very long. Hence an optimum temperature close to
400C - 450C is maintained.
3 Catalyst Porous vanadium pentoxide (V2O5) catalyst is used to speed up the
equilibrium process and high yield of SO3 is achieved in a short period.
Presence of moisture deactivates the catalyst. Only dry and pure SO2 and
O2 gases are used over the catalyst.
4 Removal of
SO3 SO3 is the anhydride of H2SO4. Therefore, SO3 from contact process along
with steam is used for the manufacture of oleum and H2SO4.
5 Yield The maximum yield of SO3 is 97%
6. Discuss the effect of temperature and pressure on the following equilibrium
N2O4 (g) ⇌ 2NO2(g) ∆ H = + 59 kJ mol-1 [Jun-2009, Sep-2012, Jun-2016]
Effect of temperature
According to Le Chatlier‟s principle, increase of temperature would shift the equilibrium in the
direction in which heat is absorbed. This means that increase of temperature favours endothermic
reaction.
N2O4 (g) 2NO2(g) H = +59 kJ mol-1endothermic
exothermic
In the above equilibrium, the forward reaction is endothermic. So, increase of temperature favours
forward reaction.
Effect of pressure
Increase in the total pressure of the system in equilibrium will decrease the volume proportionately.
According to Le Chatlier‟s principle, increase of pressure would shift the equilibrium in the direction in
which number of moles (or volume) is decreased.
N2O4 (g) ⇌ 2NO2(g)
1 mole 2 moles
1 volume P
2 volume
In the above equilibrium, increase of pressure would shift the equilibrium in the backward direction.
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7. Apply Le Chatelier’s principle for the following reaction: N2g) + O2(g) ⇌ 2NO(g) and discuss the
effect of pressure and concentration on it. [Mar-2016]
i) Effect of pressure
Increase in the total pressure of the system in equilibrium will decrease the volume proportionately.
According to Le Chatlier‟s principle, increase of pressure would shift the equilibrium in the direction in
which number of moles (or volume) is decreased.
N2 (g) + O2 (g) ⇌ 2NO (g)
2 moles 2 moles
2 volume 2 volume
In the above equilibrium, number of moles of reactants = number of moles of products
So, the equilibrium is not affected by increase or decrease of pressure.
ii) Effect of concentration
According Le Chatelier‟s principle, increasing the concentration of reactants (addition of reactants)
would shift the equilibrium in the forward direction and increasing the concentration of products
(addition of products) would shift the equilibrium in the backward direction.
N2 (g) + O2 (g) ⇌ 2NO (g)
If the concentration of reactants (N2 & O2) is purposely raised in the above equilibrium, the excess
amounts of N2 and O2 react to give products. So, addition of reactants favours forward reaction.
If the concentration of product (NO) is purposely raised in the above equilibrium, the excess amounts of
NO react in the reverse direction to produce back the reactants. So, addition of products favour
backward reaction.
8. Dissociation equilibrium constant of HI is 2.06 × 10–2
at 458C. At equilibrium, concentrations of
HI and I2 are 0.36M and 0.15M respectively. What is the equilibrium concentration of H2 at
458C. [Sep-2006]
Refer – 3 mark Q - 16
Additional questions
1 MARK
1. The condition for equilibrium is
a) kf = kr b) Rf = Rr c) Kc = Kp d) all the above
2. Kc for the equilibrium 2SO2 + O2 ⇌ 2SO3 is 50, then Kc for 2SO3 ⇌ 2SO2 + O2 is
a) 50 b) -50 c) 50
1 d) 5
3. 3H2(g) + N2(g) ⇌ 2NH3(g), the value of Kp and Kc are related as
a) Kp = Kc b) Kp > Kc c) Kp < Kc d) Kp = Kc =0
4 . Favourable conditions to get maximum yield of NO2 according to the reaction
N2O4 (g) ⇌2NO2 (g) ∆H = +59 kJ mol–1
a) high temperature b) low pressure c) addition of N2O4 d) all the above
5. If the reaction quotient of a reaction is less than equilibrium constant, then
a) forward reaction is favoured b) reverse reaction is favoured
c) reaction is at equilibrium d) none of these
Note: If Q = Kc then the reaction is at equilibrium If Q > Kc then the reverse reaction is favoured
If Q < Kc then the forward reaction is favoured
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6. Equilibrium constant can be calculated by the formula,
a) K= kf – kr b) K = kf + kr c) K =
r
f
k
k d) K =
f
r
k
k
7. Catalyst is added in equilibrium reactions in order to
a) increase the amount of product b) increase the values of Kp and Kc
c) attain the equilibrium quickly d) decrease the rate of reverse reaction 8. The mole ratio of H2 and N2 in Haber‟s process is
a) 1 : 2 b) 2 : 1 c) 3 : 1 d) 1 : 3
3 MARK 1. Define degree of dissociation
Degree of dissociation (x) is defined as the fraction of total moles of reactant dissociated.
x = initiallypresentmolesofnumberTotal
ddissociatemolesofNumber
5 MARK
1. Discuss the effect of concentration, pressure and temperature on the following reaction.
N2g) + O2(g) ⇌ 2NO(g)
i) Effect of concentration & ii) Effect of pressure (Refer –5 mark- Q 7)
iii) Effect of temperature
According to Le Chatlier‟s principle, increase of temperature would shift the equilibrium in the
direction in which heat is absorbed. This means that increase of temperature favours endothermic
reaction.
N2(g) + O2(g) 2NO(g) H0f = +43.2 kCal mol-1
endothermic
exothermic
In the above equilibrium, the forward reaction is endothermic. So, increase of temperature favours
forward reaction.
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11. CHEMICAL KINETICS – II
Blue print 1 Mark = 1 3 Mark = 2 5 Mark = 1 Total marks = 12
Govt. Examination questions 1 MARK
1. Hydrolysis of an ester by dilute HCl is an example for [June-2007, Mar-2010, Sep-2016]
a) second order reaction b) zero order reaction
c) pseudo first order reaction d) first order reaction
2. The excess energy which a molecule must possess to become active is known as
[Mar-2009, June-2010, Mar-2015, Sep-2015]
a) kinetic energy b) threshold energy c) potential energy d) activation energy
3. Arrhenius equation is [Sep-2012, June-2014, Sep-2014]
a) k = Ae–1/RT
b) k = Ae–RT/Ea
c) k = Ae–Ea/RT
d) k = AeEa/RT
4. The term A in Arrhenius equation is called as [Mar-2012, June-2015]
a) Probability factor b) Activation of energy c) Collision factor d) Frequency factor
5. The sum of the powers of the concentration terms that occur in the rate equation is called
[Sep-2007, June-2009]
a) molecularity b) order c) rate d) rate constant
6. Reactions in which the reacting molecules react in more than one way yielding different set of products
are called [Sep-2009]
a) consecutive reactions b) parallel reactions c) opposing reactions d) chain reactions
7. The half life period of a first order reaction is 10 minutes. Then its rate constant is [June-06, Mar-17]
a) 6.93 102 min
–1 b) 0.693 10
–2 min
–1 c) 6.932 10
–2 min
–1 d) 69.3 10
–1 min
–1
8. For a reaction : aA → bB, the rate of reaction is doubled when the concentration of A is increased by
four times. The rate of reaction is equal to [Sep-2010, Jun-2016]
a) k [A]a b) k [A]
½ c) k [A]
1/a d) k [A]
9. For a reaction, Ea = 0 and k = 4.2 105 sec
–1 at 300 K, the value of k at 310 K will be
[Mar-2006, June-2011, Mar-2013, Sep-2013, Mar-2016]
a) 4.2 105 sec
–1 b) 8.4 10
5 sec
–1 c) 8.4 10
5 sec
–1 d) unpredictable
10. The first order rate constant of a reaction is 0.0693 min–1
. Then the time required for 50% completion of
a reaction is [Sep-2006]
a) 10 min b) 1min c) 100 min d) 50 min
11. 50% of a first order reaction is completed in 20 minutes. The time required for 75% completion is
[Mar-2007]
a) 60 minutes b) 10 minutes c) 40 minutes d) 80 minutes
12. Half life period of a first order reaction is 20 min. The time taken for the completion of 99.9% of the
reaction is [Mar-2008]
a) 20 min b) 2000 min c) 250 sec d) 200 min
[ For a first order reaction time required for 99.9% completion is 10 times that required for 50%
completion t99.9% = 10 t50%
For a first order reaction time required for 99% completion is twice the time required for 90%
completion of the reaction. t99% = 2 t90% ] 13. In a first order reaction the concentration of the reactant is increased by 2 times. The rate of the reaction
is increased by [June-2008]
a) 2 times b) 4 times c) 10 times d) 6 times
14. Decomposition of nitrogen pentoxide in CCl4 is an example for [Sep-2008, June-2012]
a) second order reaction b) third order reaction
c) zero order reaction d) first order reaction
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15. The t ½ of a first order reaction is 100 minutes. Then its rate constant is [Mar-2011]
a) 6.93 103 min
–1 b) 0.693 10
–3 min
–1 c) 6.93 10
–3 min
–1 d) 69.3 10
–2 min
–1
16. If the activation energy is high, then the rate of the reaction is [Sep-2011]
a) high b) moderate c) low d) cannot be predicted
17. The rate constant of a first order reaction is 1.54 10–3
s–1
. Its half life period is [June-2013]
a) 540 seconds b) 450 seconds c) 45 seconds d) 54 seconds
18. The unit of zero order rate constant is [Mar-2014]
a) litre mol – 1
sec – 1
b) mol litre – 1
sec – 1
c) sec – 1
d) litre2 sec
– 1
Unit of k = mole 1–n
litre n–1
sec–1
[ n = order ]
Unit of zero order rate constant mol litre–1
sec–1
Unit of first order rate constant sec–1
Unit of second order rate constant mol–1
litre sec–1
3 MARK 1. Define order of a reaction
[June-2007, June-2008, Mar-2009, Sep-2010, Sep-2011, Sep-2013, Sep-2014, June-2015]
Order is defined as the sum of the powers of the concentration terms that occur in the rate equation.
aA + bB products
rate = k[A]p [B]
q
The total order of the reaction = p + q
2. Define half life period [June-2013, Sep-2014]
Half life period, ‘t½’, of a reaction is defined as the time required to reduce the concentration of a
reactant to one half of its initial value.
t ½ = 1k
0.693
3. Give three examples for first order reaction [Sep-2012, Sep-2016]
1. All radioactive transformations follow first order kinetics.
92U238
90Th234
+ 2He4
2. Decomposition of sulphuryl chloride in the gas phase
SO2Cl2(g) SO2(g) + Cl2(g)
3. Decomposition of nitrogen pentoxide in CCl4 medium
N2O5 2NO2 + ½ O2
4. What are the characteristics of first order reaction? [June-2013]
1. When the concentration of the reactant is increased by ‘n’ times, the rate of reaction is also
increased by n times. That is, if the concentration of the reactant is doubled, the rate is doubled.
2. The unit of rate constant of a first order reaction is sec–1
or time–1
.
k1 = 1
11
litmol
seclitmol
)xa(
rate
= sec–1
3. The time required to complete a definite fraction of reaction is independent of the initial
concentration
t ½ = )1u(
ulog
k
303.2
1
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5. Derive the relationship between half-life period and rate constant for a first order reaction.
[June-2008, June-2012, Mar-2012]
Half life period, ‘t½’, of a reaction is defined as the time required to reduce the concentration of a
reactant to one half of its initial value.
For first order reaction,
k1 = xa
alog
t
2.303
If amount reacted x = 2
a then t = t ½
t ½ =
2aa
alog
k
2.303
1
t ½ =
1k
2log2.303
t ½ =
1k
0.693 sec
Thus half life period of a first order reaction is independent of the initial concentration of the reactant
and also, inversely proportional to the rate constant of the reaction.
6. What is a pseudo first order reaction? Give an example
[Mar-2008, June-2009, Sep-2009, June-2010, Mar-2011, Mar-17]
A pseudo first order reaction is one whose stoichiometric equation indicates that more than one reactant
is taking part in the reaction but only the concentration of one reactant determines the reaction rate. The
other is present in such a large excess that there is no change in the concentration in the course of the
reaction.
Eg- Acid catalysed hydrolysis of ester
CH3COOCH3 + H2O H
CH3COOH + CH3OH
7. Write the Arrhenius equation and explain the terms
[Mar-2007, Mar-2009, Mar-2011, June-2011, Mar-2013, June-2015, Sep-2015, Mar-17]
k = A e –Ea / RT
where k = rate constant, Ea = activation energy, A = frequency factor,
R = gas constant, T = temperature in Kelvin.
8. What are simple and complex reactions? [Sep-2007, Sep-2008, Mar-2015]
No Simple reactions Complex reactions
1 Occurs in single step Occurs in many steps.
2 No side reactions Many side reactions are present
3 Products are formed directly from
the reactants
Products are not formed directly from
the reactants.
9. What is activation energy?
[Sep-2007, Mar-2008, Sep-2008, Mar-2010, Mar-2014, June-2014, Sep-2015]
The additional energy required by the reactant molecules to attain the threshold energy in addition to the
energy of colliding molecules is called as activation energy ‘Ea’.
Activation energy = Threshold energy – Energy of colliding molecules.
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P.E of reactants
P.E of products
Reaction coordinate
Ener
gy
Threshold energy
Ea
= activation energyEa
10. What is threshold energy? [Mar-2016]
The minimum energy possessed by all colliding molecules for making effective and successful
collisions is called threshold energy.
Threshold energy = Energy of colliding molecules + Activation energy
11. What are consecutive reactions?
Refer – 5 mark – Q – 7 [Jun-2006, Sep-2006, Sep-2009, Marr-2012, Sep-2012, Mar-2014, Jun-2016 ]
12. What are opposing reactions? [Mar-2006, Jun-2007, Jun-2009, Mar-2013, Jun-2014, Sep-2016]
Refer – 5 mark – Q – 7
13. What are parallel reactions? [Mar-2007, Mar-2010,June-2011, Sep-2011,June-2012, Sep-2013]
Refer – 5 mark – Q – 7
14. The initial rate of a first order reaction is 5.2 10–6
mol.lit–1
.s–1
at 298 K. When the initial
concentration of reactant is 2.6 10–3
mol.lit–1
, calculate the first order rate constant of the
reaction at the same temperature. [June-2006]
Rate = k1 [A] 1
k1 = 1]A[
Rate =
3
6
102.6
105.2
= 2 × 10
–3 sec
–1
15. The half life period of a first order reaction is 20 minutes. Calculate the rate constant [Mar-2006]
k =
21t
0.693 =
20
0.693 = 0.03465 min
–1
16. The rate constant of a first order reaction is 1.54 10–3
s–1
. Calculate the half life period
[Sep-2010, Mar-2015, Jun-2016]
t ½ = k
0.693 =
3-101.54
0.693
= 450 sec
17. Show that for a first order reaction time required for 99% completion is twice the time required
for 90% completion of the reaction [Sep-2006]
For first order reaction, k1 = xa
alog
t
2.303
t 99% = 99100
100log
k
2.303
1 = 100log
k
2.303
1
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t 90% = 90100
100log
k
2.303
1 = 10log
k
2.303
1
10log
100log
t
t
%90
%99 = 1
2 = 2
t 99% = 2 t 90%
18. Write any three characteristics of a simple reaction [Mar-2016]
Refer – 3 mark – Q – 8
5 MARK
1. What are the characteristics of order of a reaction?
[Mar-2006, Sep-2008, Mar-2010, June-2010, Mar-2011, Mar-2013, Mar-17]
1. The magnitude of order of a reaction may be zero, or fractional or integral values. For an elementary
reaction, its order is never fractional since it is a one step process.
2. Order of a reaction should be determined only by experiments. It cannot be predicted interms of
stoichiometry of reactants and products.
3. Simple reactions possess low values of order like n = 0,1,2, Reactions with order greater than or
equal to 3 are called complex reactions. Higher order reactions are rare.
4. Some reactions show fractional order depending on rate.
5. Higher order reactions may be experimentally converted into simpler order (pseudo) reactions by
using excess concentrations of one or more reactants.
2. What are the characteristics of first order reaction? [Sep-2006, Mar-2015]
Refer – 3 mark – Q – 4
3. Write the differences between simple and complex reactions
[Jun-07, Sep-10, Sep-13, Jun-16, Sep-16]
No. Simple reactions Complex reactions
1 Occurs in single step Occurs in multi (or) many steps.
2 Overall order values are small. Total and
pseudo order values lie between
0,1,2 and 3.
Overall order values are large and greater than
3. Sometimes fractional orders such as 1/2, 1/3,
3/2 etc. are seen.
3 No side reactions Many side reactions are present.
4 Products are formed directly from the
reactants
In some complex reactions products are not
formed in steps directly involving the reactants.
5 Experimental rate constant values agree with
the calculated values. Theories of reaction
rates apply well on simple reactions.
Experimental overall rate constant values differ
from the calculated values. Theories of reaction
rates do not agree well on complex reactions.
4. Derive an equation for the rate constant of a first order reaction
[June-2009, June-2011, Mar-2014, Sep-2015]
The reactions in which, the overall rate of the reaction is proportional to the first power of concentration
of one of the reactants only are called as first order reactions.
Consider the reaction
A 1k
products
Rate of reaction = dt
d[A] = k1 [A]
1
where k1 is the rate constant of the first order reaction.
Concentration of A at the beginning of the reaction (time ‘t’ = 0) = a mole lit –1
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Concentration of A that has reacted after time ‘t’ = x mole lit –1
Concentration of unreacted A remaining at time ‘t’ = (a – x) mole lit –1
For a first order reaction,
Rate = dt
dx = k1 (a – x)
upon integrating,
)( xa
dx
= k1 dt
–ln(a – x) = k1t + c
c = integration constant
at time, t = 0, x = 0.
– ln (a – 0) = k1 0 + C
or C = –ln a.
Substituting C value in the above equation
–ln (a – x) = k1t – ln a
Rearranging, k1 = xa
a
ln
t
1
k1 = xa
a
log
t
2.303
Unit of k1 is sec–1
.
5. How will you determine the rate constant of decomposition of H2O2 in aqueous solution?
[June-2006, Sep-2007, June-2012]
H2O2 Pt
H2O + ½ O2
1. The decomposition of H2O2 in aqueous medium in the presence of Pt catalyst follows a first order
reaction.
2. The progress of the reaction is followed by titrating equal volumes of the reaction mixture at regular
time intervals against standard KMnO4 solution.
3. Volume of KMnO4 consumed at t = 0 is ‘Vo’ which is proportional to ‘a’, the initial concentration
of H2O2.
4. Volume of KMnO4 consumed after time‘t’ is ‘Vt’ which is proportional to (a – x). , the unreacted
H2O2
5. The first order rate constant ‘k1’ of the reaction is, k1 =
tV
Vlog
t
2.303 0
6. How would you determine the rate constant of acid catalysed hydrolysis of an ester (methyl
acetate)? [Mar-2007, Sep-2011, Sep-2012, June-2013]
The acid catalysed hydrolysis of an ester follows pseudo first order kinetics.
CH3COOCH3 + H2O H
CH3COOH + CH3OH
Methyl acetate
The overall rate of the reaction depends on the concentrations of reactants and also on the catalyst
concentration.
Rate = k3 [ester] [H2O] [H+]
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k3 = rate constant of the third order reaction.
Therefore the true order of the reaction is 3. Since water is used as the solvent, its concentration is
excess. Therefore change in the concentration of water considered as negligible and concentration of
water is assumed to be constant.
Since acid acts the catalyst, there will be no change in the catalyst concentration before initial and after
the completion of the reaction. Hence [H+] is considered as a constant value.
Hence the above expression can be rewritten as
Rate = k3' [ester]
Procedure
Initially to a definite volume of (100 ml) hydrochloric acid (0.5 N), 10 ml of ester is added and the start
of the reaction corresponds to time of addition of ester. The rate of the reaction is followed by
withdrawing a definite volume of the reaction mixture consisting of the ester and acid at various time
intervals and arresting the further progress of reaction by adding ice. The whole cold mixture is titrated
with standard NaOH (0.1 N) using phenolphthalein as the indicator.
Let the volume of alkali consumed at t = 0 be Vo cc which is equivalent to the amount of hydrochloric
acid present in the definite volume of the reaction mixture drawn out at regular intervals of time.
If Vt cc is the volume of alkali consumed for the same definite volume of the reaction mixture drawn out
after reaction time ‘t’, then (Vt –Vo) cc is equivalent to the acetic acid produced by the hydrolysis of
ester in time ‘t’.
A final titration is done after about 8 hours or after refluxing the solution for 45 minutes to complete the
hydrolysis which is Vcc. (V–Vo) cc is equivalent to acetic acid produced from complete hydrolysis
of ester.
Calculation
The initial concentration of ester = a (V– Vo) cc
Concentration of ester reacted at ‘t’ = x (Vt – Vo) cc
Concentration of ester remaining at time ‘t’ = (a – x) (V– Vt)
)V(V
)V(V
xa
a
t
0
The first order rate expression for the hydrolysis of ester can be written as
k =)V(V
)V(Vlog
2.303
t
0
t
By substituting Vt values for various ‘t’ values, k is determined. These values are found to be constant
indicating k as the rate constant of the reaction.
7. Explain different types of complex reaction [Mar-2008, June-2008, Sep-2014, Mar-2016]
Reactions which do not take place in a single step but take place in a sequence of a number of
elementary steps are called as complex reactions.
a) Consecutive reactions
The reactions in which the reactant forms an intermediate and the intermediate forms the product in one
or many subsequent reactions are called as consecutive or sequential reactions. In such reactions the
product is not formed directly from the reactant.
A 1k B 2k
C
A = reactant ; B = intermediate ; C = product.
Eg-Saponification of a diester in presence of an alkali
R’OOC-(CH2)n-COOR 1k R’OOC-(CH2)n-COOH 2k
HOOC-(CH2)n-COOH
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b) Parallel reactions In these reactions, one or more reactants react simultaneously in two or more pathways to give two or
more products. The parallel reactions are also called as side reactions.
A
B
C
D
k1
k2
k3
Eg- i) Bromination of bromobenzene
Br
Br
Br
Br
Br
+
+
HBr
HBr
Bromo benzene
o-dibromobenzene
p-dibromobenzene
ii) Dehydration of 2-methyl-2-butanol
CH3
C CH2
OH
2-methyl-2-butanol
2-methyl-2-butene
2-methyl-1-butene
CH3
CH3
CH3
C CH CH3
CH3
C CH3
CH3
CH2
CH2
c) Opposing reactions In opposing reactions the products formed react back simultaneously to form the reactants. These
reactions are also called as reversible reactions.
i) Reaction between CO and NO2 gases
CO (g)
+ NO2 (g)
CO2 (g)
+ NO (g)
kf
kr
ii) Isomerisation of cyclopropane to propene
CH3-CH=CH
2
kf
krCH
2
CH2
CH2
iii) Dissociation of hydrogen iodide in gas phase
2HI (g) ⇌ H2 (g) + I2 (g)
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8. Compound A reacts by first order kinetics. At 25°C, the rate constant of the reaction is 0.45 sec–1
.
What is the half life of A at 25°C? What is the time required to have 12.5% unreacted A for first
order reaction? [Mar-2009]
t ½ =
1k
0.693 = ssec54.1
0.45
0.693
No. of t½ Amount unreacted from 100%
1 50%
2 25%
3 12.5%
Time of three half-life periods = 3 1.54 = 4.62 sec
9. A first order reaction is 75% complete in 100 minutes. What are the rate constant and half life
period of the reaction? [Sep-2009]
k1 = 12 min10386.1
75100
100log
100
303.2log
t
2.303
xa
a
t ½ = minutes5010386.1
693.0
k
0.6932
1
10. Show that for a first order reaction, the time required for 99.9% completion of the reaction is 10
times that required for 50% completion. [Mar-2012, June-2014, June-2015]
For first order reaction, k1 = xa
alog
t
2.303
t = xa
alog
k
2.303
1
t 99.9 % = 99.9100
100log
k
2.303
1 =
1.0
100log
k
2.303
1
= 1000logk
2.303
1
t 50% = 50100
100log
k
2.303
1
=
50
100log
k
2.303
1
= 2logk
2.303
1
2logk
303.2
1000logk
303.2
1
1
%50
%9.99 t
t =
2log
1000log =
3010.0
3=10
t 99.9% = 10 t50%
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Additional questions
1 MARK
1. 2N2O5 4NO2 + O2, –dt
]Od[N 52 = k1[N2O5], dt
]d[NO2 = k2 [N2O5] and dt
]d[O2 = k3 [N2O5], the
relation between k1, k2 and k3 is
a) 2k1 = 4k2 = k3 b) k1 = k2 = k3 c) 2k1 = k2 = 4k3 d) 2k1 = k2 = k3
2. For a reaction, rate = k [A]p [B]
q . Order of the reaction is
a) p+q b) p-q c) A+B d) A-B
3. For a reaction, A product, when the concentration of A is doubled, the rate is increased by four
times the initial rate. Then the order of the reaction is
a) 1 b) 2 c) 3 d) 0
4. For a reaction, rate = k [HCrO4–] [I
–]2 [H
+]2
.Order of the reaction is
a) 1 b) 2 c) 3 d) 5
5. For a reaction, rate = k [H2O2] [I–]
.Order of the reaction is
a) 1 b) 2 c) 3 d) 5
6. Arrhenius plot of log k against 1/T values gives a straight line with slope value equal to
a) –Ea/2.303R b) R/2.303Ea c) –2.303R/Ea d) none
7. The intercept of a plot of log k against 1/T is equal to
a) log A b) –log A c) –Ea/2.303R d) none
8. The energy which is needed to make the collisions effective and successful is
a) activation energy b) threshold energy c) kinetic energy d) none
9. Which of the following is correct?
a) If Ea = 0, then k = A b) If T = , then k = A
c) Higher the value of Ea, slower is the rate of the reaction d) all are correct
10. Half life of -------- order reaction does not depend on initial concentration of the reactant
a) 0 b) 1 c) 2 d) 3
5 MARK
1. How will you determine the rate constant of decomposition of nitrogen pentoxide in CCl4?
N2O5 1k
2NO2 + ½ O2
At time t = 0, the volume of oxygen liberated is zero. Let Vt and Vbe the measured volumes of oxygen
liberated after the reactant has reacted in ‘t’ time and at completion (t = ). Initial concentration of
N2O5 is proportional to total volume of oxygen liberated (i.e.,) (V).
(V - Vt) is proportional to undecomposed N2O5 at time ‘t’.
k1 = )V(V
Vlog
t
2.303
t
sec–1
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12. SURFACE CHEMISTRY
Blue print 1 Mark = 3 3 Mark = 1 5 Mark = 1 Total marks = 11
Govt. Examination questions 1 MARK
1. The migration of colloidal particles under the influence of an electric field is known as
[Sep-2010, Sep-2011, Sep-2012, Mar-2016]
a) electro osmosis b) cataphoresis c) electrodialysis d) electrophoresis
2. Which one is the correct factor that explains the increase of rate of reaction by a catalyst?
a) shape selectivity b) particle size [Jun-2008, Jun-2016]
c) increase of free energy d) lowering of activation energy
3. Fog is a colloidal solution of [Mar-2007, Sep-2008, Sep-2012]
a) gas in liquid b) liquid in gas c) gas in solid d) solid in gas
4. Curd is a colloidal solution of [Mar-2006]
a) liquid in liquid b) liquid in solid c) solid in liquid d) solid in solid
5. An example of gel is [June-2006]
a) paint b) pumice stone c) milk d) curd
6. Smoke is a colloidal solution of [Sep-2006]
a) gas in solid b) solid in gas c) gas in liquid d) liquid in gas
7. Which type of colloid is a sol? [June-2007, Sep-2010]
a) solid in liquid b) liquid in solid c) solid in solid d) gas in solid
8. An emulsion is a colloidal solution of [Mar-2006, June-2010, June-2011, Sep-2013, Mar-15, Sep-15]
a) two solids b) two liquids c) two gases d) one solid and one liquid
9. Ruby glass is a colloidal solution of [June-2013]
a) Solid-sol b) Gel c) Emulsion d) Sol
10. Haze is a colloidal solution of [Sep-2015]
a) gas in liquid b) liquid in gas c) gas in solid d) solid in gas
11. The phenomenon of Tyndall’s effect is not observed in
[June-2006, Sep-2007, June-2009, June-2012, Mar-2013, June-2013, June-2014, Sep-2015]
a) emulsion b) colloidal solution c) true solution d) None
12. The Tyndall’s effect associated with colloidal particles is due to [Mar-08, Mar-10, Mar-12, Sep-16, Mar-17]
a) presence of charge b) scattering of light c) absorption of light d) reflection of light
Dispersed
phase
Dispersion
medium Name Example
Solid Solid Solid-sol Ruby glass, Alloys
Solid Liquid Sol Paint, Ink
Solid Gas Aerosol Smoke, Haze
Liquid Solid Gel Curd, Cheese
Liquid Liquid Emulsion Milk, Cream
Liquid Gas Liquid Aerosol Cloud, Mist, Fog
Gas Solid Solid foam Cork, Pumice stone
Gas Liquid Foam Froths of air
13. In case of physical adsorption, there is desorption when
[Sep-2006, Mar-2007, Mar-2009, Sep-2010, June-2011, Mar-2015, June-2015]
a) temperature increases b) temperature decreases
c) pressure increases d) concentration increases
14. Colloidal medicines are more effective because [Sep-06, Jun-07, Jun-08, Mar-11, Sep-14, Jun-16]
a) they are clean b) they are easy to prepare
c) the germs more towards, them d) they are easily assimilated and adsorbed
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15. For chemisorption, which is wrong? [Mar-2011, Sep-2011, June-2012, June-2013]
a) irreversible b) it requires activation energy
c) it forms multimolecular layers on adsorbate d) surface compounds are formed
16. Colloids are purified by [Mar-2007, Mar-2008, Sep-2008, Sep-2009, Mar-2010, Mar-2012, Mar-2014]
a) precipitation b) coagulation c) dialysis d) filtration
17. Decomposition of hydrogen peroxide is retarded in the presence of [Jun-06, Sep-09, June-12, Mar-14]
a) alcohol b) glycerine c) MnO2 d) Mo
18. The decomposition of hydrogen peroxide in the presence of colloidal platinum is [Mar-2011]
a) positive catalysis b) negative catalysis c) auto catalysis d) induced catalysis
19. The oxidation of sodium sulphite by air is retarded by [June-2011, June-2015]
a) MnO2 b) H2S c) alcohol d) As2O3
Process Positive catalyst Negative catalyst or
inhibitors or retardant
Decomposition of H2O2
2H2O2 2H2O + O2 Colloidal Pt Glycerine
Oxidation of sodium sulphite
2Na2SO3 + O2 2Na2SO4
alcohol
20. The iron catalyst used in Haber’s process is poisoned by [Mar-2006, Sep-2012]
a) Pt b) H2 c) H2S d) As2O3
21. The platinum catalyst used in the oxidation of SO2 by contact process is poisoned by [June-07, Sep-13]
a) As2O3 b) V2O5 c) Fe2O3 d) CuCl2
22. Coconut charcoal has a great capacity of the ------- of gases [Sep-2007]
a) adsorption b) absorption c) desorption d) all of these
23. Emulsifying agent is used for [Sep-2007, Mar-2013]
a) precipitation of an emulsion b) coagulation of an emulsion
c) stabilization of an emulsion d) none of these
24. Which one of the following factors is false regarding catalyst? [Mar-2008]
a) small quantity is enough b) initiate the reaction
c) remains unchanged in mass and chemical composition d) specific in its action
25. The function of FeCl3 in the conversion of Fe(OH)3 precipitate into a colloid is [June-2008]
a) peptising agent b) emulsifying agent c) reducing agent d) precipitating agent
26. An example of lyophilic colloid is [Sep-2008, Mar-17]
a) sulphur in water b) phosphorus in water c) starch d) all of these
Lyophobic colloid (solvent hating) Colloidal solutions of metals, sulphur in water
Lyophilic colloid (solvent loving) Gelatin, protein, starch, glue
27. In the reaction between oxalic acid and potassium permanganate in the presence of dil.H2SO4 ---------
acts as an auto catalyst [Mar-2009, June-2010, Mar-2016, Sep-2016]
a) K2SO4 b) MnSO4 c) MnO2 d) Mn2O3
28. The Blue colour of the sky is due to [Mar-2009]
a) Tyndall effect b) Brownian movement c) electrophoresis d) electro-osmosis
29. The sky looks blue due to [June-2010]
a) adsorption b) dispersion c) reflection d) scatering of light
30. The emulsifying agent used in O/W emulsion is [Sep-2009, Mar-2014]
a) long chain alcohol b) lamp black c) protein d) glycerol
31. Theprincipal emulsifying agent for W / O emulsion is [Mar-2015]
a) protein b) gum c) lamp black d) synthetic soaps
Emulsifying agents for O/W emulsion Proteins, gums, natural and synthetic soaps
Emulsifying agents for W/O emulsion Heavy metal salts of fatty acids, long chain
alcohols and lampblack.
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32. The colloid used for stomach disorder is [Mar-2010, June-2015]
a) colloidal silver b) colloidal antimony c) colloidal gold d) milk of magnesia
33. Argyrol is [June-2009]
a) colloidal silver b) colloidal antimony c) colloidal gold d) milk of magnesia
34. Medicine used as an eye lotion is [Mar-2012, Mar-2016]
a) silver sol b) colloidal gold c) colloidal antimony d) milk of magnesia
35. Colloidal milk of Magnesia is used [Sep-2014]
a) as germ killer b) as fertilizer c) for stomach disorders d) as tonic
Colloidal medicines Uses
Argyrol [Silver sol] Eye lotion
Colloidal antimony kalazar
Colloidal gold Intramuscular injection
Milk of magnesia Stomach disorder
36. Silica gel is utilized for the ------------ of the number of gases [Mar-2013]
a) adsorption b) absorption c) desorption d) all of these
37. Electrophoresis is a ------------ property of a colloid [Sep-2013]
a) optical b) kinetic c) electrical d) magnetic
38. When an oil soluble dye is mixed with emulsion and emulsion remains colourless then, the emulsion is
[June-2014]
a) O / W b) W / O c) O / O d) W / W
39. A substance which increases the activity of a catalyst [Sep-2014]
a) Positive catalyst b) Negative catalyst c) Promotors d) catalytic poison
40. The catalyst used for the decomposition of KClO3 is [Sep-2011, Sep-2014]
a) MnO2 b) Cl2 c) V2O5 d) Pt
41. Catalyst used in Deacon’s method of manufacture of chlorine is [June-2009]
a) NO b) CuCl2 c) Fe2O3 d) Ni
42. The intermediate compound produced in the formation of SO3 by lead chamber process is [Jun-2016]
a) NO b) NO2 c) SO2 d) O2
43. The magnitude of gaseous adsorption does not depend upon [Sep-2016]
a) temperature b) pressure c) nature of the gas d) amount of the adsorbent
44. Which one of the following processes does not involve coagulation? [Mar-2017]
a) peptisation b) formation of delta
c) purification of drinking water using alum d) tanning of leather using tannin
3 MARK
1. Write any three differences between physical adsorption and chemical adsorption [Sep-2009]
No Physical adsorption Chemical adsorption
1 It is due to intermolecular Vander
waal’s force.
It is due to chemical bond formation
2 Reversible. Irreversible.
3 Forms multimolecular layers on
adsorbent surface.
Forms unimolecular layer
2. Write the general characteristics of catalytic reactions [Mar-2011]
1. The catalyst remains unchanged in mass and in chemical composition at the end of the reaction.
2. Only a small quantity of catalyst is generally needed.
3. A catalyst cannot initiate a reaction. The function of a catalyst is only to alter the speed of the
reaction which is already occurring at a particular rate.
4. A catalyst does not alter the position of equilibrium in a reversible reaction.
5. The catalyst is generally specific in its action.
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3. What is catalysis? Give an example [Mar-2015]
A catalyst is a substance which alters the speed of a chemical reaction without itself undergoing any
chemical change and the phenomenon is known as catalysis.
Example:
2KClO3 2MnO 2KCl + 3O2
In the above reaction, MnO2 acts as a catalyst.
4. What are the two types of catalysis? [March-2008]
1. Homogeneous Catalysis:
In these reactions, the reactants and catalyst remain in the same phase.
2SO2(g) + O2(g) 2SO3(g)
NO(g)
2. Heterogeneous Catalysis:
The catalytic process in which the reactants and the catalyst are in different phases is known as
heterogeneous catalysis.
N2 (g) + 3H2 (g) (s)
Fe
2NH3 (g)
5. What is an auto catalyst? Give an example [Sep-2006, June-2014]
One of the products formed during the reaction acts as a catalyst for that reaction. Such type of catalyst
is called auto catalyst and the phenomenon is known as auto catalysis.
In the oxidation of oxalic acid by potassium permanganate, one of the products MnSO4 acts as a auto-
catalyst because it increases the speed of the reaction.
COOH
COOH5 2 KMnO
4 3 H2SO
42 MnSO
4K
2SO
410 CO
28 H
2O+ + + + +
6. What are active centers? [June-2012, Sep-2015]
The catalytic surface has unbalanced chemical bonds on it. The reactant gaseous molecules are adsorbed
on the surface by these free bonds. This accelerates the rate of the reaction. The distribution of free
bonds on the catalytic surface is not uniform. These are crowded at the peaks, cracks and corners of the
catalyst. The catalytic activity due to adsorption of reacting molecules is maximum at these spots. These
are, therefore, referred to as the active centres.
7. What are promoters? Give an example [June-2008, Mar-2010, June-2010, June-2015]
The activity of a catalyst can be increased by addition of a small quantity of a second material. A
substance which, though itself not a catalyst, promotes the activity of a catalyst is called a promoter.
Eg-In the Haber’s process for the synthesis of ammonia, traces of molybdenum increase the activity of
finely divided iron which acts as a catalyst.
N2 + 3H2 2NH3
Fe
Mo
8. What are catalytic poisons? Give an example [June-2007]
A substance which destroys the activity of the catalyst is called a poison and the process is called
catalytic poisoning.
Eg-1. The iron catalyst used in the synthesis of ammonia in Haber process is poisoned by H2S
N2 + 3H2 2NH3
Fe
Poisoned by H2S
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9. Define colloidal solution [Sep-2012]
When the diameter of the particles of a substance dispersed in a solvent ranges from about 10A° to
2000A°, the system is termed a colloidal solution.
10. Why colloidal system in gas in gas does not exist? [March-2007, Mar-2016]
A colloidal solution of gas in gas is not possible as gases are completely miscible and always form true
solutions.
11. What is peptisation? Give an example [Mar-2009, Mar-17]
The dispersion of a precipitated material into colloidal solution by the action of an electrolyte in solution
is termed as peptisation. The electrolyte used is called a peptizing agent.
Examples : i. Silver chloride can be converted into a sol by adding hydrochloric acid
ii. Ferric hydroxide yields a sol by adding ferric chloride
12. What are lyophobic and lyophilic colloids? [Sep-2011, Jun-2016]
Colloidal solutions in which the dispersed phase has very little affinity for the dispersion medium are
termed as lyophobic (solvent hating) colloids.
Eg- Colloidal solutions of metals and sulphur in water
Colloidal solutions in which the dispersed phase has considerable affinity for the dispersion medium are
called lyophilic (solvent loving) colloids.
Eg-Gelatin, protein and starch
13. What is Tyndall effect? [Sep-2010, June-2011, Mar-2014]
When a strong beam of light is passed through a sol and viewed at right angles, the path of light shows
up as a hazy beam. This is due to the fact that sol particles absorb light energy and then emit it in all
directions. This scattering of light illuminates the path of the beam. The phenomenon of the scattering of
light by the sol particles is called Tyndall effect.
14. What is Brownian movement? Give reason [June-2009, Mar-2012]
The continuous rapid zig-zag, chaotic, random and ceaseless movement executed by a colloidal particle
in the dispersion medium is called brownian movement. This is due to the unbalanced bombardment of
the particles by the molecules of the dispersion medium.
15. What is electrophoresis? [March-2006, Mar-2013, Sep-2014]
The movement of sol particles under an applied electric potential is called electrophoresis or
cataphoresis.
If the sol particles are negatively charged, they migrate toward the positive electrode.
If the sol particles are positively charged, they migrate toward the negative electrode.
16. Write a note on electrodialysis [Sep-2008]
Electrodialysis is a method of purification of sol. In this process, dialysis is carried under the influence
of electric field. Potential is applied between the metal screens supporting the membranes. This speeds
up the migration of ions to the opposite electrode. In this way the electrolytes are removed from sol.
+ -
+-
Membrane Metal Screen
Impure Sol WaterWater
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17. Formation of delta is a colloidal phenomenon. Explain [Sep-2013]
River water is colloidal solution of clay. Sea water contains a number of electrolytes. When river water
meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay which
gets deposited with the formation of delta.
18. What is tanning? [Sep-2007]
Animal hides are colloidal in nature. When a hide, which has positively charged particles, is soaked in
tannin, which contains negatively charged colloidal particles, mutual coagulation takes place, which
results in the hardening of leather. The process is termed as tanning.
19. What are emulsions? Give two examples [June-2006, June-2013]
Emulsions are liquid-liquid colloidal systems.
If a mixture of two immiscible liquids is shaken, a coarse dispersion of one liquid in the other is
obtained which is called emulsion.
Eg: Milk, Butter
20. Give the dispersed phase and dispersion medium for a gel. Give example for gel [Sep-2016]
Dispersed
phase
Dispersion
medium Name Example
Liquid Solid Gel Curd, Cheese
5 MARK
1. Distinguish between physical adsorption and chemical adsorption
[June-2009, June-2013, Sep-2015, Mar-17]
No Physical adsorption Chemical adsorption
1 It is due to intermolecular Vander waal’s
force.
It is due to chemical bond formation
2 Heat of adsorption is small. Heat of adsorption is large.
3 Reversible. Irreversible.
4 Increase of pressure increases adsorption. Change of pressure has no effect.
5 Forms multimolecular layers on
adsorbent surface.
Forms unimolecular layer
2. Discuss the factors affecting adsorption [Sep-2012]
1. Temperature
Adsorption is invariably accompanied by evolution of heat. Therefore, in accordance with
Le chatelier’s principle, the magnitude of adsorption increases with decrease in temperature.
2. Pressure Since adsorption of a gas leads to decrease of pressure, the magnitude of adsorption increases with
increase in pressure.
3. Nature of the gas
It is observed that the more readily soluble and easily liquefiable gases such as ammonia, chlorine
and sulphur dioxide are adsorbed more than the hydrogen, nitrogen and oxygen. The reason is that
Vander waal’s or intermolecular forces which are involved in adsorption are more predominant in
the former than in the latter.
4. Nature of the adsorbent
Adsorption is a surface phenomenon. Therefore, the greater the surface area per unit mass of the
adsorbent, the greater is its capacity for adsorption under the given conditions of temperature and
pressure.
3. Write the general characteristics of catalytic reactions [Sep-2011, Jun-2016]
Refer – 3 mark – Q - 2
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4. Write briefly about Intermediate Compound Formation Theory of catalysis. (or)
Explain the theory of homogeneous catalysis
[Mar-2007, Sep-2008, June-2011, June-2012, June-2015]
Intermediate compound formation theory explains the mechanism of homogeneous catalytic reactions.
According to this theory, the catalyst first forms an intermediate compound with one of the reactants.
The compound is formed with less energy consumption than needed for the actual reaction. The
intermediate compound being unstable combines with other reactant to form the desired product and the
catalyst is regenerated.
For example, a reaction of the type
A + B C AB
which occurs in presence of a catalyst C, may take place as
A + C AC
Catalyst Intermediate compound
AC + B AB + C
Product catalyst
Eg: The catalytic oxidation of SO2 to SO3 in the lead chamber process.
2NO + O2 2NO2
Catalyst Intermediate compound
NO2 + SO2 SO3 + NO
Product catalyst
5. Write on Adsorption Theory of catalysis (or)
Explain the theory of heterogeneous catalysis [Mar-2006, Mar-2009, Mar-2013, Sep-16]
This theory explains the mechanism of heterogeneous catalysis. Here, the catalyst functions by
adsorption of the reacting molecules on its surface.
A B
C D C D
step-1 step-2
step-3 step-4
In general, there are four steps involved in the heterogeneous catalysis.
A(g) + B(g) )solid(Catalyst
C(g) + D(g)
Step - 1. Adsorption of reactant molecules
The reactant molecules A and B strike the surface of the catalyst. They are held up at the
surface by weak vanderwaal’s forces or by partial chemical bonds.
Step - 2. Formation of Activated complex
The particles of the reactants adjacent to one another join to form an intermediate complex (A-
B). The activated complex is unstable.
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Step - 3. Decomposition of Activated complex
The activated complex breaks to form the products C and D. The separated particles of the
products hold to the catalyst surface by partial chemical bonds.
Step - 4. Desorption of Products
The particles of the products are desorbed or released from the surface.
6. Write briefly about the preparation of colloids by dispersion methods.
[June-2006, Mar-2008, Sep-2013, Sep-2014, Mar-2016]
Dispersion methods: By splitting coarse aggregates of a substance into a colloidal size.
1. Mechanical dispersion using colloidal mill
The solid along with the liquid is fed into a colloidal mill. The colloidal mill consists of two steel
plates nearly touching each other and rotating in opposite directions with high speed. The solid
particles are ground down to colloidal size and then dispersed in the liquid. Colloidal graphite and
printing inks are made by this method.
Suspension
Metal disc
Discharge
Driving belt
Discharge
2. Electro-dispersion method: (Bredig’s Arc Method)
This method is suitable for the preparation of colloidal solution of metals like gold, silver, platinum
etc. An arc is struck between the metal electrodes under the surface of water containing some
stabilising agent such as trace of alkali. The water is cooled by immersing the container in a cold
bath. The intense heat of the arc vapourises some of the metal which condenses under cold water.
+ -Metal rods
Water + KOH
Ice
3. Ultra-sonic dispersion
The sound waves of high frequency are usually called ultra-sonic waves. Ultrasonic waves are
passed through the solution containing larger particles. They break down to form colloidal solution.
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4. Peptisation
The dispersion of a precipitated material into colloidal solution by the action of an electrolyte in
solution is termed as peptisation. The electrolyte used is called a peptizing agent.
Examples
i. Silver chloride can be converted into a sol by adding hydrochloric acid
ii. Ferric hydroxide yields a sol by adding ferric chloride
7. Write briefly about the preparation of colloids by chemical methods
[Sep-2006, June-2007, Mar-2010, June-2010, Sep-2010,Mar-2012, Mar-2014, June-2014]
1. Double decomposition
An Arsenic sulphide sol is prepared by passing a slow stream of hydrogen sulphide gas through a
cold solution of arsenious oxide. This is continued till the yellow colour of the sol attains maximum
intensity.
As2O3 + 3H2S As2S3 + 3H2O
Yellow
Excess hydrogen sulphide is removed by passing in a stream of hydrogen
2. Oxidation
A colloidal solution of sulphur is obtained by passing H2S into a solution of sulphur dioxide.
2H2S + SO2 2H2O + 3S
3. Reduction
Silver sols and gold sols can be obtained by treating dilute solution of silver nitrate or gold chloride
with organic reducing agents like tannic acid or formaldehyde.
AgNO3 + tannic acid Ag Sol
AuCl3 + tannic acid Au Sol
4. Hydrolysis
A colloidal solution of ferric hydroxide is obtained by boiling a dilute solution of ferric chloride.
FeCl3 + 3H2O Fe(OH)3 + 3HCl
Red Sol
8. Write notes on i) Auto catalyst ii) Promoters [Sep-2009]
Refer – 3 mark – Q – 5 & 7
9. How are colloids purified by dialysis? [Mar-2011]
The process of removing ions (or molecules) from a sol by diffusion through a permeable membrane is
called Dialysis.
Animal membranes (bladder) or those made of parchment paper and cellophane sheet, have very fine
pores. These pores permit ions (or small molecules) to pass through but not the large colloidal particles.
When a sol containing dissolved ions (electrolyte) or molecules is placed in a bag of semi permeable
membrane dipping in pure water, the ions diffuse through the membrane. By using a continuous flow of
fresh water, the concentration of the electrolyte outside the membrane tends to be zero. Thus diffusion
of the ions into pure water remains brisk all the time. In this way, practically all the electrolyte present
in the sol can be removed easily.
+
+
+
+
-
-
-
-
Water
Colloidal
particles
Dialysis bag
Molecules
Ions
Water
+
+-
-
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10. Write a note on Electrophoresis or How can you determine the charge of the sol particles? [Mar-15]
1. The movement of sol particles under an applied electric potential is called electrophoresis
or cataphoresis. 2. If the sol particles have negatively charged, they migrate toward the positive electrode.
3. If the sol particles have positively charged, they move toward the negative electrode.
4. From the direction of movement of the sol particles, we can determine the charge of the sol particles.
5. The phenomenon of electrophoresis can be demonstrated by placing a layer of As2S3 sol under two
limbs of a U-tube. When a potential difference of about 100 volts is applied across the two platinum
electrodes dipping in deionised water, it is observed that the level of the sol drops on the negative
electrode side and rises on the positive electrode side. This shows that As2S3 sol has migrated to the
positive electrode, indicating that the particles are negatively charged.
+ -
Deionized waterOriginal
Sol levels
Platinum electrode
As2S3 SOL
(Negative)
11. What is electro osmosis? Explain [Sep-2007, June-2008]
In a sol, the dispersion medium carries an equal but opposite charge to that of the dispersed particles.
Thus, the medium will move in opposite direction to the dispersed phase under the influence of applied
electric potential. The movement of the dispersion medium under the influence of applied potential is
known as electro-osmosis.
The phenomenon of electro osmosis can be demonstrated by using a U-tube in which a plug of wet clay
(a negative colloid) is fixed. The two limbs of the tube are filled with water to the same level. The
platinum electrodes are immersed in water and potential applied across them. It will be observed that
water level rises on the cathode side and falls on anode side. This movement of the medium towards the
negative electrode, shows that the charge on the medium is positive
+++
+ -
Wetclay(Negative)
Deionized water
Original
Sol levels
Platinum electrode
+++
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Additional questions
1 MARK
1. For selective hydrogenation of alkynes into alkene the catalyst used is
a) Ni at 250°C b) Pt at 25°C c) Pd, partially inactivated by quinoline d) Raney nickel
2. Which one of the following gases is adsorbed more?
a) H2 b) N2 c) O2 d) Cl2
3. The size of colloidal particles ranges from
a) 10 Å to 2000 Å b) 1 Å to 10 Å c) 100 nm to 200 nm d) 1 nm to 2000 nm
4. The migration of dispersion medium under the influence of an electric field is known as
a) electro osmosis b) cataphoresis c) electrodialysis d) electrophoresis
5. Brownian movement is due to the unbalanced bombardment of the colloidal particles by
a) other colloidal particles b) molecules of dispersion medium c) both d) none
6. Artificial rain can be caused by throwing ------- from an aeroplane
a) starch b) ice c) electrified sand d) all of these
7. Which of the following food articles are colloidal in nature?
a) milk, butter b) halwa, ice cream c) fruit juice d) all the above
8. Which of the following can coagulate blood? [styptic action]
a) alum b) ferric chloride c) both a and b d) none
9. Which of the following is added in the purification of drinking water?
a) Alum b) FeCl3 c) Tannic acid d) all of these
10. Formation of delta is due to
a) coagulation of colloidal solution of clay b) Brownian movement
c) electrophoresis d) electro-osmosis
11. In the process of tanning, animal hide is soaked in
a) tannin b) chromium salts c) both are correct d) none
12. Photographic plates are prepared by coating an emulsion of light sensitive ------- in gelatin
a) chromium salt b) tannin c) silver bromide d) none of these
13. The function of HCl in the conversion of AgCl precipitate into a colloid is
a) peptising agent b) emulsifying agent c) reducing agent d) precipitating agent
14. A colloidal solution of gas in gas is not possible as gases
a) are completely miscible and form true solution b) are immiscible
c) form suspension d) none of these
15. Which is used as the stabilising agent in the preparation of colloidal solution of metals like gold, silver
and platinum by Bredig’s arc method?
a) traces of alkali (KOH) b) clay c) AgCl d) FeCl3
16. Which is O/W type colloid?
a) milk b) vanishing cream c) both a and b d) none
17. Which is W/O type colloid?
a) butter b) cream c) both a and b d) none
18. Emulsions can be broken into constituent liquids by
a) heating b) freezing c) centrifuging d) all of these
19. In O/W colloid the dispersion medium is
a) oil b) water c) both d) none
20. Which is used for intramuscular injection?
a) Milk of magnesia b) colloidal antimony c) Argyrol d) colloidal gold
3 MARK
1. Define adsorption.
The condition in which concentration of a substance in the interfacial layers between two phases is
greater than in the bulk of either phase, then the substance is said to be adsorbed at the interface and the
phenomenon is known as adsorption.
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2. What is meant by physical adsorption?
This adsorption is due to the operation of forces between solid surface and the adsorbate molecules that
are similar to vander waal’s forces between molecules. These forces are generally undirected and
relatively non specific. Physical adsorption can also be defined as that type of adsorption where physical
forces hold the gas molecules to the solids.
3. What is meant by chemical adsorption?
Chemical adsorption or chemisorption is defined as a type of adsorption in which chemical bonds serve the
function of holding gas molecules to the surface. It occurs due to the stronger binding forces, comparable
with those leading to formation of chemical compounds. It is generally an irreversible process.
4. What is positive catalyst? Give an example
A catalyst which enhances the speed of the reaction is called positive catalyst and the phenomenon is
known as positive catalysis.
Decomposition of H2O2 increases in presence of colloidal platinum
2H2O2 Pt
2H2O + O2
5. What is negative catalyst or inhibitor? Give an example
There are certain substances which, when added to the reaction mixture, retard the reaction rate instead
of increasing it. These are called negative catalysts or inhibitors and the phenomenon is known as
negative catalysis.
The decomposition of hydrogen peroxide decreases in presence of glycerine.
2H2O2 Glycerine
2H2O + O2
6. What is an induced catalyst?
When one reactant influences the rate of other reaction, which does not occur under ordinary conditions,
the phenomenon is known as induced catalysis.
Sodium arsenite solution is not oxidised by air. If, however, air is passed through a mixture of the
solution of sodium arsenite and sodium sulphite, both of them undergo simultaneous oxidation. Thus
sulphite has induced the arsenite and hence is called induced catalyst.
7. Define dispersed phase and dispersion medium
A colloidal system is made up of two phases. The substance distributed as the colloidal particles is
called the dispersed phase. The second continuous phase in which the colloidal particles are dispersed
is called the dispersion medium.
8. How are lyophobic sols prepared by ultrasonic dispersion method?
The sound waves of high frequency are usually called ultra-sonic waves. Ultrasonic waves are passed
through the solution containing larger particles. They break down to form colloidal solution.
9. What is meant by Helmholtz double layer?
The surface of colloidal particle acquires a positive charges by selective adsorption of a layer of positive
ions around it. This layer attracts counterions from the medium which form a second layer of negative
charges. The combination of the two layers of charges around the sol particle is called Helmholtz
double layer.
Colloidal
particle
++
+
++
+
+
+
-
-
-
-
-
-
-
-
Fixed layer
Movable layer
Liquid medium
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10. Define electro osmosis.
The movement of the dispersion medium under the influence of applied potential is known as electro-osmosis.
11. Why does sky appear blue?
Colloidal particles scatter blue light. Dust particles along with water suspended in air scatter blue light
which reaches our eyes and the sky looks blue to us.
12. Formation of rain is a colloidal phenomenon. Explain
Clouds are aerosols having small droplets of water suspended in air. On account of condensation in the
upper atmosphere, the colloidal droplets of water grow bigger and bigger in size, till they come down in
the form of rain. Sometimes, the rainfall occurs when two oppositively charged clouds meet.
13. How is artificial rain caused?
It is possible to cause artificial rain by throwing electrified sand or spraying a sol carrying charge
opposite to the one on clouds from an aeroplane.
14. Formation of fog and mist is a colloidal phenomenon. Explain When a large mass of air, containing dust particles, is cooled below its dewpoint, the moisture from the
air condenses on the surfaces of these particles forming fine droplets. These droplets being colloidal in
nature continue to float in the air in the form of mist or fog.
15. Explain the use of colloids in the purification of drinking water
The water obtained from natural sources often contains bacteria and suspended impurities. Alum is
added to such water so as to destroy the bacteria as well as to coagulate the suspended impurities and
make water fit for drinking purposes.
5 MARK
1. Write a note on ultra filtration
Sols pass through an ordinary filter paper. Its pores are too large to retain the colloidal particles.
However, if the filter paper is impregnated with collodion or a regenerated cellulose such as cellophane
or visking, the pore size is much reduced. Such a modified filter paper is called an ultrafilter. The
separation of the sol particles from the liquid medium and electrolytes by filtration through an ultrafilter
is called ultrafiltration.
Ultrafiltration is a slow process. Gas pressure (or suction) has to be applied to speed it up. The colloidal
particles are left on the ultrafilter in the form of slime. The slime may be stirred into fresh medium to
get back the pure sol.
Gas Pressure
Sol + Electrolyte
Ultrafilter supported
on wiremesh
Water + Electrolyte
+
+
-
-
+
+-
-
2. Explain the use of colloids in medicine
1. Most of the medicines in use are colloidal in nature.
2. Argyrol is a silver sol used as an eye lotion.
3. Colloidal antimony is used in curing kalazar.
4. Colloidal gold is used for intramuscular injection.
5. Milk of magnesia, an emulsion, is used for stomach disorders.
Colloidal medicines are more effective because these are easily assimilated.
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3. Write a note on emulsions These are liquid-liquid colloidal systems. If a mixture of two immiscible liquids is shaken, a coarse
dispersion of one liquid in the other is obtained which is called emulsion.
There are two types of emulsions
(i) Oil dispersed in water (O/W type)
In this system water acts as dispersion medium.
Examples: milk and vanishing cream. In milk, liquid fat is dispersed in water.
(ii) Water dispersed in oil (W/O type)
In this system oil acts as dispersion medium. Examples: butter and cream.
Emulsions of oil and water are unstable and sometimes they separate into two layers on standing. For
the stabilization of an emulsion, a third component called emulsifying agent is usually added. The
emulsifying agent forms an interfacial film between suspended particles and the medium.
The principal agents for O/W emulsions are proteins, gums, natural and synthetic soaps.
For W/O emulsions, the principal emulsifying agents are heavy metal salts of fatty acids, long chain
alcohols and lampblack.
4. Write a note on Cottrell smoke precipitator
Electrical precipitation of smoke: Smoke is a colloidal solution of solid particles such as carbon,
arsenic compounds, dust, etc., in air. The smoke, before it comes out from the chimney, is led through a
chamber containing plates having a charge opposite to that carried by smoke particles. The particles on
coming in contact with these plates lose their charge and get precipitated. The particles thus settle down
on the floor of the chamber. The precipitator is called Cottrell precipitator.
Charged knob
Chimney
Dust free gases
Precipitated
carbon or dust
Smoke
************
Process Positive catalyst Promoter
Lead chamber process for the manufacture of H2SO4 Nitric oxide NO
Contact process for the manufacture of H2SO4
Pt (Platinised asbestos) or
V2O5
Hydrolysis of methyl acetate H+ ions furnished by HCl
Haber’s process Finely divided Fe Mo
Decomposition of H2O2 Colloidal Pt
Decomposition of KClO3 MnO2
Ostwald’s process for the manufacture of HNO3 Platinised asbestos
Deacon’s process for the manufacture of Cl2 Cupric chloride CuCl2
Bosch’s process for the manufacture of H2 Ferric oxide Fe2O3 Chromic oxide
Cr2O3
Hydrogenation of vegetable oils Finely divided Ni
Bergius process for the synthesis of petrol from coal Ferric oxide Fe2O3
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13. ELECTRO CHEMISTRY –I
Blue print 1 Mark = 1 3 Mark = 1 5 Mark = 1+1* Total marks = 14
Govt. Examination questions 1 MARK
1. When one coulomb of electricity is passed through an electrolytic solution, the mass deposited on the
electrode is equal to [Mar-2006, Sep-2007, Sep-2011, Sep-2013]
a) equivalent weight b) molecular weight c) electrochemical equivalent d) one gram
2. Faraday’s laws of electrolysis are related to ............... [Mar-2009]
a) atomic number of the cation b) atomic number of the anion
c) equivalent weight of the electrolyte d) speed of the cation
3. The equivalent conductivity of CH3COOH at 25oC is 80 ohm
-1 cm
2 eq
–1 and at infinite dilution 400
ohm–1
cm2 eq
–1. The degree of dissociation of CH3COOH is........... [June-2008, June-2013]
a) 1 b) 0.2 c) 0.1 d) 0.3
4. When sodium acetate is added to acetic acid, degree of ionisation of acetic acid
[Sep-2006, June-2009, Sep-2009, June-2015, Mar-2016]
a) increases b) decreases c) does not change d) becomes zero
5. Ostwald’s dilution law is applicable in the case of the solution of ........... [Mar-2010, Mar-2012]
a) CH3COOH b) NaCl c) NaOH d) H2SO4
6. When 10–6
mole of a monobasic strong acid is dissolved in one litre of solvent, the pH of the solution is
[Sep-2008, Sep-2015]
a) 6 b) 7 c) less than 6 d) more than 7
7. When pH of a solution is 2, the hydrogen ion concentration in moles litre–1
is
[Mar-2008, Sep-2010, Sep-2012, Mar-2013]
a) 1 10–12
b) 1 10–2
c) 1 10–7
d) 1 10–4
8. The pH of a solution containing 0.1 N NaOH solution is -------------
[June-2010, June-2011, June-2012, Mar-2014]
a) 1 b) 10–1
c) 13 d) 10–13
9. If 0.2 ampere can deposit 0.1978g of copper in 50 minutes, how much of copper will be deposited by
600 coulombs? [June-2007]
a) 19.78g b) 1.978g c) 0.1978g d) 197.8g
10. The value of enthalpy of neutralisation of strong acid by strong base is [June-2014]
a) – 87.32 kJ equiv–1
b) – 57.32 kJ equiv–1
c) – 72.57 kJ equiv–1
d) – 72.23 kJ equiv–1
11. The number of moles of electrons required to discharge one mole of Al+3
is [Mar-2015]
a) 3 b) 1 c) 2 d) 4
12. For the titration between oxalic acid and NaOH, the indicator used is [June-2006, Mar-2011]
a) potassium permanganate b) phenolphthalein c) litmus d) methyl orange
13. The indicator used in the titration of NH4OH with HCl is [Mar-2007]
a) KMnO4 b) Methyl orange c) Phenolphthalein d) Litmus
14. For the titration between HCl and sodium carbonate, the indicator used is [Sep-2014, Mar-17]
a) potassium permanganate b) phenolphthalein c) litmus d) methyl orange
Titration Suitable indicator
Strong acid Vs Strong base
( HCl Vs NaOH )
Phenolphthalein
& Methyl orange
Weak acid Vs Strong base
( Oxalic acid Vs NaOH ) Phenolphthalein
Strong acid Vs Weak base
( HCl Vs Na2CO3 ) & ( HCl Vs NH4OH ) Methyl orange
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15. pH range of phenol red is [Jun-2016]
a) 6.8 – 8.4 b) 4.4 – 6.2 c) 3.1 – 4.4 d) 8.3 – 10
Indicator pH range
Methyl orange 3.1 – 4.4
Phenolphthalein 8.3 – 10 Phenol red 6.8 – 8.4 Methyl red 4.4 – 6.2
16. Which one of the following relationship is correct? [Sep-2016]
a) pH = ][H
1
b) pH = 10log [H+] c) 10log pH = [H
+] d) pH = 10log
][H
1
3 MARK
1. Why does the metallic conduction decrease with increase in temperature? [Sep-2011]
Conductivity of metal decreases with increase in temperature due to the enhanced thermal vibration of
metal atoms disrupting the movement of electrons passing through them.
2. Define specific and equivalent conductance. How are they related? [June-2008]
Specific conductance (is defined as the conductance of one metre cube of an electrolyte solution.
R
1
a
l mho m
–1
Equivalent conductance (C) is defined as the conductance of an electrolyte solution containing one
gram equivalent of the electrolyte.
Relation between specific and equivalent conductance
C =N
10 3
mho m2 (gm equiv)
–1
3. State Faraday’s first law and second law of electrolysis [Mar-2006]
Faraday’s First law: [June 2015]
The mass of the substance (m) liberated at the electrodes during the electrolysis is directly proportional
to the quantity of electricity (Q) that passes through the electrolyte.
m ∝Q
m = Z I t
Faraday’s Second law: [Mar-2013]
When the same quantity of electricity passes through solutions of different electrolytes, the amounts of
the substances liberated at the electrodes are directly proportional to their chemical equivalents.
4. Define electrochemical equivalent [June-2009, Sep-2014]
Electrochemical equivalent is the amount of a substance deposited by 1 ampere current passing for 1
second.
m = z I t
where z = Electrochemical equivalent
If I = 1 ampere and t = 1 second, then m = z
Unit of electrochemical equivalent (z) = g coulomb–1
5. State Ostwald’s dilution law [Mar-2009, Mar-2015]
Ostwald’s dilution law relates the dissociation constant of the weak electrolyte with the degree of
dissociation and the concentration of the weak electrolyte.
Ka = α1
Cα2
is known as the Ostwald’s dilution law
Where, C = concentration of weak electrolyte= the degree of dissociation Ka = dissociation constant
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6. State Kohlrausch’s law [Jun-2006 , Sep-2008, Sep-2009, Sep-2012, Mar-2016, Sep-2016]
‘‘At infinite dilution wherein the ionisation of all electrolytes is complete, each ion migrates
independently and contributes a definite value to the total equivalent conductance of the electrolyte’’.
7. What is ionic product of water? Give its value [June-2013, Sep-2013]
Water is a weak electrolyte. The dissociation equilibrium of water can be considered as,
2H2O ⇌ H3O+ + OH
–
According to law of mass action, Keq = 2
2
3
]OH[
]OH[]OH[
Since water as a solvent is always in excess and change in concentration due its dissociation is
negligible. Hence water concentration is assumed to be constant.
Keq [H2O]2 = [H3O
+] [OH
–]
Kw = [H3O+] [OH
–]
The constant Kw is called as the ionic product of water and its value is given by the product of
concentrations of hydronium (H3O+) and hydroxide (OH
–) ions.
At 298 K, Kw = 1 10–14
mol2
dm–6
.
8. What is common ion effect? Give an example
[Sep-2006, June-2007, Sep-2007, March-2008, Mar-2010, Mar-2011, Mar-2014, June-2014]
The reduction of the degree of dissociation of a salt by the addition of a common-ion is called the
Common-ion effect.
Example: In a saturated solution of silver chloride, we have the equilibrium
AgCl (s) ⇌ Ag+
(aq) + Cl–
(aq)
When sodium chloride is added to the solution, the concentration of Cl– ions will increase. Therefore,
according to Le Chatelier’s principle, the equilibrium will be shifted to the left to form more of solid
AgCl. Thus the dissociation of AgCl will decrease.
9. What are buffer solutions? What are the types of buffer solutions?
[June-2010, Sep-2010, June-2011, Mar-2012, June-2012]
A buffer solution is one which maintains its pH fairly constant even upon the addition of small amounts
of acid or base.
Two types of buffer solutions are :
1. Acid buffers : a weak acid and its salt with a strong base. e.g., CH3COOH + CH3COONa.
2. Basic buffers : a weak base and its salt with a strong acid e.g., NH4OH + NH4Cl.
10. The mass of the substance deposited by the passage of 10 ampere of current for 2 Hours 40
minutes and 50 seconds is 9.65 g. Calculate electro chemical equivalent [March-2007]
According to Faraday’s first law of electrolysis, m= z It
Given that, m = 9.65 g , I = 10 ampere,
t = (2 × 60 × 60) + (40 × 60) + 50 = 9650 seconds
z = It
m =
965010
65.9
= 0.0001 g C
–1
11. What are indicators? Give examples [Sep-2015]
An indicator is a substance which indicates the completion of a reaction by sharp colour change at the
end point without taking part in the reaction. Eg: Phenolphthalein, Methyl orange.
12. Define molar conductance. Mention its unit [Jun-2016]
Molar conductance (C) is defined as the conductance of a solution containing one mole of the
electrolyte dissolved in it.
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Unit : mho m2 mol
–1
C =
M
10 3 mho m
2 mol
–1
where M is the molarity of the electrolytic solution and is the specific conductance
13. Write the significances of Henderson equation [Mar-17]
1. The pH of a buffer solution can be calculated from the initial concentrations of the weak acid and
the salt provided Ka is given. However, the Henderson-Hasselbalch equation for a basic buffer will
give pOH and its pH can be calculated as (14 – pOH).
2. The dissociation constant of a weak acid (or weak base) can be determined by measuring the pH of
a buffer solution containing equimolar concentrations of the acid (or base) and the salt.
pH = pKa + log [acid]
[salt]
Since, [salt] = [acid], log [acid]
[salt] = log 1 = 0 & pKa = pH
The measured pH, therefore, gives the value of pKa of the weak acid. Likewise we can find the
pKb of a weak base by determining the pOH of equimolar basic buffer.
3. A buffer solution of desired pH can be prepared by adjusting the concentrations of the salt and the
acid added for the buffer.
5 MARK
1. Write the differences between Metallic conductors and Electrolytic conductors
[June-2011, Sep-2014, Sep-2015]
No. Metallic conductors
( Electronic conductors ) Electrolytic conductors
1 Pure metals and their alloys are called as
metallic conductors.
Ionic compounds in fused state or in dissolved
state are called electrolytic conductors.
2 Metallic conductors conduct electricity
due to the movement of free electrons.
Electrolytic conductors conduct electricity
due to the movement of positive and negative
ions.
3 There is no chemical change in the
material when electricity is passed.
Chemical change occurs.
4 There is only flow of electrical energy but
there is no transfer of matter
There is actual transfer of matter since ions
move towards respective electrodes.
5 Conductivity of metal decreases with
increase in temperature due to the
enhanced thermal vibration of metal
atoms disrupting the movement of
electrons passing through them.
The conductivity of electrolytes increases
with increase in temperature. This is due to
increase in ionic mobility.
2. Write the postulates of Arrhenius theory of electrolytic dissociation
[June-2009, June-2010, Mar-2012, Sep-2013]
1. When dissolved in water, neutral electrolyte molecules are split up into two types of charged
particles. The positively charged particles are called cations and those having negative charge are
called anions. The theory assumes that the ions are already present in the solid electrolyte and these
are held together by electrostatic force. When placed in water, these neutral molecules dissociate to
form separate anions and cations.
BAwater
BA
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2. The ions present in solution constantly reunite to form neutral molecules. Thus there is a state of
equilibrium between the undissociated molecules and the ions.
AB ⇌ A+ + B
–
Applying the Law of Mass Action to the ionic equilibrium we have,
K = [AB]
][B][A
where K is called the Dissociation constant.
3. The charged ions are free to move through the solution to the oppositely charged electrode. This is
called as migration of ions. This movement of the ions constitutes the electric current through
electrolytes.
4. The electrical conductivity of an electrolyte solution depends on the number of ions present in
solution.
When current is passed through the electrolytic solution, the ions migrate to the opposite electrodes.
Thus in the electrolytic solution of AgNO3, the cations (Ag+) will move to the cathode and anions
(NO3–) will move to the anode.
+
Ag+
NO3
-
CathodeAnode
Migration of ions through electrolytic solution to opposite electrodes
5. The properties of solution of electrolytes are the properties of ions. Presence of hydrogen ions (H+)
renders the solution acidic while presence of hydroxide ions (OH–) renders the solution basic.
6. There are two types of electrolytes:
i) Strong electrolytes: Completely ionised in water.Eg: NaCl, KCl, AgNO3
ii) Weak electrolytes: Partially ionised in water.Eg: CH3COOH, NH4OH
3. Write the evidences of Arrhenius theory of electrolytic dissociation. [Mar-2010, Mar-2014]
1. The enthalpy of neutralisation of strong acid by strong base is a constant value and is equal to –
57.32 kJ. gm.equiv–1
. This aspect is well explained by Arrhenius theory of electrolytic dissociation.
Strong acids and strong bases are completely ionised in water and produce H+ and OH
– ions
respectively along with the counter ions. The net reaction in the acid-base neutralisation is the
formation of water from H+ and OH
– ions.
H+ + OH
– H2O ∆ H = –57.32 kJ mol
–1
2. The colour of certain salts or their solution is due to the ions present.
For example, copper sulphate is blue due to Cu2+
ions.
Nickel salts are green due to Ni2+
ions.
Metallic chromates are yellow due to CrO42–
ions.
3. Ostwalds dilution law, common ion effect and solubility product and other such concepts are based
on Arrhenius theory.
4. Chemical reactions between electrolytes are almost ionic reactions. This is because these are
essentially the reaction between oppositely charged ions. For example,
Ag+ + Cl
– AgCl
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5. Electrolytic solutions conduct current due to the presence of ions which migrate in the
presence of electric field.
6. Colligative properties depend on the number of particles present in the solution. Electrolytic
solution has abnormal colligative properties. For example, 0.1 molal solution of NaCl has elevation
of boiling point about twice that of 0.1 molal solution of non-electrolyte. The abnormal
colligative properties of electrolytic solutions can be explained with theory of electrolytic
dissociation.
4. Explain Ostwald’s dilution law [Mar-2006, Mar-2007, Sep-2008, Sep-2011, Mar-2015, Mar-17]
Ostwald’s dilution law relates the dissociation constant of the weak electrolyte with the degree of
dissociation and the concentration of the weak electrolyte.
Consider the dissociation equilibrium of CH3COOH which is a weak electrolyte in water.
CH3COOH ⇌ CH3COO–
+ H+
Initial number of moles 1 0 0
Number of moles reacted α – –
Number of moles remaining at equilibrium 1 – α α α
Equilibrium molar concentration C (1 – α) C α C α
= degree of dissociation of CH3COOH
C = Initial concentration of CH3COOH
Ka = COOH][CH
][H]COO[CH
3
3
Ka = α)-(1C
CαCα =
α)-(1
Cα 2
If is too small, then Ka = 2C
Ka = α)-(1
Cα 2
is known as Ostwald’s dilution law.
5. What is a buffer solution? Explain buffer action with example [Sep-2009, Sep-2012, Mar-2016]
A buffer solution is one which maintains its pH fairly constant even upon the addition of small amounts
of acid or base.
Two types of buffer solutions are :
1. Acid buffers : a weak acid and its salt with a strong base. e.g., CH3COOH + CH3COONa.
2. Basic buffers : a weak base and its salt with a strong acid e.g., NH4OH + NH4Cl.
Buffer action of acidic buffer:
Consider the buffer system consisting of a solution of acetic acid and sodium acetate
(CH3COOH/CH3COONa).
1. Addition of HCl.
Upon the addition of HCl, the increase in H+ ions is counteracted by association with the excess of
acetate ions to form unionised CH3COOH. Thus the added H+ ions are neutralised and the pH of the
buffer solution remains unchanged.
2. Addition of NaOH
When NaOH is added to the buffer solution, the additional OH– ions combine with CH3COOH to give
CH3COO– and H2O. Thus pH of the buffer solution is maintained almost constant.
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CH3COOH H+CH3COO
-+
Na+
CH3COO-
+CH3COONa
Addition of OH-
H2O
Addition of H+
CH3COOH
CH3COO-
+
Mechanism of Buffer action of an acid buffer
Buffer action of basic buffer:
Consider the buffer system consisting of a solution of ammonium hydroxide and ammonium chloride
(NH4OH/NH4Cl).
1. Addition of HCl
Upon addition of HCl the H+ ions combine with NH4OH to form NH4
+ and H2O. Thus pH of the buffer
solution is maintained almost constant.
2. Addition of NaOH
When NaOH is added, the OH– ions combine with NH4
+ ions present in the buffer solution to give
NH4OH and hence pH of the buffer solution is maintained almost constant.
NH4OH OHNH4
+ -+
ClNH4
+ -+NH4Cl
Addition of H+
H2O
Addition of OH NH4OH-
NH4
++
Mechanism of Buffer action of a basic buffer
6. Derive Henderson equation [Sep-2006, Jun-2007, Mar-2008, Mar-2011, Jun-2014, Jun-15, Jun-16]
The pH of an acid buffer can be calculated from the dissociation constant, Ka of the weak acid and the
concentrations of the acid and the salt used.
Consider an acidic buffer : Weak acid HA + its salt NaA
HA ⇌ H+ + A
–
ANaNaA
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The dissociation expression of the weak acid, HA, may be represented as
HA ⇌ H+ + A
–
Ka = [HA]
][A][H
[H+] =
][A
[HA]K a
The weak acid is only slightly dissociated and its dissociation is further depressed by the addition of the
salt (NaA) which provides A– ions (Common ion effect). As a result the equilibrium concentration of
the unionised acid is nearly equal to the initial concentration of the acid.
The equilibrium concentration [A–] is equal to the initial concentration of the salt added since it is
completely dissociated. Thus we can write the above equation as
[H+] = Ka
[salt]
[acid]
where [acid] = initial concentration of the acid , [salt] = initial concentration of the salt
Taking negative logs on both sides
– log [H+] = –log Ka – log
[salt]
[acid]
But, – log [H+] = pH and – log Ka = pKa
∴ pH = pKa – log [salt]
[acid]
pH = pKa + log [acid]
[salt]
This relationship is called the Henderson-Hasselbalch equation or simply Henderson equation.
In a similar way, the Henderson-Hasselbalch equation for a basic buffer can be derived. This can be
stated as :
pOH = pKb + log [base]
[salt]
7. Write notes on Ostwald’s theory of indicators [June-2006, June-2012. June-2013]
1. It is based on Arrhenius theory
2. According to this theory, the acid-base indicator is either a weak acid or a weak base.
3. They are partially ionised in solution.
4. The ionised and unionised forms have different colours.
5. The indicator exists predominantly in one of the two forms depending on the nature of the medium
and hence there is colour change when the nature of the medium changes.
Eg-i) Phenolphthalein: It is a weak acid and its ionisation can be written as
HPh ⇌ H+ + Ph
–
Unionised form Ionised form
(colourless) (pink)
In acidic medium, excess H+ ions are present which suppress the dissociation of HPh due to
common ion effect. Hence the indicator is mostly in unionised form and it is colourless.
In basic medium, the OH– ion combines with H+ ion to form water. So the dissociation of Hph is
favoured and the indicator is mostly in ionised form and it is pink in colour.
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Eg-ii) Methyl orange: It is a weak base and its ionisation can be written as
MeOH ⇌ Me+ + OH
–
Unionised form Ionised form
(yellow) (pink)
In basic medium, excess OH– ions suppress the dissociation of MeOH due to common ion
effect. Hence in basic medium, the indicator is mostly in unionised form which is yellow.
In acidic medium the H+ ions combine with OH
– ions to form water. So the dissociation of
MeOH is favoured. Hence the indicator is mostly in ionised form and has pink colour.
8. Write on Quinonoid theory of indicators [Mar-09, Sep-07, June-08, Sep-10, Mar-13, Sep-16]
1. According to Quinonoid theory the colour change of an acid-base indicator arises as a result of
structural change.
2. An indicator exists as an equilibrium mixture of two tautomeric forms namely, benzenoid and
quinonoid forms.
benzenoid form quinonoid form
3. One form exists in acidic solution and the other form in basic solution.
4. At least one of the tautomers is a weak acid or a weak base.
5. The two forms possess two different colours and as the pH of the solution containing the indicator is
changed, the solution shows a change of colour. The colour change is due to the fact that one
tautomer changes over to the other.
For example, phenolphthalein is tautomeric mixture of the two forms.
C
O
O
C
OOH
H
OH-
H+
C
O
O
C
OOH
C
O
C
OH
O
O
Benzenoid form
Colourless
Exists predominantly in
acidic medium
Quinonoid form
Pink
Exists predominantly in
basic medium
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Additional questions 1 MARK
1. The process in which chemical change occurs on passing electricity is termed as
a) neutralization b) hydrolysis c) electrolysis d) ionisation
2. The laws of electrolysis were enunciated first by
a) Dalton b) Faraday c) Kekule d) Avogadro
3. The specific conductance of a 0.01 M solution of KCl is 0.0014 ohm–1
cm–1
at 25oC. Its equivalent
conductance is ...............
a) 14 ohm–1
cm2 eq
–1 b) 140 ohm
–1 cm
2 eq
–1 c) 1.4 ohm
–1 cm
2 eq
–1 d) 0.14 ohm
–1 cm
2 eq
–1
4. NH4OH is a weak base because ...............
a) it has low vapour pressure b) it is only partially ionised
c) it is completely ionised d) it has low density
5. Which one of the following formulae represents Ostwald’s dilution law for a binary electrolyte whose
degree of dissociation is and concentration C
a) K =
α
Cα1 b) K =
α1
Cα2
c) K =
2α
Cα1 d) K =
Cα1
Cα2
6. A solution which is resistant to changes of pH on addition of small amounts of an acid or a base is
known as .............
a) buffer solution b) true solution c) isohydric solution d) ideal solution
7. The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salt is given by ...
a) [H+] = Ka
[Salt]
[Acid]
b) [H+] = Ka [Salt] c) [H
+] = Ka [Acid] d) [H
+] = Ka
[Acid]
[Salt]
8. Indicators used in acid-base titrations are ...........
a) strong organic acids b) strong organic bases
c) weak organic acids or weak organic bases d) non-electrolysis
9. Metallic conductors or electronic conductors such as metals and alloys conduct electricity due to the
movement of
a) ions b) electrons c) atoms d) none of these
10. Electrolytic conductors such as electrovalent or ionic compounds conduct electricity due to the
movement of
a) ions b) electrons c) atoms d) none of these
11. Intrinsic semi conductors become good conductors at
a) low temperature b) high temperature c) high pressure d) low pressure
12. Extrinsic semi conduction is due to
a) low temperature b) high temperature c) addition of impurity d) none of these
13. Silicon or Germanium doped with boron (B) becomes
a) non-conductors b) p-type semiconductor c) n-type d) none of these
14. Silicon or Germanium doped with arsenic (As) becomes
a) non-conductors b) n-type semiconductor c) p-type d) none of these
15. Which is a weak electrolyte?
a) CH3COOH b) NH4OH c) both d) none
16. Which is a strong electrolyte?
a) AgNO3 b) NaCl c) KCl d) All the above
17. Ostwalds dilution law is not applicable to
a) CH3COOH b) NH4OH c) NaCl d) all of these
18. For strong electrolytes, the degree of dissociation (α) tends to
a) 0 b) 1 c) 2 d) none
19. The quantity of electricity required to liberate electrochemical equivalent of substance is
a) 1 Coulomb b) 1 F c) 96495 coulomb d) 96495 F [1 ampere current passing for 1 second = 1 coulomb]
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20. The quantity of electricity required to liberate one gram equivalent of substance is
a) 1 coulomb b) 1 F or 96495 coulomb c) 96495 F d) none of these
21. Current carried by 1 mole of electrons or 6.023 × 1023
electrons is
a) 1 Coulomb b) 6.023 x 1023
Coulomb c) 1F or 96495 Coulomb d) none
22. The quantity of electricity required to liberate 1 mole of aluminium from aluminium sulphate solution is
a) 1 Coulomb b) 1 F c) 2 F d) 3 F
[quantity of electricity required to liberate 1 mole of Ag+ = 1 F
quantity of electricity required to liberate 1 mole of Cu 2+
=2 F]
23. Current carried by 1 electron or charge of an electron is
a) 1 Coulomb b) 6.023 × 1023
Coulomb c) 1.602 × 10–19
Coulomb d) none of these
24. Specific conductance is equal to
a) cell constant × conductance b)cell constant / resistance c) both a and b d) none
25. For 1 : 1 electrolyte like NaCl
a) λc = μc b) λc > μc c) λc < μc d) none of these
26. The plot of λc versus square root of concentration of strong electrolytes such as NaCl, NaOH, KCl,
AgNO3, Al2(SO4)3 etc.. is
a) linear b) non linear c) curve d) none of the above
27. The plot of λc versus square root of concentration of weak electrolytes such as CH3COOH, NH4OH
etc… is
a) linear b) non linear or curve c) both d) none of the above
28. Addition of which of the following decreases the dissociation of AgCl?
a) NaOH b) NaCl c) HF d) NH4OH
29. The degree of dissociation is given by the formula
a) C + b) λ
λ c c) C d) none of these
30. Addition of which of the following decreases the dissociation of NH4OH?
a) NaBr b) NaCl c) HF d) NH4Cl
31. Addition of which of the following decreases the dissociation of HF?
a) NaOH b) NaCl c) NaF d) NH4OH
32. The value of ionic product of water ( KW) at 298 K is
a) 1×10–7
mol dm–3
b) 1×10–14
mol2 dm
–6 c) 1×10
14 mol
2 dm
–6 d) none
33. The value of pKW at 298 K is
a) 14 b) 7 c) 0 d) 1×10–14
34. Which is correct?
a) [H+] = 10
–pH b) [H
+] = 10
pH c) [H
+] = 10
–pOH d) none
35. [H+] of a solution is 10
–5. Then, [OH
–] of that solution is
a) 10–5
b) 10–9
c) 10–7
d) 107
36. Which is correct?
a) [H+] = 10
–pH b) [H
+] = Cα c) [H
+] = CKa d) all are correct
37. pH + pOH = --------
a) 7 b) 14 c) 0 d) 1
38. Which is not a buffer solution?
a) CH3COOH + CH3COONa b) NH4OH + NH4Cl
c) C2H5COOH + C2H5COONa d) CH3COOH + CH3COONH4
39. pH of a buffer solution prepared by mixing equimolar concentrations of acetic acid and sodium acetate
is [pKa= 4.7] a) 4.7 b) 9.3 c) 0 d) 10–4.7
40. The pH of 1M HCl is
a) 1 b) 10 c) 0 d) 0.1
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41. The pH of 0.1 M acetic acid is 3. The degree of ionisation of acetic acid is
a) 0.1% b) 10% c) 1% d) 0.01%
42. The solution with pH = 7 is
a) acidic b) basic c) neutral d) weakly basic
3 MARK
1. Why does the electrolytic conduction increase with increase in temperature?
The conductivity of electrolytes increases with increase in temperature. This is due to increase in the
ionic mobility.
2. Define specific resistance
Specific resistance ( is the resistance in ohms which one meter cube of material offers to the passage
of electricitiy through it. Unit of specific resistance is ohm-meter.
3. What is cell constant?
In a conductance cell, the ratio of length to area of cross section is a constant and it is known as cell
constant
Cell constant = a
lUnit of cell constant = m
–1
4. Explain the variation of equivalent conductance with concentration
The effect of concentration on equivalent conductance can be studied from the plots of C values versus
square root of concentration of the electrolyte. By doing so, it has been found that different types of
plots are obtained depending on the nature of electrolyte.
For strong electrolytes C decreases linearly with increase in C while for weak electrolytes, there is a
curve type of non linear decrease of C with C
C
C
Strong electrolyte
Weak electrolyte
5. Derive the Relation between pH and pOH
Ionic product of water Kw = [H+] [OH
–]
Taking log on both sides,
log Kw = log [H+] + log [OH
–]
(or) – log Kw = –log [H+] – log [OH
–]
pKw = pH + pOH
Since Kw = 1 10–14
& pKw = –log (1 10–14
) = 14
pH + pOH = 14
6. What is meant by indicator range?
Every pH indicator changes its colour specifically in a ranging pH which is called as indicator range or
pH range.
pH range of Methyl orange = 3.1 – 4.4 & pH range of Phenolphthalein = 8.3 – 10
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7. What is a titration curve?
The pH values can be plotted against the volume of the base added and the curve so obtained is called
titration curve.
The titration curves are useful in the choice of a suitable indicator in an acidbase titration.
8. Phenolphthalein is not a suitable indicator in the titration of a strong acid against a weak base.
Give reason
Phenolphthalein is not a suitable indicator in the titration of a strong acid against a weak base. The
reason is the OH–
ions produced by the weak base at the end point is too low to cause the ionisation of
phenolphthalein. Hence, the pink colour does not appear exactly at the equivalence point. The pink
colour appears only after a sufficient excess of the weak base is added.
9. Methyl orange is not a suitable indicator in the titration of a strong base against a Weak acid.
Give reason
Methyl orange is not a suitable indicator in the titration of a strong base against a weak acid. The reason
is the H+
ions produced by the weak acid at the end point is too low to cause the ionisation of methyl
orange. Hence, the colour change does not appear exactly at the equivalence point. A sufficient excess
of the weak acid has to be added to get the colour change
5 MARK
1. What are semiconductors? Explain intrinsic and extrinsic semiconductors.
Semi conductors
Certain type of solids like pure silicon and germanium which are poor conductors of electricity at
normal temperature become good conductors either at high temperatures or in the presence of impurities
like Arsenic or Boron. Such substances are called semiconductors.
There are two types of semi conductors known as intrinsic and extrinsic semiconductors.
Intrinsic semi conductors
In the intrinsic type, these solids have very low conductivity at room temperature but at high temperatures
one of the interatomic covalent bonds between Silicon (or) Germanium atoms are broken heterolytically such
that free electrons and corresponding positive holes are created. When electrical field is applied these
electrons migrate along the direction of the applied electric field causing electrical conductivity in them. The
positive holes move in opposite direction to that of the movement of electrons.
Extrinsic semi conductor
In the extrinsic type of semi conductors addition of impurities like Arsenic or Boron causes appreciable
increase in the electrical conductivity. This effect can be obtained as follows:
N-type semi conductor
In silicon and germanium crystals, each atom is covalently bonded to four neighbours so that all its four
valence electrons are tied down. Thus in the pure state these elements are poor conductors. Suppose an
atom of arsenic is introduced in place of silicon or germanium in the crystal lattice. Arsenic has five
valence electrons, four of which will be utilised in the formation of covalent bonds and the remaining
electron is free to move through the lattice. This leads to enhanced conductivity.
p-type semi conductor
Now let a Boron atom be introduced in place of Silicon atom in the crystal lattice. A Boron atom has
only three valence electrons. It can form only three of the four bonds required for a perfect lattice. Thus
it is surrounded by seven electrons (one of Si) rather than eight. Thus electron vacancy or a ‘positive
hole’ in the lattice is produced. Another electron from the bond of the adjacent Si atom moves into this
hole, completing the four bonds on the B atom. This electron also leaves a hole at its original site. In this
way electrons move from atom to atom through the crystal structure and the holes move in the opposite
direction. Therefore the conductivity of the material improves.
Semi conductors which exhibit conductivity due to the flow of excess negative electrons, are called n-
type semiconductors (n for negative). Semiconductors which exhibit conductivity due to the positive
holes, are called p-type semiconductors (p for positive).
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2. Write a note on Selection of pH indicators
Every pH indicator changes its colour specifically in a ranging pH which is called as indicator range or
pH range.
pH range of Methyl orange = 3.1 – 4.4 pH range of Phenolphthalein = 8.3 – 10
When a base is added to a solution of an acid, the H+ ions will be slowly neutalised by the OH
– ions of
the base. Hence, there is a steady decrease in the H+ ion concentration and pH value increases
uniformly. At the end point there is a steep rise in the pH value. The pH values can be plotted against
the volume of the base added and the curve so obtained is called titration curve. The titration curves
are useful in the choice of a suitable indicator in an acid- base titration.
1. Titration of a strong acid against a strong base:
(Example, HCl vs NaOH)
In this type of titration, the change in the pH value at the end point is roughly from 4 to 10. Therefore
any indicator which changes its colour within this range may be used as a suitable indicator in the
titration of strong acid against strong base and phenolphthalein can be used as indicators for this type of
titrations.
2. Titration of a weak acid against a strong base :
(Example, Oxalic acid vs NaOH)
There is a little change in the pH value at the end point in this type of titration. The pH value changes
from 6.5 to 10. Thus phenolphthalein is the suitable indicator for this titration as its working range is
8.3 – 10. Methyl orange is not a suitable indicator since it has a working range below pH 5.
3. Titration of strong acid against weak base :
(Example, HCl vs Na2CO3]
When a strong acid like HCl is titrated against a weak base like Na2CO3, the pH changes from 3.5 to 7.5
at the end point. The best indicator for this type of titration is methyl orange which changes its colour
within this pH range.
4. Titration of weak acid against weak base :
(Example, CH3COOH vs NH4OH)
In this sypte of titration there is no sharp change in the pH value at the end point. Therefore, in the
titration of a weak acid against a weak base none of the indicators shown in the table are quite
satisfactory.
Volume of base
E
pH
3
10
4
10
Strong acid Vs Strong base
Volume of base
E
pH
10 10
Weak acid Vs Strong base
Volume of base
E
pH
3
8
Strong acid Vs Weak base
66.5
3.5
7.5
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14. ELECTRO CHEMISTRY – II
Blue print 1 Mark = 0 3 Mark = 0 5 Mark = 2 Total marks = 10
Govt. Examination questions 5 MARK
1. Write a note on Daniel cell or voltaic cell [June-2007, Sep-2009]
1. Daniel cell or a galvanic cell is an example of electrochemical cell. The overall reaction taking place
in the cell is the redox reaction given as
Zn (s) + Cu2+
(aq) Zn2+
(aq) + Cu (s)
2. This overall reaction is made of the summation of two half reactions such as oxidation half reaction
and reduction half reaction.
3. The oxidation half reaction occurring at the zinc electrode in contact with the aqueous electrolyte
containing Zn2+
accumulates the electrons at the zinc rod.
Zn (s) Zn2+
(aq) + 2e–
4. The reduction half reaction occurring at the copper electrode in contact with the aqueous electrolyte
containing Cu2+
ions receives the electrons from the zinc electrode when connected externally, to
produce metallic copper according to the reaction as,
Cu2+
(aq) + 2e– Cu (s)
5. The two half cells each comprising the metal electrode and its aqueous electrolyte kept in separate
containers and can be connected externally as below:
Zn
+
SO4
-
Cu
Anode Cathode
Zn2+ SO4
2-
Cu2+
e e- -
SO4
solution solution
Salt Bridge
CuZn
( )( )
6. When the cell is set up, electrons flow from zinc electrode through the wire to the copper cathode.
As a result, zinc dissolves in the anode solution to form Zn2+
ions. The Cu2+
ions in the cathode half
cell pick up electrons and are converted to Cu atoms on the cathode.
2. Explain IUPAC conventions for writing cell diagrams or IUPAC conventions of representation of
a cell. [Mar-06, Mar-07, Mar-09, Mar-10, Mar-11, Sep-11, Mar-13, Jun-14, Jun-15, Mar-17]
IUPAC recommended the following conventions for writing cell diagrams:
1. A single vertical line represents a phase boundary between metal electrode and electrolytic solution.
Thus the two half-cells in a voltaic cell are indicated as
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Zn Zn2+Cu2+
Cu
Anode Half-Cell Cathode Half-Cell
Phase Boundary
It may be noted that the metal electrode in anode half-cell is on the left, while in cathode
half-cell it is on the right of the metal ion.
2. A double vertical line represents the salt bridge that permits ion flow while preventing the
electrolyte from mixing.
3. Anode half-cell is written on the left and cathode half-cell on the right.
4. In the complete cell diagram, the two half-cells are separated by a double vertical line (salt bridge)
in between.
The zinc-copper cell can be written as
Zn Zn2+Cu2+
Cu
Anode Half-Cell Cathode Half-Cell
Salt Bridge
5. The symbol for an inert electrode, like the platinum electrode is often enclosed in a bracket. For
example,
Mg Mg2+H+ H2 (Pt)
Anode Half-Cell Cathode Half-Cell
Inert Electrode
6. The value of emf of a cell is written on the right of the cell diagram. Thus a zinc-copper cell has emf
1.1 V and is represented as
Zn ZnSO4 CuSO4Cu E = + 1.1 V
Direction of electron flow
If the emf acts in the opposite direction through the cell circuit it is denoted as a negative
value.
Zn ZnSO4CuSO4Cu
Direction of electron flow
E = 1.1 V
3. How will you determine single electrode potential? Or emf of a half-cell Or Explain Standard
Hydrogen Electrode. Write its use [Mar-2008, Sep-2009, June-2013, Sep-2014, Sep-2015]
The potential of a single electrode in a half-cell is called the Single electrode potential.
Determination of emf of a half-cell or single electrode potential:
The Standard Hydrogen Electrode (SHE) is selected for coupling with the unknown half-cell. It consists
of a platinum electrode immersed in a 1 M solution of H+ ions maintained at 25
oC. Hydrogen gas at one
atmosphere enters the glass hood and bubbles over the platinum electrode. The hydrogen gas at the
platinum electrode passes into solution, forming H+ ions and electrons.
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H2 gas at 1 atm pressure
Glass Hood
Pt electrode covered with H2
1 M solution of H+ ions at 250C
+
The emf of the standard hydrogen electrode is arbitrarily assigned the value of zero volts. So, SHE
can be used as a standard for other electrodes. The half-cell whose potential is desired, is combined with
the hydrogen electrode and the emf of the complete cell determined with a voltmeter. The emf of the
cell is the emf of the half-cell.
For example, it is desired to determine the emf of the zinc electrode, Zn | Zn2+
. It is connected with the
SHE. The complete electrochemical cell may be represented as :
Zn | Zn2+
|| H+ | H2 (1 atm), Pt
E0 cell = E
0 right – E
0 left
E0 cell = E
0 SHE – E
0 left
0.76 V = 0 – E left
Therefore the standard reduction potential of zinc electrode = – 0.76 V
SHE is placed on the right-hand side of the zinc electrode. So, the hydrogen electron reaction is:
2H+ + 2e
– H2
4. With the help of electrochemical series, how will you predict whether a metal will displace
another metal from its salt solution or not? Give examples [June-2012]
1. Metals near the bottom of the electrochemical series are strong reducing agents and are themselves
oxidised to metal ions.
2. Metals lying higher up in the series are strong oxidizing agents and their ions are readily reduced to
the metal itself.
3. For example, zinc lying down below the series is oxidised to Zn2+
ion, while copper which is higher
up in the series is produced by reduction of Cu2+
ion.
Zn Zn2+
+ 2 e–
Cu2+
+ 2 e– Cu
4. Thus when zinc is placed in CuSO4 solution, Cu metal gets precipitated. In general we can say that a
metal lower down the electrochemical series can precipitate the one higher up in the series.
Zn + Cu SO4 ZnSO4 + Cu
5. Silver cannot precipitate Cu from CuSO4 solution, since silver metal has position higher up in the
series and is strong oxidising agent.
Ag + CuSO4 (reaction is not feasible)
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5. Write an account on Cell terminology [June-2009, Mar-2012, Mar-2014, June-2014, June-2015, Mar-2016]
1. Current is the flow of electrons through a wire or any conductor.
2. Electrode is the metallic rod which conducts electrons into and out of a solution.
3. Anode is the electrode at which oxidation occurs. It sends electrons into the outer circuit. It has
negative charge and is shown as (–) in cell diagrams.
4. Cathode is the electrode at which electrons are received from the outer circuit. It has a positive
charge and is shown as (+) in the cell diagrams.
5. Electrolyte is the salt solution in a cell.
6. Anode compartment is the compartment of the cell in which oxidation half-reaction occurs. It
contains the anode.
7. Cathode compartment is the compartment of the cell in which reduction half-reaction occurs. It
contains the cathode.
8. Half-cell. Each half of an electrochemical cell, where oxidation occurs and the half where reduction
occurs, is called the half cell.
6. Derive the relation between EMF and free energy [Sep-2006, Sep-07, Jun-09, Mar-15, Jun-16]
The maximum amount of work obtainable from the cell is the product of charge flowing per mole and
maximum potential difference, E, through which the charge is transferred.
Wmax = – n FE ... (1)
where
n = number of moles of electrons transferred and is equal to the valency of the ion
participating in the cell reaction.
F = Faraday and is equal to 96,495 coulombs
E = emf of the cell.
According to thermodynamics, the maximum work that can be derived from a chemical reaction is equal
to the free energy (G) for the reaction,
Wmax = G ... (2)
Therefore, from (1) and (2), we can write
G = – n FE ... (3)
Thus only when E has a positive value, G value will be negative and the cell reaction will be
spontaneous and the e.m.f. of the cell can be measured.
Here E refers to the Ecell. Thus, the electrical energy supplied by the cell is (nFE) equal to the free
energy decrease (–G) of the cell reaction occurring in the cell.
7. Derive Nernst equation
[Jun-06, Sep-06, Jun-08, Sep-08, Mar-09, Jun-10, Sep-10, Jun-11, Mar-12, Sep-12, Sep-13, Sep-16]
Suppose the reaction occurring in a reversible cell is represented by the equation
A + B ⇌ C + D
The decrease in free energy, –G, accompanying the process is given by the well known
thermodynamic equation
– G = – Go – RT ln J
where –Go is the decrease in free energy accompanying the same process when all the reactants and
products are in their standard states of unit activity and J stands for the reaction quotient of the activities
of the products and reactants at any given stage of the reaction.
Substituting the value of J, we have
– G = – Go – RT ln
BA
DC
aa
aa
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If E is the E.M.F. of the cell in volts and the cell reaction involves the passage of ‘n’ faradays (i.e.,) nF
coulombs, the electrical work done by the cell is in nFE volt-coulombs or Joules. Hence free energy
decrease of the system, –G, is given by the expression
–G = n FE
n FE = – Go – RT ln
BA
DC
aa
aa
n FE = n FE0 – RT ln
BA
DC
aa
aa
Divide both sides by n F
E = E0 –
Fn
RT ln
BA
DC
aa
aa
where Eo is the E.M.F. of the cell in which the activity, or as an approximation, the concentration of
each reactant and each product of the cell reaction is equal to unity. Eo is known as the standard E.M.F.
of the cell.
Replacing activities by concentrations, as an approximation, the above equation may be written as
E = E0 –
Fn
RT ln
]B[]A[
]D[]C[
where the quantities in parantheses represent the concentration of the species involved. Replacing
]B[]A[
]D[]C[ as equal to K, the equilibrium constant in the molar concentration units,
E = E0 –
Fn
RT ln K
E = E0 –
Fn
RT303.2 log K
This equation is known as Nernst equation.
where Eo = standard electrode potential
R = gas constant
T = Kelvin temperature
n = number of electrons transferred in the half-reaction
F = Faraday of electricity
K = equilibrium constant for the half-cell reaction as in equilibrium law.
8. Calculate the E.M.F. of the zinc-silver cell at 25oC when [Zn
2+] = 0.10 M and [Ag
+]=10 M. (E
o cell
at 25oC = 1.56 volt] [Mar-2006, June-2006, June-2007, June-2008, Sep-2008, Mar-2010, Sep-2010]
The cell reaction in the zinc - silver cell would be
2Ag+ + Zn ⇌ 2Ag + Zn
2+
According to Nernst equation, Ecell = Eocell
–
n
0591.0 log K
Ecell = Eocell
–
n
0591.0 log
[Zn]][Ag
][Zn[Ag]
2
22
Ecell = 1.56 – n
0591.0 log
[Zn]][Ag
][Zn[Ag]
2
22
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Since concentrations of solids are taken as unity,
Ecell = 1.56 –
n
0591.0 log
2
2
][Ag
][Zn
= 1.56 – 2
0591.0 log
2(10)
0.10
= 1.56 – 2
0591.0 log (10)
–3
= 1.56 + 2
)0591.0(3
Ecell = 1.648 V
9. Calculate the E.M.F. of the zinc-silver cell at 25oC when [Zn
2+] = 0.001 M and [Ag
+]=0.1 M.
(Eo
cell at 25oC = 1.56 volt] [Mar-2015]
The cell reaction in the zinc - silver cell would be
2Ag+ + Zn ⇌ 2Ag + Zn
2+
According to Nernst equation, Ecell = Eocell
–
n
0591.0 log K
Ecell = Eocell
–
n
0591.0 log
[Zn]][Ag
][Zn[Ag]
2
22
Ecell = 1.56 – n
0591.0 log
[Zn]][Ag
][Zn[Ag]
2
22
Since concentrations of solids are taken as unity,
Ecell = 1.56 –
n
0591.0 log
2
2
][Ag
][Zn
= 1.56 – 2
0591.0 log
2(0.1)
0.001
= 1.56 – 2
0591.0 log (10)
–1
= 1.56 + 2
0591.0
Ecell = 1.58955 V V
10. A zinc rod is placed in 0.095 M zinc chloride solution at 25oC. emf of this half cell is – 0.79V.
Calculate Eo
Zn2
/ Zn [Sep-2014]
Half cell reaction is Zn2+
+ 2e– Zn
E = Eo
– n
0591.0 log
][Zn
[Zn]2
Since concentrations of solids are taken as unity,
E = Eo
– n
0591.0 log
][Zn
12
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– 0.79 = Eo
– 2
0591.0 log
095.0
1
Eo = – 0.76 V
11. The emf of the half cell Cu2+
(aq)/Cu(s) containing 0.01 M Cu2+
solution is + 0.301 V. Calculate
the standard emf of the half cell. [Mar-2007, June-2010, Mar-17]
Half cell reaction , Cu2+
+ 2e– ⇌ Cu
E = Eo –
n
0591.0 log K
0.301 = Eo –
2
0591.0 log
][Cu
]Cu[2
Since concentrations of solids are taken as unity,
0.301 = Eo –
2
0591.0 log
][Cu
12
0.301 = Eo –
2
0591.0 log
01.0
1
0.301 = Eo –
2
0591.0 log (10)
2
0.301 = Eo– log
2
)0591.0(210
0.301 = Eo – 0.0591
Eo = 0.301 + 0.0591
Eo = 0.3601 V
12. Determine the standard e.m.f of the cell and standard free energy changes of the cell reaction,
Zn/Zn2+
// Ni2+
/Ni. The standard reduction potentials of Zn2+
/ Zn and Ni2+
/ Ni half cells are –0.76
V and –0.25 V respectively. [Sep-2007, Mar-2008, Mar-2013, Sep-2013]
Zn /Zn2+
// Ni2+
/ Ni
The standard e.m.f of the cell, Eo cell = E
o right – Eo left
= -0.25 – (– 0.76)
= 0.76 – 0.25
= 0.51 V
Standard free energy change Go = – n FE
o
= – 2 × 96495 × 0.51
= –98425 J
= –98.425 kJ
13. Calculate the equilibrium constant for the reaction,
2Ag+ + Zn ⇌ Zn
2+ + 2Ag V0.76EandV0.80E Zn/Zn
0 2Ag/Ag0 [Sep-2011]
Zn | Zn2+
|| Ag + | Ag
Eocell = E
oRight – E
oLeft
= 0.8 – (– 0.76)
= 0.8 + 0.76
= 1.56 V
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According to Nernst equation, E = Eo –
n
0591.0 log K
At equilibrium, E = 0
0 = Eo –
n
0591.0 log K
Eo
= n
0591.0 log K
1.56 = 2
0591.0 log K
log K = 52.79
∴ K = Anti log of (52.79)
= 6.19 × 1052
14. What is the potential of a half-cell consisting of zinc electrode in 0.01 M ZnSO4 solution at 25oC?
Eo = 0.763 V. [Jun-12, Sep-2012, Mar-2014, Sep-16]
The half-cell reaction is Zn Zn2+
+ 2e–
The Nernst equation for the oxidation half-cell reaction is
E = Eo
– n
0591.0 log
[Zn]
][Zn2
Since concentrations of solids are taken as unity,
E = Eo
– n
0591.0 log [Zn
2+]
= 0.763 –
2
0591.0 log 0.01
= 0.763 + 0.0591
= 0.8221 V
15. The standard electrode potentials of the half cells Ag+ / Ag and Fe
3+/ Fe
2+ (Pt) are 0.7991 V and
0.771 V respectively. Calculate the equilibrium constant of the reaction [June-2013]
Ag(s) + Fe3+
⇌ Ag+ + Fe
2+
Ag+ / Ag and Fe
3+, Fe
2+ / Pt
Eocell = E
oRight – E
oLeft
= 0.771 – 07991
= – 0.0281 V
According to Nernst equation, E = Eo –
n
0591.0 log K
At equilibrium, E = 0
0 = Eo
– n
0591.0 log K
Eo
= n
0591.0 log K
– 0.0281 = 1
0591.0 log K
log K = – 0.4751
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∴ K = Anti log of (– 0.4751)
= 0.335
16. Calculate the standard emf and standard free energy change of the following cell:
Zn | Zn2+
|| Cu2+
| Cu
V0.337EandV0.762E Cu2Cu/Zn2Zn /00 [Jun-2011]
Eocell = E
oRight – E
oLeft
= 0.337 – (– 0.762)
= 0.337 + 0.762
= 1.099 V
Go = – n FE
o
= – 2 × 96495 × 1.099
= – 212096 J = – 212.096 kJ
17. Calculate the emf of the cell.
Zn | Zn2+
(0.001 M) || Ag+ (0.1 M) | Ag
The standard reduction potential of Ag+/Ag half-cell is + 0.80 V and Zn
2+/Zn is – 0.76 V.
[Mar-2011]
Zn | Zn2+
(0.001 M) || Ag+ (0.1 M) | Ag
Eocell = E
oRight – E
oLeft
= 0.8 – (– 0.76)
= 0.8 + 0.76
= 1.56 V
Anode Hale-cell reaction: Zn Zn2+
+ 2e–
Cathode Hale-cell reaction: 2Ag+ + 2e
– 2Ag
Cell reaction: Zn + 2Ag+ ⇌ Zn
2+ + 2Ag
According to Nernst equation,
E = Eo –
n
0591.0 log K
= 1.56 – 2
0591.0 log
2
2
][Ag
][Zn
= 1.56 – 2
0591.0 log
2)1.0(
001.0
= 1.56 – 2
0591.0 log
2
3
10
10
= 1.56 – 2
0591.0 log 10
– 1
= 1.56 + 2
0591.0 log 10
= 1.58955 V
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18. Calculate the equilibrium constant for the reaction,
2Ag+ + Zn ⇌ Zn
2+ + 2Ag V56.1cell
0E [Sep-2015]
According to Nernst equation, E = Eo –
n
0591.0 log K
At equilibrium, E = 0
0 = Eo –
n
0591.0 log K
Eo
= n
0591.0 log K
1.56 = 2
0591.0 log K
log K = 52.79
∴ K = Anti log of (52.79)
= 6.19 × 1052
19. Calculate the potential of the following cell at 298K
Zn / Zn2+
(a = 0.1) // Cu2+
(a = 0.01) / Cu
V0.337EandV0.762E Cu2Cu/Zn2Zn /00 [Mar-2016, Jun-2016]
Solution:
Eocell = E
oRight – E
oLeft
= 0.337 – (– 0.762)
= 0.337 + 0.762
= 1.099 V
The cell reaction in the zinc - copper cell would be
Zn + Cu2+
⇌ Zn2+
+ Cu
According to Nernst equation, Ecell = Eocell
–
n
0591.0 log K
Ecell = Eocell
–
n
0591.0 log
2
2Zna
CuZn
Cu
aa
a
Ecell = Eocell
–
n
0591.0 log
2
2Zna
Cua
(since activity of a pure metal is unity)
Ecell = 1.099 –
2
0591.0 log
01.0
0.1
Ecell = 1.099 –
2
0591.0 log 10
Ecell = 1.099 – 0.02955
Ecell = 1.06945 V
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Additional questions
5 MARK
1. What are the uses or applications of Standard reduction potentials?
Standard reduction potentials are used,
i) for predicting cell EMF
E0 cell = E
0 right – E
0 left
Where,
E0 right = Standard reduction potential of right-hand electrode
E0 left = Standard reduction potential of right-hand electrode
ii) for predicting feasibility of redox reactions
If Eo cell = + ve, the reaction is feasible
Eo cell = – ve, the reaction is not feasible.
Zn + Cu2+
Zn2+
+ Cu Eo = + 1.1 V (reaction is feasible)
Zn2+
+ Cu Zn + Cu2+
Eo = – 1.1 V (reaction is not feasible)
iii) for predicting whether a metal will displace another metal from its salt solution or not
Refer – Q – 4 (Govt. Examinatin questions)
iv) for predicting whether a metal will displace hydrogen from a dilute solution or not
Any metal above hydrogen in the electrochemical series is a weaker reducing agent than hydrogen and
will not displace hydrogen. This explains why Zn (strong reducing agent) lying below hydrogen reacts
with dil. H2SO4 to liberate H2, while Cu (weak reducing agent) lying above hydrogen does not react.
Zn + H2SO
4 ZnSO
4 + H
2
dil
Cu + H2SO
4
dil
2. The standard reduction potential for the reaction Sn4+
+ 2e- Sn
2+ is + 0.15 V. Calculate the
standard free energy change of the reaction.
Go = – n FE
o
= – 2 × 96495 × 0.15
= –28948.5 J
= –28.948 kJ
3. How will you calculate half-cell potential using Nernst equation?
For an oxidation half-cell reaction when the metal electrode M gives Mn+
ion,
M Mn+
+ ne–
the Nernst equation takes the form
E = E0 –
Fn
RT303.2 log
[M]
][M n
The activity of solid metal [M] is equal to unity. Therefore, the Nernst equation can be written as
E = E0 –
Fn
RT303.2 log [M
n+]
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Substituting the values of R, F and T at 25oC, the quantity
F
RT303.2comes to be 0.0591. Thus the
Nernst equation can be written in its simplified form as
E = E0 –
n
0591.0 log [M
n+]
This is the equation for a half-cell in which oxidation occurs. In case it is a reduction, the sign of E will
have to be reversed.
4. A zinc rod is placed in 0.095 M zinc chloride solution at 25
oC. emf of this half cell is –
0.79V. Calculate Eo
Zn2
/ Zn
Half cell reaction is Zn2+
+ 2e– Zn
E = Eo
– n
0591.0 log
][Zn
[Zn]2
Since concentrations of solids are taken as unity,
E = Eo
– n
0591.0 log
][Zn
12
– 0.79 = Eo
– 2
0591.0 log
095.0
1
Eo = –0.76 V
5. The standard free energy change of the reaction M+ (aq) + e M(s) is –23.125 kJ. Calculate the
standard emf of the half cell.
Go
= –23.125 kJ = –23125 J
Go = – n FE
o
–23125 = – 1 × 96495 × Eo
Eo = 0.2396 V
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15. ISOMERIM IN ORGANIC CHEMISTRY
Blue print 1 Mark = 0 3 Mark = 1 5 Mark = 1 Total marks = 8
Govt. Examination questions
3 MARK
1. Identify Cis-Trans isomer from the following [Sep-2008]
C
C
Br
Br
a) b)
CH3
CH3
C
C
Br
BrCH3
CH3 Ans:
C
C
Br
Br
a) b)
CH3
CH3
C
C
Br
BrCH3
CH3
Cis Trans
2. Label the following as E, Z isomers [June-2007]
C
C
a) b)
CH3
C
C
Br
CH2OH H
H
Cl
FHOH2C
Ans:
1
1 1
12
2 2
2
ZE
C
C
a) b)
CH3
C
C
Br
CH2OH H
H
Cl
FHOH2C
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3. What are the conditions (or criteria) for a compound to be optically active?
Or Write the conditions for optical activity [Sep-2007, Mar-2009]
An organic compound will be optically active if it satisfies the following conditions:
1. It should contain atleast one chiral carbon atom ( or asymmetric carbon atom )
[ carbon atom attached to 4 different groups is called chiral carbon atom ]
C**
H OH
CH3
COOH
C = Chiral carbon
2. Its configuration is non superimposable on its mirror image.
3. “Chirality” is the essential and the sufficient condition for a molecule to be optically active.
4. What are enantiomers? [June-2008]
Enantiomers are stereo isomers which have object-mirror image relationship and they are
non-superimposable. Enantiomers possess identical properties and same magnitude of optical rotation
but they differ only in the sign of (or direction of) optical rotation.
Eg:
CH3
H OH
COOH
HO
CH3
COOH
H
d-Lactic acid l-Lactic acid
5. What is a racemic mixture? Explain with suitable example
[March-2008, June-2009, Sep-2009, Mar-2014]
When equal amounts of d-isomer and l-isomer are mixed, we get a racemic mixture and this process is
called racemisation.
d-tartaric acid (50%) + l-tartaric acid (50% ) onRacemisati
dl-tartaric acid
(racemic mixture)
Racemic mixture is optically inactive due to external compensation.
6. Trans isomer is more stable than cis isomer. Why? [Jun-2011, Sep-2016]
In cis isomer, similar groups lie on the same side and hence it experiences more Vanderwaals
repulsion and steric hindrance. So, cis isomer is less stable.
In trans isomer, similar groups lie on the opposite sides and hence it does not experience steric
interaction. So, trans isomer is more stable than cis isomer.
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7. Give the structure of Cis-Trans isomers of 2-pentene [Sep-2011]
Cis Trans 2-pentene
H
H H
H CH2CH3CH2CH3
CH3 H3C
C
C
C
C
8. Mesotartaric acid is an optically inactive compound with chiral carbon atoms. Justify
[March-2006, Sep-2006, March-2010, Mar-2013, Sep-2014, Sep-2015]
1. Meso tartaric acid contains two chiral carbon atoms but it is optically inactive. One chiral atom is
dextro rotatory and the other chiral atom is laevo rotatory. Both chiral carbons rotate the plane
polarized light to the same extent in opposite directions. The net result is that the meso tartaric acid
is optically inactive.
COOH
COOH
OH
OHH
H
symmetric plane
2. Meso tartaric acid has symmetric plane, which divides the molecule into two equal halves. One half
of the molecule looks like the mirror image of the other half. The rotation of plane polarized light
towards clockwise direction by one half of the molecule is exactly cancelled by equal rotation
towards anticlockwise direction by the other half. So, meso tartaric acid is optically inactive due to
internal compensation.
3. Meso tartaric acid has configuration which is super-imposable on its mirror image and hence the
molecule becomes “achiral” (optically inactive)
9. Give the structure of Z and E forms of cinnamic acid [Mar-2015]
CC
COOH
H H
C6H
5
CC
1 1 1
12 2 2
2
EZ
H
H
COOH
C6H
5
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10. Distinguish racemic form from Mesoform
[Mar-2007, June-2010, Sep-2010, Mar-2011, Sep-2012, June-2013, June-2014, June-2015]
No. Racemic form Meso form
1 It is a mixture that can be separated into
two optically active forms
It is a single compound and hence
cannot be separated
2 Optically inactive due to external
compensation
Optically inactive due to internal
compensation
3 The individual molecules of racemic
mixture are chiral
Molecule is achiral
Eg d-tartaric acid + l-tartaric acid Meso tartaric acid
11. Distinguish enantiomers and diasteromers [June-2006, Mar-2012, June-2012, Sep-2013, Mar-17]
No. Enantiomers Diastereomers
1 Optical isomers having the same
magnitude but different sign of rotation
Differ in the magnitude of optical
rotation
2 Enantiomers have object-mirror image
relationship.
They are non-superimposable
Diastereomers are not mirror images of
each other
3 Enantiomers are identical in all
properties except the sign of optical
rotation
Diastereomers differ in all physical
properties
4 Separation of enantiomers is a tedious
process
Separation of diastereomers is easy
Eg d-tartaric acid & l-tartaric acid d–tartaric acid & meso tartaric acid
12. Draw the structure of S-cis and S-trans form of 1,3-butadiene. [Mar-2016]
CH2= CH
CH2= CH
CH2= CH
CH = CH2
S-cis S-trans
S-cis S-trans
13. Write the formula (configuration) for d, l and meso tartaric acid [Jun-2016]
COOH
COOH
H
H
COOH
COOH
OHH HO
HO H OH
COOH
COOH
OH
H
H
OH
meso tartaric acidd-tartaric acid l-tartaric acid
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5 MARK
1. Distinguish racemic form from Mesoform [Sep-2006]
Refer –3 mark – Q - 10
2. Distinguish enantiomers and diasteromers [Mar-2006, Mar-2008, Mar-2014, Mar-2015]
Refer –3 mark – Q - 11
3. Explain geometrical isomerism or cis-trans isomerism in organic compounds with examples
[Sep-2007, June-2009, Sep-2009, Mar-2011, Sep-2012, Mar-2013, June-2013, Mar-2016]
1. Isomerism that arises due to the difference in the spatial arrangement of atoms or groups about
the doubly bonded carbon atoms is called geometrical isomerism or cis-trans isomerism.
2. This isomerism arises due to the restricted rotation about C = C bond.
3. Eg: 2 - butene
Cis form Trans form
C C
CC
H
H H
H CH3CH3
CH3 CH3
4. Cis isomer
Similar groups lie on the same side
Since similar groups are nearer to each other, cis isomer experiences more Vanderwaals
repulsion and steric hindrance. So, cis isomer is less stable.
5. Trans isomer
Similar groups lie on the opposite sides
Since similar groups are diagonally opposite to each other, trans isomer does not
experience steric interaction. So, trans isomer is more stable.
6. Cis and trans isomers do not differ much in chemical properties. They differ in physical
properties like boiling point, melting point, crystal structure and solubility.
7. At high temperature, trans isomer can be converted into cis isomer and vice-versa. Breaking of
carbon-carbon π - bond and its reformation is responsible for this interconversion.
Trans isomer Cis isomer
4. Explain optical isomerism (or optical activity) in tartaric acid.
[Jun-06, Jun-08, Sep-08, Mar-09, Jun-10, Jun-11, Mar-12, Jun-12, Jun-13, Sep-13, June-14, Sep-16]
1. Tartaric acid contains two chiral carbon atoms or asymmetric carbon atoms.
COOH
COOH
OHH
HO H
C
C
*
**
C chiral carbon
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2. d-tartaric acid and l-tartaric acid are related as object and mirror image and they are non
superimposable. They are enantiomers.
COOH
COOH
H
H
COOH
COOH
OHH HO
HO H OH
d -tartaric acid l-tartaric acid
3. In d-tartaric acid, each of the two asymmetric carbon atoms rotate the plane of the polarized light
towards right (clocwise) leading to overall dextro rotation. So, d-tartaric acid is optically active.
4. In l-tartaric acid, each of the two asymmetric carbon atoms rotate the plane of the polarized light
towards left (anticlockwise) leading to overall laevo rotation. So, l-tartaric acid is optically active.
5. d and l tartaric acids have the same magnitude of optical rotation but they differ only in the sign of
optical rotation.
6. Meso tartaric acid contains two chiral carbon atoms but it is optically inactive. One chiral atom is
dextro rotatory and the other chiral atom is laevo rotatory. Both chiral carbons rotate the plane
polarized light to the same extent in opposite directions. The net result is that the meso tartaric acid is
optically inactive.
COOH
COOH
OH
OHH
H
symmetric plane
Meso tartaric acid has symmetric plane, which divides the molecule into two equal halves. One half
of the molecule looks like the mirror image of the other half. The rotation of plane polarized light
towards clockwise direction by one half of the molecule is exactly cancelled by equal rotation
towards anticlockwise direction by the other half. So, meso tartaric acid is optically inactive due to
internal compensation.
Meso tartaric acid has configuration which is superimposable on its mirror image and hence the
molecule becomed “achiral”
7. When equal amounts of d-tartaric acid and l-tartaric acid are mixed one gets a „racemic mixture‟‟ and
this process is called racemisation. A racemic mixture becomes optically inactive. Because, in this
mixture rotation towards clockwise direction by the dextro isomers is compensated by the rotation
towards the anticlockwise direction by the laevo isomers. The optical inactivity of a racemic mixture
is said to be due to ‘external compensation’
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5. Explain internal and external compensation with suitable examples [June-2007]
Refer-Q-4 ( points 6 & 7 )
6. Describe the conformations of cyclo hexanol. Comment on their stability
[Mar-2007, Mar-2010, Sep-2010, Sep-2011, Sep-2014, June-2015, Mar-17]
Cyclo hexanol exists in two chair forms. These two forms are interconvertible and exist in equilibrium
.
Axial cyclo hexanol ( 10% ) Equatorial cyclo hexanol ( 90% )
OH - group : axial OH - group : equatorial
experience 1:3 diaxial interaction no 1:3 diaxial interaction
more energy
more stable
less energy
less stable
OH
OH
In axial cyclo hexanol, the axial –OH group experiences steric interaction with the axial H-atoms
present at the third carbon atoms. This increases the energy and decreases the stability of axial cyclo
hexanol. This is called 1:3-diaxial interaction.
OHH
H
1:3-Diaxial interaction
1
2
3
3
2
This interaction is absent in equatorial cyclo hexanol. So, equatorial cyclo hexanol is more stable than
axial cyclo hexanol.
In the equilibrium mixture the more stable conformer, equatorial cyclo hexanol is present to an extent of
90% and the less stable conformer, axial cyclo hexanol is present only to 10%.
Energy level diagram
progress of reaction
E
0.7
K cal
11
K cal
equatorial
axial cyclo hexanol
cyclo hexanol
boat form
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7. Write a note different types of hydrogen in cyclohexane [Sep-2015]
Cyclo hexane can exist in two non-planar forms.
a) Chair form
H
H H
H
H
H
H
H
H
H
H
H
axial
equatorial
Axial Equatorial
(six C-H bonds parallel to the
axis of the molecule)
(six C-H bonds subtending an angle of
about 70o and 110
o with the axis)
-axial -axial -equatorial -equatorial
(three axial bonds
facing upward)(three axial bonds facing downward)
(three equatorial bondsfacing upward) facing down ward)
(three equatorial bonds
Types of C-H bonds
b) Boat form
H
HH
H
HH
H (f)
H (S)
H (x)
(s)
(x)
( f )(s)
( f )
(x)
: flag pole hydrogens
: bowspirit hydrogens
: eclipsing hydrogens
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8. Write the position isomers of dichloro benzene and their dipolemoment order [Jun-2016]
Cl
Cl
Cl
Cl
Cl
Cl
ortho dichlorobenzene meta dichlorobenzene para dichlorobenzene
These three compounds have the same molecular formula but differ in the position of chlorine atom. So,
they are called position isomers.
DPM is the best method of distinguishing the three isomers of disubstituted benzenes.
ortho meta para
angle θ =60o θ =120
o θ = 180
o
Their DPM is in the order, ortho isomer > meta isomer > para isomer
For para dichlorobenzene DPM = 0
Additional questions
3 MARK 1. Boat form of cyclo hexane is less stable than chair form. Why?
Boat form of cyclo hexane has two pairs of carbon atoms with eclipsing bonds. This eclipsing
interaction destabilizes boat form. So, boat form is less stable than chair form.
2. What is ring flipping? Chair form of cyclo hexane is sufficiently flexible to turn upside down called ring flipping so that all
bonds which are axial originally become equatorial and viceversa.
H
R
H
R
H
R
Chair form Boat form Chair form 3. What is meant by resolution of racemic mixture?
The process of separation of optically inactive racemic mixture into two optically active forms is called
resolution of racemic mixture.
Racemic mixture solutionRe
d-isomer & l-isomer
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5 MARK
1. Explain geometrical isomerism in 1,3-butadiene. CH2=CH-CH=CH2 1,3-butadiene
This molecule can exist in two forms.
CH2= CH
CH2= CH
CH2= CH
CH = CH2
S-cis S-trans
These two forms do not arise due to the hindrance to rotation about C=C nstead the restricted rotation
about C-C (since bulky groups are attached to ingle bonded carbon atoms C-C rotation is restricted).
These isomers can also be represented as,
S-cis S-trans
S-trans form is also called „transoid form‟
S-cis form is also called „cisoid form‟.
S = indicates the restricted rotation about C-C single bond.
the energy of S-trans form is 3 K cal less than S-cis form.
S-trans form is more stable than S-cis form
these two forms differ in chemical reactivity.
they give different products in reactions.
These are easily interconvertible and exist in equilibrium.
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16. HYDROXY DERIVATIVES
Blue print 1 Mark = 1 3 Mark = 2 5 Mark = 1* Total marks = 12
Govt. Examination questions
1 MARK
1. Order of reactivity of alcohol towards sodium metal is [Sep-2010, Sep-2011, Sep-2012]
a) primary < secondary > tertiary b) primary > secondary > tertiary
c) primary < secondary < tertiary d) primary > secondary < tertiary
2. The compound that reacts fastest with Lucas reagent is
[Mar-2007, June-2007, Sep-2007, Mar-2010, June-2015]
a) butan-1-ol b) butan-2-ol c) 2-methyl propan-1-ol d) 2-methyl propan-2-ol
3. When phenol is distilled with Zn dust it gives [Sep-2009, June-2011, Sep-2013, Mar-2015]
a) benzaldehyde b) benzoic acid c) toluene d) benzene
4. A compound that undergoes bromination easily is [Mar-2008, Mar-2012]
a) benzoic acid b) benzene c) phenol d) toluene
5. Ethylene diamine is converted to ethylene glycol using [June-2006, Mar-2013]
a) Na2CO3 solution b) nitrous acid c) NaHCO3 (aqueous) d) Baeyer’s reagent
6. Glycerol is used [June-2012]
a) as a sweetening agent b) in the manufacture of good quality soap
c) in the manufacture of nitro glycerin d) in all the above
7. The number of secondary alcoholic group in glycerol is [Sep-2008, June-2010, June-2013]
a) 1 b) 2 c) 3 d) 0
8. The active component of dynamite is [Mar-2009]
a) Keiselghur b) Nitro glycerine c) Nitro benzene d) Trinitro toluene
9. The reaction of ethylene glycol with PI3 gives [Jun-2008, Sep-2016, Mar17]
a) ICH2CH2I b) CH2 = CH2 c) CH2 = CHI d) ICH = CHI
10. Oxidation of glycerol with bismuth nitrate gives [Mar-2006]
a) meso oxalic acid b) glyceric acid c) tartronic acid d) both b and c
11. The characteristic odour of lower phenol is [Sep-2006]
a) carbolic acid b) fruity c) oil of bitter almonds d) rotten fish
12. Which of the following compounds is oxidised to give ethyl methyl ketone? [June-2009]
a) 2-propanol b) 2-pentanone c) 1-butanol d) 2-butanol
13. Compound which is used as medicine for Asthma and whooping cough is [Mar-2011]
a) Benzyl acetate b) Ethyl acetate c) Benzyl benzoate d)Benzyl formate
14. Ethyl alcohol cannot be used as a solvent for CH3MgI because [Mar-2014]
a) CH3MgI reacts with alcohol giving methane
b) The reaction between them is explosive in nature
c) CH3MgI is converted to C2H5MgI
d) Alcohol is immiscible with CH3MgI
15. A compound that gives a positive iodoform test is [June-2014, Mar-2016]
a) 1- pentanol b) 2 pentanone c) 3 - pentanone d) pentanal
16. Ethylene glycol forms terylene with [Sep-2014, Jun-2016]
a) adipic acid b) Phthalic anhydride c) terephthalic acid d) oxalic acid 17. Isomerism exhibited by ethylene glycol is [Sep-2015]
a) position isomerism b) chain isomerism c) functional isomerism d) both (a) and (c)
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3 MARK
1. Alcohols cannot be used as solvent for Grignard reagents. Why? [Mar-2008, Mar-17]
Grignard reagents are decomposed by alcohol.
R–OH + CH3MgBr RO–Mg–Br + CH4
Hence alcohols cannot be used as a solvent for Grignard’s reagents.
2. Write a note on ‘Breath analysis test’ or How can the consumption of alcohol by a person be
detected? [Mar-2006]
The breath analysis test for the detection of ethanol involves the oxidation of alcohol in the breath of a
person who has consumed alcohol, by acidic potassium dichromate and observing the change in the
colour of the chromium ion (CrVI
) from yellow orange to CrIII
which is blue green colour.
CrVI
alcohol
CrIII
yellow orange blue green
3. How are 1-propanol and 2-propanol distinguished by oxidation method? [June-2012]
(1o alcohol)
K2Cr2O7/H+
propanaldehyde
CCH3 CH3
OH
H(2o alcohol)
CCH3CH3
O
acetone
+ (O)
+ (O)K2Cr2O7/H
+
- H2O
- H2O
(O)
(O)
propanoicacid
acetic acid
OH
1-propanol
2-propanol
CCH3
O
CH3-CH
2-CH
2OH CH
3-CH
2-CHO CH
3-CH
2-COOH
4. Give chemical tests to distinguish between propan-2-ol and 2-methylpropan-2-ol [Mar-2015]
Test Propan-2-ol
(2o alcohol)
2-methylpropan-2-ol
(3o alcohol)
Lucas test
Con.HCl + anhyd. ZnCl2
Turbidity appears after
5-10 minutes at room
temp.
Turbidity appears immediately
at room temp.
Victor Meyer’s test Gives blue colour No colur
Catalytic
dehydrogenation
Cu/573 K
Acetone is formed
It undergoes dehydration to
give isobutylene
Iodoform test I2/KOH Gives iodoform Does not give iodoform
CCH3 CH3
OH
H
CCH3
CH3
CH3
OH
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5. How will you convert 2-methyl-2-propanol into 2-methyl propene? [June-2009, Mar-2011]
Or What happens when the vapour of tertiary butyl alcohol is passed over heated Cu at 573 K ?
[Sep-2014]
CCH3
CH3
CH3
OHdehydration
CCH3
CH3
CH2
Cu/573K
isobutylene
+ H2O
(2-methyl-2-propanol) (2-methylpropene)
tert.butyl alcohol
6. Give the Victor Meyer’s test for secondary alcohol [Sep-2016]
CCH3
CH3
OH
2o alcohol
P / I2CCH3
CH3
IAgNO2
CCH3
CH3
NO2
H H H
HO-N=OCCH3
CH3
NO2
N=OPseudo nitrol (Blue colour)
KOH
No reaction but blue colour remains
- H2O
7. Give the Victor Meyer’s test for tertiary alcohol [Mar-2014]
CCH3
CH3
CH3
OH
3o alcohol
P / I2CCH3
CH3
CH3
IAgNO2
CCH3
CH3
CH3
NO2
HNO2No reaction
Nitro compound (Colourless)
Absence of α-H in tertiary alcohol makes it inactive to nitrous acid.
8. What happens when excess of ethyl alcohol is treated with con.H2SO4 at 410 K? [June-2007]
When excess of ethyl alcohol is treated with con.H2SO4 at 410 K it undergoes intermolecular
dehydration to give diethyl ether.
C2H5 HSO4 + H2O410 K
C2H5OH + H HSO4
C2H5 HSO4 C2H5+ H O C2H5O C2H5
+ H2SO4
Conexcess
diethyl ether
9. Give a chemical test to distinguish between ethanol and methanol [Sep-2008]
Ethyl alcohol undergoes iodoform reaction with iodine and potassium hydroxide.
CH3CH2OH /KOHI
2 CH3CHO 2I
CI3CHO KOH
CHI3 + HCOOK Ethyl alcohol yellow crystals
of iodoform
Methyl alcohol does not undergo iodoform reaction.
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10. Give reason for high viscosity, high boiling point and polymeric structure of glycol [Jun-06, Mar-16]
Because of the presence of two hydroxyl groups the intermolecular hydrogen bonding is made much
stronger. Hydrogen bond can be formed between both OH groups resulting in a polymeric structure.
This leads to high viscosity and high boiling point.
CH2
CH2
O
O
H
H
H O
CH2CH2
O
H
11. How will you convert ethylene into ethylene glycol? [Sep-2009]
Or What happens when ethylene reacts with alkaline KMnO4 solution? [June-2014]
When ethylene is treated with cold dilute alkaline potassium permanganate solution (Baeyer’s reagent)
glycol is formed
CH2=CH2 + H2O + (O) ReagentsBaeyer'
HO-CH2-CH2-OH
Glycol
12. Write the equation for action of heat on ethylene glycol with conc. phosphoric acid [Jun-2016]
When heated with conc. phosphoric acid, ethylene glycol eliminates a molecule of water forming
diethylene glycol.
CH2CH2HO
CH2CH2HO
O H
OH
H3PO4
CH2CH2HO
CH2CH2HO
O
diethylene glycol
13. How will you convert glycol into dioxan? [Mar-2008, Sep-2008, June-2012, Sep-2013]
Concentrated sulphuric acid being more powerful acid and dehydrating agent removes two molecules of
water forming dioxan.
CH2CH2
HO
CH2CH2HO
O H
OH
con.H2SO
4 con.H2SO
4
CH2CH2
CH2CH2
O
H O
HO
CH2CH2
CH2CH2
OO
dioxan
14. How is terylene prepared? Write its use [Mar-2009, June-2010, Mar-2011, June-2013]
Ethylene glycol reacts with terephthalic acid to form condensation polymer ‘Terylene’ (also known as
dacron or terene), which is extensively used as a synthetic fibre.
C-O-]n H
O
HO-[-CH2-CH2-O-C
O
+ (2n-1) H2O
C-O-H
OO
Terylene or Dacron or terene
n HO-CH2-CH2-OH + H-O-Cn
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15. How will you convert propylene into glycerol? [June-2008]
CH2
CH
CH3
Cl2
CH2
CH
CH2Cl
Na2CO
3
CH2
CH
CH2OH
HOCl
CH2
CH
OH
Cl
CH2OH
NaOH
CH2
CH
OH
CH2OH
OH + NaCl
propylene allyl chloride allyl alcohol glycerol
chlorohydringlycerol
16. What is TNG? How is it prepared? Write its use [Sep-2011, June-2015]
With a well cooled mixture of concentrated sulphuric acid and fuming nitric acid, glycerol forms
trinitroglycerine (TNG). It is used as an explosive.
con. H2SO4
CH2
CH
OH
CH2OH
OH
glycerol
+ 3 H ONO2
CH2
CH
CH2
+ 3 H2O
ONO2
ONO2
ONO2
Trinitroglycerine or Glyceryl trinitrate
17. How is acrolein or acrylic aldehyde prepared from glycerol?
[Mar-2007, Mar-2010, Sep-2010, June-2011, Mar-2012, Mar-2016]
When glycerol is heated with potassium bisulphate or conc. Sulphuric acid or phosphorus pentoxide
dehydration takes place. Two -elimination reaction takes place to give acrolein or acrylic aldehyde.
C
C
H
H
KHSO4
2H2O
OH
C OH
OH
H
H
H
C
C
H
H
OH
CH
C
C
CH2
H
HO
unstable
Acrolein
18. How will you convert glycerol into allyl alcohol? [Sep-2007]
(OR) What is the action of oxalic acid on glycerol at 533K? [Jun-2016]
At 533 K glycerol reacts with oxalic acid to give allyl alcohol
CH2
CH
CH2OH
+
C
C
O
O
O H OH533 K
O H HO
CH2
CH
CH2OH
C
C
O
O
O
O2 CO2
CH2
CH
CH2OH
allyl alcoholunstable
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19. How will you convert glycerol into formic acid? [Sep-2012]
At 383 K glycerol reacts with oxalic acid to give glycerol monoformate. This on hydrolysis gives formic
acid.
CH2
CH
CH2OH
OH +
C C
O O
O HO H OH
383 K
CH2
CH
CH2OH
OH
C
O
O C
O
O H
O
CH2
CH
CH2OH
OH
CO H
glycerol monoformate
CO2
CH2
CH
CH2OH
OH
O
CO HH O H
CH2
CH
CH2OH
OH
OH
+ HCOOH
formic acid
20. What is glycerose? How is it prepared from glycerol? [Mar-2013]
‘Glycerose’ is a mixture of glyceraldehyde and dihydroxy acetone.
It is prepared by the oxidation of glycerol using Fenton’s reagent [FeSO4 + H2O2].
CH2
CH
CH2OH
OH
O H
Glycerol
Glyceraldehyde
Dihydroxyacetone
Fenton's reagent
FeSO4 + H
2O
2
CH
CH
CH2OH
OH
O
CH2
C
CH2OH
O
O H
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21. How is benzyl alcohol prepared by Grignard’s synthesis? [Sep-2011, Mar-2015]
Benzyl alcohol is prepared by the action of phenyl magnesium bromide on formaldehyde
C6H5MgBr C6H5CH2OH MgBr
OHHC6H5
+ H C H
O
+H C H
ether
O MgBr H-O-H
+
22. How is benzyl alcohol prepared from toluene? [Mar-17]
C6H5-CH3
Cl2
HClC6H5-CH2Cl C6H5-CH2OH NaCl+
Toluene Benzyl chloride Benzyl alcohol
NaOH
23. Write a note on Dow’s process [June-2007, Mar-2010, June-2013, June-2014]
C6H5Cl + NaOH623 K
300 atmC6H5OH + NaCl
Chlorobenzene Phenol
This reaction is an example of aromatic nucleophilic substitution. The replacement of nuclear halogen is
difficult because of the stronger C–Cl bond compared to that in alkyl halides.
24. Phenol is soluble in alcohol but insoluble in water. Why? [June-2010]
Solubility of phenol in alcohol is due to intermolecular hydrogen bonding. Insolubility in water is due to
the presence of hydrophobic phenyl group.
25. Phenol is insoluble in NaHCO3 solution but acetic acid is soluble. Give reason [Mar-2007]
Phenol is weaker acid than acetic acid and hence phenol is insoluble in sodium bicarbonate (weaker
base).
Acetic acid is stronger acid than phenol. So, it is soluble in NaHCO3
26. Write three tests for identifying phenol [Mar-2009, Sep-2014]
1. Phenol gives violet colour with neutral ferric chloride
2. Phenol decolourises bromine water with the formation of white precipitate
3. Dye test: Phenol couples with benzene diazonium chloride in alkaline medium to form p-
hydroxy azobenzene which is a red orange dye.
27. What is picric acid? How is it prepared from phenol? [Sep-2010, Mar-2014]
2,4,6-trinitro phenol is known as picric acid.
OH OH
NO2
NO2O2Ncon.HNO
3 + con.H
2SO
4+ 3 H2O
2,4,6-trinitro phenol
(Picric acid)
(nitrating mixture)
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28. Write a note on coupling reaction or write a note on dye test for phenol
[Sep-2006, Sep-2007, Sep-2009, June-2011, Mar-2012, Sep-2012, Sep-2013, June-2015]
Phenol couples with benzene diazonium chloride in alkaline medium to form p-hydroxy azobenzene.
OH
benzene diazonium
chloridep-hydroxy azobenzene
(Red orange dye)Phenol
N=N-Cl + H OH N=N OH
29. Explain Kolbe’s reaction or Kolbe Schmidt reaction [June-2008]
When sodium phenoxide is heated with carbon dioxide at 400 K under pressure, sodium salicylate is
formed. This is decomposed by dilute hydrochloric acid, when salicylic acid is formed. CO2 is the
electrophile in this reaction.
ONa
H
OH
COONaCO
2
sodium phenoxide sodium salicylate
400 K , 4-7 atm
dil. HCl
OH
COOH
salicylic acid
+ NaCl
30. Give the uses of benzyl alcohol [Sep-2006]
Benzyl alcohol is used
1. as a local anaesthetic in intravenus subcutaneous injections.
2. as an antiseptic in ointments.
3. as esters in perfumery. (Benzyl acetate has fragrance of Jasmine)
4. as benzyl benzoate in the treatment of asthma and whooping cough.
31. Write a note on phthalein fusion or convert phenol into phenolphthalein
[Mar-2006, June-2006, June-2009, Mar-2013]
Phenol is heated with phthalic anhydride and con. H2SO4 to give phenolphthalein. This can be tested by
the formation of pink colour when it is treated with sodium hydroxide.
con. H2SO
4
Phenolphthalein
C
C
O
O
O
OHOH
HH
C
C
O
O
OHOH
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32. Explain Riemer-Tiemann reaction [Sep-2015]
This reaction is an example of formylation reaction. When phenol is refluxed with chloroform and
sodium hydroxide, a formyl group –CHO is introduced at the ortho or para position to –OH group.
OH
CHCl3
phenol o-hydroxy
benzaldehyde
NaOH
OH
CHO
OH
CHO
p-hydroxy
benzaldehyde
+
33. Write a note on Schotten-Baumann reaction [Sep-2015]
The reaction of phenols with benzoyl chloride in presence of sodium hydroxide to form phenyl
benzoates is known as Schotten-Baumann reaction.
C6H5O H Cl
O
C C6H5+
phenol benzoyl chloride
NaOHC6H5O
O
C C6H5
phenyl benzoate
34. Give the oxidation reaction of phenol [Sep-2016]
Phenol undergoes oxidation to quinone on treatment with chromyl chloride (CrO2Cl2)
OH + 2 (O)CrO2Cl2
OO
quinone
Additional questions
1 MARK
1. Which has the highest boiling point?
a) CH3CH3 b) CH3OH c) C2H5OH d) C3H8
2. Which is soluble in H2O?
a) Phenol b) Alkanes c) Alcohols d) Alkenes
3. The boiling point of ethyl alcohol should be less than that of
a) propane b) formic acid c) dimethyl ether d) None of the above
4. When alcohols are converted to alkyl chlorides by thionyl chloride in presence of pyridine the
intermediate formed is
a) sulphonium ion b) chlorosulphonic acid c) alkyl chlorosulphite d) chlorosulphite
5. On oxidation of an alcohol gives an aldehyde having the same number of carbon atoms as that of
alcohol. The alcohol is
a) 1o alcohol b) 2
o alcohol c) 3
o alcohol d) None
6. Among the following compounds strongest acid is
a) HC CH b) C6H6 c) C2H6 d) CH3OH
7. The ionization constant of phenol is higher than that of ethanol because
a) phenoxide ion is bulkier than ethoxide b) phenoxide ion is stronger base than ethoxide
c) phenoxide ion is stablized through delocalization d) phenoxide is less stable than ethoxide
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8. The most unlikely representation of resonance structures of p-nitrophenoxide ion is
a) b) c) d)
O
N
O O
O
N
O O+
O
N
O+
O
O
N
O O+
9. p-nitrophenol is having lower pKa value than phenol because
a) phenol is more acidic than p-nitro phenol
b) anion of p-nitrophenol is more stabilised by resonance than that of phenol
c) degree of ionisation of p-nitro phenol is less than that of phenol
d) anion of p-nitrophenol is less stable than that of phenol
10. The reaction of Lucas reagent is fast with
a) (CH3)3COH b) (CH3)2CHOH c) CH3(CH2)2OH d) CH3CH2OH
11. Isomerism exhibited by ethylene glycol is
a) position isomerism b) chain isomerism c) functional isomerism d) both (a) and (c)
12. 1-propanol and 2-propanol can be best distinghished by
a) oxidation with KMnO4 followed by reaction with Fehling solution
b) oxidation with acidic dichromate followed by reaction with Fehling solution
c) oxidation by heating with copper followed by reaction with Fehling solution
d) oxidation with concentrated H2SO4 followed by reaction with Fehling solution
13. The alcohol obtained by the hydrolysis of oils and fats is
a) pentanol b) propanol c) glycerol d) glycol
14. 2-pentanol and 3-pentanol exhibit ------- isomerism
a) chain b) position c) functional d) metamerism
15. 1-butanol and 2-methyl-1-propanol exhibit -------- isomerism
a) chain b) position c) functional d) metamerism
16. n-propyl alcohol and ethylmethyl ether exhibit ------- isomerism
a) chain b) position c) functional d) metamerism
17. Wood spirit is
a) methyl alcohol b) ethyl alcohol c) benzyl alcohol d) phenol
18. Which is called as grain alcohol?
a) methyl alcohol b) ethyl alcohol c) benzyl alcohol d) phenol
19. Ethanol which contains 5% methanol is known as ----------
a) methylated spirit or denatured spirit b) oleum c) wood spirit d) none
20. Addition of water to propylene follows -----------
a) Markovnikoff’s addition b) anti-Markovnikoff’s addition c) peroxide effect d) none
21. Alcohols are ---------- substances
a) neutral b) basic c) strongly acidic d) none
22. Which of the following has high boiling point?
a) CH3OH b) CH3CH3 c) C2H5OH d) C3H8
23. Solubility of alcohols decreases with increase in molecular weight. This is due to increase in the size of
a) hydrophobic alkyl group b) OH group c) hydrophilic group d) none
24. Alcohols are weaker acid than water because of
a) +I effect of alkyl groups b) -I effect of alkyl groups c) resonce effect d) none
25. The acidic strength of alcohols decrease in the order
a) 3o > 2
o >1
o b) 1
o > 2
o > 3
o c) 2
o >3
o > 1
o d) none of these
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26. Lucas reagent is
a) Zn/Hg + con. HCl b) HNO2 + HCl c) con.HCl + anhydrous ZnCl2 d) none
27. Reactivity of alcohols with Lucas reagent follows the order
a) 3o > 2
o >1
o b) 1
o > 2
o > 3
o c) 2
o >3
o > 1
o d) none of these
28. The alcohol which gives immediate turbidity with Lucas reagent at room temperature is
a) (CH3)3COH b) (CH3)2CHOH c) CH3CH2OH d) CH3OH
29. There is no separation problem when alkyl chloride is prepared from alcohols by using
a) Con.HCl / ZnCl2 b) SOCl2/pyridine c) PCl3 d) PCl5
30. Ethyl alcohol (excess) when treated with con.H2SO4 at 410K gives
a) diethyl ether b) ethylene c) dimethyl ether d) none of these
31. Ethyl alcohol when treated with excess con.H2SO4 at 440K gives
a) diethyl ether b) ethylene c) dimethyl ether d) none of these
32. Iodoform (or haloform) test is answered by alcohols containing --------- group
a) CH3CHOH b) -OCH3 c) C2H5CHOH d) none of these
33. Which of the following compounds undergo iodoform reaction?
a) CH3CH2OH b) CH3CH(OH)CH3 c) CH3CH2CH(OH)CH3 d) all of these
34. Which of the following does not undergo iodoform reaction?
a) CH3CH2OH b) CH3CH(OH)CH3 c) CH3CH2CH(OH)CH3 d) CH3OH
35. tert.butyl alcohol vapours when passed over heated copperat at 573K it undergoes dehydration to give
a) acetone b) acetaldehyde c) isobutyl alcohol d) isobutylene
36. Which of the following gives red colour in Victor Meyers test?
a) CH3CH2OH b) CH3CH(OH)CH3 c) (CH3)3COH d) all of these
37. Which of the following is used as an antifreeze in automobile radiators?
a) methyl alcohol b) glycol c) glycerol d) all of these
38. Which is used as a preservative for biological specimens?
a) ethyl alcohol b) phenol c) glycerol d) all of these
39. Baeyer’s reagent is
a) FeSO4/ H2O2 b) cold dilute alkaline KMnO4 c) AgNO3/NH4OH d) none
40. Aldehydes or ketones or esters can be reduced to alcohol by sodium and ethanol. This is,
a) Bouveault-Blanc reduction b) Rosenmund reduction
c) Clemmenson reduction d) none
41. High boiling point, high viscosity and polymeric structure of glycol is due to
a) intramolecular H-bonding b) intermolecular H-bonding
c) dative bonding d) none
42. An explosive prepared from glycol is
a) ethylene dinitrate b) glycol diacetate c) Dacron d) none of these
43. An explosive prepared from glycerol is
a) glycerol triacetate b) nitroglycerine or glyceryl trinitrate (TNG) c) acrolein d) none
44. Polymer (synthetic fibre) produced when glycol reacts with terephthalic acid is
a) Terylene or Dacron or Terene b) bakelite c) silicones d) none of these
45. When glycol is heated with con.H2SO4 it gives
a) ethylene epoxide b) acetaldehyde c) diethylene glycol d) dioxan
46. Which of the following oxidizes glycol into formic acid by cleaving C-C bond?
a) acidified K2Cr2O7 b) acidified KMnO4 c) periodic acid (HIO4) d) all of these
47. Which is used as a coolant in aeroplane engines?
a) methanol b) ethanol c) glycol d) glycerol
48. Glycerol contains
a) two primary alcoholic groups & one secondary alcoholic group
b) two secondary alcoholic groups & one primary alcoholic group
c) three secondary alcoholic groups
d) three primary alcoholic groups
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49. Hydrolysis of esters using alkali is called
a) esterification reaction b) saponification reaction c) trans esterification d) none
50. Hydrolysis of oils or fats by alkali gives soap and -------- as byproduct
a) glycol b) glycerol c) phenol d) all of these
51. Sodium or potassium salts of higher fatty acids are known as
a) soaps b) detergents c) wax d) none of these
52. IUPAC name of benzyl alcohol is
a) phenyl carbinol b) phenyl methanol c) hydroxyl benzene d) cresol
53. Benzyl alcohol is isomeric with
a) cresols b) anisole c) both a and b d) none of these
54. Benzyl alcohol and phenol are sparingly soluble in water because of the presence of
a) hydrophobic phenyl group b) resonance
c) hydrophilic phenyl group d) none of these
55. Preparation of phenol by heating aryl halide with caustic soda (Dow’s process) is
a) aromatic nucleophilic substitution b) aromatic electrophilic substitution
c) aromatic nucleophilic addition d) aromatic electrophilic substitution
56. Phenol turns pink or red on exposure to air and light due to
a) oxidation b) reduction c) polymerization d) sublimation
57. Phenol gives violet colour with
a) NaOH b) NaHCO3 c) neutral ferric chloride d) bromine water
58. Phenol is weakly acidic and so it dissolves in
a) NaOH b) NaHCO3 c) both d) none of these
59. Which of the following is used to separate phenol from carboxylic acid?
a) NaOH b) NaHCO3 c) HCl d) H2SO4
60. The correct order of acidic strength is
a) alcohol < phenol < carboxylic acid b) alcohol > phenol > carboxylic acid
c) phenol < alcohol < carboxylic acid d) alcohol < carboxylic acid < phenol
61. Which of the following is weaker acid?
a) phenol b)alcohol c) carboxylic acid d) water
62. Phenol is more acidic than alcohol due to
a) resonance in phenolate ion b) resonance in alkoxide ion c) OH group d) none
63. The correct order of acidic strength is
a) p-nitro phenol > Phenol > p-cresol b) p-nitro phenol < Phenol < p-cresol
c) p-cresol > p-nitro phenol > Phenol d) p-nitro phenol > p-cresol > Phenol
64. C6H5OH + Zn C6H6 + ZnO
This reaction is useful in the identification of ---------- in natural products
a) aromatic ring b) resonance c) alkyl groups d) all the above
65. Bromination of phenol in the presence of polar solvent (H2O) gives white ppt of
a) 2,4,6-tribromo phenol b) ortho bromo phenol c) para bromo phenol d) none
66. Bromination of phenol at low temperature and in the presence of non polar solvent (CCl4, CS2) gives
a) 2,4,6-tribromo phenol b) ortho bromo phenol c) para bromo phenol d) both b and c
67. With nitration mixture (con. HNO3 + con.H2SO4) phenol gives
a) 2,4,6-trinitro phenol (picric acid) b) o-nitro phenol c) p-nitro phenol d) none
68. Riemer-Tiemann reaction is an example of
a) decarboxylation b) formylation c) hydration d) alkylation
69. The three dimensional polymer obtained when phenol reacts with excess of formaldehyde
a) BH C b) bakelite c) terylene d) nylon
70. In the nitration of phenol (phenol+con.HNO3+con.H2SO4) the electrophile is
a) nitronium ion (NO2+) b) Nitrosonium ion (NO
+) c) H
+ d) NO3
+
71. In the nitrosation of phenol (phenol+HNO2) the electrophile is
a) nitronium ion (NO2+) b) Nitrosonium ion (NO
+) c) H
+ d) NO3
+
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72. Which of the following does not give Iodoform reaction?
a) CH3CH2OH b) CH3CH(OH)CH2CH3 c)CH3CH(OH)CH3 d) (CH3)3COH
3MARK
1. How will you convert propylene into isopropyl alcohol?
Markovnikoff’s addition of water to propylene in the presence of dilute sulphuric acid give isopropyl
alcohol.
CH3-CH=CH2 + H-OH CH3-CH-CH3
H2SO4
OH
2. How will you convert ethylene into ethyl alcohol?
CH2=CH2 + H HSO4 CH3-CH2-HSO4 HOH
CH3-CH2OH + H2SO4
Conc. Ethyl hydrogen sulphate Ethyl alcohol
3. With suitable examples explain isomerism in alcohol
i. Chain isomerism : arising out of the difference in the carbon chain structure of the parent
hydrocarbon
CH3-CH2-CH2-CH2-OH CH3-CH-CH2-OH
CH3
1-butanol 2-methyl-1-propanol
(linear chain) (branched chain)
ii. Position isomerism : arising out of the difference in the position of –OH group, the hydrocarbon
chain being the same.
CH3-CH
2-CH
2-CH
2-CH
2-OH CH
3-CH
2-CH
2-CH-CH
3
OH
CH3-CH
2-CH-CH
2-CH
3
OH
1-pentanol 2-pentanol 3-pentanol
iii. Functional isomerism : Alcohols are isomeric with ethers, having the same formula with different
functional groups.
CH3-CH2-CH2-OH
n-propyl alcohol ethyl methyl ether
CH3-O-CH2-CH3
4. Explain iodoform test for alcohols
When alcohols containing CH3CHOH- group is treated with iodine and potassium hydroxide, yellow
crystals of iodoform is formed.
CH3CH2OH KOHI /
2 CH3CHO 2I
CI3CHO KOH
CHI3 + HCOOK
Ethyl alcohol iodoform
The formation of haloform by this procedure is called haloform reaction. Since methanol does not
contain CH3CHOH- group, it does not undergo haloform reaction. Ethyl alcohol, isopropyl alcohol and
sec.butyl alcohol undergo iodoform or haloform reaction.
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5. How will you distinguish 1o, 2
o and 3
o alcohols using Lucas test?
C
C
CCH3
CH3
CH3
OH
CH3
CH3
OH
OH
H
CH3
H
H
3o alcohol
2o alcohol
1o alcohol
con.HCl
anhydrous ZnCl2
con.HCl
anhydrous ZnCl2
con.HCl
anhydrous ZnCl2
CCH3
CH3
CH3
Cl
CCH3
CH3
H
Cl
CCH3
H
H
Cl
Turbidity appears immediately at
room temperature
Turbidity appears after 5-10 minutes
at room temperature
No turbidity appears easily
but only on heating
6. How will you distinguish 1o, 2
o and 3
o alcohols by catalytic dehydrogenation method?
C OHCH3
H
H
1o alcohol
Cu/573K CCH3
H
O + H2
acetaldehyde
CCH3
CH3
OH
H
2o alcohol
CCH3
CH3
Cu/573K O + H2
acetone
CCH3
CH3
CH3
OH
3o alcohol
dehydrationCCH3
CH3
CH2
Cu/573K
isobutylene
+ H2O
7. Give the Victor Meyer’s test for primary alcohol
CCH3 OH
1o alcohol
P / I2CCH3 I
AgNO2CCH3
NO2
H H H
CCH3NO2
KOH
- H2O
H H HO=N-OH
N-OH
CCH3NO2
N-OK
Nitrolic acid
Potassium salt of nitrolic
acid (Red colour)
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8. What happens when ethyl alcohol is treated with excess conc.H2SO4 at 440 K?
When ethyl alcohol is treated with excess of con.H2SO4 at 440 K it undergoes intramolecular
dehydration to give ethylene.
CH3 HSO4+ H2O
440 KOH + H HSO4
Con(excess)
CH2 CH3CH2
CH2 CH2
H HSO4
CH2CH2 + H2SO4
ethylene
9. Explain isomerism in diols or dihydric alcohols
i. Position Isomerism
Diols having the same molecular formula differ in the position of the hydroxyl groups.
CH3CH2CH
OHOH
CH2CH2
CH2
OHOH
propane-1,2-diol propane-1,3-diol
ii. Functional Isomerism: Diols are isomeric with ethers or hydroxy ethers.
CH3CH2CH
OHOH
propane-1,2-diol
is isomeric with CH3-O-CH2-O-CH3 CH3-O-CH2-CH2-OHand
dimethoxy methane 2-methoxy ethanol
10. How will you prepare glycol from i) ethylene dibromide and ii) ethylene diamine?
i) from ethylene dibromide
CH2Br
CH2Br
H2O CH2OH
CH2OH
+ 2 NaBr + CO2
ethylene dibromide glycol
Na2CO
3
ii) from ethylene diamine
CH2NH2
CH2NH2
2HNO2
CH2OH
CH2OH
+ 2 N2 + 2H2O
ethylene diamine glycol
11. Write a note on Bouveault-Blanc reduction.
Ester can be reduced to alcohol by sodium and ethanol.
COOC2H5
COOC2H5
+ 8(H)Na + Ethanol
CH2OH
CH2OH
+ 2 C2H5OH
diethyl oxalate glycol
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12. Give reason for hygroscopic nature of glycol.
In glycol intermolecular hydrogen bond can be formed between both OH groups resulting in a
polymeric structure.
CH2
CH2
O
O
H
H
H O
CH2CH2
O
H
Water forms stronger H-bonding with glycol, breaking the polymeric structure. This explains the
hygroscopic nature of glycol.
13. How will you prepare ethylene dinitrate? Give its use
On heating with nitric acid in presence of catalytic amount of con. sulphuric acid, ethylene glycol forms
ethylene dinitrate. It is used as an explosive.
con.H2SO
4CH2ONO2
CH2ONO2
+ 2 H2O
ethylene dinitrate
(explosive liquid)
CH2 OH
CH2 OH
glycol
2 H ONO2+
14. What is the action of PI3 on glycol?
With PI3, glycol forms ethylene diiodide, which being unstable decomposes to form ethylene.
PI3
ethylene
CH2 OH
CH2 OH
glycol ethylene diiodide
CH2I
CH2I
- I2CH2
CH2
15. What happens when glycol is i) heated alone and ii) heated with anhydrous ZnCl2?
Intramolecular dehydration :
i) When heated alone up to 773 K glycol forms ethylene oxide or ethylene epoxide. This is an
intramolecular dehydration reaction in which a water molecule is eliminated from the two –OH groups
of same molecule.
+CH2 OH
CH2 OH
glycol
773 KCH2
CH2
ethylene oxide or
ethylene epoxide
O H2O
ii) When heated with anhydrous zinc chloride, glycol forms acetaldehyde.
C
C
H
H
H
OHanhydrous
ZnCl2
H2OOH
H C
C
H
H
H
O H C
C
H
H
H
O
H
Acetaldehyde
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16. Write a note on saponification reaction Glycerol is prepared in a large scale by the hydrolysis of oils or fats either by using alkali (in soap
industry) or by super heated steam (in candle industry). Hydrolysing with alkali, forms sodium or
potassium salt of higher fatty acids as solids. These are called soaps. In general hydrolysis of
esters using alkali is called ‘Saponification’ reaction.
CH2
CH
CH2
O
O
O
C
C
C
O
O
O
R
R
R
+ 3Na OH
CH2
CH
CH2
O
O
O
+
H
H
H
3 RCOONa
Soap
Glycerol
Oil or Fat
17. What is the action of HI or PI3 or I2 on glycerol?
With hydrogen iodide or phosphorous triiodide glycerol forms allyl iodide. An unstable triiodide is
formed as an intermediate which loses iodine to give allyl iodide. With excess of the reagent allyl iodide
further reacts giving isopropyl iodide.
CH2
CH
OH
CH2OH
OH3 HI
CH2
CH
CH2I
I
I
I
I2
CH2
CH
CH2
HI
CH3
CH
CH2
I
I
I2
CH3
CH
CH2
I HI
CH3
CH
CH3
glycerol unstable allyl iodide propylene isopropyl iodideunstable
18. Write a note on Cannizzaro reacton
Aldehydes with no α –hydrogen undergo Cannizzaro reacton in the presence of 50% caustic soda
solution.
This reaction involves oxidation of one aldehyde molecule to acid, while the other aldehyde molecule is
reduced to alcohol.
Simple Cannizzaro reacton
C6H5CHO
H2O50% NaOH
C6H5CH2OH C6H5COOH
C6H5CHO
+ +
Benzyl alcohol Benzoic acid
Crossed Cannizzaro reacton
HCHO
HCOOH
C6H5CHO
H2O50% NaOH
C6H5CH2OH+ +
Benzyl alcohol Formic acid
19. From benzyl alcohol prepare i) toluene ii) C6H5CHO and iii) C6H5COOH
i) On heating with phosphorous and hydriodic acid, benzyl alcohol is reduced to toluene.
C6H5 C6H5-CH3
CH2OH
H I
H2O C6H5 CH2I
H I
+ I2
Toluene
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ii) With mild oxidising agents like copper nitrate or lead nitrate, benzyl alcohol is converted to
benzaldehyde
C6H5 CH2OH
Pb(NO3)2
or Cu(NO3)2
C6H5 CHO + H2O+ (O)
iii) When benzyl alcohol is oxidised with acidified K2Cr2O7 or alkaline KMnO4, it gives benzoic acid.
Benzoic acid
C6H5 CH2OH
Na2Cr
2O
7 /H+
or KMnO4/OH-
C6H5 CHO+ (O) C6H5 COOH(O)
Benzaldehyde
20. How is phenol prepared from i) benzenediazonium chloride and ii) sodium salicylate?
i) from benzenediazonium chloride
C6H5N2Cl + H2O H
C6H5OH + HCl +N2
ii) from sodium salicylate (decarboxylation reaction)
OH
COONa
NaOH/CaO
(Sodalime)
OH
+ Na2CO3
21. Give the industrial preparation of phenol from cumene or convert benzene into phenol.
Industrially phenol is prepared from cumene. Cumene is prepared from benzene and propylene in
presence of Lewis acid like anhydrous aluminium chloride.
+ CH3-CH=CH2
anhydrous
AlCl3
cumene cumene hydro peroxidebenzene
acetonephenol
CH3-CH-CH3
O2
(air)
CH3-C-CH3
O
O
H
aqueous HCl
OH
+ CH3-CO-CH3
propene
22. Give two reactions that differentiate phenol from alcohols
i) Phenol gives violet colour with neutral ferric chloride It is a characteristic test for phenol.
ii) Acidic character
Phenol is more acidic than alcohols and so, it dissolves in sodium hydroxide
C6H5OH + NaOH C6H5ONa + H2O
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23. What happens when phenol is distilled with zinc dust? Write the use of this reaction
Phenol on distillation with zinc dust gives benzene (aromatic hydrocarbon)
C6H5OH + Zn C6H6 + ZnO
This reaction is useful in the identification of aromatic ring present in natural products.
24. Convert phenol into i) chloro benzene ii) anisole iii) phenetole and iv) aniline
i) Phenol to chlorobenzene
C6H5 O H
Cl PCl3 Cl
C6H5Cl + POCl3 + HCl
ii) Phenol to anisole
C6H5OH + CH2N2 OH
C6H5OCH3 + N2
Diazomethane Anisole or methoxy benzene
iii) Phenol to phenetole
C6H5OH + C2H5Br OH
C6H5OC2H5 + HBr
Ethyl bromide phenetole or ethoxy benzene
iv) Phenol to aniline
C6H5 OH H NH2
anhydrous ZnCl2
+ C6H5 NH2 + H2O
Aniline
25. What happens when phenol is refluxed with CCl4 and NaOH?
Similarly with CCl4 and NaOH hydroxy benzoic acid is formed
OH
CCl4
phenol o-hydroxy
benzoic acid
NaOH
OH
COOH
OH
COOHp-hydroxy benzoic acid
+
26. From phenol prepare cyclohexanol
Phenol on hydrogenation in presence of nickel forms cyclohexanol.
OH
Ni+ 3 H2
OH
cyclohexanol
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27. Explain Lederer-Manasse reaction
Phenol, when treated with formaldehyde and sodium hydroxide, undergoes condensation reaction. This
is called ‘Lederer-Manasse reaction’.
+
OH
NaOHHCHO
OH
CH2OH
p-hydroxy phenyl methanol
28. Write the uses of methyl alcohol
1. industrial solvent 2. antifreeze in automobile radiators 3. manufacture of formaldehyde
29. Write the uses of ethyl alcohol
1. industrial solvent 2. in alcoholic beverage 3. preservative for biological specimens
30. What is methylated spirit or denatured spirit? Write its use
Ethyl alcohol which contains 5% methyl alcohol is known as methylated spirit or denatured spirit. It is
used as a fuel in spirit lamp and as a solvent for wood polish.
31. Write the uses of glycol
1. antifreeze in automobile radiators
2. coolant in aeroplane engines 3. in the preparation of, terylene
32. Write the uses of glycerol
1. in the manufacture of explosive (TNG)
2. antifreeze in automobile radiators 3. in copying inks and stamp pad inks
33. Write the uses of phenol
1. manufacture of dyes, plastics, pesticides and explosives
2. antiseptic and germicide 3. in soaps and lotions
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17. ETHERS
Blue print 1 Mark = 2 3 Mark = 0 5 Mark = 1 Total marks = 7
Govt. Examination questions 1 MARK
1. Which one of the following is simple ether? [Mar-2010, Mar-2013, Sep-2014]
a) CH3 – O – C2H5 b) C2H5 – O – CH3 c) C2H5 – O – C2H5 d) C3H7 – O – C2H5
2. Diethyl ether can be decomposed with [Mar-2009, Sep-2011, Mar-2012, Jun-2012, Sep-15, Sep-16]
a) HI b) KMnO4 c) NaOH d) H2O
3. Oxygen atom of ether is [Sep-2006, June-2008, Mar-2014, June-2014, Mar-2015]
a) very active b) Replaceable c) oxidising d) Comparatively inert
4. According to Lewis concept of acids and bases, ethers are
[Mar-2006, Mar-2007, Jun-2007, Sep-2007, Sep-2010, Jun-2016]
a) Neutral b) Acidic c) Basic d) Amphoteric
5. Intermolecular hydrogen bonds are not present in [Sep-2012, June-2013]
a) CH3COOH b) C2H5OC2H5 c) CH3CH2OH d) C2H5NH2
6. When ethyl Iodide is treated with dry silver oxide it forms [Sep-2008, Sep-2009]
a) Ethyl alcohol b) diethylether c) silver ethoxide d) ethylmethyl ether
7. Williamson's synthesis is an example of [June-2009, June-2011]
a) nucleophilic addition b) electrophilic addition
c) electrophilic substitution d) Nucleophilic substitution reaction
8. When ether is exposed to air for sometime an explosive substance produced is
[June-2006, Mar-2008, June-2010, Mar-2011, Sep-2013, Sep-2014, June-2015, Mar-2016]
a) peroxide b) oxide c) TNT d) superoxide
9. Ether is formed when alkyl halide is treated with sodium alkoxide. This method is known as
[Mar-2008, Mar-2014]
a) Hoffmann reaction b) Williamson's synthesis c) Wurtz synthesis d) Kolbe's reaction
10. The compound mixed with ethanol to serve as substitute for petrol is [Mar-2006]
a) methoxymethane b) ethoxyethane c) methanol d) ethanal
11. The isomerism exhibited by 1-propanol and methoxyethane is [June-2006]
a) chain b) position c) functional d) metamerism
12. Diethylether behaves as a [Sep-2006, June-2011]
a) Lewis acid b) Lewis base c) neutral compound d) Bronsted acid
13. Zeisel’s method of detection and estimation of alkoxy group in alkaloids involves the reaction of ether
with [Mar-2007]
a) HI b) Cl2 c) PCl5 d) AlCl3
14. The number of ether isomers possible for C4H10O is [June-2008, Mar-2011]
a) 7 b) 5 c) 4 d) 3
15. How many alcohol isomers are possible for the molecular formula C4H10O? [Mar-2012]
a) 4 b) 2 c) 3 d) 7
16. In the formation of oxonium salt when etherial oxygen reacts with strong mineral acid is called
[June-2009, Mar-2013]
a) electronation b) protonation c) deprotonation d) dehydration
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17. Ethers should never be evaporated to dryness because [June-2010]
a) they form explosive peroxide b) they are insoluble in water
c) they are inert d) they are lighter than water
18. Higher ethers can be prepared from lower members by the action of [Sep-2010]
a) con. H2SO4 b) AgOH c) sodium alkoxide d) Grignard reagent
19. Which of the following produces ether when heated with con. H2SO4 at 413 K? [Sep-2011]
a) organic acid b) aldehyde c) alcohol d) keteone
20. The solvent for Grignard reagent is [June-2012, June-2015]
a) ethyl alcohol b) diethyl ether c) acetone d) benzene
21. When diethyl ether reacts with Cl2 in the presence of sunlight, it gives [Sep-2012]
a) -chloro diethyl ether b) ,’-dichloro diethyl ether
c) perchloro diethyl ether d) both a & b
22. The ether used in perfumery is [June-2007, Sep-2009]
a) diethyl ether b) dimethyl ether c) methylphenylether (Anisole) d) ethyl methyl ether
23. Which among the following is unsymmetrical ether? [Sep-2007]
a) C6H5OC6H5 b) C2H5OC2H5 c) CH3OCH3 d) C6H5OCH3
24. Anisole on bromination yields [Mar-2009, June-2014]
a) m-bromo anisole b) o-bromo anisole c) o & p-bromo anisole d) benzoic acid
25. The IUPAC name of phenetole is [Mar-2010, Sep-2013]
a) ethyl phenyl ether b) methyl phenyl ether c) diethyl ether d) ethoxy benzene
CH
CH3
OCH3CH326. IUPAC name of is [June-2013]
a) 1-methoxy propane b) Methyl isopropylether
c) isopropylmethylether d) 2-methoxy propane
27. Which one of the following does not form peroxide easily? [Mar-2015]
a) Diethyl ether b) Ethylmethyl ether c) Dimethyl ether d) Anisole
28. Anisole is used in [Sep-2015]
a) refrigerant b) anaesthetic c) perfumery d) substitute for petrol
29. The number of ether isomers possible for the molecular formula C5H12O [Mar-2016]
a) 5 b) 6 c) 4 d) 7
30. Diethyl ether is used as a solvent for [Jun-2016]
a) Lucas reagent b) Fenton’s reagent c) Tollen’s reagent d) Grignard reagent
31. -------------- is used as refrigerant in refrigerator [Sep-2016]
a) methoxy benzene b) ethoxy ethane c) methoxy methane d) ethoxy benzene
32. The isomerism exhibited by C2H
5-O-C
2H
5 and CH
CH3
OCH3CH3 is [Mar-2017]
a) Functional b) metamerism c) position d) chain
33. With a mixture of conc.HNO3 and conc.H2SO4 anisole gives [Mar-17]
a) o-nitro anisole b) p-nitro anisole c) o & p-nitro anisole d) m-nitro anisole
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5 MARK
1. Write a note on the type of isomerism exhibited by ethers [June-2006, Sep-2007, Mar-2014]
1. Functional Isomerism
Ethers are functional isomers of alcohols as both have the same general formula CnH2n+2O.
Molecular
formula Ethers Alcohols
C2H6O CH3–O– CH3
Dimethyl ether
CH3CH2OH
Ethyl alcohol
C3H8O CH3–O–CH2–CH3
Ethyl methyl ether
CH3–CH2–CH2–OH
n-propyl alcohol
CCH3 CH3
OH
H
Isopropyl alcohol
C4H10O
CH3–CH2–O–CH2–CH3
Diethyl ether
CH3– O–CH2–CH2–CH3
Methyl n-propyl ether
CH
CH3
OCH3CH3
Methyl isopropyl ether
CH3–CH2–CH2–CH2–OH
n-butyl alcohol
CCH3OHH CH2
CH3
Isobutyl alcohol
CCH3 CH3
OH
HCH2
sec.butyl alcohol
CCH3
CH3
CH3
OH
tert.butyl alcohol
2. Metamerism
It is a special isomerism in which molecules with same formula, same functional group, differing only
in the nature of the alkyl group attached to oxygen.
CH3
OCH3 CH3OCH3CH3
OCH3 CH2 CH
2CH
2 CH2
CH3 CH
methyl n-propylether diethylether methyl isopropylether
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2. Write all possible isomers with the molecular formula C4H10O and name them [June-2014]
Isomers (C4H10O) Common name IUPAC name
CH3–CH2–CH2–CH2–OH
n-butyl alcohol
1-butanol
CCH3OHH CH2
CH3
Isobutyl alcohol
2-methyl-1-propanol
CCH3 CH3
OH
HCH2
sec.butyl alcohol
2-butanol
CCH3
CH3
CH3
OH
tert.butyl alcohol
2-methyl- 2-propanol
CH3–CH2–O–CH2–CH3
Diethyl ether
Ethoxy ethane
CH3– O–CH2–CH2–CH3
Methyl n-propyl ether
1-methoxy propane
CH
CH3
OCH3CH3
Methyl isopropyl ether 2-methoxy propane
3. Give the methods of preparing diethyl ether
[June-2007, Mar-2008, Mar-2010, Mar-2012, Mar-2013, Sep-2014, Jne-2012]
1. Intermolecular dehydration of alcohol
When excess of ethyl alcohol is heated with con. H2SO4 two molecules condense losing a molecule
of water forming diethyl ether.
C2H
5-O-H + H-O-C
2H
5 C
2H
5-O-C
2H
5 + H
2O
413 K
conc. H2SO
4
2. Williamson’s synthesis
Diethyl ether is prepared by heating ethyl bromide with sodium ethoxide.
C2H
5-Br + Na-O-C
2H
5 C
2H
5-O-C
2H
5 + NaBr
3. From ethyl iodide
Diethyl ether is prepared by heating ethyl iodide with dry silver oxide.
C2H
5 - I
C2H
5 - I
Ag2 O C
2H
5 - O - C
2H
5 + 2AgI
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4. How does ether react with the following? a) Air or O2 / long contact b) dil.H2SO4 c) PCl5 [Mar-2007, Jun-2011, Jun-2016]
a) Air or O2 / long contact
C2H5–O–C2H5 )O(
(C2H5)2O2
Diethyl peroxide
Diethyl peroxide is unstable and decomposes violently with explosion on heating. Hence ether
should not be evaporated to dryness.
b) dil.H2SO4 ( Hydrolysis )
Diethyl ether on boiling with water in presence dilute H2SO4 is hydrolysed to form ethyl alcohol.
C2H
5-O-C
2H
5
H - OH
dil. H2SO
42 C
2H
5-O-H
c) PCl5
C2H
5 - O - C
2H
5
Cl - PC3 - Cl
2 C2H
5-Cl + POCl
3
Ethyl chloride
5. How does diethyl ether react with PCl5, one equivalent of HI and excess of HI? [Sep-2010]
a) PCl5 ( Refer – Q – 4 c )
b) With one equivalent of HI
C2H
5-O-C
2H
5
H - I
C2H
5-O-H + C
2H
5I
Ethyl alcohol Ethyl iodide
c) With excess of HI ( excess hot concentrated hydroiodic acid )
C2H
5-O-C
2H
5
H - I
I - H
2 C2H
5-I + H
2O
Ethyl iodide
6. How does diethyl ether react with the following reagents?
a) O2 / long contact b) HI in excess c) PCl5 [Mar-2015]
Ans: Refer – Q – 4 & 5
7. How do ethers react with HI? Give the significance of the reaction [Mar-2006]
With one equivalent of HI
C2H
5-O-CH
3
H - I
C2H
5-O-H + CH
3I
Ethyl alcohol Methyl iodide
Halogen (I) prefers to attack the carbon atom of the smaller alkyl group.
With excess hot concentrated hydroiodic acid
C2H
5-O-CH
3
H - I
I - H
C2H
5-I + CH
3-I + H
2O
Ethyl iodide Methyl iodide
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This reaction is used in the Zeisel’s method of detection and estimation of alkoxy (especially
methoxy) group in natural products like alkaloids.
8. Write three methods of preparation of anisole.
[June-2008, Mar-2009, June-2012, Sep-2012, June-2013, Sep-2015]
1. Williamson’s synthesis
C6H5ONa + CH3–I C6H5–O–CH3 + NaI
sodium phenoxide Anisole
2. Using diazomethane
C6H5 –O–H + CH2 –N2 C6H5–O–CH3 + N2
Phenol diazomethane Anisole
3. Using dimethyl sulphate
C6H5 –O–H + (CH3)2 –SO4 NaOH C6H5–O–CH3 + CH3HSO4
Phenol dimethyl sulphate Anisole
9. Give any two methods of preparation of anisole and explain the reaction of HI with anisole. [Jun-08]
Preparation: Refer-Q-8
Anisole reacts with hydroiodic acid to give phenol and methyl iodide
C6H5 _ O _ CH3
H _ I
C6H5OH + CH3I
Because of the strong C–O bond in anisole [bond between carbon atom of benzene ring and oxygen],
cleavage of the C–O bond does not occur and iodobenzene, methyl alcohol are never formed.
10. Give the differences between aromatic ether and aliphatic ether
[Sep-2006, Sep-2008, June-2009, Sep-2009, June-2010, Mar-2011, Sep-2011, Sep-2013, Mar-17]
No Aromatic ether
(Anisole)
Aliphatic ether
(Diethyl ether)
1 Comparatively high boiling liquid Volatile liquid
2 Used in perfumery Used as anaesthetic
3 Not used as solvent Used as a solvent
4
Cannot be used as a substitute for
petrol
Mixed with alcohol, used as a
substitute for petrol
5 Does not form peroxide easily Forms peroxide in air.
11. How are the following reactions effected?
i) diethyl ether with Grignard reagent
ii) diethyl ether with mineral acid
iii) ethyl methyl ether with excess hot conc. HI acid [Mar-2016]
i) Grignard reagent forms the following complex with diethyl ether. Thus the Grignard reagent is
stabilised in dry ether.
2 C2H5_O_C2H5 + CH3MgI
CH3
Mg
I
O(C2H5)2
O(C2H5)2
Hence ether is used as a solvent for Grignard’s reagent.
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ii) Strong mineral acids protonate the ethereal oxygen forming oxonium salts. In this reaction diethyl
ether acts as Lewis base.
C2H5_O_C2H5 + HCl C2H5
_O_C2H5
H
Cl +
diethyl oxonium chloride
C2H5_O_C2H5
C2H5_O_C2H5
H
+H2SO4
HSO4+
diethyl oxonium hydrogensulphate
conc.
iii) With excess hot concentrated hydroiodic acid:
C2H
5-O-CH
3
H - I
I - H
C2H
5-I + CH
3-I + H
2O
Ethyl iodide Methyl iodide
12. How does diethyl ether react with the following?
i) conc. H2SO4 ii) dil. H2SO4 iii) HI [Sep-2016]
i) conc. H2SO4 Refer-Q-11-ii
ii) dil. H2SO4 Refer-Q-4-b
iii) HI Refer-Q-5-b & c
Additional questions 1 MARK
1. Ethers are functional isomers of
a) alcohols b) aldehydes c) thioethers d) esters
2. The general formula ethers and alcohols
a) CnH2n+2O b) CnH2n c) CnH2n-2O d) CnH2nO
3. Ethers are insoluble in water due to the absence of
a) hydrogen bonding b) functional group c) non-polar group d) none of these
4. Ethers do not react with
a) acids b) alkalis c) sodium d) all the above
5. Ether oxygen is capable of forming a coordinate covalent bond with
a) Oxygen b) BF3 c) AlCl3 d) all the above
6. Diethyl ether acts as Lewis base in the reaction with
a) con.HCl , con.H2SO4 b) BF3 c) AlCl3 d) all the above
7. Ether is used as a solvent for Grignard reagent because
a) it forms stable complex b) Grignard’s reagent is soluble only in ether
c) ether forms H-bond with Grignard reagent d) all are correct
8. Diethyl ether is used as
a) Refrigerant b) medium for the preparation of Grignard reagent
c) Anaesthetic d) all the above
9. Substitute for petrol is the mixture of
a) diethyl ether + ethanol b) diethyl ether + anisole
c) dimethyl ether + ethanol d) dimethyl ether + methanol
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10. Which is not a phenolic ether or aromatic ether?
a) C6H5OC6H5 b) C6H5OCH3 c) C6H5OC2H5 d) C2H5OCH3
11. C6H5-O-CH3 + HI X. X is
a) CH3OH & C6H5I b) C6H5OH & CH3I c) CH3I & C6H5I d) o-iodo anisole
12. Anisole can be prepared from
a) C6H5ONa + CH3I b) C6H5I+CH3ONa c) C6H5Br+CH3ONa d) C6H5OH+CH3ONa
13. Which is ortho, para directing group?
a)-OH b)-OCH3 c) both a and b d)-CHO
5 MARK
1. How does diethyl ether react with the following?
1. Cl2 / dark
CH3CH2
O CH2 CH3
Cl2 / dark
CH3CH O CH2 CH3
Cldiethylether
CH3CH O CH3
Cl
CH
Cl
Cl2 / dark
- chloro diethylether
' - dichloro diethylether
2. Cl2 / sunlight In sunlight, substitution is complete
C2H5–O–C2H5 + 10 Cl2 Light
C2Cl5–O–C2Cl5 + 10HCl perchloro diethyl ether
2. Explain electrophilic substitution reactions of anisole. Write its uses
Electrophilic substitution reactions:
–OCH3, group increase the reactivity of the benzene ring with respect to electrophilic attack
and is ortho, para –directing.
a) With a mixture of con.HNO3 and con.H2SO4 it gives a mixture of ortho and para nitro anisole
OCH3
con.HNO3
con.H2SO
4
OCH3 OCH3
NO2
NO2
+
b) Bromination yields ortho bromoanisole and para bromoanisole
OCH3 OCH3OCH3
+Br2
Br
Br
Uses of anisole
1. It is used in perfumery.
2. It is used as a starting material in organic synthesis.
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18. CARBONYL COMPOUNDS
Blue print 1 Mark = 1 3 Mark = 1 5 Mark = 1+1* Total marks = 14
Govt. Examination questions 1 MARK
1. Schiffs reagent gives pink colour with [Mar-2007, Sep-2007, Sep-2009, June-2010, Mar-2012]
a) acetone b) acetaldehyde c) ethyl alcohol d) methyl acetate
2. Methyl ketones are usually characterised by [Sep-2011, Mar-2015, Sep-2015]
a) the Fehling’s solution b) the iodoform test c) the Schiff’s test d) the Tollen’s reagent
3. The compound that does not undergo Cannizzaro reaction is [Mar-2009, Mar-2013, Sep-2013]
a) formaldehyde b) acetaldehyde c) benzaldehyde d) trimethyl acetaldehyde
4. The formation of cyanohydrin from a ketone is an example of [Sep-2006, Jun-2016]
a) electrophilic addition b) nucleophilic addition
c) nucleophilic substitution d) electrophilic substitution
5. Hydrogenation of benzoyl chloride in the presence of Pd on BaSO4 gives [Mar-2008]
a) phenol b) benzoic acid c) benzyl alcohol d) benzaldehyde
6. Aldol is [Mar-2006, June-2011, Sep-2014]
a) 2-hydroxy butanol b) 3-hydroxy butanol c) 3-hydroxy butanal d) 2-hydroxy butanal
7. A cyanohydrin of a compound X on hydrolysis gives lactic acid. The X is [Mar-2011]
a) HCHO b) CH3CHO c) (CH3)2 CO d) C6H5CH2CHO
8. Which of the following does not give iodoform test? [June-2006]
a) aceto phenone b) benzophenone
c) d) CH3
CH3 CH
CH3
OH CH
OH
CH2
CH2 CH
3
9. The compound which does not reduce Fehling solution is [June-2008, June-2012]
a) formaldehyde b) acetaldehyde c) benzaldehyde d) propionaldehyde
10. CH3COCH3 4SO2Hconc. The product is [Sep-2012]
a) mesitylene b) mesityl oxide c) phorone d) paraldehyde
11. The compound used in the preparation of the tranquilizer, sulphonal is [Mar-2010, June-2013]
a) acetone b) acetophenone c) isopropyl alcohol d) glycol
12. Calcium acetate + Calcium benzoate ondistillati gives [Sep-2010]
a) benzophenone b) benzaldehyde c) acetophenone d) phenyl benzoate
13. Tertiary butyl alcohol can be prepared by treating CH3MgI with --------- [Jun-2007, Sep-2016]
a) Formqaldehyde b) acetaldehyde c) acetone d) CO2
14. Propanone is usually characterized by [Sep-2008]
a) Fehling’s test b) iodoform test c) Schiff’s test d) Tollen’s test
15. Compound which undergoes iodoform test is [June-2009]
a) 1-pentanol b) 2-pentanone c) 3-pentanone d) pentanal
16. Tollen's Reagent is [Mar-2014]
a) Ammoniacal cuprous chloride b) Ammoniacal cuprous oxide
c) Ammoniacal silver nitrate d) Ammoniacal silver chloride
17. Formaldehyde polymerises to give [June-2014, Mar-17]
a) Paraldehyde b) paraformaldehyde c) Formalin d) Formic acid
18. Which of the following compounds is oxidised to give ethyl methyl ketone? [June-2015]
a) 2-propanol b) 2-pentanone c) 1-butanol d) 2-butanol
19. During reduction of aldehydes with hydrazine and C2H5ONa the product formed is [Mar-2016]
a) R–CH = N–NH2 b) R–C N c) R – CO – NH2 d) R–CH3
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3 MARK
1. Give IUPAC name of i) crotonaldehyde ii) methyl n-propyl ketone iii) phenyl acetaldehyde
iv) Benzaldeyde [Mar-2006, Mar-2013]
Formula Common Name IUPAC Name
CH3CH=CHCHO Crotonaldehyde 2-butenal
CH3COCH2CH2CH3 Methyl n-propyl ketone 2-pentanone
C6H5CH2CHO Phenyl acetaldehyde Phenyl ethanal
C6H5CHO Benzaldehyde Phenyl methanal
2. Give the IUPAC names of [Sep-2015]
i) CH3-CH-C-CH-OCH
2CH
3
OCH3
O
CH3
ii) CH3-CO-CH-CH
2-CH
2-Cl
C2H
5
i) 2-ethoxy-4-methoxy-3-pentanone ii) 3-ethyl-5-chloro-2-pentanone
3. Give three tests for aldehydes [June-2008, Mar-2012, Sep-2013]
1. Aldehydes reduce ammonical silver nitrate (Tollen’s reagent) to metallic silver
CH3CHO + Ag2O CH3COOH + 2Ag
2. Aldehydes reduce Fehling’s solution (copper sulphate, sodium potassium tartrate) to red cuprous oxide.
CH3CHO + 2CuO CH3COOH + Cu2O
(blue) (red precipitate)
3. Aliphatic aldehydes restore the original colour (red-pink) of the Schiff's reagent. (When SO2 is
passed through intensely pink coloured Schiff’s reagent in water, it forms a colourless solution. This
colourless solution is used for this test).
4. What is formalin? Give its use [June-2007, June-2014]
40% aqueous solution of formaldehyde is known as formalin.
It is used as a preservative for biological specimens and in leather tanning.
5. What is urotropine? How is it prepared? Give its structure and use. [June-2006, Sep-2006, Mar-
2008, June-2009, Sep-2009, Mar-2010, Sep-2010, June-2011, Sep-2011, June-2012, Mar-14, Sep-14]
Formaldehyde reacts with ammonia to form hexamethylene tetramine which is known as ‘Urotropine’
6CH2O + 4NH3 (CH2)6 N4 + 6H2O
Hexa methylene tetramine
It is used as Urinary antiseptic in medicine.
Structure of urotropine:
N
N N
N
CH2
CH2
CH2CH
2
CH2CH
2
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6. What is Rosenmund’s reduction ? What is the purpose of adding BaSO4 in it?
[Sep-2007, June-2015, Mar-17]
Acetyl chloride is reduced to acetaldehyde by hydrogen in presence of palladium catalyst and barium
sulphate as catalytic poison.
CH3
C
O
Cl +Pd
BaSO4
HH2
CH3
C
O
+ HCl
Acetyl chloride Acetaldehyde
BaSO4 is used as a catalytic poison, to stop the reduction at the stage of aldehyde. Otherwise, the
aldehyde formed will be further reduced to primary alcohol.
7. Write a note on Clemmenson reduction [June-2010]
Aldehydes and ketones can be reduced to hydrocarbons by zinc amalgam and con.HCl
CH3 C
O
4 ( H )CH3 +
Zn / Hg
Con. HCl
CH3 C CH
3H2
Acetone Propane
+ H2O
8. What happens when benzaldehyde is heated with concentrated NaOH? [Sep-2012]
Benzaldehyde undergoes Cannizaro reaction when heated with concentrated NaOH because of the
absence of α-hydrogen. It involves self oxidation and reduction.
C6H5CHO + C6H5CHO NaOHC6H5COONa + C6H5CH2OH
Benzaldehyde Benzaldehyde Sodium benzoate Benzyl alcohol
9. Write on Perkins reaction [Mar-2015]
When benzaldehyde is heated with sodium acetate in presence of acetic anhydride, it forms cinnamic acid.
C6H
5-C = O
H
+ H2 CH-CO
O
CH3-CO H
OH
CH3COONa
C6H
5-C =
H
CH-COOH + CH3COOH
Acetic anhydride
Cinnamic acid
10. Write a note on haloform reaction [Sep-2008]
The compounds having CH3CHOH– or CH3CO– group undergoes haloform reaction.
CH3–CO–CH3 + 3Cl2 CCl3 – CO – CH3 + 3HCl
CCl3 – CO – CH3 + NaOH CH3COONa + CHCl3
Chloroform
The formation of haloform by this procedure is called haloform reaction.
11. From acetone prepare chloretone [June-2013]
With chloroform acetone forms an addition product
KOHCH
3 C
CH3
+ H-CCl3
O
CH3 C
CH3
CCl3
OH
Chloretone
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12. How is acetophenone prepared by Friedel-Crafts method? [Mar-2007, Mar-2009, Mar-2011]
In presence of anhydrous Aluminium chloride acetylation of benzene takes place with the formation of
acetophenone. CH3CO+ is the electrophile.
anhy. AlCl3
+ CH3
C
O
Cl
H
+ HCl
CH3
C
O
Benzene Acetyl Chloride Acetophenone
13. How is benzophenone prepared by Friedel-Crafts reaction? [Mar-2016]
In presence of anhydrous Aluminium chloride benzoylation of benzene takes place with the formation
of benzophenone. Benzoyl cation (C6H5CO+) is the electrophile.
anhy. AlCl3
+ C
O
Cl
H
+ HCl
C
O
Benzene Benzoyl Chloride Benzophenone
C6H
5
C6H
5
14. Write Knoevenagal reaction [Sep-2016]
Refer – 5 mark – Q – 10 -b
5 MARK 1. Explain the mechanism of aldol condensation of acetaldehyde
[Sep-2006, Mar-2007, Jun-2007, Sep-2011, Mar-2012 , Mar-2014, Jun-2014, Jun-2015, Jun-2016]
This reaction is catalysed by base. The carbanion generated is nucleophilic in nature. Hence it can bring
about nucleophilic attack on carbonyl group.
Step 1: The carbanion is formed as the -hydrogen atom is removed as a proton by the base.
O
+HCH3 C
OH
O
HCH2 C H
2O
Carbanion
Step 2: The carbanion attacks the carbonyl carbon of another unionised aldehyde molecule.
O
H
CH3 C
O
HCH2 C+
O
H
CH3 C
O
HCH2 C
Alkoxide ion
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Step 3: The alkoxide ion formed is protonated by water to give ‘aldol’.
OH+
HO
H
CH3 C
O
HCH2 C
H
CH3 C
O
HCH2 C
OH
OH
Aldol
2. Explain the mechanism of aldol condensation of acetone [Sep-2008]
This reaction is catalysed by base. The carbanion generated is nucleophilic in nature. Hence it can bring
about nucleophilic attack on carbonyl group.
Step 1: The carbanion is formed as the -hydrogen atom is removed as a proton by the base.
O
+CH3 C
OH
O
CH2 C H
2O
Carbanion
CH3
CH3
Step 2: The carbanion attacks the carbonyl carbon of another unionised ketone molecule.
O
CH3 C
O
CH2 C+
O
CH3 C
O
CH2 C
Alkoxide ionCH
3
CH3
CH3
CH3
Step 3: The alkoxide ion formed is protonated by water
OH+
HO
CH3 C
O
CH2 C CH
3 C
O
CH2 C
OH
OH
CH3
CH3
CH3
CH3
3. Explain the mechanism of crossed aldol condensation [Jun-06, June-10, Jun-12, Sep-14, Mar-15]
In this reaction an aldehyde and a ketone undergo condensation. This reaction is catalysed by base. The
carbanion generated is nucleophilic in nature. Hence it can bring about nucleophilic attack on carbonyl
group.
Step 1: The carbanion is formed as the -hydrogen atom is removed as a proton by the base.
O
+HCH3 C
OH
O
HCH2 C H
2O
Carbanion
Step 2: The carbanion attacks the carbonyl carbon of unionised acetone molecule.
O
CH3 C
O
CH2 C+
O
CH3 C
O
CH2 C
Alkoxide ionCH
3CH
3
H H
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Step 3: The alkoxide ion formed is protonated by water to give ‘aldol’.
H
OH+
HO
CH3 C
O
CH2 C CH
3 C
O
CH2 C
OH
OH
CH3 CH
3
H
4. Explain the mechanism of Cannizzaro reaction [Mar-06, Jun-08, June-09, Mar-10, Jun-11, Sep-13]
Benzaldehyde undergoes Cannizaro reaction because of the absence -hydrogen. It involves self
oxidation and reduction of benzaldehyde when heated with concentrated NaOH.
C6H5CHO + C6H5CHO NaOH
C6H5COONa + C6H5CH2OH
Benzaldehyde Benzaldehyde Sodium benzoate Benzyl alcohol
The mechanism involves the transfer of hydride ion from one molecule of benzaldehyde to the other
molecule.
Step 1: Nucleophilic attack by OH– ion on carbonyl carbon.
C6H
5 C O
H
OH
C6H
5 C O
H
OH
Step 2: Transfer of hydride ion from the anion to carbonyl carbon of another molecule.
+C6H
5 C O
H
OH
C6H
5 C O
H
C6H
5 C O
OH
+ C6H
5 C O
H
H
Step 3: The benzyloxide ion picks up the acidic proton from benzoic acid to give benzyl
alcohol.
C6H
5 C O + C6H
5 C O
H
HO H
C6H
5 C O + C6H
5 C
H
HO
OH
5. Explain the mechanism of Claisen-Schmidt reaction [Sep-2007, Sep-2010]
Benzaldehyde reacts with aliphatic aldehydes or ketones in presence of NaOH forming α, β unsaturated
aldehyde or ketone.
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NaOH
-elimination
H2O
CH3
CHO CH2
CHO
C6H
5C
O
H
+ CH2
CHO C6H
5C
H
CH2
CHO
OH+
C6H
5C
H
CH2
CHO
OH
CHOC6H
5C
H
OH
CH
H
CHOC6H
5CH
Carbanion
NaOH
Nucleophilic attack
Cinnamaldehyde
CH
Similarly,
C6H
5-C = O
H
+ H2 CH-C-CH
3
NaOHC
6H
5-C =
H
Acetone
+ H2O
O
CH-C-CH3
O
The carbanion formed from acetaldehyde or acetone brings forth a nucleophilic attack on the carbonyl
carbon of benzaldehyde. The aldol type of product undergoes β-elimination (of water). This reaction
is known as Claisen Schmidt reaction.
6. Explain Popott’s rule with an example [Mar-2009, Jun-2016 (3 mark)]
Popott’s Rule :
During oxidation of unsymmetric ketones with oxidising agent which brings about the cleavage of C–C
bond, the smaller alkyl group goes preferentially with the carbonyl group resulting in the carboxylic acids.
CH3
C CH2
O
CH2
CH3
a b
'a' cleavage 'b' cleavage
HCOOH CH3-CH
2-CH
2-COOH CH
3-COOH CH
3-CH
2-COOH
Acetic acid Propionic acidn-butyric acidCO
2 + H
2O
Minor products Major products
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7. Write the differences between acetaldehyde and acetone [Mar-2008, Sep-2015]
Reaction Acetaldehyde Acetone
1. With Fehling’s solution. gives a red precipitate does not react.
2. With Tollen’s reagent gives silver mirror no silver mirror.
3. With Schiff’s reagent pink colour appears in cold no pink colour in cold
4. Oxidation gives acetic acid gives acetic acid with loss of
one carbon atom.
5. Polymerisation forms paraldehyde forms condensation
products
8. How will you distinguish between acetaldehyde and benzaldehyde? [Mar-2013, Sep-2016]
Reactions Acetaldehyde CH3CHO Benzaldehyde C6H5CHO
1. Heating with Fehling’s
solution.
gives a red precipitate
no reaction
2.With NaOH undergoes Aldol condensation undergoes Cannizzaro reaction
3. Polymerisation undergoes polymerisation does not polymerise.
4. Electrophilic substitution. does not undergo. undergoes at the meta position
5. With primary amines does not form Schiff’s base forms Schiff’s base
9. Write a note on i) Stephen’s reaction and ii) Perkin’s reaction [Sep-2009]
i) Stephen’s reaction Acetaldehyde can be prepared by Stephen’s reaction, by the reduction of methyl cyanide dissolved in
ether with Stannous chloride and hydrochloric acid.
CH3 C N
HH
SnCl2
HClCH
3CH NH HCl
CH3
CH NH HCl
O H2
HydrolysisCH
3CH NH
4ClO +
Methyl cyanide Iminimum hydrochloride
Acetaldehyde
ii) Perkins reaction: Refer – 3 mark – Q - 9
10. Write note on a) Clemmenson reduction b) Knoevenagal reaction [June-2013]
a) Clemmenson reduction: Refer – 3 mark – Q - 7
b) Knoevenagal reaction
Benzaldehyde condenses with malonic acid in presence of pyridine forming cinnamic acid, pyridine is
the basic catalyst here.
C6H
5-C = O
H
+ H2 C Pyridine
COOH
COOH
C6H
5-C =
H
COO H
COOH
C C6H
5-C =
H
CH-COOH
Cinnamic acid
- CO2
Malonic acid
11. How is acetone converted to i) Mesityl oxide and ii) Mesitylene? [Mar-2011]
i) Mesityl oxide
With dry hydrogen chloride acetone forms mesityl oxide
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CH3 C O
CH3
+ H2 CH C
O
CH3
Dry. HClCH
3 C
CH3
CH C
O
CH3
Mesityl oxide
(4-methylpent-3-ene-2-one)
ii) Mesitylene
In presence of con. H2SO4 three molecules of acetone condense to give mesitylene
(1, 3, 5 trimethyl benzene).
3 CH3-CO-CH
3
Con. H2SO
4
CH3
CH3CH
3
+ 3 H2O
Mesitylene
12. Illustrate the reducing properties of acetaldehyde with examples [Sep-2012]
1. Aldehydes reduce Tollen’s reagent ( ammonical silver nitrate ) to metallic silver
CH3CHO + Ag2O CH3COOH + 2Ag
2. Aldehydes reduce Fehling’s solution ( copper sulphate + sodium potassium tartrate ) to red cuprous oxide.
CH3CHO + 2CuO CH3COOH + Cu2O
(blue) (red precipitate)
3. Aliphatic aldehydes restore the original colour (red-pink) of the Schiff's reagent. (When SO2 is
passed through intensely pink coloured Schiff’s reagent in water, it forms a colourless solution. This
colourless solution is used for this test). This is characteristic test for aldehydes.
13. Explain i) benzoin condensation and ii) Knoevenagal reaction [Mar-2016]
i) benzoin condensation
When benzaldehyde is refluxed with aqueous alcoholic potassium cyanide -hydroxy ketone called
benzoin is formed. Cyanide ion (CN–) is the specific catalyst in this reaction.
Benzoin can be considered as dimer of benzaldehyde.
C6H
5-C
H
+
O
C6H
5 C
H
O
Alcoholic KCNC
6H
5-C
H
O
C6H
5 C
H
O
Benzoin
ii) Knoevenagal reaction Refer – 5 mark – Q – 10 -b
14. Explain i) Clemmenson’s reduction and ii) Perkins reaction [Mar-2017]
Refer-3 mark-Q-7, 9
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Additional questions 1 MARK
1. The chain isomer of 2-methyl propanal is
a) 2-butanone b) butanal c) 2-methyl propanol d) but-3-ene-2-ol
2. Isopropyl alcohol vapours with air over silver catalyst at 520 K give
a) tert.butyl alcohol b) acetaldehyde c) acetone d) 2-propanol
3. When acetaldehyde is heated with Fehling solution, it gives a precipitate of
a) Cu2O b) CuO c) CuO + Cu2O d) Cu
4. From which of the following, tertiary butyl alcohol is obtained by the action of methyl magnesium
iodide?
a) HCHO b) CH3CHO c) CH3COCH3 d) CO2
5. In the reduction of acetaldehyde using LiAlH4 the hydride ion acts as
a) electrophile b) nucleophile c) both (a) and (b) d) a free radical
6. Which of the following statement is wrong?
a) 2-pentanone and 3-pentanone are position isomers
b) aqueous solution of formaldehyde is known as formalin
c) aldehydes and ketones undergo nucleophilic substitution
d) aldehydes act as reducing agents
7. The IUPAC name of CH3
C
CH3
CH C
O
CH3 is
a) 4-methylpent-3-en-2-one b) 2-methylpent-3-en-2-one
c) 3-methyl pent-2-en-1-one d) None of these
8. Which compound on strong oxidation gives propionic acid?
a) CH3 – CH(OH) – CH3 b) CH3 – CO – CH3
c) (CH3)3COH d) CH3CH2CH2OH
9. Bakelite is a product of reaction between
a) formaldehyde and NaOH b) phenol and methanal
c) aniline and NaOH d) phenol and chloroform
10. Butanal and 2-methylpropanal exhibits --------- isomerism
a) chain b) position c) functional d) metamerism
11. Pentanal and 2,2-dimethylpropanal exhibits ------- isomerism
a) chain b) position c) functional d) metamerism
12. Acetone, propanaldehyde and 2-propen-1-ol exhibit ------- isomerism
a) chain b) position c) functional d) metamerism
13. Aldehydes are functional isomers of ---------
a) ketones b) unsaturated alcohols c) both a and b d) none of these
14. 2-pentanone and 3-methyl-2-butanone exhibits ------- isomerism
a) chain b) position c) functional d) metamerism
15. 3-pentanone and 2-pentanone exhibits ------- isomerism
a) position b) chain c) functional d) none of these
16. Dry distillation of calcium acetate gives --------
a) acetaldehyde b) acetone c) formaldehyde d) none
17. In Rosenmund reduction, acid chlorides are reduced to aldehydes in the presence of
a) Zn/HCl b) Pt/ZnSO4 c) Pd/BaSO4 d) N2H4/C2H5ONa
18. Formaldehyde cannot be prepared by Rosenmund reduction because
a) Formyl chloride is unstable at room temperature b) Formaldehyde is unstable
c) Formyl chloride cannot be reduced d) none of these
19. In Rosenmund reduction, the catalyst and catalytic poison respectively are
a) BaSO4, Pd b) Pd, BaSO4 c) Zn,Hg d) N2H4, C2H5ONa
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20. In Rosenmund reduction, BaSO4 is used as a catalytic poison to stop the reduction at the stage of
a) alcohol b) aldehyde c) carboxylic acid d) none of these
21. If BaSO4 is not added in Rosenmund reduction, then the aldehyde formed will be further reduced to
a) alkane b) carboxylic acid c) primary alcohol d) none of these
22. Ozonolysis of alkenes containing double bond at the terminal carbon give ------- as one of the products
a) formaldehyde b) acetaldehyde c) acetone d)none of these
23. Ozonolysis is used to
a) prepare aldehyde b) prepare ketone
c) locate the position of double bond d) all the above
24. The correct reactivity order towards nucleophilic addition reaction is
a) HCHO > CH3CHO > CH3COCH3 b) HCHO < CH3CHO < CH3COCH3
c) CH3CHO > HCHO> CH3COCH3 d) none of these
25. On heating aldol gives
a) Crotonaldehyde b) Acetaldehyde c) ethyl alcohol d) none of these
26. In medicine urotropine is used as
a) urinary antiseptic b) anaesthetic c) pain killer d) all of these
27. Which of the following compounds do not undergo aldol condensation?
a) HCHO b) C6H5CHO c) C6H5COC6H5 d) all of these
[Carbonyl compounds having α-hydrogen undergo aldol condensation
CH3CHO, CH3COCH3 and C6H5COCH3 have α-hydrogens
HCHO, C6H5CHO and C6H5COC6H5 do not have α-hydrogens]
28. Which of the following does not undergo iodoform (haloform) reaction?
a) CH3CHO b) CH3COCH3 c) C6H5COCH3 d) HCHO
[Aldehydes & ketones which contain CH3
C
O
group answer iodoform test
CH3CHO, CH3COCH3, CH3CH2COCH3, CH3CH2CH2COCH3 and C6H5COCH3 undergo iodoform
reaction. HCHO, C6H5CHO, C6H5COC6H5 and CH3CH2COCH2CH3 do not undergo iodoform reaction]
29. Aldehydes and ketones are reduced to hydrocarbons by
a) Zn/Hg and con.HCl b) N2H4 and C2H5ONa c) both a and b d) none of these
30. Which of the following reduces Tollen’s reagent and Fehling’s solution?
a) acetaldehyde b) acetone c) acetophenone d) all of these
[Aliphatic aldehydes (HCHO, CH3CHO) reduce Tollen’s reagent and Fehling’s solution and restore the
original red-pink colour of Schiff’s reagent where as ketones do not. Benzaldehyde reduces Tollen’s
reagent but it does not reduce Fehling’s solution]
31. Which of the following does not undergo polymerization?
a) HCHO b) CH3CHO c) C6H5CHO d) all of these
32. When aqueous solution of formaldehyde is evaporated to dryness it forms a white crystalline polymer called
a) paraldehyde b) paraformaldehyde c) bakelite d) none of these
33. When a drop of con.H2SO4 is added to acetaldehyde it polymerises to a cyclic compound
a) paraldehyde b) paraformaldehyde c) bakelite d) none of these
34. Paraldehyde is used as
a) hypnotic b) urinary antiseptic c) antipyretic d) none of these
35. 40% aqueous solution of formaldehyde is known as
a) para formaldehyde b) formic acid c) formalin d) paradehyde
36. Which is used for silvering of mirrors?
a) formaldehyde b) acetaldehyde c) acetone d) none of these
37. IUPAC name of benzaldehyde is
a) phenyl methanal b) phenyl carbinol c) phenyl methanol d) none of these
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38. Toluene is oxidized by air in the presence of V2O5 to give
a) benzaldehyde b) aniline c) benzyl alcohol d) peroxide
39. Which is known as oil of bitter almonds?
a) formaldehyde b) acetaldehyde c) benzaldehyde d) aniline
40. Benzaldehyde can be reduced to toluene by
a) Clemmenson reduction b) Wolff Kishner reduction c) both a and b d) none of these
41. Benzaldehyde reacts with primary amines to form ‘aldimine’ type of compound called
a) hydro benzamide b) phenol c) Schiffs base d) dye
42. Benzoin can be considered as dimer of
a) benzaldehyde b) benzoic acid c) phenol d) all of these
43. Benzoin prepared from benzaldehyde is used as ‘tincture benzoin’ in medicine for
a) throat infection b) hypnotic c) urinary antiseptic d) none of these
44. Which of the following undergoes Cannizzaro reaction?
a) benzaldehyde b) acetaldehyde c) acetone d) acetophenone
[Aldehydes with no α-hydrogen undergo Cannizzaro reaction HCHO and C6H5CHO do not have α-
hydrogen & undergo Cannizzaro reaction CH3CHO has α-hydrogen & does not undergo Cannizzaro
reaction]
45. Bezaldehyde condenses with two moleculesof N,N-dimethyl aniline forming triphenyl methane dye
called
a) ajobenzene b) Malachitegreen c) benzanilide d) benzoin
46. In presence of con. H2SO4 three molecules of acetone condense to give
a) phorone b) mesitylene c) mesityl oxide d) chloretone
47. Acetone is used in the preparation of
a) Tranquilizers like sulphonal, in medicine b) cordite c) both a and b d) none of these
48. Acetophenone is used as a hypnotic (sleep inducing) by name
a) urotropine b) hypnone c) acetaldoxime d) mesitylene
49. Diphenyl carbinol is also called
a) diphenyl methane b) malachite green c) benzhydrol d) benzoin
50. Which is used in perfumery?
a) benzaldehyde b) acetophenone c) benzophenone d) all of these
51. A new carbon to carbon linkage is set up in
a) aldol condensation b) cannizzaro reaction c) Stephen’s reaction d) all of these
3 MARK
1. Write a note on polymerization reaction formaidehyde
When aqueous solution of formaldehyde is evaporated to dryness it forms a white crystalline polymer
called paraformaldehyde.
C
H
On
H
C
H
O
H
n
2. Convert a] acetylene into acetaldehyde and b] propyne into acetone
a) Acetaldehyde is obtained by the hydration of acetylene in 42% sulphuric acid containing HgSO4 as a
catalyst.
CH CH
H2 O
H+
Hg2+CH
3-CHO
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b) Acetone is obtained by the hydration of propyne in 42% sulphuric acid containing HgSO4 as a
catalyst.
CH3-C CH
H2O
H+
Hg2+CH
3-CO-CH
3
3. How is paraldehyde prepared? Mention its use
Acetaldehyde polymerises to a cyclic structure called paraldehyde when a drop of concentrated
sulphuric acid is added to it.
3 CH3-CHO
Con. H2SO
4O O
CH-CH3
CH3-CH
CH3
CH
OParaldehyde
Paraldehyde is used as hypnotic.
4. Ho will you convert acetaldehyde into lactic acid?
O
H
CH3 C + HCN
O
H
CH3 C
H
CNH
2O / H+
O
H
CH3 C
H
COOH
Acetaldehyde Acetaldehyde cyanohydrin Lactic acid
5. From acetaldehyde prepare a] ethyl alcohol b] ethane
a) acetaldehyde to ethyl alcohol
CH3-CHO 2 [H]
LiAlH4
+ CH3-CH
2-OH
b) acetaldehyde to ethane Wolff-Kishner reduction
CH3 C
ON
2H
4
C2H
5ONa
CH3
CH3
Acetaldehyde Ethane
+ H2OH + N
2
6. Ethanal is more reactive towards nucleophilic addition reaction than propanone. Why?
Or Aldehydes are more reactive than ketones. Explain
Aldehydes are more reactive than ketones for both steric and electronic reasons.
The presence of alkyl group increases the crowding near carbonyl group and also increases the
electron density at the carbonyl carbon by Inductive effect. (+I effect)
C = O
Most reactive
H
H
C = O
CH3
HC = O
Less reactive Least reactive
CH3
CH3
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7. How is benzaldehyde prepared from toluene?
Toluene is oxidized by air in the presence of V2O5 at 773 K
C6H5CH3 + 2(O) 5O
2Vair/
C6H5CHO + H2O
8. From benzaldehyde prepare a] benzyl alcohol b] toluene and c] benzoic acid
a] benzaldehyde to benzyl alcohol
C6H
5-CHO 2 [H]
LiAlH4
+ C6H
5-CH
2-OH
b] benzaldehyde to toluene Clemmenson reduction
C6H
5 C
O
C6H
5CH
3HZn / Hg + Con. HCl
c] benzaldehyde to benzoic acid
Benzaldehyde is oxidized to benzoic acid by alkaline KMnO4
C6H
5-CHO [O]
KMnO4
+ C6H
5-COOH
9. How is hydro benzamide prepared?
Benzaldehyde undergoes condensation reaction with ammonia to form hydrobenzamide.
C OC6H
5
H
C OC6H
5
H
H2 N
+
H
H2 N H
CO C6H
5
H CC6H
5
H
CC6H
5
H
N
N
C C6H
5
H
Hydrobenzamide
10. How will you prepare Malachite green?
In presence of con. H2SO4 benzaldehyde condenses with two molecules of N,N-dimethyl aniline
forming triphenyl methane dye called Malachite green.
CH= O
H
H
N(CH3)2
N(CH3)2
Con. H2SO
4CH
N(CH3)2
N(CH3)2
Triphenyl methane dyeN,N-dimethyl aniline
11. Explain the reactions of benzaldehyde with Cl2
a] in the absence of catalyst b] in the presence of FeCl3 as Lewis acid catalyst
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a] in the absence of catalyst
CC6H
5
O
+H ClCl CC6H
5
O
Cl + HCl
Benzaldehyde Benzoyl chloride
b] in the presence of FeCl3 as Lewis acid catalyst
CHO
Cl2
CHO
Cl
m-chloro benzaldehyde
FeCl3
12. How is Schiff’s base formed from benzaldehyde? Benzaldehyde reacts with primary amines to form ‘aldimine’ type of compound called Schiffs base.
Benzaldehyde
CC6H
5
O
+
H
N
H
H
C6H
5CC
6H
5
O
H
N
H H
C6H
5CC
6H
5
H
N C6H
5
Aniline Schiffs base
-H2O
13. From acetone prepare a] isopropyl alcohol and b] propane
2 [H]LiAlH
4+CCH
3
O
CH3 CCH
3 CH3
OH
H
Acetone isopropyl alcohol
+ + H2OCCH
3
O
CH3 CH
3 CH3
Acetone Propane
CH2
Zn / Hg + Con. HCl4 [H]
14. From acetone prepare mesityl oxide and phorone
With dry hydrogen chloride first acetone forms mesityl oxide and then phorone.
CH3 C O
CH3
+ H2 CH C
O
CH3
Dry. HClCH
3 C
CH3
CH C
O
CH3
Mesityl oxide
(4-methylpent-3-ene-2-one)
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CH3 C
CH3
CH C
O
CH H2
CH3
C CH3O+ CH
3 C
CH3
CH C
O
CH
CH3
C CH3
Phorone
(2,6-dimethylhept-2,5-diene-4-one)
15. From acetophenone prepare a] phenyl methyl carbinol and b] ethyl benzene
CC6H
5
O
CH3
2 [H]LiAlH
4+ C CH
3H
Phenyl methyl carbinol
C6H
5
OH
Acetophenone
CC6H
5
O
CH3
4 [H]+ CH3
Ethyl benzene
C6H
5
Acetophenone
Zn / Hg + Con. HClCH
2+ H
2O
16. How will you prepare benzophenone from phosgene?
By the reaction of excess of benzene with carbonyl chloride (phosgene) in presence of anhydrous
aluminium chloride as a catalyst, Benzophenone is formed.
C6H6 + Cl-CO-Cl + C6H6 3AlClanhydrous
C6H5 -CO-C6H5
17. From benzophenone prepare a] diphenyl carbinol b] diphenyl methane and c] benzene
a] diphenyl carbinol
CC6H
5
O
2 [H]LiAlH
4+ CH
Diphenyl carbinol
C6H
5
OH
Benzo phenone
C6H
5 C6H
5
(Benzhydrol)
b] diphenyl methane
Zn / Hg + Con. HClCH
2CC6H
5
O
4 [H]+
Diphenyl methane
C6H
5
Benzo phenone
C6H
5 C6H
5
c] benzene
On fusion with potassium hydroxide, it undergoes disproportionation reaction.
CC6H
5
O
Potassium benzoate
C6H
6
Benzo phenone
C6H
5 + H-O-K + CC6H
5
O
O-K
Benzene
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18. Write the uses of formaldehyde
1. 40% aqueous solution of formaldehyde is known as formalin and is used as a preservative for
biological specimens and in leather tanning
2. Urotropine is used in medicine for urinary infection
3. To decolourise vat dyes
19. Write the uses of acetaldehyde
1. For silvering of mirror 2. Its trimer ‘paraldehyde’ is used in hypnotic
20. Write the uses of benzaldehyde
1. Benzoin prepared from benzaldehyde is used as ‘tincture benzoin’ in medicine for throat infection.
2. It is used for the preparation of triphenyl methane dyes.
3. It is used in perfumary.
21. Write the uses of acetone 1. In the preparation of Tranquilizers like sulphonal, in medicine.
2. In the manufacture of cordite.
22. Write the uses of acetophenone
1. Used as a hypnotic (sleep inducing) by name hypnone 2. In perfumary.
23. Write the uses of benzophenone 1. Benzophenone is used in perfumery
2. In the preparation of benzhydrol drop and diphenyl methane
5 MARK 1. How will you convert
a] formaldehyde into primary alcohol (ethyl alcohol)
b] acetaldehyde into secondary alcohol (iso propylalcohol)
c] acetone into tertiary alcohol (tertiary butyl alcohol)
H C
O
CH3-MgI
H OH
H + H C
O
H
MgI
CH3
HydrolysisCH
3-CH
2-OH + Mg
I
OH
FormaldehydeEthyl alcohol
( 1o )
C
O
CH3-MgI
H OH
H + C
O
H
MgI
CH3
Hydrolysis+ Mg
I
OH
Acetaldehydeisopropyl alcohol
( 2o )
CH3CH
3CH
3-CH-CH
3
OH
C
O
CH3-MgI
H OH
+ C
O MgI
CH3
Hydrolysis+ Mg
I
OH
Acetonetert.butyl alcohol
( 3o )
CH3CH
3CH
3-C-CH
3
OH
CH3
CH3
CH3
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2. Explain isomerism in aldehydes
Aldehydes exhibit (i) chain isomerism and (ii) functional isomerism.
i) chain isomerism:
Chain Isomerism arises due to changes in carbon chain.
CH3
CH2
CH2
CHO CH3 CHOCHand
Butanal
CH3
2-methylpropanal
ii) Functional isomerism:
Aldehydes are functional isomers of ketones and unsaturated alcohols.
CH3
CH2
CHO
Propanaldehyde Acetone 2-propen-1-ol
CH3
CH3C
O
CH2
CH CH2
OH
3. Explain isomerism in ketones
Ketones exhibit i) chain isomerism ii) functional isomerism and iii) Positional Isomerism
i) Chain isomerism:
CH3C
O
CH2
CH2
CH3
CH3C
O
CH3
and CH
CH3
2-pentanone 3-methyl-2-butanone
ii) Functional isomerism:
Ketones are functional isomers of aldehydes and unsaturated alcohols.
CH3
CH2
CHO
Propanaldehyde Acetone 2-propen-1-ol
CH3
CH3C
O
CH2
CH CH2
OH
iii) Positional Isomerism:
The carbonyl group may occupy different positions in the carbon chain to give Positional isomers.
C
O
CH2CH
2CH
3
3-pentanone
CH3C
O
CH2
CH2
CH3
2-pentanone
CH3
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4. How will you prepare the following by dry distillation of calcium salt of fatty acids?
a] HCHO b] CH3CHO c] C6H5CHO
a] HCHO
Calcium formate
Formaldehyde
Dry distillation+ CaCO
3
C
O
O
C
O
OCa C
O
H
H
H H
b] CH3CHO
Calcium acetate
Dry distillation
Calcium formate
C
O
O
C
O
OCa
CH3
Ca+
O
O C
C
O
O
+ 2 CaCO3
C
O
2
Acetaldehyde
H
H
H
CH3
CH3
c] C6H5CHO
Calcium benzoate
Dry distillation
Calcium formate
C
O
O
C
O
OCa
C6H
5
C6H
5
Ca+
O
O C
C
O
O
+ 2 CaCO3
C
O
C6H
52
Benzaldehyde
H
H
H
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5. How will you prepare the following by dry distillation of calcium salt of fatty acids?
a] CH3COCH3 b] C6H5COCH3 c] C6H5COC6H5
a] CH3COCH3
Dry distillation+ CaCO
3
Calcium acetate
Acetone
CH3
C
O
O
C
O
CH3
OCa C
O
CH3
CH3
b] C6H5COCH3
Calcium benzoate
Dry distillation
Calcium acetate
C
O
O
C
O
OCa
C6H
5
C6H
5
CH3
Ca+
O
O C
C
O
O
CH3
+ 2 CaCO3
C
O
C6H
5CH
32
Acetophenone
c] C6H5COC6H5
Calcium benzoate
Benzophenone
Dry distillation+ CaCO
3
C
O
O
C
O
OCa C
OC
6H
5
C6H
5
C6H
5 C6H
5
6. How will you distinguish between formaldehyde and acetaldehyde
Reactions Formaldehyde HCHO Acetaldehyde CH3CHO
1. Iodoform test does not undergo gives iodoform
2. With alkaline solution
(NaOH)
undergoes Cannizzaro
reaction
undergoes aldol
condensation.
3. With phenol forms thermosetting
plastic-bakelite does not form resin
4. With Ammonia forms ‘urotropine’ gives simple addition
product
5. With RMgX forms primary alcohol forms secondary alcohol
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19. CARBOXYLIC ACIDS
Blue print 1 Mark = 1 3 Mark = 1 5 Mark = 2 Total marks = 14
Govt. Examination questions 1 MARK
1. Which of the following is least acidic [Mar-2009]
a) C2H5OH b) CH3COOH c) C6H5OH d) ClCH2COOH
2. The acid which reduces Tollen's reagent is [Jun-2010, Sep-2011, Sep-2016]
a) acetic acid b) benzoic acid c) formicacid d)oxalicacid
3. The Isomerism exhibited by CH3CH2COOH and CH3COOCH3 is [Sep-2006, Mar-2010, June-2011]
a) metamerism b) position c) chain d) functional
4. The acid that cannot be prepared by Grignard reagent is
[Mar-2007, Sep-2008, Sep-2009, Sep-2010, Sep-2012, Sep-2013, Mar-2015]
a) acetic acid b) formic acid c) butyric acid d) benzoic acid
5. Which order of arrangement is correct in terms of the strength of the acid [Sep-2007]
a) CH3–CH2COOH > CH3COOH < HCOOH < ClCH2COOH
b) ClCH2COOH < HCOOH < CH3COOH < CH3CH2COOH
c) CH3–CH2COOH < CH3COOH < HCOOH < ClCH2COOH
d) HCOOH > CH3CH2COOH < CH3COOH > ClCH2COOH
6. Among the following the strongest acid is [Jun-2007, Jun-2012, Jun-2016]
a) ClCH2COOH b) Cl3CCOOH c) CH3COOH d) Cl2CHCOOH
7. Which of the following compound is optically active? [Mar-2012, Sep-2014, Mar-17]
a) CH3CH2COOH b) HOOC–CH2–COOH c) CH3CH(OH)COOH d) Cl2CHCOOH
8. CH3CH(OH)COOH 2Fe/
2O
2H
? The product is [Mar-2006]
a) CH3COCOOH b) CH3CH2COOH c) CH3CHOHCHO d) COOHCH2COOH
9. The compound found in some stony deposit in kidneys is
[June-2006, Mar-2011, June-2014, Sep-2015, Mar-2016]
a) potassium oxalate b) oxalic acid c) potassium succinate d) calcium oxalate
10. Ethylene dicyanide on hydrolysis using acid gives [June-2008, June-2009]
a) oxalic acid b) succinic acid c) adipic acid d) propionic acid
11. Concentrated solution of sodium acetate on electrolysis gives [Mar-2008]
a) ethane b) propane c) methane d) butane
12. Among the following the strongest acid is [Mar-2013]
a) CH3COOH b) HCOOH c) ClCH2COOH d) CH3CH2COOH
13. Aspirin is [June-2013]
a) Salicylic acid b) Acetyl salicylic acid c) Salicylaldehyde d) Methyl salicylate
14. The order of reactivity of carboxylic acid derivatives is [Mar-2014]
a) Acid chloride > Ester > Amide > Acid anhydride
b) Acid chloride > Acid anhydride > Ester > Amide
c) Acid chloride > Amide > Acid anhydride > Ester
d) Acid anhydride > Ester >Amide > Acid chloride
160 0CThe product is?C = O
H
HO
15.
[June-2015]
a) CO + H2O b) HCOOH c) H2 + CO2 d) HCHO + O2
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3 MARK
1. Give the source and trivial names of i) HCOOH ii) CH3COOH iii) C3H7COOH
iv) C4H9COOH v) C11H23COOH [June-2007]
Formula Source Trivial name
HCOOH Red ant Formic acid
CH3COOH vinegar Acetic acid
C3H7COOH Butter Butyric acid
C4H9COOH Root of valarion plant Valeric acid
C11H23COOH Laurel oil Lauric acid
2. Give the IUPAC name of the following [June-2014]
COOH
COOHCH3COOHa) b) c) (CH2)4
COOH
COOH
Ethanoic acidEthane dioic acid Hexane dioic acid
3. Write a note on esterification reaction with an example [June-2006]
Carboxylic acids react with alcohols in the presence of mineral acid as catalyst and form esters. This
reaction is called esterification.
CH3 CO OH + C2H5H OH
+CH3
COOC2H5 + H2O
acetic acid ethyl alcohol ethyl acetate
4. Write the tests for carboxylic acid [Jun-2009, Sep-2012, Mar-2013, Jun-2015, June-2016]
1. Aqueous solution of carboxylic acids turn blue litmus into red colour.
2. Carboxylic acids give brisk effervescence with sodium bi-carbonate due to the evolution of carbon-
di-oxide.
3. On warming carboxylic acid with alcohol and concentrated sulphuric acid it forms ester which is
identified from its fruity odour.
5. Formic acid reduces Tollen's reagent, but acetic acid does not-Give reason
[Sep-2007, Sep-2008, Sep-2009, Sep-2010, June-2012, Sep-2015]
Formic acid is unique because it contains both an aldehyde group and carboxyl group. Hence it can act
as a reducing agent. It reduces Tollens reagent and Fehling’s solution.
H C OH
O
H C OH
O
aldehyde group carboxlic acid group
Since acetic acid does not contain aldehyde group, it does not reduce Tollen's reagent.
6. Write a note on HVZ reaction [Sep-2006, Sep-2014]
Refer – 5 mark – Q – 27 a
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7. What is the action of dilute sulphuric acid with lactic acid? [Mar-2008]
Refer – 5 mark – Q – 12 c
8. Write a note on trans esterification [Mar-2006, Mar-2014, Sep-2016]
Refer – 5 mark – Q – 27 b
9. What happens when lactic acid is treated with PCl5? Write the equation [Mar-2012]
Refer – 5 mark – Q – 12 d
10. Write the tests for salicylic acid [Mar-2010]
1. An aqueous solution of salicylic acid gives violet colour with neutral ferric chloride.
2. It gives effervescence with sodium bicarbonate.
3. It is soluble in sodium hydroxide and reprecipitated on acidification.
4. With Bromine water the colour is discharged with the formation of white precipitate.
11. Give the structures of lactyl chloride and lactide [Mar-2015]
O=C
CHCH3
O
O
C=O
CH CH3
Lactide
CH3 C COClH
ClLactyl chloride
12. How is methyl cyanide obtained from acetamide? [Mar-2017]
P2O5CH3CN
Methyl cyanide
CH3CO NH2 - H2O
13. How is methyl salicylate prepared? [June-2008]
On heating salicylic acid with methyl alcohol in presence of con.H2SO4 methyl salicylate is formed.
+
O H
COOCH3
H2O
Methyl salicylate
H2SO4
O H
CO OH
+ H OCH3
conc.
Methyl salicylate is present in the oil of winter green.
14. What is aspirin? Write its preparation and use [Mar-2011]
Acetyl salicylic acid is known as aspirin
Salicylic acid undergoes acetylation by heating with acetic anhydride to form aspirin which is used as an
analgesic and antipyretic.
O
COOH
+
C
O
CH3
Acetyl salicylic acid or Aspirin
O H
COOH
CH3 C
O
O C
O
CH3+
CH3COOH
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15. Compare the strength of mono, di, trichloro acetic acid [Sep-2013]
Cl is electron withdrawing group (–I effect). So, it decreases the strength of the –O–H bond and hence
the release of hydrogen becomes easy. Three chlorine atoms in trichloro acetic acid withdraw the
electrons more powerfully making the acid very strong. Thus the strength of mono, di, trichloro acetic
acid varies in the order,
CCl3COOH > CHCl2COOH > CH2ClCOOH
16. Write the uses of benzoic acid [June-2010]
1. Benzoic acid is used as an urinary antiseptic
2. Sodium benzoate is used as food preservative
3. Benzoic acid vapours are used to disinfect bronchial tube.
4. It is used for the manufacture of dyes.
17. Mention the uses of oxalic acid [Mar-2007, Mar-2009, June-2011, Sep-2011]
1. for removing ink stains and iron stains.
2. as mordant in dyeing and calico printing.
3. in manufacture of ink and metal polishes.
4. Redox titration
18. Write the uses of lactic acid [June-2013]
1. In tanning industry.
2. In soft drinks.
3. In the treatment of digestive disorder in children.
4. Silver lactate as an antiseptic and astringent.
19. Write the uses of formic acid [Mar-2016]
1. In Textile Industry for preparing mordants
2. In leather tanning for removing lime from the hides.
3. In coagulating rubber latex.
4. As an antiseptic and in preservation of fruits.
5. Nickel formate as hydrogenation catalyst
6. Stimulant for the growth of yeast
7. Treatment of gout
5 MARK
1. Give the mechanism involved in the esterification of a carboxylic acid with alcohol.
[Mar-2006, June-2007, Mar-2008, Sep-2009, Mar-2011, Sep-2012, Mar-2013, Sep-2015, Mar-17]
Step 1. Protonation of carboxylic acid
H+
O
HCH3 C O
O
CH3 C O+
H
H
Step 2. Attack by nucleophile.
O
+H
H
..
CH3 C O
C2H
5O H
O
+H
H
CH3 C O
C2H
5O H
+
CH3 C O C2H
5
O
+ H2O + H+
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2. Write the mechanism of bromination of salicylic acid [Sep-07, Sep-08, Jun-13, Mar-16, Sep-16]
OH
OH
OH
Br
OH
BrBr
Br
C
O
O
C
O
O
C
O
O
BrBr
O+..
Br Br
H
+
Br
+OH
Br
2Br2
+ CO2
3. Write the mechanism of Kolbe’s reaction [Mar-2009, Mar-2012]
In Kolbe’s reaction sodium phenoxide is heated with CO2 at 403 K under pressure to form sodium
salicylate. This on treatment with dil.HCl gives salicylic acid.
OH ONa
OH
COONaOH
COOH
H
C
O
O
OO
NaOH
+
HCl+ NaCl
OH
C
O
OC
O
O
4. Explain the isomerism exhibited by carboxylic acids [Sep-2006]
1. Chain isomerism
This arises due to the difference in the carbon chain of alkyl group attached to carboxyl group.
CH3 CH
2 CH
2 CH
2 COOH CH
3 CH CH
2 COOH
CH3
Pentanoic acid 3-methyl butanoic acid
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2. Functional isomerism
Carboxylic acids are functional isomers of esters.
CH3– CH2– COOH CH3– COOCH3 H–COOC2H5
propanoic acid methyl acetate ethyl formate
5. Account for reducing nature of Formic acid. [June-2006, June-2009, Mar-2010, Mar-2013]
Formic acid is unique because it contains both an aldehyde group and carboxyl group. Hence it can act
as a reducing agent. It reduces Tollens reagent and Fehling’s solution.
H C OH
O
H C OH
O
aldehyde group carboxlic acid group
1. Formic acid reduces ammoniacal silver nitrate solution (Tollen’s reagent) to metallic silver.
H-COOH + Ag2O CO2 + H2O + 2Ag ↓ (Metallic silver)
2. Formic acid reduces Fehling’s solution. It reduces blue coloured cupric ions to red coloured cuprous
ions.
H-COOH + 2CuO CO2 + H2O + Cu2O
Blue Red
3. Formic acid decolourises pink coloured KMnO4 solution.
6. How do you distinguish formic acid from acetic acid? [Mar-2007, June-2008, June-2015]
No. Formic acid (HCOOH) Acetic acid (CH3COOH)
1 It contains both aldehyde group and
carboxylic acid group
It contains only carboxylic acid group
2 It reduces Tollen’s reagent & Fehling’s
solution
It does not reduce Tollen’s reagent &
Fehling’s solution
3 On dehydration with con.H2SO4 it
gives CO & H2O
It undergoes inter molecular dehydration with
P2O5 to give acetic anhydride
4 It does not contain α-hydrogen and
hence it does not react with Br2/P
It contains α-hydrogen and hence it reacts
with Br2/P to give bromo acetic acid
5 With PCl5 it gives formyl chloride.
Formyl chloride being unstable
decomposes to CO & HCl
With PCl5 it gives acetyl chloride.
7. How is benzoic acid obtained from the following? [Sep-2006, Sep-2009, Mar-2014]
a) From methyl benzene (Toluene)
acidified KMnO4C
6H
5CH
3 + ( O ) C
6H
5COOH
b) From ethyl benzene
acidified KMnO4C
6H
5CH
2CH
3 + ( O ) C
6H
5COOH
c) From phenyl cyanide
C6H5CN
H/OH2 C6H5CONH2
H/OH2 C6H5COOH
Benzamide
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d) From CO2 & Grignard reagent
C6H5MgBr + O=C=O
C6H5
O MgBr MgBr
O=CH-O-H
+
OHC
6H
5COOH
e) From benzyl chloride
C6H5CH2ClNaOH
C6H5CH2OH(O)
KMnO4
C6H5COOH
f) From benzaldehyde
C6H5CHOKMnO4/H
+
C6H5COOH
8. How is oxalic acid manufactured from sodium formate? [Sep-2007, Sep-2010, Mar-2014, Sep-2015]
Oxalic acid is made industrially by heating sodium formate
HCOONa
HCOONa
COONa
COONaH2++
673 K
Sodium formate Sodium oxalate
The sodium oxalate thus formed is dissolved in water and calcium hydroxide is added to precipitate
calcium oxalate. The solution is filtered and the precipitate is treated with calculated quantity of dilute
sulphuric acid to liberate the oxalic acid.
Sodium oxalate
COONa
COONa
+ Ca(OH)2
COO
COO
Ca + 2NaOH
COO
COO
Ca + H2SO4
COOH
COOH+ CaSO4
Oxalic acid
Calcium oxalate
Calcium sulphate is precipitates and oxalic acid is crystallised as dihydrate (COOH)2.2H2O.
9. How is lactic acid manufactured in large scale (fermentation method) ? How can it be converted
into cyclic diester? [Mar-2006, Jun-2014, Sep-2014, Jun-2016]
To a dilute solution of cane sugar a little of sour milk is added. Temperature is maintained at 40 – 45o C
for six days. The Bacillus acidi lacti brings forth fermentation. Methyl glyoxal forms as intermediate
compound. Acid is removed by the addition of CaCO3 which precipitates calcium lactate. It is filtered
and decomposed with dilute sulphuric acid. The filterate is distilled under reduced pressure.
C12H22O11 2H O
C6H12O6 + C6H12O6
Cane sugar glucose fructose
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C6H12O6
BAL2 CH3
C COOH
OH
C
O
CHO 2 CH3
HMethyl glyoxal Lactic acid
Lactic acid to lactide: Refer-Q-12
10. How is lactic acid synthesized from acetylene? How can it be converted into cyclic diester [Jun-11]
Acetylene is prepared by striking an electric arc using carbon electrodes in an atomsphere of hydrogen.
This is passed through dilute sulphuric acid containing mercuric ion catalyst. Acetaldehyde is formed. It
is converted to cyanohydrin on treatment with HCN, which is then hydrolysed to get lactic acid.
2C + H2
electric arc CH CH
CH CHdil.H
2SO
4
Hg2+catalystH2
O
CH3C H
OHCN
CH3C H
OH
CN
dil. HCl
H2OC H
OH
CH3
COOH
acetylene
acetaldehydeacetaldehyde
cyanohydrin
Lactic acid
Cyclic diester: Refer -5 mark-Q-12-e
11. How can salicylic acid be converted into
a) Aspirin b) Methyl salicylate c) 2,4,6-tribromophenol [June-2006, Sep-2011]
Or
How does salicylic acid react with the following ?
a) (CH3CO)2O b) CH3OH c) Br2 / H2O [Sep-2014]
a) Aspirin : Refer – 3 mark – Q – 14
b) Methyl salicylate: Refer – 3 mark – Q – 13
c) 2,4,6-tribromophenol
OH
BrBr
Br
3Br2
OH
COOH
OH
Br2
- CO2
2,4,6-tribromo phenol
/ H2O / H
2O
(white precipitate)
12. Give the reactions of lactic acid with the following [Sep-2013]
a) Fenton's reagent b) dilute acidified KMnO4 c) dilute H2SO4 d) PCl5 e) Heat
a) Fenton's reagent
Mild oxidising agent like Fenton’s reagent Fe2+
/H2O2 forms pyruvic acid with lactic acid
H2O2 / Fe2+
CH3 C
O
+H COOH (O)
H
CH3 C
O
COOH
Pyruvic acid
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b) dilute acidified KMnO4
With dilute acidified permanganate, lactic acid decomposes to form acetaldehyde.
KMnO4 / H+
CH3CHO + CO2 + H2OCHCH3
O H
COO H + (O)
c) dilute H2SO4
With dil. H2SO4 acid lactic acid dissociates to acetaldehyde and formic acid
dil.H2SO4CHCH3
O H
COOH CH3CHO + HCOOH
d) PCl5
CH3 C
O
H COOH
H
CH3 C COClPCl5 H
Cl
Lactyl chloride
e) Heat with conc.H2SO4
When lactic acid is heated with con.H2SO4 it gives cyclic diester called lactide
CH
CH3O H
OH
HO
CH
CH3H O
+con.H2SO4
Lactide
+ 2H2O
C
O
CO CH
CH3
CH
CH3
C
O
CO
O
O
13. What happens when lactic acid is [June-2010]
a) treated with dilute H2SO4 b) oxidized with Fenton’s reagent c) added to PCl5
(Refer-5 mark-Q-12)
14. How are the following conversions carried out? [Jun-2012, Sep-2013, Mar-17]
a) Salicylic acid Aspirin (Refer – 3 mark – Q – 14)
b) Salicylic acid Methyl salicylate (Refer – 3 mark – Q – 13)
c) Acetamide Methyl amine (Refer -5 mark-Q-32-c)
d) Formic acid Formamide
HCOOH + NH3 HCOONH4 HCONH2 + H2O
Ammonium formate Formamide
15. How are the following conversions carried out? [Mar-2011]
a) Lactic acid Lactide (Refer -5 mark-Q-12)
b) Salicylic acid Methyl salicylate (Refer – 3 mark – Q – 13)
16. How are the following conversions carried out? [June-2010]
a) Salicylic acid Aspirin (Refer – 3 mark – Q – 14)
b) Salicylic acid Methyl salicylate (Refer – 3 mark – Q – 13)
c) Lactic acid Lactide (Refer-5 mark-Q-12)
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17. How are the following conversions carried out? [Mar-2009]
a) Salicylic acid Aspirin (Refer – 3 mark – Q – 14)
b) Methyl acetate Ethyl acetate (Refer -5 mark-Q-27 - b)
c) Lactic acid Pyruvic acid (Refer-5 mark-Q-12)
18. What happens when lactic acid is [Mar-2007, June-2009]
a) treated with dilute H2SO4 (Refer-5 mark- Q-12)
b) heated (Refer -5 mark-Q-12)
c) oxidised with dilute acidified KMnO4 (Refer -5 mark-Q-12)
19. How are the following conversions carried out? [June-2013]
a) Lactic acid to Pyruvic acid (Refer -5 mark-Q-12)
b) Methyl acetate to Ethyl acetate (Refer -5 mark-Q-27-b)
c) Oxalic acid to Oxamide
COOH 2NH3 2 H2O
COOH
+
COONH4
COONH4
+
CONH2
CONH2
20. How are the following conversions carried out? [June-2007]
a) Lactic acid to lactide (Refer Q-12)
b) Salicylic acid to aspirin (Refer – 3 mark – Q – 14)
c) Succinic acid to succinimide
CH2COOH
2NH3
CH2COOH
+
CH2COONH4
CH2COONH4
CH2CO
CH2CO
NH
Succinic acid Ammonium succinate Succinimide
21. How are the following conversions carried out? [Mar-2008]
a) salicylic acid to methyl salicylate (Refer – 3 mark – Q – 13)
b) Lactic acid to pyruvic acid (Refer Q-12)
c) Methyl cyanide to acetamide
Partial hydrolysis of methyl cyanide with alkaline H2O2 gives acetamide
CH3 CNH2O2
NaOHCH3CONH2
22. Bring out the following conversions [Sep-2008]
a) Salicylic acid to aspirin (Refer – 3 mark – Q – 14)
b) Lactic acid to lactide (Refer-5 mark- Q-12)
c) benzoic acid to benzyl alcohol
LiAlH4C6H5 C O H
O
[C6H5CHO]LiAlH4
C6H5CH2OH
Benzyl alcohol
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23. What happens when benzoic acid reacts with [June-2011]
a) conc.HNO3 / conc.H2SO4 b) Cl2 / FeCl3 c) PCl5 d) sodalime
a) conc.HNO3 / conc.H2SO4
COOHCOOH
NO2
HNO3
H2SO4
m-nitro benzoic acid b) Cl2 / FeCl3
COOHCOOH
Cl
Cl2
FeCl3
m-chloro benzoic acid c) PCl5
C6H
5-CO O H
Cl PCl3 Cl
C6H
5-COCl + POCl
3 + HCl
Benzoyl chloride
d) Sodalime
C6H
5COOH C
6H
6 + CO
2
NaOH / CaO
SodalimeBenzene
24. How are the following compounds obtained from benzoic acid? [Mar-2015]
a) Ethyl benzoate b) Benzyl alcohol c) Benzene
a) Ethyl benzoate
H+
C6H5COOC2H5 + H2O
ethyl benzoate
C6H5CO OH + H OC2H5
b) Benzyl alcohol Refer –-5 mark-Q – 22 (c)
c) Benzene Refer –-5 mark-Q – 23 (d)
25. What happens when [June-2008]
a) oxalic acid is treated with ammonia (Refer -5 mark-Q-19-c)
b) benzoic acid is treated with PCl5 (Refer-5 mark-Q-23 - c)
26. Give the equation for the action of heat on
a) Formic acid b) Oxalic acid c) Succinic acid [Sep-2011, Mar-2016]
a) Formic acid
H - COO H160 0C
H2 + CO
2
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b) Oxalic acid
473 KCOO H
COOH
HCOOH + CO2
Formic acid
c) Succinic acid
CH2-CO
CH2-CO
+ H2O
succinic anhydride
300 0C
CH2-COO H
CH2-CO OH
O
(a cyclic compound)
27. Write short notes on the following
a) HVZ – reaction b) Trans – esterification
c) Kolbe’s electrolytic reaction d) decarboxylation [Mar-2010, Mar-2012, Jun-2014]
a) HVZ – reaction
Carboxylic acids having α-hydrogen atoms can be converted to α-halo acids by halogen and
phosphorus trihalide. This reaction is known as Hell-Volhard Zelinsky reaction.
R OHBr
2 PBr3
C
H
H
C
O
RBr
2 PBr3
C
H
H
C
O
Br R C
H
C
O
Br
BrH OH
R C
H
C
OBr
OH
b) Trans – esterification
In presence of a little acid, methyl acetate is cleaved by ethyl alcohol to form ethyl acetate. This is
called ‘trans esterification’.
Methyl acetate Ethyl alcohol Ethyl acetate Methyl alcohol
CH3-COO CH
3 + HO - C
2H
5 CH
3-COOC
2H
5 + CH
3OH
H+
c) Kolbe’s electrolytic reaction
Electrolysis of concentrated aqueous solution of sodium acetate gives ethane.
CH3 COO Na2CO2 + 2Na
CH3 COO Na
ElectrolysisCH3
CH3
+
sodium acetate ethane
d) decarboxylation
When anhydrous sodium acetate is heated with sodalime, carboxyl group is removed with the
formation of methane.
CH3COONa NaOH/CaO
CH4 + Na2CO3
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28. Explain the following reactions [June-2012]
a) Friedel-craft’s acetylation b) Trans - esterification
a) Friedel-craft’s acetylation
C
anhydrous AlCl3
O
CH3
+ HCl
Acetophenone
H
O
+ CH3CCl
Benzene Acetyl chloride
b) Trans – esterification (Refer-5 mark- Q-27 - b)
29. Explain the following [Sep-2012]
a) HVZ – reaction (Refer-5 mark- Q-27 - a)
b) Claisen ester condensation
In presence of strong bases like sodium ethoxide, it undergoes condensation forming aceto acetic ester.
C
O
O CH3
C2H5ONaC
O
O CH3CH3CH2 + H+
C
O
O CH3CH3
C
O
O CH3CH2
C
O
CH3C
O
O CH3CH2+ OCH3
Methyl acetoacetate
30. Explain the following reactions [Mar-2015]
a) HVZ reaction (Refer -5 mark- Q-27-a)
b) Trans esterification (Refer -5 mark- Q-27-b)
c) Methyl salicylate formation Refer – 3 mark – Q – 13
31. Write short notes on the following [Jun-2016]
a) HVZ – reaction b) Kolbe’s electrolytic reaction c) Friedel-Crafts acylation
a) HVZ – reaction (Refer -5 mark-Q-27-a)
b) Kolbe’s electrolytic reaction (Refer -5 mark- Q-27-c)
c) Friedel-Crafts acylation (Refer -5 mark-Q-28-a)
32. Explain the reactions of acetamide with the following [Sep-2010]
a) Hydrolysis by an acid
H+
CH3COOH + NH3
Acetic acid
CH3CO NH2 + H OH
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b) P2O5
P2O5CH3CN
Methyl cyanide
CH3CO NH2 - H2O
c) Br2 / NaOH (Hoffmann’s reaction)
Br2CH3NH2
Methyl amine
CH3CONH2 NaOH+ CO2
Acetamide
33. How is lactic acid synthesized from acetylene? Explain the reaction of lactic acid with con. H2SO4
[June-2015]
Refer - Q-10 & 12-e
34. How do succinic acid reacts with the following? [Sep-2016] a) NaOH b) NH3 c) PCl5
a) NaOH
CH2COOH
CH2COOH
+
Succinic acid
NaOH
CH2COOH
CH2COONa
NaOHCH
2COONa
CH2COONa
mono sodium succinate disodium succinate
b) NH3
CH2COOH
2NH3
CH2COOH
+
CH2COONH4
CH2COONH4
CH2CO
CH2CO
NH
Succinic acid Ammonium succinate Succinimide c) PCl5
PCl5
CH2COOH
CH2COOH
+
CH2COCl
CH2COCl
POCl3 + H2O+
succinoyl chloride
Additional questions
1 MARK
1. Weakest acid among the following is
a) Acetic acid b) Phenol c) Water d) Acetylene
2. Ester formation involves the reaction of
a) an aldehyde and a ketone b) An alcohol with RMgX
c) Two molecules of an acid with dehydrating agent d) An acylhalide with an alcohol
3. Heating a mixture of sodium acetate and soda lime gives
a) methane b) ethane c) aceticacid d) benzene
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4. The IUPAC name of CH3 CH
2 CH COOH
CH3
is
a) -methyl butric acid b) 3-methyl butanoic acid c) 2-methyl butanoic acid d) Iso pentanoic acid
5. The compound which undergoes intramolecular dehydration with P2O5 is
a) acetic acid b) formic acid c) propionic acid d) Butyric acid
7. When chlorine is passed through acetic acid in presence of red P, it forms.
a) acetyl chloride b) Trichloro acetaldehyde c) Trichloro acetic acid d) Methyl chloride
8. Which of the following compounds will react with NaHCO3 solution to give sodium salt and CO2?
a) acetic acid b) n-hexanol c) phenol d) both (a) and (c)
9. When propanoic acid is treated with aqueous sodium-bicarbonatate, CO2 is liberated. The "C" of CO2
comes from
a) methyl group b) carboxylic acid group c) methylene group d) bicarbonate
10. Carboxylic acids are more acidic than phenol and alcohol because of
a) inter molecular hydrogen bonding b) formation of dimers
c) highly acidic hydrogen d) greater resonance stabilisation of their conjugate base
11. Pentanoic acid and 3-methyl butanoic acid exhibit --------- isomerism
a) functional b) chain c) position d) metamerism
12. Propanoic acid and methyl acetate exhibit --------- isomerism
a) chain b) functional c) position d) metamerism
13. Propanoic acid and ethyl formate exhibit --------- isomerism
a) chain b) functional c) position d) metamerism
14. The functional isomers of propanoic acid are
a) ethyl formate b) methyl acetate c) both a and b d) none of these
15. Carboxylic acids are the functional isomers of ------
a) esters b) aldehydes c) ethers d) alcohols
16. Formic acid cannot be prepared from Grignard reagent because
a) formic acid is unstable b) it has α-hydrogen
c) formic acid contains only one carbon atom d) none of these
17. Formic acid reacts with PCl5 to give
a) CO2 + HCl b) CO + HCl c) CO2 + PCl3 d) none
18. The reagent used for decarboxylation reaction is
a) soda lime [NaOH + CaO] b) Pt/ZnSO4 c) Pd/BaSO4 d) N2H4/C2H5ONa
19. The carboxylic acid which contains both aldehyde group and carboxyl group and reduces Tollen’s
reagent and Fehling’s solution is
a) acetic acid b) benzoic acid c) formic acid d) none of these
20. Which of the following decolourise pink coloured KMnO4?
a) CH3COOH b) HCOOH c) CH3OH d) none of these
21. The main constituent of sour milk is
a) oxalic acid b) lauryl alcohol c) lactic acid d) none of these
22. The fermentation of cane sugar or glucose or maltose into lactic acid is brought by
a) Bacillus acidi lacti b) invertase c) zymase d) none of these
23. Which is hygroscopic?
a) formic acid b) acetic acid c) lactic acid d) all the above
24. Which of the following undergoes haloform reaction?
a) formic acid b) acetic acid c) lactic acid d) all the above
25. Which of the following is optically active?
a) formic acid b) acetic acid c) lactic acid d) all the above
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26. Oxalic acid occurs as ---------- in the wood sorrel and tomatoes
a) sodium oxalate b) oxalyl chloride c) potassium hydrogen oxalate d) none of these
27. Which acid is an active poison depressing the central nervous system?
a) formic acid b) acetic acid c) lactic acid d) oxalic acid
28. CH2ClCOOH is more acidic than CH3COOH due to --------- effect of Cl
a) +M effect b) +I effect c) –I effect d) none of these
29. CH3COOH is weaker than HCOOH due to ---------- effect of methyl group
a) -M effect b) +I effect c) –I effect d) none of these
30. The correct order of acidic strength is
a) CH3CH2COOH < CH3COOH < HCOOH b) CH3CH2COOH >CH3COOH >HCOOH
c) CH3COOH < CH3CH2COOH < HCOOH d) CH3CH2COOH < HCOOH < CH3COOH
31. The correct order of acidic strength is
a) Cl3CCOOH > Cl2CHCOOH > ClH2COOH > CH3COOH
b) Cl3CCOOH < Cl2CHCOOH < ClH2COOH < CH3COOH
c) Cl3CCOOH > Cl2CHCOOH < ClH2COOH < CH3COOH
d) Cl2CHCOOH > ClH2COOH > CH3COOH > Cl3CCOOH
32. Which of the following gives violet colour with neutral FeCl3?
a) benzoic acid b) acetic acid c) benzyl alcohol d) salicylic acid
33. Which of the following discharge the colour of bromine water and gives white precipitate?
a) butyric acid b) acetic acid c) propionic acid d) salicylic acid
34. Chemical name of aspirin is
a) methyl salicylate b) acetyl salicylic acid c) o-hydroxy benzoic acid d) none of these
35. Oil of winter green contains
a) methyl salicylate b) Aspirin c) salol d) none of these
36. An analgesic and antipyretic prepared from salicylic acid is
a) Aspirin b) methyl salicylate c) sodium salicylate d) none of these
37. Relative reactivity of carboxylic acid derivatives towards nucleophilic substitution reaction is
a) Acid chloride > acid anhydride > carboxylic acid > ester > amide
b) Acid chloride < acid anhydride < carboxylic acid < ester < amide
c) Acid chloride > carboxylic acid > acid anhydride > ester > amide
d) acid anhydride > carboxylic acid > ester > amide> acid chloride
38. Acetyl chloride fumes in moist air due to hydrolysis producing --------- gas
a) Chlorine b) hydrogen chloride c) COCl2 d) none of these
39. The catalyst (Lewis acid) used in Friedel Craft’s reaction is
a) anhydrous AlCl3 b) ZnCl2 c) Pd d) none of these
40. Acetyl chloride is used in the detection and estimation of ---------
a) alcoholic and amino groups b) Chlorine c) COOH group d) none of these
41. Acetamide exists as a dimer due to
a) intermolecular hydrogen bonding b) covalent bonding c) both d) none of these
42. Cyclic diester formed on heating lactic acid is
a) epoxide b) lactyl acetate c) lactide d) none of these
43. The Lewis acid AlCl3 generates -------- electrophile from acetyl chloride in the following reaction
+ CH3COClanhy.AlCl3
COCH3
+ HCl
a) acetyl cation CH3–C
+ =O b) CH3
+ c) OCH3
+ d) none of these
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3 MARK
1. Explain dehydration reactions of acetic acid and formic acid
CH3C O H
O
CH3C O H
O
P2O5
CH3C
O
CH3C
O
O
Acetic anhydride
con.H2SO4
H2O + COH C
O
OH
2. Write on decarboxylation reaction.
When anhydrous sodium acetate is heated with sodalime, carboxyl group is removed with the formation
of methane.
CH3COONa NaOH/CaO
CH4 + Na2CO3
3. Acetyl chloride fumes in moist air. Why?
Acetyl chloride fumes in moist air due to hydrolysis producing hydrogen chloride gas.
CH3C
O
Cl H OH+ CH3C
O
OH + HCl
4. Write the uses of salicylic acid
1. It is an antiseptic and disinfectant
2. as a food preservative
3. as a medicine for rheumatic pain
4. in the manufacture of aspirin, salol, methyl salicylate
5. in the preparation of azo dyes
5. Write the preparation and uses of succinic acid
Preparation
H2O CH2-COO H
CH2-CO OH
CH2-CN
CH2-CN
CH2-Br
CH2-Br
2 NaCN
HCl
ethylene dibromide ethylene dicyanide succinic acid
Uses
1. in the manufacture of lacquers and dyes
2. It is a very important laboratory reagent
6. Write the uses of acetyl chloride.
1. As an acetylating agent.
2. In the preparation of acetic anhydride.
3. In the detection and estimation of alcoholic and amino groups.
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7. Write the uses of acetic anhydride.
1. As an acetylating agent for the manufacture of dyes and cellulose acetate
2. In the manufacture of aspirin and some drugs.
8. Write the uses of methyl acetate.
1. It is used as a very good laboratory and industrial solvent.
2. It is used for the preparation of acetoaceticester a compound of synthetic importance.
9. Write the uses of acetamide.
1. For the preparation of methyl cyanide.
2. In leather tanning.
3. As soldering flux.
4. As a plasticiser in cloth.
5 MARK 1. How will you prepare acetic acid from the following?
a) From methyl cyanide or acetonitrile
CH3-CN + H2O H
CH3-CONH2 OH-H
CH3-COOH + NH3
b) From trichloro ethane
CH3C Cl
Cl
Cl
+ 3 K OH CH3C
OH
O H
OH
- H2O CH3-COOH
trichloro ethane
c) From Grignard reagent
CH3MgI + O=C=O
CH3
O MgI OH MgI
O=CH-O-H
O=C +
OHCH3
2. From acetic acid prepare-
a) ethyl acetate
H+
CH3COOC2H5 + H2O
ethyl acetate
CH3CO OH + H OC2H5
b) acetic anhydride: Intermolecular dehydration
CH3C O H
O
CH3C O H
O
P2O5
CH3C
O
CH3C
O
O
Acetic anhydride
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c) acetyl chloride
POCl3 + HClCH3 C O H
O
PCl3 ClCl
C
O
Cl +
Acetyl chloride
CH3
d) Methane
CH3 C O H
O
HO-Na+ CH3 C O
O
Na
sodalimeCH3 C O
O
Na
NaO H + CaOCH4 + Na2CO3
methane
e) Ethane
CH3–COOH P/HI
CH3–CH3
f) Ethyl alcohol
CH3COOH 4LiAlH
CH3CH2OH
g) Acetone
Calcium acetate
CH3 C O
O
CH3 C O
O
Ca CaCO3+CH3C
O
CH3
Acetone
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20. ORGANIC NITROGEN COMPOUNDS
Blue print 1 Mark = 3 3 Mark = 1 5 Mark = 1 Total marks = 11
Govt. Examination questions 1 MARK
1. The isomerism exhibited by CH3CH2NO2 and CH3 CH2 – O – N = O is [Mar-06, Sep-11, Mar-14]
a) position b) chain c) functional d) tautomerism
2. In nitro alkanes –NO2 group is converted to –NH2 group by the reaction with
[Sep-2006, Mar-2008, June-2009, Mar-2013]
a) Sn/HCl b) Zn dust c) Zn/NH4Cl d) Zn/NaOH
3. When nitromethane is reduced with Zn dust + NH4Cl in neutral medium, we get
[Sep-2011, June-2012, Mar-17]
a) CH3NH2 b) C2H5NH2 c) CH3NHOH d) C2H5COOH
4. The compound that is most reactive towards electrophilic nitration is [Mar-2007]
a) Toluene b) benzene c) benzoic acid d) nitrobenzene
5. Nitromethane condenses with acetaldehyde to give [June-2007, Mar-2011, June-2014, Sep-2014]
a) nitro propane b) 1-nitro-2-propanol c) 2-nitro-1-propanol d) 3-nitro propanol
6. Which of the following compounds has the smell of bitter almonds?
[June-2007, Sep-2009, June-2011, June-2015]
a) aniline b) nitro methane c) benzene sulphonic acid d) nitrobenzene
7. Nitrobenzene on electrolytic reduction in con. sulphuric acid, the intermediate formed is
[June-2006, Sep-2012, Sep-2014]
a) C6H5NH – NHC6H5 b) C6H5 – NHOH c) C6H5 – N = N – C6H5 d) C6H5.HSO4
8. Electrophile used in the nitration of benzene is [June-2010, Sep-2011, Sep-2013, Mar-2016]
a) hydronium ion b) sulphonic acid c) nitronium ion [NO2+] d) bromide ion
9. The basic character of amines is due to the [June-2007, Mar-2010, Mar-17]
a) tetrahedral structure b) presence of nitrogen atom
c) lone pair of electrons on nitrogen atom d) high electronegativity of nitrogen
10. The organic compound that undergoes carbylamine reaction is
[Sep-2006, Sep-2007, Mar-2008, June-2012, Mar-2013, June-2013, June-2015]
a) (C2H5)2 NH b) C2H5NH2 c) (C2H5)3 N d) (C2H5)4 N+ I
–
11. Primary amine acts as [Jun-2008, Sep-2008, Jun-2010, Jun-2014, Sep-2015, Sep-2016]
a) Electrophile b) Lewis base c) Lewis acid d) Free radical
12. Oxidation of aniline with acidified potassium dichromate gives [Mar-2009, Mar-2015]
a) p-benzo quinone b) benzoic acid c) benzaldehyde d) benzyl alcohol
13. Which one of the following is a secondary amine? [Mar-2011, Mar-2014, Jun-2016]
a) aniline b) diphenyl amine c) sec.butylamine d) tert.butylamine
14. C6H5NH2 HCl/
2NaNO
X. Identify X. [June-2008, Sep-2008, June-2013, Sep-2013]
a) C6H5Cl b) C6H5NHOH c) C6H5N2Cl d) C6H5OH
15. Which of the following will not undergo diazotisation?
[June-2006, Sep-2009, Sep-2010, Mar-2012, Sep-2012, Jun-2016]
a) m-toluidine b) aniline c) p-amino phenol d) benzyl amine
16. Aniline differs from ethylamine by the reaction with [Mar-2007]
a) metallic sodium b) an alkyl halide c) chloroform and caustic potash d) nitrous acid
17. When aqueous solution of benzene diazonium chloride is boiled the product formed is
[Mar-2009, Mar-2011, June-2012]
a) benzyl alcohol b) benzene + N2 c) phenol d) phenyl hydroxylamine
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18. Conversion of benzene diazonium chloride to chloro benzene is called [Mar-2006, Mar-2013]
a) Sand Meyer’s reaction b) Stephen’s reaction
c) Gomberg reaction d) Schotten-Baumann reaction
19. The compound used in the preparation of sulpha drugs is [Sep-2006]
a) methyl amine b) nitro methane c) amino benzene d) nitro benzene
20. The compound that does not show tautomerism is [June-2006, Sep-2013]
a) nitro benzene b) nitro methane c) nitro ethane d) 2-nitropropane
21. Nitro-acinitro tautomerism is exhibited by [Mar-2007, June-2011]
a) nitro methane b) nitro benzene c) CCl3NO2 d) o-toluidene
22. Chloropicrin is [June-2013]
a) CCl3CHO b) CCl3NO2 c) CHCl3 d) CH3NO2
23. Chloropicrin ( CCl3NO2 ) is used as [Sep-2007, Mar-2008, June-2010, June-2011]
a) soil sterilizing agent b) organic synthesis c) good solvent d) antioxidant
24. Which compound is used as soil sterilizing agent? [Mar-2015]
a) Nitrobenzene b) Nitroethane c) Aniline d) Chloropicrin
25. Oil of mirbane is [June-2008, Mar-2012]
a) nitro benzene b) benzaldehyde c) methyl salicylate d) Aspirin
26. The reaction between benzene diazonium chloride and benzene in the presence of NaOH is [Sep-2007]
a) Perkins reaction b) Gatterman’s reaction
c) Sand Meyer’s reaction d) Gomberg Bachmann reaction
27. Which one of the following will not undergo Hoffman’s bromamide reaction? [Sep-2008]
a) ethanamide b) propanamide c) methanamide d) phenyl methanamide
28. Which among the following is a tertiary amine? [Mar-2009]
a) (CH3)3CNH2 b) (CH3)2CHNHCH3 c) (CH3)2-N-C2H5 d) (C2H5)2C(CH3)NH2
29. The isomerism exhibited by [Sep-2010]
CH3-N
O
O
Nitroform
CH2=N
O
OH
Acinitroform
and is
a) Position b) Chain c) Functional d) Tautomerism
30. Which of the following nitro compounds behaves as an acid in the presence of strong alkali [Mar-2006]
a) primary b) secondary c) tertiary d) both a and b
31. The product obtained when nitrobenzene is treated with Zn / NaOH is [Sep-2010]
a) aniline b) azoxybenzene c) azobenzene d) hydrozobenzene
32. The correct order of basic strength is [Sep-2009]
a) NH3 < CH3NH2 < (CH3)2NH b) NH3 > CH3NH2 > (CH3)2NH
c) CH3NH2 < NH3 < (CH3)2NH d) CH3NH2 > NH3 > (CH3)2NH
33. Which one of the following is the most basic? [Sep-2012]
a) Ammonia b) Methylamine c) Dimetylamine d) Aniline
34. The tertiary nitro compound is [June-2009]
a) 2-nitropropane b) 1-nitropropane
c) 1-nitro-2,2-dimethylpropane d) 2-nitro-2-methylpropane
35. The intermediate formed in the nitration of benzene is [June-2009]
a) Arrenium ion b) Carbanion c) Oxonium ion d) Nitrite ion
36. Aniline reacts with benzoyl chloride in the presence of sodium hydroxide and gives benzanilide. This
reaction is known as [Mar-2010]
a) Gattermann reaction b) Sandmeyer’s reaction
c) Schotten-Baumann reaction d) Gomberg-Bachmann reaction
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37. C6H5N2Cl HCl/
2Cl
2Cu
X. The compound X is [Mar-2010]
a) C6H5NH2 b) C6H5NHNH2 c) C6H5-C6H5 d) C6H5Cl
38. C6H5N2Cl Cu/
2PO
3H
X + N2 + HCl. The compound X is [Mar-2012]
a) C6H5Cl b) C6H5NHNH2 c) C6H6 d) C6H5NO2
39. Methyl isocyanide on reduction using LiAlH4 gives [Mar-2014]
a) methyl amine b) ethyl amine c) dimethyl amine d) trimethyl amine
40. The reagent that cannot be used for the conversion of C6H5NO2 C6H5NH2 [June-2014]
a) Sn / HCl b) LiAlH4 c) H2 / Ni d) Zn / NaOH
41. The reaction between a primary amine, chloroform and alcoholic KOH is known as [Sep-2014]
a) Gabriel reaction b) Mustard oil reaction c) Carbylamine reaction d) Hoffmann reaction
42. CH3CONH2 + 4 (H) EthanolNa X. the compound X is [Mar-2015]
a) Methylamine b) Ethylamine c) Dimethylamine d) Nitromethane
43. Which of the following does not react with benzene diazonium chloride? [June-2015]
a) phenol b) benzene c) aniline d) benzaldehyde
44. Carbylamine reaction is characterized by [Sep-2015]
a) tertiary amine b) primary amine c) secondary amine d) all the above 45. Nitration of nitrobenzene at 373 K results in [Sep-2015]
a) o-dinitrobenzene b) 1,3,5 trinitrobenzene c) p-dinitrobenzene d) m-dinitrobenzene 46. The organic compound that does not undergoes carbylamine test [Mar-2016]
a) ethyl methyl amine b) isobutylamine c) n-propyl amine d) isopropyl amine 47. When benzene diazonium chloride reacts with N,N-dimethyl aniline at ice cold condition gives [Mar-16]
a) p-aminoazobenzene b) diazoaminobenzene
c) p-dimethyl aminoazobenzene d) p- methyl aminoazobenzene 48. Which of the following will not dissolve in NaOH? [Jun-2016]
CH3-NO
2CH
3 CH CH3
NO2
a) b) c) d) CH3-CH
2-CH
2-NO
2CH3
CH3
NO2
C
CH3
49. -------------- nitro compounds behave as acids in presence of strong alkali. [Sep-2016]
a) primary b) secondary c) (a) and (b) d) tertiary
50. C6H5N2+ is more stable than CH3N2
+ because of [Sep-2016]
a) resonance b) steric effect c) inductive effect d) both (b) and (c)
51. The IUPAC name of dimethyl sec.butylamine is [Mar-2017]
a) 2-amino-3-methylbutane b) 2-(N-methylamino)butane
c) 2-(N,N-dimethylamino)butane d) 2-(N,N-dimethylamino)propane
3 MARK
1. How is chloropicrin prepared? Write its use [Sep-2014]
CH3NO2 + 3Cl2 NaOHCCl3 NO2 + 3HCl
Nitromethane Chloropicrin
(trichloro nitro methane)
Chloropicrin is used as soil sterilizing agent
2. From acetamide prepare a) ethyl amine b) methyl amine [Mar-2006]
a) CH3CONH2 + 4[H] 4LiAlH
CH3CH2NH2 + H2O
b) CH3CONH2 KOH/
2Br
CH3NH2 + CO2
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3. From benzylamine prepare a) benzyl alcohol b) benzoic acid and c) N-benzyl acetamide [Jun-13]
a) benzyl alcohol
Benzyl amine reacts with nitrous acid to form benzyl alcohol.
C6H
5CH
2 N H
2
HO N= O
C6H
5CH
2OH + N
2 + H
2O
Benzyl alcohol
b) benzoic acid
On oxidation with permanganate the side chain with the amino group is oxidised to give benzoic acid.
CH2NH
2COOH
KMnO4
[O]
Benzyl amine Benzoic acid
c) N-benzyl acetamide
C6H
5N
H
H C
O
Cl
Benzyl amine
CH3
+
Acetyl chloride N-benzyl acetamide
+ HClCH2
C6H
5N
H
CH2
C
O
CH3
4. What is Gabriel phthalimide synthesis ? [Sep-06, Mar-07, Sep-07, Sep-08, Jun-10, Jun-11, Mar-17]
CO
CO
N HKOH
H2O
CO
CO
N KRX
KX
CO
CO
N R
aqueous KOH
CO
CO
R
OK
OK
+ NH2
Phthalimide Potassium phthalimide N-alkyl phthalimide
Potassium phthalate
Alkylamine (1o)
This involves the treatment of phthalimide with potassium hydroxide to form potassium salt. The salt
is then heated with an alkyl halide to give N-alkyl phthalimide, which in turn reacts with potassium
hydroxide to form a potassium phthalate salt and a pure primary amine.
5. Write a note on diazotisation reaction [June-2007, June-2009]
Aniline reacts with nitrous acid and HCl in ice cold condition to give benzene diazonium chloride. This
reaction is known as ‘diazotisation’.
Nitrous acid is formed by mixing aqueous solutions of sodium nitrite and HCl.
NaNO2 + HCl HNO2 + NaCl
C6H
5N H
2O=N OH+ + H Cl
273 KC
6H
5N
2Cl 2H
2O+
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6. When benzamide is treated with Br2 /alkali gives (A). When it is reduced by LiAlH4
compound (B) is formed. Identify A and B and explain the reactions [Mar-2008]
C6H5CONH2 KOH/
2Br
C6H5NH2 + CO2
Benzamide Aniline (A)
C6H5CONH2 + 4 [H] 4LiAlH
C6H5CH2NH2 + H2O
Benzyl amine (B)
7. An organic compound (A) of M.F C2H7N is treated with nitrous acid to give (B) of M.F
C2H6O which answers iodoform test. Identify A and B and explain the reactions [June-2006]
CH3CH
2N H
2 O=N OH+ CH3CH
2N=N OH[ ] CH
3CH
2OH
UnstableEthyl amine (A) Ethyl alcohol (B)
8. An organic compound (A) of molecular formula C2H5ON treated with bromine and KOH gives
B of molecular formula CH5N. Identify A and B. Write the equation involved. [Mar- 2011]
CH3CONH2 KOH/
2Br
CH3NH2 + CO2
Acetamide (A) Methyl amine (B)
9. C6H5CH2NH2 2HNO
A [O}
B HClHg/Zn
C
Identify A, B and C and explain the reactions [June-2008]
C6H
5CH
2 N H
2
HO N= O
C6H
5CH
2OH + N
2 + H
2O
C6H
5CH
2OH
(O)C
6H
5CHO
C6H
5CHO
Zn/Hg + HClC
6H
5CH
3
Benzyl alcohol
(A)
Benzaldehyde (B)
Toluene (C)
10. An aromatic primary amine A with M.F C6H7N undergoes diazotization to give B. B when
treated with H3PO2 gives C. Identify A, B and C and explain the reactions [Mar-2009]
Diazotization
C6H
5N H
2O=N OH+ + H Cl
273 KC
6H
5N
2Cl 2H
2O+
Aniline (A) Benzene diazonium chloride (B)
C6H
5 N
2 Cl
H H
H3PO
2
Cu+
C6H
6 + N
2 + HCl
Benzene
(C)
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11. An aromatic simplest nitro compound A on reduction using Sn and HCl gives B. B undergoes
carbylamines reaction. Identify A and B. Give any one use of compound A [Sep-2009]
Or
Compound A is yellow coloured liquid and it is called oil of mirbane. A on reduction with tin
andHCl gives B. B answers carbylamines test. Identify A and B. [Mar-2010, Mar-2016]
C6H5NO2 + 6[H] HCl/Sn
C6H5NH2 + 2H2O
Nitrobenzene (A) Aniline (B)
( Oil of mirbane )
Use of nitrobenzene: It is used to prepare explosives like TNT, 1,3,5-trinitro benzene
12. An organic compound (A) with molecular formula C6H7N gives (B) with HNO2 / HCl at 273 K.
The aqueous solution of (B) on heating gives (C) which gives violet colour with neutral FeCl3.
Identify the compounds A, B and C. [Sep-2010, Sep-2015]
Or
An aromatic primary amine (A) with molecular formula C6H7N undergoes diazotization to give
(B). When the aqueous solution of (B) is boiled it gives (C). Identify A, B and C. [Mar-2014]
C6H
5N H
2O=N OH+ + H Cl
273 KC
6H
5N
2Cl 2H
2O+
Aniline (A) Benzene diazonium chloride (B)
C6H
5 N
2 Cl
H
C6H
5OH + N
2 + HCl
HOPhenol
(C)
13. An organic compound (A) of molecular formula C2H5NO on treatment with Na / C2H5OH gives
B (C2H7N) and with Br2 / KOH gives C (CH5N). Identify A, B and C. [Sep- 2011]
CH3CONH2 + 4[H] OH5H
2C/Na
CH3CH2NH2 + H2O
Acetamide (A) Ethyl amine (B)
CH3CONH2 KOH/
2Br
CH3NH2 + CO2
Acetamide (A) Methyl amine (C)
14. CH3NO2 HCl/Sn
A KOHalcoholic/
3CHCl
B Pt/
2H
C. Identify A, B and C [Mar-2012]
CH3NO2 + 6[H] HCl/Sn
CH3NH2 + 2H2O
Nitromethane Methylamine (A)
CH3NH2 + CHCl3 + 3KOH CH3NC + 3KCl + 3H2O
Methylisocyanide (B)
CH3NC + 4 [H] Pt/
2H
CH3NHCH3
Dimethylamine (C)
15. Identify A, B and C [Sep-2013]
HNO3
400oCCH
4A
Sn / HClB
CS2
HgCl2
C
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HNO3
400oCCH
4
Sn / HCl CS2
HgCl2
CH3NO
2 CH3NH
2CH
3N=C=S
(A) (B) (C)
Nitro methane Methyl amine Methyl isothiocyanate
16. Identify A,B and C
CH5N KOHalcoholic/
3CHCl
B Pt/
2H orLiAlH
4 C [Jun-2014, Sep-2016]
(A)
CH3NH2 + CHCl3 + 3KOH CH3NC + 3KCl + 3H2O
(A) Methylisocyanide (B)
CH3NC + 4 [H] Pt/
2H
CH3NHCH3
Dimethylamine (C)
17. An organic compound ‘A’ C7H7NO when treated with Br2 / alkali gives ‘B’ C6H7N.
‘B’ undergoes diazotization reaction. Identify ‘A’ & ‘B’ [Sep-2012]
C6H5CONH2 KOH/
2Br
C6H5NH2 + CO2
Benzamide (A) Aniline (B)
C6H
5N H
2O=N OH+ + H Cl
273 KC
6H
5N
2Cl 2H
2O+
Benzene diazonium chloride
18. An aromatic hydrocarbon A on nitration gives B which is known as oil of mirbane. B on
warming with conc. H2SO4 gives compound C. Identify A, B and C. [June-2012]
+ HNO3
Conc.
Conc. H2SO
4
330 K
NO2
+ H2O
Benzene (A) Nitrobenzene (B)(Oil of mirbane)
NO2
+ con. H2SO
4
NO2
SO3H
m-nitrobenzene sulphonic acid (C)
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19. An organic compound (A) of molecular formula C2H5NO reacts with Br2 / NaOH to give
compound (B) of molecular formula CH5N. (A) is reduced by LiAlH4 to give compound (C) of
formula C2H7N. Identify (A) , (B) and (C) [Mar- 2013]
CH3CONH2 NaOH/
2Br
CH3NH2 + CO2
Acetamide (A) Methyl amine (B)
CH3CONH2 + 4[H] 4LiAlH
CH3CH2NH2 + H2O
Acetamide (A) Ethyl amine (C)
20. C6H5CONH2 KOH/
2Br
X HCl
2HNO
Y Cu/
2NaNO
Z. Identify X, Y and Z.
[Mar-2015]
C6H5CONH2 KOH/
2Br
C6H5NH2 + CO2
Benzamide Aniline (X)
C6H
5N H
2O=N OH+ + H Cl
273 KC
6H
5N
2Cl 2H
2O+
Benzene diazoniumchloride (Y)
C6H
5 N
2 Cl + NaNO
2
Cu+
C6H
5NO
2 + NaCl + N
2
Nitrobenzene (Z)
21. C6H5NH2 KHClHNO 273
2 A
2)(
2CNCuKCN
B
OH3 C. Identify A,B and C
[June-2015]
C6H
5N H
2O=N OH+ + H Cl
273 KC
6H
5N
2Cl 2H
2O+
Benzene diazoniumchloride (A)
C6H5N2Cl + KCN 2)(
2CNCu
C6H5CN + KCl + N2
Phenyl cyanide (B)
C6H5CN
OH3 C6H5COOH
Benzoic acid (C)
22. C2H
3N
LiAlH4
etherB
HNO2
C
A
Identify A, B and C [Jun-2016]
CH3-C N
LiAlH4
etherCH
3-CH
2-NH
2
A B
Methyl cyanide Ethyl amine
CH3-CH2NH2 + O = N – OH [CH3-CH2–N=N – OH] CH3-CH2OH + N2
B unstable Ethyl alcohol ( C )
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5 MARK
1. Distinguish between primary, secondary and tertiary amines [Mar-2006, June-2006, Sep-2007,
June-2009, Sep-2009, Mar-2010, Sep-2010, June-2011, Mar-2013, Sep-2014, Mar-2015]
Primary amine
RNH2
Secondary amine
R2NH
Tertiary amine
R3N
1. With HNO2 forms alcohol forms N-nitroso amine forms salt
2. With CHCl3 / KOH forms
carbylamine
No reaction No reaction
3. With acetyl chloride forms
N-alkyl acetamide
form N,N-dialkyl acetamide No reaction
4. With CS2 and HgCl2 alkyl
isothiocyanate is formed
(Mustard oil reaction)
No reaction No reaction
5. With three molar proportion of
alkyl halide, quaternary
ammonium salt- crystalline
compound is formed
RNH2 + 3RX R4N+X
–
With two molar proportion of
alkyl halide, quaternary
ammonium salt is formed
R2NH + 2RX R4N+X
–
With only one molar
proportion of alkyl halide
quaternary ammonium salt is
formed.
R3NH + RX R4N+X
–
2. How do primary, secondary and tertiary amines react with nitrous acid ? [June-2008, June-2012, Sep-2013, Mar-17]
1. Primary amines react with nitrous acid to form alcohols and nitrogen gas.
CH3NH2 + O = N – OH [CH3–N=N – OH] CH3OH + N2
Methyl amine unstable methyl alcohol
2. Secondary amines react with nitrous acid to form N-nitroso dialkyl amines which are water
insoluble yellow oils.
(CH3)2 N H + HO – N = O (CH3)2 N – N = O
Dimethyl amine N-nitroso dimethyl amine - yellow oil
(insoluble in water)
3. Tertiary amines react with nitrous acid to form trialkyl ammonium nitrite salts which are soluble in
water.
(CH3)3 N + HONO (CH3)3 NH+ NO2
–
Trimethyl amine trimethyl ammonium nitrite
(salt - soluble in water)
3. Write a note on the reduction of nitro benzene under different conditions [Mar-2007, Mar-2014, Sep-2015, Mar-2016]
1. Strongly acidic medium
C6H5NO2 + 6[H] HCl/Sn
C6H5NH2 + 2H2O
Nitrobenzene Aniline
2. Neutral medium
C6H5NO2 + 4[H] Cl
4NH/Zn
C6H5NHOH + H2O
Nitrobenzene phenyl hydroxylamine
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3. Alkaline medium
C6H
5NO
2
Zn / NaOH
SnCl2 + NaOH
or Glucose + NaOH
Na3AsO
3 / NaOH
C6H
5-NH-NH-C
6H
5
azoxy benzene
azo benzene
hydrazo benzene
C6H
5-N=N-C
6H
5
O
C6H
5-N=N-C
6H
5
4. Catalytic reduction
Lithium Aluminium hydride reduces nitro benzene to aniline. This reduction can also be carried out
by H2/Ni.
C6H5NO2 4LiAlH
C6H5NH2
Nitrobenzene Aniline
5. Electrolytic Reduction :
When nitro benzene is reduced electrolytically in presence of concentrated sulphuric acid, phenyl
hydroxylamine is first produced which rearranges to give p-amino phenol.
4. How are the following conversions carried out? [Sep-2006]
a) nitro benzene to phenyl hydroxyl amine b) aniline to phenyl isocyanide
c) benzene diazonium chloride to biphenyl
a) C6H5NO2 + 4[H] Cl
4NH/Zn
C6H5NHOH + H2O
Nitrobenzene phenyl hydroxylamine
b) C6H5NH2 + CHCl3 + 3KOH C6H5NC + 3KCl + 3H2O
Aniline Phenyl isocyanide
+ N2 + HClC
6H
5N
2Cl + C
6H
6
NaOHC
6H
5C
6H
5
Biphenyl
c)
Benzene
5. How are the following conversions carried out? [June-2007]
a) CH3NO2 CH3NH2 b) CH3NH2 CH3NC
c) benzene diazonium chloride biphenyl
a) CH3NO2 + 6[H] HCl/Sn
CH3NH2 + 2H2O
Nitromethane Methylamine
b) CH3NH2 + CHCl3 + 3KOH CH3NC + 3KCl + 3H2O
Methyl amine Methylisocyanide
c) Refer Q-4 c
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6. How are phenol, chlorobenzene and biphenyl prepared from benzene diazonium chloride?
[Mar-2008]
C6H
5 N
2 Cl
H
C6H
5OH + N
2 + HCl
HO Phenol
a)
C6H
5 N
2 Cl C
6H
5Cl + N
2
HCl
Cu
Chlorobenzene
b)
c) Refer Q-4 c
7. Explain the reduction of nitro benzene [June-2010]
i) in alkaline medium ii) by catalytic reduction
Refer Q – 3
8. Explain the following reactions of aniline [Sep-2008]
a) Coupling reaction b) Schotten Baumann reaction c) Carbylamine reaction
a) Coupling reaction
N=N-Cl +
Benzene diazonium chloride
H HN
Aniline
N=N HN
diazo amino benzene
N=N HN
(Rearrangement)
HClN=N NH
2
p-amino azobenzene
b) Schotten Baumann reaction
C6H
5 N
H
H C
O
Cl
Aniline
C6H
5+
Benzoyl chloride
NaOH
Benzanilide
C6H
5 N
H
C
O
C6H
5 + HCl
c) Carbylamine reaction
Aniline reacts with chloroform and alcoholic KOH to give an offensive smelling liquid, phenyl
isocyanide.
C6H5NH2 + CHCl3 + 3KOH C6H5NC + 3KCl + 3H2O
Aniline Phenyl isocyanide
9. Explain mustard oil reaction and Gabriel’s phthalimide synthesis [Mar-2009]
Mustard oil reaction When primary amines are warmed with carbondisulphide and mercuric chloride, alkyl
isothiocyanate, having a pungent mustard like odour is obtained.
CH3NH2 + S=C=S 2HgCl
CH3N=C=S + H2S
Methyl amine Methyl isothiocyanate
Gabriel phthalimide synthesis ( Pure primary amine ) : Refer – 3 mark – Q - 4
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10. Write note on the following reactions [Mar-2011]
a) Carbylamine reaction b) Gabriel’s phthalimide synthesis
a) Carbylamine reaction Primary amines on heating with chloroform and alcoholic KOH forms a foul smelling substance
called carbylamine or alkyl isocyanide.
Methyl amine
(1o)
Chloroform Methyl isocyanide
CH3 N H C 3 K OH CH
3NC + 3KCl + 3H
2O
H
H+
Cl
Cl
Cl
+
Gabriel’s phthalimide synthesis: Refer – 3 mark – Q - 4
11. Write note on the following reactions [Sep-2011]
a) Mustard oil reaction b) Diazotisation reaction c) Gomberg reaction a) Mustard oil reaction: Refer Q – 9
b) Diazotisation reaction: Refer – 3 mark – Q - 5
c) Gomberg reaction or Gomberg Bachmann Reaction
Decomposition of diazonium salts in presence of sodium hydroxide and benzene, results in the
formation of Biphenyl.
+ N2 + HClC
6H
5N
2Cl + C
6H
6
NaOHC
6H
5C
6H
5
Biphenyl
12. Write notes on the following: [Mar-2012]
a) Mustard oil reaction b) Formation of Schiff’s base
a) Mustard oil reaction: Refer Q – 9
b) Formation of Schiff’s base Aniline is Primary aromatic amine. It reacts with aldehydes to form aldimines or Schiff’s base.
C6H
5N H
2 + RCO
H
C6H
5N +RC
H
Schiff's base
H2O
13. Write any three methods of preparing benzyl amine [Sep-2012]
a) C6H5CN + 4[H] 4LiAlH
C6H5CH2NH2
Benzonitrile Benzyl amine
b) C6H5CONH2 + 4[H] 4LiAlH
C6H5CH2NH2 + H2O
Benzamide Benzyl amine
c) C6H5CH2 –Br + H–NH2 C6H5CH2NH2 + HBr
Benzyl bromide Alcoholic Benzyl amine
ammonia
14. How can the folloing conversions be effected? [Jun-2016]
a) chlorobenzene aniline
Chlorobenzene reacts with NH3 at high temperature and pressure in the presence of copper salts to give aniline.
C6H5Cl + 2NH3 2CuCl
C6H5NH2 + NH4Cl
b) Aniline Schiff’s base (Refer-Q-12-b)
c) Benzene diazonium chloride Biphenyl (Refer-Q-11-c)
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15. Write a note on a) Carbylamine reaction b) Mustard oil reaction c) Sand Meyer reaction
[June-2013]
a) Carbylamine reaction: Refer Q – 10
b) Mustard oil reaction : Refer Q – 9
c) Sand Meyer reaction:
When aqueous solution of benzene diazonium chloride is warmed with Cu2Cl2 or Cu2Br2 in
halogen acid, halogenated benzene is formed.
C6H
5 N
2 Cl C
6H
5Cl + N
2
HCl
Cu2Cl
2
Chlorobenzene
C6H
5 N
2 Cl C
6H
5Br + N
2
HBr
Cu2Br
2
Bromobenzene
16. Starting from benzene diazonium chloride how will you obtain the following:
a) Phenol b) Anisole c) p- hydrory azobenzene [June-2014]
a) Phenol
C6H
5 N
2 Cl
H
C6H
5OH + N
2 + HCl
HO Phenol
b) Anisole
Anisole is prepared by warming benzene diazonium chloride with methyl alcohol.
C6H
5 N
2 Cl
H
C6H
5OCH
3 + N
2 + HCl
CH3O
Anisole
c) p- hydrory azobenzene
Phenol couples with benzene diazonium chloride in alkaline medium to form p-hydroxy
azobenzene.
OH
benzene diazonium
chloridep-hydroxy azobenzene
(Red orange dye)Phenol
N=N-Cl + H OH N=N OH
17. Write note on carbylamines reaction and formation of benzanilide [Jun – 2015]
Refer : 10-a, 8-b
18. How do you prepare the following compounds from benzene diazonium chloride [Sep-16]
a) nitrobenzene b) chloro benzene c) phenyl hydrazine
a) nitrobenzene
C6H
5 N
2 Cl C
6H
5NO
2Cu
Nitrobenzene
NaNO2
+
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b) chloro benzene
C6H
5 N
2 Cl C
6H
5Cl + N
2
HCl
Cu
Chlorobenzene
c) phenyl hydrazine
C6H
5 N
2 Cl C
6H
5NHNH
2
4[H]
SnCl2 / HCl phenyl hydrazine
Additional questions 1 MARK
1. Bromo ethane reacts with silver nitrite to give
a) C2H5NO2 b) C2H5–O–NO c) C2H5Ag + NaBr d) C2H5NC
2. Nitration of nitrobenzene at 433 K results in
a) o-dinitrobenzene b) 1,3,5 trinitrobenzene c) p-dinitrobenzene d) m-dinitrobenzene
3. 1-nitro butane and 1-nitro-2-methyl propane exhibit --------- isomerism
a) Position b) Chain c) Functional d) Tautomerism
4. 1-nitro propane and 2-nitro propane exhibit --------- isomerism
a) Position b) Chain c) Functional d) Tautomerism
5. Nitro methane (CH3NO2) and Methyl nitrite (CH3– O – N = O) exhibit ------- isomerism
a) Position b) Chain c) Functional d) Tautomerism
6. The reduction of CH3 – CH2 – CN with sodium and alcohol results in the formation of
a) (CH3)2CHNH2 b) CH3 – CH2 – CH2 – OH + N2
c) CH3 – CH2 – CH2 – NH2 d) CH3 – CH2 – NH2
7. Oxidation state of nitrogen in nitro group is
a) -3 b) +3 c) +2 d) -2
8. Which of the following does not exhibit tautomerism?
CH3-CH
2-CH
2-NO
2CH
3 CH CH3
NO2
CH3
CH3
NO2CH
2C
CH3
CH3
CH3
NO2
C
CH3
a) b) c) d)
[Both primary and secondary nitro compounds contain α-hydrogen and so they exhibit tautomerism. But
tertiary nitro compounds do not contain α-hydrogen and so they do not exhibit tautomerism]
9. Which is insoluble in NaOH and does not react with nitrous acid?
a) nitro methane b) nitro ethane c) 1-nitropropane d) 2-methyl-2-nitro propane
[2-methyl-2-nitro propane is a teriary nitro compound and it has no α-hydrogen and hence it is insoluble
in NaOH and does not react with nitrous acid]
10. Explosive prepared from nitrobenzene
a) TNT b) 1,3,5-trinitrobenzene c) both a and b d) none of these
11. Which of the following amines does not form intermolecular H-bond?
a) primary b) secondary c) tertiary d) all
[in tertiary amines H atom is not connected to electronegative N]
12. The boiling points of amines are in the order
a) Secondary amine > Primary amine > Tertiary amine
b) Secondary amine < Primary amine < Tertiary amine
c) Primary amine <Secondary amine <Tertiary amine
d) none of these
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13. The order of the basic strength of amines is
a) 2o amine > 1
o amine > 3
o amine b) 2
o amine < 1
o amine < 3
o amine
c) 1o amine > 2
o amine > 3
o amine d) none of these
14. Secondary amine is more basic than primary amine due to
a) +I effect of methyl groups b) -I effect of methyl groups c) resonance d) none
15. Tertiary amine is less basic than 1o and 2
o amines due to
a) +I effect b)-I effect c) steric effect d) none of these
16. The correct order of basic strength is
a) CH3NH2 < (CH3)2NH < C6H5NH2 b) CH3NH2 > (CH3)2NH > C6H5NH2
c) C6H5NH2 < CH3NH2 < (CH3)2NH d) C6H5NH2 > CH3NH2 > (CH3)2NH
17. Amines are
a) Lewis acids b) Lewis bases c) Arrhenius acids d) Arrhenius bases
18. Amines have
a) Garlic odour b) fishy odour c) pleasant odour d) none of these
19. Methyl cyanide on reduction with LiAlH4 gives --------
a) methyl amine b) ethyl amine c) nitro methane d) nitro ethane
20. Methyl isocyanide on reduction with LiAlH4 gives --------
a) methyl amine b) dimethylamine c) nitro methane d) nitro ethane
21. CH3CONH2 reacts with Br2/KOH to give
a) CH3CH2NH2 b) CH3NH2 c) CH3NHCH3 d) CH3COOH
22. Gabriel’s phthalimide synthesis is used for the preparation of pure
a) primary amine b) secondary amine c) tertiary amine d) nitro compounds
23. Which of the following reaction is characteristic of primary amine?
a) Carbylamine reaction b) Mustard oil reaction c) both a and b d) none of these
24. Which of the following does not undergo Carbylamine reaction?
a) isopropyl amine b) aniline c) ethyl methyl amine d) isobutyl amine
25. Primary amines condense with aromatic aldehydes forming
a) Schiff’s bases b) aldimines c) aldoximes d) none of these
26. Which of the following does not react with diethyl oxalate?
a) methyl amine b) dimethyl amine c) trimethyl amine d) iso propyl amine
27. On reduction with LiAlH4 benzonitrile gives ---------
a) phenyl cyanide b) benzamide c) aniline d) benzylamine
28. On reduction with LiAlH4 or H2/Pd benzamide gives ---------
a) phenyl cyanide b) benzaldehyde c) aniline d) benzylamine
29. When benzamide is treated with bromine and alkali it gives
a) aniline b) benzyl amine c) phenazyl bromide d) bromo benzene
30. Which is the weak base among the following?
a) methyl amine b) ethyl amine c) benzyl amine d) dimethyl amine
31. An organic compound of MF C7H9N answers carbylamine test and undergoes diazotization reaction.
The compound is
a) benzyl amine b) N-methyl aniline c) toluidine d) none of these
32. Which is used for the manufacture of anti oxidants in rubber industry and for preparing sulpha drugs
a) aniline b) benzyl amine c) benzaldehyde d) nitro benzene
33. ---------- is used in the manufacture of synthetic polymers like PAN.
a) Acrylo nitrile (CH2=CH–CN) b) acetonitrile c) benzonitrile d) formonitrile
34. Which of the following diazonium salts can be dried without decomposition?
a) ArN2Cl b) ArN2Br c) ArN2NO2 d) ArN2BF4
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3 MARK 1. From aniline prepare a) s-diphenyl urea and b) s-diphenyl thiourea
a) s-diphenyl urea
C6H
5 N
H
H
C6H
5 N
H
H
C O
Cl
Cl
C6H
5 N
H
C6H
5 N
H
C O + 2HCl
S-diphenyl ureaAnilineCarbonyl chloride
('S' stands for symmetric)
b) s-diphenyl thiourea
('S' stands for symmetric)
C6H
5 N
H
C6H
5 N
H
C +
Aniline
H2S
Carbondisulphide
S-diphenyl thiourea
C6H
5 N
H
H
C6H
5 N
H
H
C SS+ S
2. Aromatic amine is less basic than aliphatic amines. Why?
Aniline (Aromatic amine) is less basic than aliphatic amines. This is because, the lone pair of
electrons on the nitrogen atom in aniline is involved in resonance and is not easily available for
donation to protons. Because of positive charge on nitrogen protonation becomes difficult.
H2N H
2N
+H
2N
+H
2N
+
3. Write a note on Hoffmann’s bromamide reaction
Hoffman’s hypobromite reaction or Hoffman’s bromamide reaction:
When an amide is treated with bromine and alkali, the amide is converted into primary amine
containing one carbon less than that of amide.
CH3CONH2 KOH/
2Br
CH3NH2 + CO2
Acetamide Methyl amine
4. Write a note on Gattermann reaction When the benzene diazonium chloride solution is warmed with copper powder and the hydrogen
halide, the corresponding halobenzene is obtained.
Iodo benzene cannot be prepared by this procedure.
C6H
5 N
2 Cl C
6H
5Cl + N
2
HCl
CuChlorobenzene
C6H
5 N
2 Cl C
6H
5Br + N
2
HBr
Cu Bromobenzene
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5. Write the uses of nitro benzene
1. Nitro benzene is used to prepare aniline
2. It is used to prepare explosives like TNT, 1,3,5-trinitro benzene.
3. It is used in making dye stuffs and pharmaceuticals.
6. Write the uses of aniline
1. For preparing dyes and dye intermediates.
2. For the manufacture of anti oxidants in Rubber industry.
3. For preparing drugs (e.g.,) sulpha drugs.
4. For making isocyanates required for polyurethane plastics.
5 MARK
1. With examples explain different types of isomerism in nitro compounds
1. Chain isomerism
Chain isomerism arises due to the difference in the arrangement of carbon atoms.
CH3– CH2– CH2– CH2– NO2 (linear chain)
1-nitro butane
CH3 CH
CH3
NO2CH
2 (branched chain)
1-nitro-2-methylpropane
2. Position isomerism
Position isomerism arises due to the difference in the position of nitro group.
CH3– CH2– CH2– NO2
1-nitro propane
CH3 CH CH
3
NO2
2-nitropropane
3. Functional isomerism
Functional isomerism is due to the difference in the nature of functional group. Nitro alkanes
are functional isomers of alkyl nitrites.
CH3
NO
O
CH3
O N O
Methyl nitriteNitromethane
4. Tauto merism
Nitromethane exhibits tautomerism.
CH3
NO
O
CH2
NO
OH
Nitro form Acinitro form
These two functional isomers exist in equilibrium.
This is said to exhibit ‘‘Nitro-acinitro’’ tautomerism.
Both primary and secondary nitro compounds exhibit these kinds of tautomerism.
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21. BIOMOLECULES
Blue print 1 Mark = 2 3 Mark = 0 5 Mark = 1 Total marks = 7
Govt. Examination questions 1 MARK
1. Identify the reducing sugar [June-2006, Mar-2007, Mar-2009, Sep-2011, Mar-2012, Mar-2013]
a) Sucrose b) Cellulose c) Starch d) Glucose
[Reducing sugars : reduce Tollen’s reagent and Fehling’s solution eg: glucose, fructose, lactose]
2. Sucrose contains glucose and fructose linked by [Jun-2007, Jun-2017, Sep-2016]
a) C1 – C1 b) C1 – C2 c) C1 – C4 d) C1 – C6
3. Glucose is not oxidised to gluconic acid by [Sep-2008, June-2013, Mar-2015]
a) Br2/H2O b) Fehling solutions c) Tollen’s reagent d) Conc. HNO3
4. Inversion of sucrose refers to [June-2008, June-2009]
a) oxidation of sucrose b) reduction of sucrose
c) hydrolysis of sucrose to glucose and fructose d) polymerisation of sucrose
5. Glucose forms -----------with acetic anhydride and sodium acetate
[Sep-2007, Sep-2010, June-2012, Sep-2014, Sep-2015]
a) di acetate b) tetra acetate c) penta acetate d) hexa acetate
6. The amino acid without chiral carbon is [June-2006, Mar-2008, Mar-2009, Mar-2010, June-2010,
Mar-2011, Sep-2011, Mar-2012, Mar-203, June-2014]
a) Glycine b) Alanine c) Proline d) Thyrosine
7. The building block of proteins are [June-2007, Sep-2012, Mar-17]
a) -hydroxy acids b) -amino acids c) -hydroxy acids d) -amino acids
8. Which is not true of amino acid? [Mar-2007, June-2012, Sep-2014]
a) amino acid forms Zwitter ion b) has isoelectric point
c) dual behaviours d) amino acid is insoluble in NaOH solution
9. A di peptide does not have [June-2009, Sep-2009, Sep-2010]
a) two peptide units b) portions of two amino acids
c) an amino group d) salt like structure
10. Ultimate products of hydrolysis of proteins is [Sep-2006, June-2008, Mar-2016]
a) aniline b) aliphatic acid c) amino acid d) aromatic acid
11. Proteins are [Mar-2006, Sep-2013]
a) polypeptides b) poly acids c) poly phenols d) poly esters
12. Important constituent of cell wall is [Sep-2009, June-2010, Mar-2014]
a) lipid b) cellulose c) protein d) vitamin
13. The solution having equal molecules of D (+) glucose and D (–) fructose is termed as
[Mar-2006, Mar-2011, June-2011, June-2014]
a) invert sugar b) fruit sugar c) racemic mixture d) cane sugar or non-sugar
14. When starch is heated to 200–250oC, it changes into [Mar-2010]
a) caramel b) dextrin c) cellulose d) barley sugar
15. ------------ occur in the white matter of the brain and of all nervous tissue. [June-2013]
a) Lecithins b) Cephalin c) Galactolipids d) Amino acid
16. Sorbitol and mannitol are [Mar-2008]
a) isomers b) polymers c) epimers d) dimmers
17. Optically inactive amino acid is [Sep-2008]
a) glycine b) alanine c) proline d) phenyl alanine
18. An example for reducing disaccharide is---------------- [Sep-2010]
a) Glucose b) Fructose c) Sucrose d) Lactose
19. ------------- is involved in the process of blood coagulation [June-2011]
a) Fats and oils b) Cephalin c) Glycolipids d) Lecithins
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20. Cephalins have been implicated in the process of [Sep-2013]
a) metabolism b) organization of the body c) blood purification d) blood coagulation 21. ----------------- act as protective agent on the surface of animals and plants. [Mar-2014]
a) Carbohydrates b) Vitamins c) Nucleic acids d) Waxes
22. Which one of the following is a polysaccharide? [Mar-2015]
a) Sucrose b) Cellulose c) Maltose d) Raffinose 23. Sucrose is not [June-2015]
a) a disaccharide b) a non-reducing sugar.
c) hydrolysed to only glucose d) hydrolysed to glucose & fructose 24. When cellulose is boiled with dilute H2SO4 it gives [June-2015]
a) D-glucose b) D-fructose c) Dextrine d) Starch 25. The number of secondary alcohol groups present in fructose is [Sep-2015]
a) 4 b) 2 c) 0 d) 3 Note: Glucose contains one primary alcohol group and four secondary alcohol groups.
Fructose contains two primary alcohol groups and three secondary alcohol groups. 26. Which is a monosaccharide among the following? [Mar-2016]
a) Sucrose b) Cellulose c) Maltose d) Glucose
27. What are the three monosaccharides obtained during hydrolysis of raffinose? [June-2016, Mar-17]
a) sucrose, fructose, lactose b) glucose, fructose, galactose
c) fructose, lactose, galactose d) maltose, lactose, sucrose
Monosaccharide C6H12O6 Glucose, Fructose, Galactose
Disaccharide C12H22O11 Sucrose, Lactose, Maltose
Trisaccharide C18H32O16 Raffinose
Polysaccharide (C6H10O5)n Strach, Cellulose, Inulin
28. Which of the following contains a lipid? [Sep-2016]
a) starch b) mineral oil c) edible oil d) peptide
5 MARK
1. Outline the classification of carbohydrates giving example for each
[June-2007, Mar-2008, June-2008, Mar-2010, Mar-2011, Sep-2013, Mar-2016]
Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones. They are classified into two broad
groups: 1. Sugars and 2. Non-sugars or polysaccharides.
1. Sugars : Sugars are sweet crystalline substances and soluble in water.
a) Monosaccharides:
The monosaccharides cannot be hydrolysed into simpler sugars.
Eg- Glucose, Fructose.
They may again be classified according to the nature of carbonyl group.
(a) Aldoses, which contain an aldehyde group
(b) Ketoses, which contain a keto group
b) Oligosaccharides :
Oligosaccharides are sugars that yield two to ten monosaccharide molecules on hydrolysis.
i) Disaccharides : The disaccharides are sugars which on hydrolysis give two molecules
of the same or different mono saccharides. Eg-Sucrose
C12H22O11 + H2O H
C6H12O6 + C6H12O6
Sucrose Glucose Fructose
ii) Trisaccharides : These give three molecules of mono saccharides on hydrolysis.
Eg-Raffinose
C18H32O16 + 2H2O H
C6H12O6 + C6H12O6 + C6H12O6
Raffinose Glucose Fructose Galactose
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2. Non-sugars (or) Polysaccharides
These are carbohydrates which involve a large number of monosaccharide units linked to each other
by oxide bridges. These linkages are called glycosidic linkages. They get hydrolysed to give
monosaccharides. Eg- Starch, Cellulose and inulin
(C6H10O5) n + n H2O H
n C6H12O6
Starch Glucose
Thus whole of the classification of the carbohydrates may be summarized as below.
Carbohydrates
Sugars Non-sugars(Polysaccharides)
Monosaccharides Oligosaccharides
Aldoses Ketoses
Trisaccharides TetrasaccharidesDisaccharides
Homo
polysaccharides
Hetero
polysaccharides
... ... ...
2. Elucidate the structure of glucose
[Jun-06, Mar-09, Jun-10, Jun-11, Jun-12, Mar-13, Mar-14, Jun-14, Jun-15, Mar-17]
1. Elemental analysis and molecular weight determination show that the molecular formula of glucose
is C6H12O6.
2. Complete reduction of glucose with concentrated hydriodic acid in the presence of red phosphorous
produces n-hexane as the major product.
This indicates that the six carbon atoms in the glucose molecule form an unbranched chain of six
carbon atoms.
Glucose HI/PCH3 – CH2 – CH2 – CH2 – CH2 – CH3
n-hexane
3. Glucose readily dissolves in water to give a neutral solution. This indicates that the glucose
molecule does not contain a carboxyl group.
4. Glucose reacts with hydroxylamine to form a monoxime or adds only one mole of HCN to give a
cyanohydrin. This reaction indicates the presence of either an aldehyde or a ketone group.
5. Mild oxidation of glucose with bromine water gives gluconic acid. This indicates the presence of
an aldehyde group since only the aldehyde group can be oxidised to an acid, containing same
number of carbon atoms. Since the six carbon atoms in glucose form a consecutive unbranched
chain, the aldehyde group, must occupy one end of this chain.
6. Further oxidation of gluconic acid with nitric acid gives saccharic acid. This indicates the presence
of a primary alcoholic group.
GlucoseBr
2 / H
2O
Mild
oxidation
HNO3
Strong
oxidation
Gluconic acid Saccharic acid
COOH
(CHOH)4
CH2OH
COOH
(CHOH)4
COOH
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7. Glucose reduces an ammoniacal solution of silver nitrate (Tollen’s reagent) to metallic silver or a
basic solution of cupric ion (Fehling’s solution) to red cuprous oxide. These reactions further
confirm the presence of an aldehyde group.
8. Glucose reacts with acetic anhydride in the presence of pyridine to form a penta acetate. This
reaction indicates the presence of five hydroxyl groups in a glucose molecule.
From the above evidences we conclude that glucose is a penta hydroxyl hexanal (an
aldohexose) and can be represented by the following structure.
CHOH
CH2OH
CHOH
CHOH
CHOH
C
H O
Glucose
(2,3,4,5,6-penta hydroxy hexanal)
3. How is the structure of fructose determined? [Mar-2006, Mar-2007, Sep-2007, Sep-2008, June-
2009, Sep-2009, Sep-2010, Mar-2012, June-2013, Sep-2014, Mar-2015]
1. Elemental analysis and molecular weight determination show that the molecular formula of fructose
is C6H12O6.
2. Complete reduction of fructose with concentrated hydriodic acid in the presence of red phosphorous
produces n-hexane as the major product.
This indicates that the six carbon atoms in the fructose molecule form an unbranched chain of six
carbon atoms.
Fructose HI/PCH3 – CH2 – CH2 – CH2 – CH2 – CH3
n-hexane
3. Fructose readily dissolves in water to give a neutral solution. This indicates that the fructose
molecule does not contain a carboxyl group.
4. Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction
indicates the presence of five hydroxyl groups in a fructose molecule.
5. Fructose reacts with hydroxylamine to form a monoxime or adds only one mole of HCN to give a
cyanohydrin. This reaction indicates the presence of either an aldehyde or a ketone group.
6. Fructose is not oxidized by bromine water indicating the absence of aldehyde group. Oxidation of
fructose with conc.HNO3 yields a mixture of glycolic acid and tartaric acid. Since this oxidation
occurs with the rupture of carbon chain, the carbonyl group must be present as ketone group.
HNO3
Fructose + 4 (O)
COOH
(CHOH)2
COOH
+COOH
CH2OH
+ 3 H2O
Tartaric acid Glycollic acid
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7. Partial reduction of fructose with sodium amalgam and water produces a mixture of two epimeric
alcohols, sorbitol and mannitol, because a new asymmetric centre is being created at C–2. This
confirms the presence of a ketone group.
8. When fructose is treated with HCN, it forms an addition product which upon hydrolysis and
subsequent reduction with hydriodic acid and red phosphorous gives 2-methyl-hexanoic acid. This
indicates that the ketone group is adjacent to one of the terminal carbon atoms.
HCNFructose
COOH
(CHOH)3
CH2OH
CCN
OH
CH2OH
Hydrolysis
(CHOH)3
CH2OH
COH
CH2OH
COOH
(CH2)3
CH3
COHHI / P
CH3
2-methyl hexanoic acid
From the above evidences we conclude that fructose is a pentahydroxyhexanone
(a ketohexose) and can be represented by the following structure.
CH2OH
CHOH
CHOH
CHOH
Fructose
(1,3,4,5,6-penta hydroxy-2- hexanone)
CH2OH
C = O
4. What is peptide bond? Illustrate the formation of peptide bond in glycylalanine. Draw the
structure of glucose and fructose. [Sep-2006, Sep-2011, Jun-2016, Sep-16]
The bond formed between two amino acids by the elimination of a water molecule is called a peptide
linkage or peptide bond.
Glycine and alanine combine to give a dipeptide called glycylalanine
H2O
C
O
NOHH2N CH
2 + C
O
OHCHH
H
CH3
C
O
NH2N CH
2 C
O
OHCH
H
CH3
Glycine Alanine Glycylalanine
Peptide bond
Structure of glucose and fructose : Refer Q-2, 3
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5. What are lipids? Mention the biological importance of lipids [Sep-2012, Sep-2015]
Lipids form a group of organic compounds which are widely distributed in living system. Lipids are
mainly divided into three types, namely simple, compound and derived, depending on the basis of the
nature of products obtained on hydrolysis.
Functions of lipids in biosystems:
1. Fats and oils Fats and oils act as storage of energy in plants and animals.
Fat gives about 2¼ times as much energy as carbohydrates or proteins.
Fat is a poor conductor of heat. Hence, fat layer under skin serves to prevent losses of heat from the body.
2. Wax
Wax acts as a protective agent on the surfaces of animals and plants.
Waxy coating on the surface of plants and fruits protects them from excessive loss of moisture and
becoming infected with fungi and bacteria.
3. Phospholipids: a) Lecithins are required for normal transport and utilisation of other lipids, especially in the liver.
Lecithin aids in the organisation of the cell structure.
b) Cephalins are found in the brain.
Cephalins are involved in the process of blood coagulation.
4. Galactolipids occur in considerable amount in the white matter of the brain and of all nervous
tissue.
5. The presence of galactose in the glycolipids suggests the importance of milk sugar in the diet of
infants and children during the development of the brain and nervous system.
Additional questions
1 MARK 1. Hydrolysis of raffinose gives
a) glucose b) fructose c) galactose d) All the above
2. Two amino acids say A, B- react to give
a) two dipeptides b) three dipeptides c) four dipeptides d) only one
3. Proteins are not sensitive to
a) acids b) bases c) elevated temperature d) water
4. Denaturation does not involve
a) breaking up of H– bonding in proteins b) the loss of biological action of enzyme
c) the loss of secondary structure d) loss of primary structure of proteins
5. Which among the following contains triglyceride?
a) Wax b) Cooking oil c) Essential oil d) Albumin
6. Which contains a long chain ester?
a) wax b) cooking oil c) turpentine oil d) cellulose
7. An example of a fatty acid obtained from a cooking oil is
a) acetic acid b) stearic acid c) benzoic acid d) oxalic acid
8. Which is not a saturated fatty acid?
a) Palmitic acid b) Stearic acid c) Oleic acid d) Glyceric acid
9. Alkaline hydrolysis of cooking oil gives
a) soap b) glycerol c) fatty acid d) both (a) and (b)
10. Hair and nail contains
a) cellulose b) fat c) keratin d) lipid
11. In polysaccharides, monosaccharide units are linked to each other by oxide bridges. These linkages are called
a) glycosidic linkages b) peptide linkages c) hydrogen bonds d) none of these
12. Oligosaccharides are sugars that yield ---- monosaccharide molecules on hydrolysis
a) 2 b) 15 c) 2 to 10 d) 1 to 20
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13. The reagent used to differentiate glucose from fructose is ----
a) Tollen’s reagent b) Fehling’s solution c) HI/P d) Br2/H2O
14. When reduced with HI/P, glucose and fructose give ----
a) n-hexane b) sorbitol c) mannitol d) cyclohexane
15. With Br2/H2O glucose gives ----
a) gluconic acid b) organic synthesis c) good solvent d) antioxidant
16. Hydrolysis of cane sugar in the presence of sulphuric acid is known as
a) racemisation b) inversion of sucrose c) decarboxylation d) none of these
17. Aqueous solution of glucose or fructose is -----------
a) acidic b) basic c) neutral d) pH=0
18. With acetic anhydride glucose and fructose give
a) octa acetate b) hexa acetate c) penta acetate d) tetra acetate
19. Partial reduction of fructose with sodium amalgam and water produces a mixture of two epimeric
alcohols. They are,
a) glycine and alanine b) sorbitol and mannitol c) glycol and gltcerol d)none of these
20. In the partial reduction of fructose with sodium amalgam and water, new asymmetric centre is created at
a) C-2 b) C-1 c) C-3 d) C-4
21. Sorbitol and mannitol are epimers and they differ only at
a) C-2 b) C-1 c) C-3 d) C-4
22. Oxidation of fructose with concentrated nitric acid yields
a) tartaric acid b) glycollic acid c) both a and b d) none of these
23. IUPAC name of glucose is
a) 2,2,3,5,6-penta hydroxy hexanal b) 3,3,4,5,6-penta hydroxy hexanal
c) 2,3,3,6,6-penta hydroxy hexanal d) 2,3,4,5,6-penta hydroxy hexanal
24. IUPAC name of fructose is
a) 1, 3, 4, 5, 6-pentahydroxy-2-hexanone b) 1, 3, 4, 5, 6-pentahydroxy-3-hexanone
c) 2, 3, 4, 5, 6-pentahydroxy-2-hexanone d) 1, 3, 4, 5, 6-entahydroxy-2-hexanal
25. Sucrose, on heating slowly, melts and when allowed to cool, it solidifies to pale yellow glassy mass
called
a) fruit sugar b) cane sugar c) invert sugar d) Barley Sugar
26. When heated to 200oC, sucrose loses water to form brown amorphous mass called
a) caramel b) dextrin c) cellulose d) none of these
27. Starch is converted into maltose in the presence of -------------- enzyme
a) zymase b) invertase c) diastase d) none of these
28. Starch solution gives blue colour with
a) FeCl3 b) AgNO3 c) iodine d) none of these
29. The two fractions present in starch
a) amylose & amylopectin b)dextrin & glycogen c)glucose &fructose d) none of these
30. Proteins are made up of ----
a) β-aminoacids b) fatty acids c) α-aminoacids d) none of these
31. Amino acids exist in an ionic form called
a) carbanion b) zwitter ion c) carbonium ion d) none of these
32. The pH at which the amino acid shows no tendency to migrate when placed in an electric field is known as
a) end point b) pH = 0 c) isoelectric point d) all are correct
33. A tri peptide contains
a) three amino acid molecules b) two peptide bonds c) three peptide bonds d)both a & b
34. Which is a lipid?
a) Fats b) Oils c) Wax d) All the above
35. Fats give about ------- times as much energy as carbohydrates or proteins
a) 1¼ b) 2¼ c) ¼ d) none of these
36. Which of the following acts as a protective agent on the surfaces of animals and plants?
a) carbohydrates b) proteins c) wax d) none of these
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37. Lecithins and cephalins are
a) carbo hydrates b) proteins c) phospholipids d) galactolipids
38. Which of the following is peptide bond?
C NHC C O
O
NN
H
a) b) c)
O
H2 H
2 d)none of these
39. The lecithins are required for
a) normal transport and utilisation of other lipids b) the organisation of the cell structure
c) both a & b d) none of these
40. ---------------- is a non-reducing sugar
a) Glucose b) Fructose c) Sucrose d) All of these
41. When starch is boiled with dilute acid it yields [June-2015]
a) glucose b) fructose c) raffinose d) maltose
5 MARK
1. Write note on a) Zwitter ion b) Isoelectric point
a) Zwitter ion
Due to the presence of an acidic and a basic group in the same molecule, amino acids exist in an
ionic form called a Zwitter ion where the proton of –COOH group is transferred to the –NH2 group.
NH2
COOH
NH3
COOR CH R CH
+
Amino acid Zwitter ion
b) Isoelectric point
Depending on the pH of the solution, the amino acid can donate or accept a proton.
NH3
COOR CH
+
Zwitter ion
H+ OHCOOR CH
High pH (anion)
NH3
R CH
+
Low pH (cation)
COOH
NH2
Acidic solution Basic solution
III III
When an ionised form of amino acid is placed in an electric field, it will migrate towards theposite
electrode. In acidic solution (low pH), the cation (II) move towards cathode. In basic solution (high pH),
the anion (III) move towards anode. The zwitter ion does not move towards any of the electrodes.
The pH at which the amino acid shows no tendency to migrate when placed in an electric field is known
as isoelectric point.
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22. CHEMISTRY IN ACTION
Blue print 1 Mark = 0 3 Mark = 1 5 Mark = 1 Total marks = 8
Govt. Examination questions 3 MARK
1. Define chemotherapy [Sep-2014]
Treatment of certain diseases by destroying the invading organism without damaging the cells of the
host, by the use of certain organic compounds is known as chemotherapy.
2. Write the characteristics of dyes [Mar-2011, Mar-17]
Refer – 5 mark – Q – 2
3. What are anaesthetics ? Give example [June-2007, June-2008, Sep-2010, Sep-2012]
Refer – 5 mark – Q – 4
4. Why Iodoform and phenolic solutions are called antiseptic? [Sep-2006, Mar-2008, Sep-2016]
Or Write a brief note on antiseptic
Antiseptic is a substance that renders micro organisms innocuous by killing them or preventing their
growth. This term is used particularly for preparations applied to living tissues.
Iodoform CHI3 and phenolic solutions (0.2 percent solution of phenol) are called antiseptic because they
kill micro organisms or prevent their growth
5. What are antibiotics? [June-2010, June-2013]
Refer – 5 mark – Q – 5
6. What are antipyretics? Give an example [Mar-2013]
Antipyretics are the compounds which are used for the purpose of reducing fever.
They lower the body temperature to the normal.
Examples: i) Aspirin ii) Paracetamol
7. What are antacids? Give examples [Sep-2007, Sep-2011, June-2012, Mar-2014]
Antacids are drugs which are used for balancing acidity in stomach and for providing relief from
burning sensation. They are available in tablets and syrup.
Antacids contain magnesium hydroxide and aluminium hydroxide, in addition to flavouring agents and
colour.
8. What are food preservatives? Give examples [Mar-2015]
A chemical substance which prevents the spoilage of food material by destroying the food-spoiling
microorganisms in it is called a food preservative.
Examples: i) Sodium benzoate ii) Potassium metabisulphite
9. What are artificial sweetening agents? Give two examples. [Sep-2009, Sep-2015]
Certain organic compounds which have been synthesized in laboratories are known to be many times
sweeter than cane sugar. Such compounds are called artificial sweetening agents or artificial sweeteners.
Examples : i) Saccharin, ii) Dulcin, iii) Cyclamate, iv) Nectarin and v) Sucralose
10. What are antioxidants? Give examples [Mar-2012, Mar-2016]
The substances that act against oxidants are called antioxidants. Antioxidants thus minimise the
damage caused by oxidants. Antioxidants protect us against cardiovascular disease, cancer and cataract
and they slow down the effect of ageing. Antioxidants act as radical inhibitors.
Examples: vitamin C, vitamin E and -carotene.
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11. What are chromophores ? Give two examples [Mar-2007, June-2009, Mar-2010, June-2014]
An organic compound appears coloured due to the presence of certain unsaturated groups (the groups
with multiple bonds) in it. Such groups with multiple bonds are called chromophores.
Examples :
N N = N
O
C = O
O
O
Nitro Azoxy Carbonyl
12. What are auxochromes? Give an example [Jun-2016]
The presence of certain groups which are not chromophores themselves, but deepen the colour of the
chromogen. Such supporting groups are called auxochromes.
Eg: –OH, –NH2, –NHR, NR2.
13. How is Terylene (or Dacron) prepared? Write its use. [Mar-2006]
Glycol reacts with Terephthalic acid to form the condensation polymer ‘Terylene’ (also known as
dacron or terene), which is used as a synthetic fibre.
n HO-CH2-CH2-OH + n HO-C
HO-[-CH2-CH2-O-C
O
(2n-1) H2OC-O-]n H
O
C-OH
O
O
+
13. Write the preparation and use of Buna-S rubber [June-2006, June-2011]
Refer – 5 mark – Q – 6
14. Write the preparation and use of Buna-N rubber [Sep-2008, Sep-2013]
Refer – 5 mark – Q – 6
15. How is nylon-66 prepared? Give its use [Mar-2009, June-2015]
Refer – 5 mark – Q – 7
5 MARK
1. Explain briefly on characteristics of rocket propellants. [March-2006, June-2006, June-2007, Sep-
2007, June-2008, Mar-2009, June-2010, Mar-2011, Sep-2011, Mar-2014, Mar-2016]
1. Rocket motors are used both in space vehicles and in offensive weapons such as missiles.
2. The propulsion system in most space vehicles consists of rocket engines powered by chemical
propellants. These are also called rocket propellants.
3. Propellants are combustible compounds which on ignition undergo rapid combustion to release
large quantities of hot gases.
4. A propellant is a combination of an oxidiser and a fuel.
5. Examples for propellants are : Hydrazine, Liquid hydrogen, Polyurethane, etc.
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6. Working of a propellant:
When a propellant is ignited, it burns to produce a large quantity of hot gases. These gases then
come out through the nozzle of the rocket motor. The passage of gases through the nozzle of the
rocket motor, provides the necessary thrust for the rocket to move forward according to the
Newton’s Third law of Motion (to every action, there is an equal and opposite reaction).
2. What are dyes? How are they classified? Write the characteristics of a dye. [June-2009]
Dyes are coloured compounds used in imparting colour to textiles, food stuffs etc.
Classification of dyes:
Natural dyes: These are obtained from vegetable sources.
Synthetic dyes: These are prepared from aromatic compounds obtained from coal tar.
Characteristics of a dye
A dye should have the following characteristics:
1. It should have a suitable colour.
2. It should be able to fix itself or be capable of being fixed to the fabric.
3. It should be fast to light.
4. It should be resistant to the action of water, dilute acids and alkalies
3. Explain chromophore and auxochrome theory (Otto Witt theory) of dyes.
Or Write note on colour and structure of dyes
[Sep-2008, Sep-2009, Sep-2010, June-2011, June-2012, Sep-2012, Sep-2013]
1. Chromophores:
An organic compound appears coloured due to the presence of certain unsaturated groups (the groups
with multiple bonds) in it. Such groups with multiple bonds are called chromophores.
Example:
N C=O
O
O
Nitro Carbonyl
N=N
Azo Quinonoid
2. Chromogen:
The compound containing the chromophore group is called chromogen. The colour intensity increases
with the number of chromophores or the degree of conjugation. For example, ethene (CH2 = CH2) is
colourless, but the compound CH3 – (CH = CH)6 – CH3 is yellow in colour.
3. Auxochromes:
The presence of certain groups which are not chromophores themselves, but deepen the colour of the
chromogen. Such supporting groups are called auxochromes. Auxochromes may be acidic (phenolic)
or basic. Some important auxochromes are –OH, –NH2, –NHR, NR2.
4. The presence of an auxochrome in the chromogen molecule is essential to make it a dye. However, if
an auxochrome is present in the meta position to the chromophore, it does not affect the colour.
5. For example, in the compound p-hydroxyazobenzene (a bright red dye),
N=N OH
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N=Na) azobenzene, is the chromogen
b) diazo group, – N = N – is the chromophore
c) hydroxyl group, –OH is auxochrome
4. What are anaesthetics ? Give example [Sep-06, Mar-10, Jun-13, Jun-14, Jun-15, Mar-17]
The drugs which produce loss of sensation are called anaesthetics.
They are classified into two types.
i) General anaesthetics are the agent, which bring about loss of all modalities of sensation, particularly
pain along with ‘reversible’ loss of consciousness.
ii) Local anaesthetics prevent the pain sensation in localised areas without affecting the degree of
consciousness.
Examples
i) Nitrous oxide N2O : It is the safest anaesthetic agent. It used after mixing general aneaesthetic like ether.
ii) Chloroform CHCl3 : With oxygen it forms a toxic carbonyl chloride. Hence it is not used now.
iii) Ether : C2H5–O–C2H5 This is mixed with stabilizer 0.002% propyl halide. After absorption by
tissues it attacks the central nervous system and makes unconscious
5. Write briefly on antibiotics. In what way antispasmodics are helpful? [Mar-2015]
Antibiotics: Many microorganisms (bacteria, fungi and moulds) produce certain chemicals which inhibit the growth
or metabolism of some other microorganism. Such chemical compounds are known as antibiotics.
Example: Penicillin C9H11N2O4S-R
Penicillin is used for rheumatic fever, narrowing of heart wall, bronchitis and pneumonia.
Antispasmodics:
Antispasmodics are medicines which are used to relieve cramps, spasms of the stomach, intestines and
bladder. Some are used with antacid or other medicine in the treatment of peptic ulcer. These medicines
prevent nausea, vomiting and motion sickness. In certain surgical and emergency procedure, these are
used to help relax stomach and intestine for certain types of examination or diagnosis.
Example: Anisotropine, Atropine, Dicyclomine, Cindinium cap
6. Write short note on synthetic rubbers. Or Write briefly on Buna rubbers
[Mar-2007, Sep-2014]
i) Buna-S rubber It is obtained by the copolymerization of butadiene and styrene in presence of sodium metal.
+
butadiene
Buna-S
Nan CH=CH2
Styrene
C6H5
CH2CH CH2CHCH2( )nCH
C6H5
n CH2=CH CH=CH2
In the name Buna-S, Bu stands for butadiene, na for sodium (acting as polymerization initiator), and S-
for styrene.
Buna-S is also called General Purpose Rubber Styrene (GRS)
Properties and Uses : Buna-S is extremely resistant towards wear and tear and used for the manufacture of tyres, rubber tubes
and other mechanical rubber goods.
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i) Buna-N rubber
It is obtained as a result of copolymerisation of two parts of butadiene with one part of acrylonitrile in
the presence of sodium metal.
CH2=CH
CH
CH2
+
CH=CH2
CN
Acrylonitrile
butadiene Buna-N
CH2=CH
CH
CH2
+
Na
CH
CH2
CH2CH
CH
CH2
CH2CH CH CH2
CN
( )n
Properties and Uses
Buna-N rubber is hard and extremely resistant to the swelling action of oils (petrol), solvents, heat etc.
Therefore, it is used for the manufacture of storage tanks for the solvents.
7. Write the preparation, properties and uses of Nylon-66 [March-2008, Mar-2012, Mar-2013] Nylon-66 is obtained by condensing adipic acid with hexamethylenediamine with the elimination of
water molecule. The chain length depends upon the temperature and time for which the process
is carried out.
n H2N(CH2)6NH2 n HO-C-(CH2)4-C-OH -(-N-(CH2)6-N-C-(CH2)4-C-)n-+
O O H H O O
Hexamethylene diamine Adipic acid Nylon-66
The polyamides are identified by numbers. These numbers refer to the number of carbon atoms in
diamine and in the dibasic acid. As in the above case, the carbon atoms are 6 in each case, therefore, the
product is described as nylon-66.
Properties and Uses:
Nylon-66 is a linear polymer, and has very high tensile strength. It shows good resistance to abrasion.
Nylon-66 is usually fabricated into sheets, bristles for brushes and in textile. Crinkled nylon fibres are
used for making elastic hosiery.
8. How will you prepare ‘Buna-N’ and ‘Nylon 66’? Mention their uses [Sep-2015] Refer – Q-6 & 7
9. Write short notes on a) antacids b) analgesics [Jun-2016] a) Antacids Antacids are drugs which are used for balancing acidity in stomach and for providing relief from
burning sensation. Antacids contain magnesium hydroxide and aluminium hydroxide, in addition to
flavouring agents and colour.
b) Analgesics Analgesics are the compounds which relieve all sorts of pains without the loss of consciousness.These
are also called as pain killers, or pain relievers. These are effective in headaches, myalgia and arthalgia.
Certain Narcotics are also used as Analgesics.
Examples : Aspirin, Novalgin
10. Write note on chemicals in food [Sep-2016]
1. Foodpreservative 2. Artificial Sweetening agents 3. Antioxidants
Refer – 3 mark – Q – 8, 9, 10
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Additional questions
3 MARK
1. What are antiprotozoals or antimalarials? Give examples
Malaria causes shivering and fever. The body temperature rises to 103-106 oF. It causesphysical
weakness with the side-effects in lever and also causes aneamia. Cinchona bark which gives rise to
quinine acts as antimalarial. Examples: i) Quinine ii) pyrimethamine iii) chloroquine
2. Write note on penicillins
Penicillins is the name given to the mixture of natural compounds having the molecular formula
C9H11N2O4 S-R, and differing only in the nature of R. Depending upon the nature of R, there are at least
six natural penicillins.
Uses. Penicillin is extensively used for rheumatic fever, narrowing of heart wall, bronchitis, and
pneumonia.
3. Write note on formaldehyde Resins Formaldehyde resins are typical thermosetting plastics. This class of plastics include phenol-
formaldehyde, urea-formaldehyde and melamineformaldehyde resins. On heating, these resins become
highly cross-linked thereby forming infusible and insoluble product.
4. How is polythene prepared? Polyethylene, (polyethene) obtained from ethylene (ethene). Ethene polymerizes under high pressure
and high temperature to give polyethene (also called polyethylene). This polymerization is catalysed by
traces of oxygen or organic peroxides.
n CH2 = CH2 pressurehighe,temperaturhighperoxide,
-(-CH2-CH2-)-n
Ethene Poly Ethene
5. Write the preparation and use of polystyrene Polystyrene is obtained from the monomer styrene
n C6H5-CH = CH2-(-CH2-CH-)n-
C6H5Styrene
polystyrene
Properties and Uses: Polystyrene is good transparent polymer. It is used for hot-drink cups, toys,
household articles, etc.
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(Q-70) HYDROXY DERIVATIVES
Hints:
1. General formula of aliphatic monohydric alcohols: CnH2n+2O
2. Alcohols liberate hydrogen gas with metallic sodium
3. Primary alcohols
No turbidity with Lucas reagent at room temperature
Give red colour in Victor Meyer’s test
Give aldehydes on oxidation and then carboxylic acid (same number of carbons)
Give aldehydes with Cu at 573 K (dehydrogenation)
4. Secondary alcohols
Give turbidity with Lucas reagent at room temperature after 5 – 10 minutes
Give blue colour in Victor Meyer’s test
Give ketones on oxidation and then carboxylic acid
(Carboxylic acid contains one carbon less than alcohol)
Give ketones with Cu at 573 K (dehydrogenation)
5. Tertiary alcohols
Give turbidity with Lucas reagent at room temperature immediately
Give no colour in Victor Meyer’s test
Give ketones on oxidation and then carboxylic acid
(Ketone contains one carbon less than alcohol & Carboxylic acid contains two carbons less than
alcohol)
Give alkene with Cu at 573 K (dehydration)
6. Alcohols containing CH3CHOH- group answer haloform test
(form iodoform with I2 / NaOH)
CH3CH
2OH CCH3 CH3
OH
H
Ethyl alcohol iso propyl alcohol
CCH3 CH3
OH
HCH2
sec.butyl alcohol
7. Benzyl alcohol (C6H5CH2OH) is a colourless pleasant smelling liquid.
8. Benzyl acetate (CH3COO CH2C6H5) has fragrance of jasmine and it is used in perfumery
9. Phenol (C6H5OH) gives violet colour with neutral ferric chloride
10. Phenol decolourises Br2 / H2O and gives white precipitate
11. Phenol gives red orange dye with benzene diazonium chloride (Dye test)
12. Salicylic acid gives violet colour with neutral ferric chloride and gives brisk effervescence with
NaHCO3
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1. An organic compound (A) C3H8O with Lucas reagent gives turbidity after 5 – 10 minutes. (A) with
P / I2 gives (B). Compound (B) on treatment with silver nitrite gives (C) which gives blue colour
with nitrous acid. Identify (A),(B) and (C) and explain the reactions. [Sep-2013]
i) (A) turbidity after 5-10 minutes
2o alcohol iso propyl alcohol
Lucas reagent
C3H
8O
CH3 CH
OH
CH3CH3 CH CH3
P/I2
I
AgNO2
2-nitropropane
HO-N=O
CH3 CH CH3
NO2
iso propyl iodideiso propyl alcohol
(A) (B) (C)
CH3C CH3
NO2
N=O
Pseudo nitrol (Blue colour)
ii)
2. An organic compound (A) of molecular formula C3H8O gives turbidity within 5-10 minutes on
reaction with anhydrous ZnCl2/HCl. Compound (A) on treatment with sodium hypochlorite gives
a carbonyl compound (B) which on further chlorination gives compound (C) of molecular formula
C3H3OCl3. Identify (A), (B) and (C) and explain the reactions. [Mar-2006]
i) (A) Turbidity within 5 - 10 minutesanhydrous ZnCl
2 / HCl
C3H
8O (A) secondary alcohol isopropyl alcohol
ii)
CCH3
CH3
OH
H
+ (O)Sodium hypochlorite
CCH3
CH3
O
acetone
+ H2O
isopropyl alcohol (A) (B)
iii)
CH
3-CO-CH
3 + 3Cl
2 CCl
3-CO-CH
3 + 3HCl
Acetone Trichloro acetone(B) (C)
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3. An organic compound (A) of molecular formula C2H6O on treatment with PCl5 gives compound
(B). Compound (B) reacts with KCN to give a compound (C) of molecular formula C3H5N which
undergoes acid hydrolysis to give compound (D) which on treatment with sodalime gives a hydro
carbon. Identify (A), (B),(C) and (D) and explain the reactions. [June-2006]
i) (A)
C2H
6O
CnH
2n+2O aliphatic alcohol Ethyl alcohol
ii)
iii)
CH
3CH
2Cl + KCN CH
3CH
2CN + KCl
Ethyl chloride Ethyl cyanide(B) (C)
iv)
CH3CH
2OH + PCl
5 CH
3CH
2Cl + POCl
3 + HCl
Ethyl alcohol Ethyl chloride(A) (B)
CH3CH
2CN CH
3CH
2COOH
H2O / H+
(C) (D)Propanoic acidEthyl cyanide
CH3CH
3
NaOH / CaO
Ethane
4. An organic compound (A) of molecular formula CH2O reacts with CH3MgI to give compound (B).
Compound (B) liberates hydrogen with metallic sodium. Compound (B) in the presence of
con.H2SO4 at 410 K on dehydration to give compound (C) of molecular formula C4H10O. Identify
(A), (B) and (C). Explain the reactions. [Sep-2016]
H C
O
CH3-MgI
H OH
H + H C
O
H
MgI
CH3
HydrolysisCH
3-CH
2-OH + Mg
I
OH
FormaldehydeEthyl alcohol
i)
(A) (B)
ii) C2H5 – O – H + H – O – C2H5 4SO
2Hcon.
C2H5 – O – C2H5 + H2O
Diethyl ether (C)
5. Two isomers (A) and (B) have the same molecular formula C4H10O. (A) when treated with copper
at 573 K gives an alkene (C) of molecular formula C4H8. (B) on heating with copper at 573 K gives
(D) of molecular formula C4H8O which does not reduce Tollen’s reagent but answers iodoform
test. Identify (A),(B),(C) and (D) and explain the reactions. [Mar-2009]
i) (A) & (B) isomers
C4H
10O
CnH
2n+2O aliphatic alcohol
CCH3
CH3
CH3
OHdehydration
CCH3
CH3
CH2
Cu/573K
isobutylene
+ H2O
tert. butyl alcohol (A) (C)
iii)
CH
3CH
3CH
2 CH
OH
Cu, 573KCH
3CH
3CH
2 C
O
+ H2
sec.butylalcohol ethylmethyl ketone(B) (D)
ii)
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6. Compound (A) of molecular formula C3H8O liberates hydrogen with sodium metal. (A) with P/I2
gives (B). Compound (B) on treatment with silver nitrite gives (C) which gives blue colour with
nitrous acid. Identify (A), (B), and (C) and explain the reactions. [Sep-2009]
i) (A) liberates H2 gas (A) is an alcohlol
Na
CH3 CH
OH
CH3CH3 CH CH3
P/I2
I
AgNO2
2-nitropropane
HO-N=O
C3H
8O
CH3 CH CH3
NO2
iso propyl iodideiso propyl alcohol (A) (B) (C)
CH3 C CH3
NO2
N=O
Pseudo nitrol (Blue colour)
ii)
7. An organic compound (A) C3H8O answers Lucas test within 5-10 minutes and on oxidation forms
(B) C3H6O. This on further oxidation forms (C) C2H4O2 which gives effervescence with Na2CO3.
(B) also undergoes iodoform reaction. Identify A, B, and C. Explain the conversion of A to B and
C. [June-2007, June-2009]
i) (A) Answers Lucas test within 5-10 minutes
C3H
8O
Secondary alcohol
(isopropyl alcohol)
ii)
CH3 C
O
CH3CH3 C
O
CH3
H
H (O) + H2OK
2Cr
2O
7 / H+
+
isopropyl alcohol Acetone(A)
(B)
iii)
(O)
K2Cr
2O
7 / H+
+CH3 C
O
CH3CH
3COOH
Acetone Acetic acid
(B) (C)
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8. An organic compound (A) of molecular formula C2H6O liberates hydrogen gas with metallic
sodium. Compound (A) on heating with excess of con.H2SO4, at 440 K gives an alkene (B).
Compound (B) when oxidised by Bayer’s reagent gives compound (C). Identify (A), (B) and (C)
and explain the reactions. [June-2009, Mar-2014]
i) (A) liberates H2 gas (A) is an alcohol
Na
C2H
6O
ethyl alcohol
ii)
CH3 HSO4 + H2O
440 KOH + H HSO4
Con(excess)
CH2 CH3CH2
CH2 CH2
H HSO4
CH2CH2 + H2SO4
ethylene
(A)ethyl alcohol
(B)
iii)
CH2=CH
2 + H
2O + (O) HO-CH
2-CH
2-OH
Bayer's reagent
ethylene ethylene glycol(B) (C)
9. An organic compound (A) of molecular formula C2H6O reacts with Al2O3 at 620 K and gives (B)
of molecular formula C2H4. (B) reacts with Bayer’s reagent to give compound (C) of molecular
C2H6O2. Identify (A), (B) and (C) and explain the reactions. [June-2010]
i) (A) (B) Al
2O
3 / 620 K
C2H
6O C
2H
4
CH3CH
2OH CH
2=CH
2 + H
2O
Al2O
3 / 620 K
Ethyl alcohol Ethylene
(A) (B)
ii) (B) (C)Bayer's reagent
C2H
6O
2
CH2=CH
2 + H
2O + (O) HO-CH
2-CH
2-OH
Bayer's reagent
Ethylene
(B)
Ethylene glycol
(C)
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10. An organic compound (A) (C2H6O) liberates hydrogen with sodium metal. (A) when heated with
alumina at 620 K gives an alkene (B) which when passed through Bayer’s reagent gives (C)
(C2H6O2). (C) reacts with PI3 and gives back (B). Identify (A), (B) & (C) and explain the reactions.
[Mar-2011]
i) (A) H2 gas (A) is an alcohol ethyl alcohol
Na
C2H
6O
ii) CH3CH
2OH CH
2=CH
2 + H
2O
Al2O
3 / 620 K
Ethyl alcohol Ethylene(A) (B)
iii) CH2=CH
2 + H
2O + (O) HO-CH
2-CH
2-OH
Bayer's reagent
cold dilute alkaline KMnO4Ethylene Ethylene glycol(B) (C)
iv)
PI3
ethylene
CH2 OH
CH2 OH
ethylene diiodide
CH2I
CH2I
- I2CH2
CH2
Ethylene glycol (C) (B)
11. An organic compound (A) (C2H6O2) liberates hydrogen with metallic sodium. Compound (A)
when heated with anhydrous ZnCl2 ultimately gives (B) (C2H4O) whereas when heated with conc.
Phosphoric acid gives (C) (C4H10O3). (A) on oxidation with acidified K2Cr2O7 gives compound (D)
(CH2O2). Identify (A),(B),(C), (D) and explain the reactions. [Sep-2012]
i) (A) H2 gas (A) is an alcohol ethylene glycol
Na
C2H
6O
2
ii)
C
C
H
H
H
OHanhydrous
ZnCl2
H2OOH
H C
C
H
H
H
O H C
C
H
H
H
O
H
AcetaldehydeEthylene glycol (A) (B)
iii)
CH2CH2HO
CH2CH2HO
O H
OH
H3PO4
CH2CH2HO
CH2CH2HO
O
diethylene glycolEthylene glycol (A) (C)
iv)
CH2 OH
CH2 OH+ 3(O)
acidified K2Cr
2O
7 HCOOH
HCOOH
+
formic acid
+ H2O
Ethylene glycol (A) (D)
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12. An organic compound (A) of molecular formula C2H4 reacts with alkaline KMnO4 and gives
compound (B) of molecular formula C2H6O2. Compound (B) when heated with anhydrous ZnCl2
forms (C) of molecular formula C2H4O. Identify A, B and C and explain the reactions. [Mar-2015]
i)
CH
2=CH
2 + H
2O + (O) HO-CH
2-CH
2-OH
Ethylene Ethylene glycol(A) (B)
ii)
alkaline KMnO4
C
C
H
H
H
OHanhydrous
ZnCl2
H2OOH
H C
C
H
H
H
O H C
C
H
H
H
O
H
AcetaldehydeEthylene glycol (B) (C)
13. Compound A with molecular formula C3H6 is obtained from petroleum. When A is treated with
chlorine at 773 K compound B of molecular formula C3H5Cl is obtained. When B is treated with
Na2CO3 solution at 773 K /12 atm. it gives the compound C with molecular formula C3H6O. C on
treatment with HOCl followed by hydrolysis with NaOH gives D having molecular formula
C3H8O3. Identify (A),(B),(C) and (D) and explain the reactions. [Mar-2008]
i) (A) obtained from petroleum Propylene
C3H
6
ii)
CH2
CH
CH3
Cl2
CH2
CH
CH2Cl
Na2CO
3
CH2
CH
CH2OH
HOCl
CH2
CH
OH
Cl
CH2OH
NaOH
CH2
CH
OH
CH2OH
OH
propylene allyl chloride allyl alcoholglycerol
chlorohydringlycerol
(A) (B) (C) (D)
solution
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14. An organic compound (A) of molecular formula C3H8O3 is obtained as by-product in the
manufacture of soap. Compound (A) on heating with P2O5 gives an unsaturated compound (B) of
molecular formula C3H4O. Compound (A) with well cooled mixture of Conc.H2SO4 and
fumingHNO3 form compound (C) with is an explosive. Identify A, B and C and explain the
reaction. [Sep-2014]
i) (A) obtained as by-product in the manufacture of soap Glycerol
C3H
8O
3
ii)
C
C
H
H
2H2O
OH
C OH
OH
H
H
H
C
C
H
H
OH
CH
C
C
CH2
H
HO
unstable
P2O5
AcroleinGlycerol (B)(A)
iii)
con. H2SO4
CH2
CH
OH
CH2OH
OH
Glycerol
+ 3 H ONO2
CH2
CH
CH2
+ 3 H2O
ONO2
ONO2
ONO2
Trinitroglycerine (A) (C)
15. Compound A of molecular formula C7H8 is treated with chlorine and then with NaOH to get
compound B of molecular formula C7H8O. B on oxidation by acidified K2Cr2O7 gives compound C
of molecular formula C7H6O. Compound C on treatment with 50% cautic soda gives the
compound B and also D. Identify (A),(B),(C) and (D) and explain the reactions. [June-2008]
i)
C6H
5CH
3 HCl
Cl2
C6H
5CH
2Cl C
6H
5CH
2OH
NaCl
NaOH
Toluene Benzyl chloride Benzyl alcohol(A) (B)
ii)
C
6H
5CH
2OH + (O) C
6H
5CHO + H
2O
K2Cr
2O
7 / H+
Benzyl alcohol (B) Benzaldehyde (C)
iii)
C6H5CHO
H2O50% NaOH
C6H5CHO
+ C6H5CH2OH C6H5COOH+
Benzyl alcohol Benzoic acid
Benzaldehyde (C)
(B) (D)
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16. Compound (A) of molecular formula C7H8 is treated with chlorine and then with NaOH to get
compound B of molecular formula C7H8O. B liberates H2 with metallic sodium. Compound B on
reaction with CH3COOH in the presence of conc. H2SO4 forms compound C of molecular formula
C9H10O2 which has the fragrance of jasmine. Identify (A),(B),(C) and explain the reactions. [Mar-2012]
i) (A) (B)Cl
2 NaOH
C7H
8C
7H
8O
NaH
2 gas (B) is an alcohol benzyl alcohol
C6H5-CH3
Cl2
HClC6H5-CH2Cl C6H5-CH2OH NaCl+
Toluene Benzyl chloride Benzyl alcohol
NaOH
(A) (B)
C6H
5CH
2O H HO C
O
CH3+
conc. H2SO
4C
6H
5CH
2O C
O
CH3
Benzyl acetate (C)Benzyl alcohol (B)
ii)
iii)
17. An organic compound (A) C7H8O liberates hydrogen with metallic sodium. (A) on treatment with
acidified K2Cr2O7 gives (B) C7H6O. Compound (B) when treated with conc.NaOH gives A. B with
acetic anhydride in the presence of sodium acetate gives (C) of molecular formula C9H8O2. Find
A, B and C. Explain the reactions. [Sep-2008]
i) (A) liberates H2 gas with metallic Na
C7H
8O
So, (A) is benzyl alcohol C6H
5CH
2OH
C6H
5CH
2OH + (O) C
6H
5CHO + H
2O
K2Cr
2O
7 / H+
Benzyl alcohol Benzaldehyde(A) (B)
C6H5CHO
H2Oconc. NaOH
C6H5CHO
+ C6H5CH2OH C6H5COOH+
Benzyl alcohol Benzoic acid
Benzaldehyde (B)
(A)
C6H
5-C = O
H
+ H2 CH-CO
O
CH3-CO H
OH
CH3COONa
C6H
5-C =
H
CH-COOH + CH3COOH
Cinnamic acidBenzaldehyde
(B) (C)
ii)
iii)
iv)
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18. An aromatic hydrocarbon (A) reacts with propene in the presence of anhydrous AlCl3 to give a
compound (B) with the molecular formula C9H12. Further compound (B) undergoes oxidation in
the presence of air to give hydroperoxide (C). Compound (C) decomposes in HCl acid solution to
give compound (D) and acetone. Identify (A),(B),(C), (D) and explain the reactions.[Jun-12,Jun-13]
+ CH3-CH=CH2
anhydrous
AlCl3
cumene cumene hydro peroxidebenzene
CH3-CH-CH3
O2
(air)
CH3-C-CH3
O
O
H
HCl
propene
(A) (B) (C)
acetonephenol
OH
+ CH3-CO-CH3
(D)
19. An organic compound (A) of molecular formula C6H6O gives violet colouration with neutral
FeCl3. Compound (A) on treatment with metallic Na gives compound (B). Compound (B) on
treatment with CO2 at 400 K under pressure gives (C). This product on acidification gives
compound (D) (C7H6O3) which is used in medicine. Identify (A), (B),(C) and (D) and explain the
reactions. [Sep-2006, Mar-2016]
i) (A) Violet colourneutral FeCl
3
C6H
6O
(A) Phenol
2C6H
5OH + 2Na 2C
6H
5ONa + H
2
Phenol Sodium phenoxide(A) (B)
iii)
ii)
ONa
H
OH
COONaCO2
sodium phenoxide sodium salicylate
400 K , 4-7 atm
dil. HCl
OH
COOH
salicylic acid
+ NaCl
(D)(C)(B)
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20. An organic compound A (C6H6O) gives maximum of two isomers B and C when an alkaline
solution of A is refluxed with chloroform at 333 K. B on oxidation gives an acid D. The acid D is
also obtained by treating sodium salt of A with CO2 under pressure. Identify (A), (B), (C) and (D)
and explain the reactions. [Mar-2007, Sep-2010]
i) (A) (B) + (C)CHCl
3 / NaOH
C6H
6O
333 K(B) & (C) Isomers Riemer-Tiemann reaction
OH
CHO
CHO
CHCl3
phenol o-hydroxy benzaldehyde
NaOH p-hydroxy benzaldehyde +
(A) (B)
(C)
iii)
Salicylic acid
(O)
o-hydroxy benzaldehyde
COOH
(B) (D)
CO2
sodium phenoxide sodium salicylate
400 K , 4-7 atm
dil. HCl
salicylic acid
+ NaCl
ONa OH
COONaH
OH
COOH
(D)
OHOH
CHO
OH OH
ii)
iv)
21. An organic compound (A) of molecular formula C6H6O gives violet colour with neutral FeCl3. (A)
gives maximum of two isomers (B) and (C) when an alkaline solution of (A) is refluxed with CCl4.
(A) also reacts with C6H5N2Cl to give the compound (D) which is red orange dye. Identify (A), (B),
(C) and (D) and explain the reactions. [Sep-2007]
i) (A) Violet colour Phenolneutral FeCl
3
ii)
C6H
6O
OH OH
COOH
OH
COOH phenol o-hydroxy
benzoic acid
NaOH
p-hydroxy benzoic acid
+
CCl4
(A)
(B) (C)
iii)
OH
benzene diazonium
chloridep-hydroxy azobenzene
(Red orange dye)Phenol
N=N-Cl + H OH N=N OH
(A)(D)
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22. An organic compound A of molecular formula C6H6O gives violet colouration with neutral FeCl3.
It reacts with CHCl3 and NaOH and gives two isomers (B) and (C) with molecular formula
C7H6O2. Compound (A) reacts with ammonia at 473 K in the presence of ZnCl2 and gives
compound (D) with molecular formula C6H7N. Compound (D) undergoes carbylamine test.
Identify (A), (B), (C) and (D) and explain the reactions [Mar-2010, Mar-2013]
i) (A) violet colourneutral FeCl
3
C6H
6O
(A) is phenol
OH OH
CHO
OH
CHO
CHCl3
phenol o-hydroxy
benzaldehyde
NaOH
p-hydroxy benzaldehyde
+
(A) (B) (C)
C6H
5OH + NH
3 C
6H
5NH
2 + H
2O
ZnCl2
phenol aniline
(A) (D)
ii)
iii)
23. An organic compound (A) of molecular formula C6H6O gives violet colouration with neutral
FeCl3. It reacts with CHCl3 and NaOH and gives two isomers (B) and (C) with molecular formula
C7H6O2. Compound (A) reacts with diazomethane in alkaline medium gives an ether (D). Identify
(A), (B), (C) and (D) and explain the reactions [Mar-2017]
i) (A) violet colourneutral FeCl
3
C6H
6O
(A) is phenol
OH OH
CHO
OH
CHO
CHCl3
phenol o-hydroxy
benzaldehyde
NaOH
p-hydroxy benzaldehyde
+
(A) (B) (C)
C6H
5OH + CH
2N
2 C
6H
5OCH
3 + N
2
phenol anisole
(A) (D)
iii)NaOH
ii)
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24. An organic compound (A) C6H6O gives violet colour with neutral FeCl3. (A) gives maximum of
two isomers (B) and (C) when an alkaline solution of (A) is refluxed with CCl4. (A) also reacts
with formaldehyde and NaOH to give compound (D). Identify (A), (B), (C) and (D) and explain
the reactions. [June-2011]
i) (A) violet colour (A) is phenolneutral FeCl
3
C6H
6O
ii)
OH OH
COOH
OH
COOH phenol o-hydroxy benzoic acid
NaOH
p-hydroxy benzoic acid
+
CCl4
(A) (B)(C)
iii)
phenol
+NaOH
HCHOp-hydroxy phenyl methanol
OH OH
CH2OH(A)
(D)
25. An organic compound (A) C6H6O gives violet colour with neutral FeCl3. (A) gives two isomers (B)
and (C) when treated with CCl4 and NaOH. (A) on oxidation with CrO2Cl2 gives (D) of molecular
formula C6H4O2. Identify (A), (B), (C) and (D) and explain the reactions. [June-2014]
i) (A) violet colour (A) is phenolneutral FeCl
3
C6H
6O
ii)
isomers
OH OH
COOH
OH
COOH phenol o-hydroxy benzoic acid
NaOHp-hydroxy benzoic acid
+
CCl4
(A) (B)
(C)
iii)
phenol
2 (O)
Quinone
OH
(A) (D)
CrO2Cl2+ OO
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26. Compound A (C6H6O) gives violet colouration with neutral FeCl3. Sodium salt of (A) reacts with
CO2 at 400 K / 4 to 7 atm followed by acidification with HCl gives B (C7H6O3). B also gives violet
colouration with neutral FeCl3 and gives effervescence with NaHCO3 solution. Compound A
reacts with NH3 at 473 K in the presence of anhydrous ZnCl2 to give compound C (C6H7N) which
undergoes carbylamines test. Identify A, B, C and explain the reactions. [Sep-2011]
i) (A) violet colour (A) is phenolneutral FeCl
3
iii)
C6H
6O
C6H
5OH
ii)
ONa
H
OH
COONaCO
2
sodium phenoxide sodium salicylate
400 K , 4-7 atm
dil. HCl
OH
COOH
salicylic acid
+ NaCl
(B)
C6H5 OH H NH2
anhydrous ZnCl2
+ C6H5 NH2 + H2O
Aniline (C)(A)Phenol
27. An organic compound (A) C6H6O gives violet colour with neutral FeCl3 solution. With NH3 in the
presence of anhydrous ZnCl2, (A) gives (B) (C6H7N). (A) with dimethyl sulphate gives (C)
(C7H8O). What are (A), (B) and (C)? Explain the reactions [June-2015]
i) (A) violet colour (A) is phenolneutral FeCl
3
C6H
6O
ii) C6H
5OH + NH
3 C
6H
5NH
2 + H
2O
iii) C6H
5OH + (CH
3)
2SO
4 C
6H
5OCH
3 + CH
3 HSO
4
anhud. ZnCl2
Aniline (B)
Anisole (C)
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28. An organic compound of molecular formula C6H5ONa is heated with CO2 at 400 K gives
compound (A) of molecular formula C7H5O3Na. Compound (A) on treating with HCl gives (B).
(B) on further reaction with NaOH / CaO gives (C) of molecular formula C6H6O which on
treatment eith nitrous acid at 200 K gives (D). What are (A), (B),(C) and (D)? Explain the
reactions [Sep-2015]
ONa
H
OH
COONaCO
2
sodium phenoxide sodium salicylate
400 K , 4-7 atm
dil. HCl
OH
COOH
salicylic acid
+ NaCl
(B)
(C)
(A)
i)
OH
COONa
NaOH/CaO
(Sodalime)
OH
+ Na2CO3ii)
salicylic acid (B) phenol
iii)
OH
phenol (C)
HNO2
200 K
OH
paranitrso phenol (D)
NO
29. An organic compound ‘A’ is a sodium salt of phenolic acid with molecular formula C7H5O3Na.
‘A’ on heating with sodalime gives compound ‘B’ of molecular formula C6H6O. ‘B’ gives violet
colour with neutral ferric chloride. ‘B’ on treatment with C6H5COCl in the presence of NaOH
gives an ester ‘C’ Identify ‘A’, ‘B’ and ‘C’. Explain the reactions. [Jun-2016]
OH
COONa
sodium salicylate
(B)
(C)
(A)
i) NaOH/CaO
(Sodalime)
OH
+ Na2CO3
ii)
phenol
C6H5O H Cl
O
C C6H5+
phenol benzoyl chloride
NaOHC6H5O
O
C C6H5
phenyl benzoate
(B)
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(Q-70) d-BLOCK ELEMENTS
Hints:
Element Group Period Colour
Cr 6 4 Silvery white metal
Cu 11 4 Reddish brown metal
Ag 11 5 White lustrous metal
Au 11 6 Lustrous yellow metal
Zn 12 4 Bluish white metal
Metallic compound Colour
Potassium dichromate K2Cr2O7 Orange red crystal
Potassium chromate K2CrO4 Yellow
Coppersulphate pentahydrate CuSO4.5H2O
(Blue Vitriol) Blue
Silver nitrate AgNO3 (Lunar Caustic) White
Zinc carbonate ZnCO3 (Calamine) White powder
Zinc oxide ZnO (Philosopher’s wool) White cloud
1. An element (A) belongs to group number 11 and period number 4. (A) is a reddish brown metal.
(A) reacts with HCl in the presence of air and gives compound (B). (A) also reacts with conc.
HNO3 to give compound (C) with the liberation of NO2. Identify (A),(B) and (C). Explain the
reactions. [Mar-2006, June-2011]
a) Element (A) is copper: Group = 11 and period = 4. It is a reddish brown metal.
b) 2Cu + 4HCl + O2 2CuCl2 + 2H2O
Compound (B) is CuCl2
c) Cu + 4HNO3 (con) Cu(NO3)2 + 2NO2 + 2H2O
Compound (C) is Cu(NO3)2
A Cu copper
B CuCl2 Cupric chloride
C Cu(NO3)2 Cupric nitrate
2. An element A occupies group number 11 and period number 4. This metal is extracted from its
mixed sulphide ore B. A reacts with dil. H2SO4 in the presence of air and forms C which is
colourless. With water C gives a blue compound D. Identify A,B,C and D and explain the
reactions. [Mar-2007]
a) Element (A) is copper : Group = 11 and period = 4.
b) Copper is extracted from its mixed sulphide ore CuFeS2. Compound (B) is CuFeS2
c) 2Cu + 2H2SO4 + O2 2CuSO4 + 2H2O
Compound (C) is anhydrous CuSO4 which is colourless.
d) CuSO4 + 5 H2O CuSO4.5H2O
Compound (D) is CuSO4.5H2O which is blue in colour.
A Cu copper
B CuFeS2 Copper pyrite
C CuSO4 Anhydrous copper sulphate
D CuSO4.5H2O Copper sulphate pentahydrate
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3. An element A belonging to group No. 11 and period No.4 is extracted from the pyrite ore. A reacts
with oxygen at different temperatures forming compounds B and C. A also reacts with conc.
HNO3 to give D with the evolution of NO2. Identify A,B,C and D and explain the reactions.
[Sep-2007]
a) Element (A) is copper: Group=11 and period=4. It is extracted from the pyrite ore
CuFeS2
b) 2 Cu + O2 K1370Below
2 CuO [black cupric oxide]
Compound (B) is CuO
c) 4 Cu + O2 K1370Above
2 Cu2O [red cuprous oxide]
Compound (C) is Cu2O
d) Cu + 4HNO3 (con) Cu(NO3)2 + 2NO2 +2H2O
Compound (D) is Cu(NO3)2
A Cu Copper
B CuO Cupric oxide
C Cu2O Cuprous oxide
D Cu(NO3)2 Cupric nitrate
4. Compound (A) is a sulphate compound of group 11 element. This compound is also called as blue
vitriol. The compound undergoes decomposition at various temperatures.
A o
100 C B
o230 C
C o
720 C D
Identify the compounds A, B, C and D and give equations. [June-2009, June-2014]
a) Compound (A) is CuSO4.5H2O : It is a sulphate compound of group 11 element (Cu). This
compound is also called as blue vitriol.
b) Decomposition of CuSO4.5H2O at various temperatures
CuSO4.5H
2O CuSO
4.H
2O CuSO
4CuO + SO
3
Blue White
100oC 230oC 720oC
- H2O- 4H
2O
Compound (B) is CuSO4.H2O, Compound (C) is CuSO4 and (D) is CuO
A CuSO4.5H2O Copper sulphate pentahydrate
B CuSO4.H2O Copper sulphate monohydrate
C CuSO4 Anhydrous Copper sulphate
D CuO Cupric oxide
5. [A] is a reddish brown metal. It belongs to group 11 and period 4 of the periodic table. When
heated below 1370 K, [A] gives a black compound [B]. When heated above 1370 K, [A] gives a red
compound [C]. With concentrated nitric acid. [A] liberates NO2 gas and gives compound [D].
Identify [A], [B], [C] and [D]. Explain the reactions. [Mar-2010, Mar-2013]
a) Metal (A) is copper : Group = 11 and period = 4. It is a reddish brown metal.
b) 2Cu + O2 K1370Below
2CuO [black cupric oxide]
Compound (B) is CuO
c) 4Cu + O2 K1370Above
2Cu2O [red cuprous oxide]
Compound (C) is Cu2O
d) Cu + 4HNO3 (con) Cu(NO3)2 + 2NO2 + 2H2O
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Compound (D) is Cu(NO3)2
A Cu Copper
B CuO Cupric oxide
C Cu2O Cuprous oxide
D Cu(NO3)2 Cupric nitrate
6. An element (A) occupying group number 11 and belonging to the 4th
period, is reddish brown in
colour. (A) is extracted from its mixed sulphide ore (B). (A) reacts with dilute H2SO4 in the
presence of air to form (C). On treatment with conc. Nitric acid (A) gives compound (D). Identify
(A), (B), (C) and (D). Explain the reactions. [June-2010]
a) Element (A) is copper: Group =11 and period = 4. It is a reddish brown in colour
b) Copper is extracted from its mixed sulphide ore, copper pyrite CuFeS2 (B)
c) 2Cu + 2H2SO4 (dil) + O2 2CuSO4 + 2H2O
Compound (C) is CuSO4
d) Cu + 4HNO3 (conc) Cu(NO3)2 + 2NO2 + 2H2O
Compound (D) is Cu(NO3)2
A Cu Copper
B CuFeS2 Copper pyrite
C CuSO4 Copper sulphate
D Cu(NO3)2 Cupric nitrate
7. (A) occupies group number 11 and period number 4. This metal is extracted from its mixed
sulphide ore (B). (A) reacts with conc. H2SO4 to form (C). Identify (A), (B) and (C). Explain the
reactions [June-2015]
a) Metal (A) is copper: Group =11 and period = 4.
b) Copper is extracted from its mixed sulphide ore, copper pyrite CuFeS2 (B)
c) Cu + 2H2SO4 (conc) CuSO4 + SO2 + 2H2O
Compound (C) is CuSO4
A Cu Copper
B CuFeS2 Copper pyrite
C CuSO4 Copper sulphate
8. An element (A) belongs to group number 11 and period number 4 reacts with conc.H2SO4 to give
its salt (B) with the liberation of SO2 gas. Compound (B) reacts with hydrogen sulphide gas gives
compound (C) which is black in colour. Identify (A), (B) and (C) and explain the reactions.
[Mar-2016]
a) Element (A) is copper: Group =11 and period = 4.
b) Cu + 2H2SO4 (conc) CuSO4 + SO2 + 2H2O
Compound (B) is CuSO4
c) CuSO4 + H2S CuS + H2SO4
Black precipitate
Compound (C) is CuS
A Cu Copper
B CuSO4 Copper sulphate
C CuS Copper sulphide
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9. Compound (A) also known as blue vitriol can be prepared by dissolving cupric oxide in dil.
H2SO4. ‘A’ on heating to 230˚C gives compound ‘B’ which is colourless. ‘A’ reacts with excess
NH4OH and gives ‘C’ which is a complex salt. ‘A’ also reacts with H2S and gives compound ‘D’, a
precipitate which is black in colour. Find out A, B, C and D. Expain the reactions. [Sep-16]
a) Compound (A) is copper sulphate pentahydrate, CuSO4. 5H2O
CuO + H2SO4 CuSO4 + H2O
b) Compound (B) is anhydrous copper sulphate, CuSO4
CuSO4.5H
2O CuSO
4.H
2O CuSO
4
Blue White
100oC 230oC
- H2O- 4H
2O
Anhydrous coppersulphate (B)
c) CuSO4 + 4NH4OH [Cu(NH3)4]SO4 + 4H2O
Tetraamminecopper(II)sulphate (C)
d) CuSO4 + H2S CuS + H2SO4
Copper sulphide (D)
10. Silver reacts with dil. HNO3 to give compound (A) which on heating at 723K gives (B). (B) on
further heating gives (C). Further compound (A) reacts with KBr and gives (D) which is highly
useful in photography. Identify (A),(B),(C) and (D) and explain the reactions. [June-2006 , Mar-2009]
a) 3Ag + 4HNO3 (dil) 3AgNO3 + NO + 2H2O
Compound (A) is AgNO3
b) 2AgNO3 K723
2AgNO2 + O2
Compound (B) is AgNO2
c) AgNO2 K980
Ag + NO2
Element (C) is Ag
d) AgNO3 + KBr AgBr + KNO3
Compound (D) is AgBr which is highly useful in photography.
A AgNO3 Silver nitrate
B AgNO2 Silver nitrite
C Ag Silver
D AgBr Silver bromide
11. The metal B is extracted from the ore A. On treatment with dil. HNO3 metal B gives a compound
C, which is also known as Lunar Caustic. The compound C on treatment with KI gives a yellow
precipitate D. Identify A, B, C and D and explain the reactions. [Mar-2008]
a) Compound (A) is Ag2S which is the ore of silver
b) The metal (B) is Silver
c) 3Ag + 4HNO3 (dil) 3AgNO3 + NO + 2H2O
Compound (C) is AgNO3 which is also known as Lunar Caustic
d) AgNO3 + KI AgI + KNO3
Compound (D) is AgI which is yellow precipitate
A Ag2S Argentite or silver glance
B Ag Silver
C AgNO3 Silver nitrate
D AgI Silver iodide
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12. An element (A) belongs to group number 11 and period number 5 and is a lustrous white metal.
(A) reacts with dil. HNO3 to give (B). (B) with KI gives (C) which is bright yellow coloured
precipitate. Identify (A), (B), and (C). Explain the reactions. [Sep-2010, Sep-2015]
a) Element (A) is silver: Group =11 and period = 5. It is a lustrous white metal
b) 3Ag + 4HNO3 (dil) 3AgNO3 + NO + 2H2O
Compound (B) is AgNO3
c) AgNO3 + KI AgI + KNO3
Compound (C) is AgI which is bright yellow precipitate
A Ag Silver
B AgNO3 Silver nitrate
C AgI Silver iodide
13. The metal (A) is extracted from its sulphide ore. On treatment with dilute nitric acid metal A
gives a compound (B), which is also known as Lunar caustic. (B) on heating at 723 K gives (C) and
O2. Identify A, B and C and explain the reactions. [Mar-2014]
a) The metal (A) is silver.
b) 3Ag + 4HNO3 (dil) 3AgNO3 + NO + 2H2O
Compound (B) is AgNO3 (Lunar caustic)
c) 2AgNO3 K723
2AgNO2 + O2
Compound (C) is AgNO2
A Ag Silver
B AgNO3 Silver nitrate
C AgNO2 Silver nitrite
14. The sulphide ore of an element of group 12 when roasted gave compound A which on reduction
with carbon gave the element B. The carbonate C of the element is used for skin diseases. Identify
A, B and C and write the required reactions. [Sep-2006]
a) Sulphide ore of group 12 element (Zn) is ZnS
b) 2 ZnS + 3O2
2 ZnO + 2 SO2
Compound (A) is ZnO
c) ZnO + C K1673
Zn + CO
Element (B) is Zn
d) Carbonate of zinc is ZnCO3. Compound (C) is ZnCO3 which is used for skin diseases
.
A ZnO Zinc oxide
B Zn Zinc
C ZnCO3 Zinc carbonate
15. A bluish white metal (A) present in 4th
period and 12th
group on heating in air gives a white cloud
(B). Metal (A) on treatment with conc. H2SO4 gives the compound (C) and SO2 gas. Find A, B and
C and explain the reactions. [June-2008]
a) Metal (A) is zinc : Group = 12 and period = 4. It is a bluish white metal
b) 2Zn + O2 K773
2 ZnO
Compound (B) is ZnO [white cloud, Philosopher’s wool]
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c) Zn + 2H2SO4 (con) ZnSO4 + SO2 + 2H2O
Compound (C) is ZnSO4
A Zn Zinc
B ZnO Zinc oxide
C ZnSO4 Zinc sulphate
16. An element (A) belonging to Group 12 and Period 4 is bluish white in colour. A reacts with hot
conc. H2SO4 forming B with liberation of SO2. A also reacts with dil.HNO3 forming C with
liberation of N2O. Identify A,B and C and explain the reactions. [Sep-2008]
a) Element (A) is zinc : Group = 12 and period = 4. It is a bluish white metal
b) Zn + 2H2SO4 (con) ZnSO4 + SO2 + 2H2O
Compound (B) is ZnSO4
c) 4Zn + 10 HNO3 (dil) 4Zn(NO3)2 + N2O + 5H2O
Compound (C) is Zn(NO3)2
A Zn Zinc
B ZnSO4 Zinc sulphate
C Zn(NO3)2 Zinc nitrate
17. A bluish white metal when treated with dilute nitric acid gives (A) along with zinc nitrate and
water. With very dilute nitric acid it gives (B) along with zinc nitrate and water. The metal when
heated with air gives (C). What are A,B and C ? Explain the reactions. [Sep-2009]
a) Bluish white metal is zinc
b) 4Zn + 10 HNO3 (dil) 4Zn(NO3)2 + N2O + 5H2O
Compound (A) is N2O
c) 4Zn + 10 HNO3 (very dilute) 4Zn(NO3)2 + NH4NO3 + 3H2O
Compound (B) is NH4NO3
d) 2Zn + O2 K773
2ZnO Compound (C) is ZnO [Philosopher’s wool]
A N2O Nitrous oxide
B NH4NO3 Ammonium nitrate
C ZnO Zinc oxide
18. An element (A) in group number 12, period number 4 is extracted from its sulphide ore. (A)
reacts with O2 present in the air at 773 K to give compound (B) which is called Philosopher’s
wool. (A) reacts with hot NaOH to give compound (C). (A) also reacts with dilute nitric acid and
forms compound (D) with the liberation of N2O. Find out (A), (B), (C) and (D). Explain the
reactions. [Sep-2011]
a) Element (A) is zinc : Group = 12 and period = 4.
Zinc is extracted from zinc blende, ZnS (sulphide ore)
b) 2Zn + O2 K773
2ZnO
Compound (B) is ZnO [Philosopher’s wool]
c) Zn +2NaOH Na2ZnO2 + H2
Compound (C) is Na2ZnO2
d) 4Zn + 10 HNO3 (dil) 4Zn(NO3)2 + N2O + 5H2O
Compound (D) is Zn(NO3)2
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A Zn Zinc
B ZnO Zinc oxide
C Na2ZnO2 Sodiumzincate
D Zn(NO3)2 Zinc nitrate
19. An element (A) belonging to Group 12 and Period 4 is bluish white metal. (A) reacts with conc.
H2SO4 forming (B) with liberation of SO2. With NaOH (A) gives compound (C). Identify A, B and
C. Explain the reactions involved. [Mar-2012]
a) Element (A) is zinc : Group = 12 and period = 4. It is a bluish white metal
b) Zn + 2H2SO4 (con) ZnSO4 + SO2 + 2H2O
Compound (B) is ZnSO4
c) Zn +2NaOH Na2ZnO2 + H2
Compound (C) is Na2ZnO2
A Zn Zinc
B ZnSO4 Zinc sulphate
C Na2ZnO2 Sodiumzincate
20. An element (A) present in period number 4 and Group number 12 on treatment with dil. HNO3
forms (B) with the liberation of N2O. (A) when hated with air at 773 K gives (C) which is known
as philosopher’s wool. Identify (A), (B) and (C). Explain the reactions involved. [Sep-2012]
a) Element (A) is zinc : Group = 12 and period = 4.
b) 4Zn + 10 HNO3 (dil) 4Zn(NO3)2 + N2O + 5H2O
Compound (B) is Zn(NO3)2
c) 2Zn + O2 K773
2ZnO Compound (C) is ZnO [Philosopher’s wool]
A Zn Zinc
B Zn(NO3)2 Zinc nitrate
C ZnO Zinc oxide
21. The chief ore of Zinc, on roasting gave a compound A, which on reduction by carbon, gives B. (B)
reacts with conc.H2SO4 to give compound (C) and SO2 gas. Identity A, B and C. Explain the
reactions. [June-2013]
a) The chief ore of Zinc is zinc blende (ZnS)
b) Roasting
2ZnS + 3O2
2 ZnO + 2SO2
Compound (A) is ZnO
c) Reduction
ZnO + C K1673
Zn + CO
Element (B) is zinc
d) Zn + 2H2SO4 (conc) ZnSO4 + SO2 + 2H2O
Compound (C) is ZnSO4
A ZnO Zinc oxide
B Zn Zinc
C ZnSO4 Zinc sulphate
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22. An element (A) belongs to group number 12 and period number 4, bluish white in colour reacts
with conc.H2SO4 to give its salt (B) with the liberation of SO2 gas. Compound (B) reacts with
NaHCO3 to give (C) which is used as ointment for curing skin diseases. Identity A, B and C.
Explain the reactions. [Mar-2015]
a) Element (A) is zinc : Group = 12 and period = 4.
b) Zn + 2H2SO4 (con) ZnSO4 + SO2 + 2H2O
Compound (B) is ZnSO4
c) ZnSO4 + 2NaHCO3 ZnCO3 + Na2SO4 + H2O + CO2
Compound (C) is ZnCO3 which is used as ointment for curing skin diseases.
A Zn Zinc
B ZnSO4 Zinc sulphate
C ZnCO3 Zinc carbonate
23. An element (A) belongs to group number 6 and period number 4 is silvery white in colour. A
decomposes steam liberating hydrogen and forming (B). (A) also reacts with conc.H2SO4 forming
(C) with evolution of SO2. Identify A,B, and C and explain the reactions. [June-2007]
a) Element (A) is chromium : Group = 6 and period = 4. It is silvery white in colour.
b) 2Cr + 3H2O (steam) Cr2O3 + 3H2 Compound (B) is Cr2O3
c) 2Cr + 6H2SO4 Cr2(SO4)3 + 3SO2 +6H2O Compound (C) is Cr2(SO4)3
A Cr Chromium
B Cr2O3 Chromic oxide
C Cr2(SO4)3 Chromic sulphate
24. A compound of chromium, in which chromium exists in +6 oxidation state. Its chief ore (A) on
roasting with molten alkali (sodium carbonate) gives compound (B).This compound on
acidification with conc.H2SO4 gives compound C. Compound C on treatment with KCl gave
compound D. Identify the compounds A,B,C and D. Explain the reactions.
[Mar-2011, June2012, Sep-2013]
a) K2Cr2O7 is the compound of chromium in which chromium exists in +6 oxidation state
b) Compound (A) is FeO.Cr2O3 which is the chief ore of Cr
c) 4FeO.Cr2O3 + 8Na2CO3 + 7O2 8Na2CrO4 + 2Fe2O3 + 8CO2 Compound (B) is Na2CrO4
d) 2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O Compound (C) is Na2Cr2O7
e) Na2Cr2O7 + 2KCl K2Cr2O7 + 2NaCl Compound (D) is K2Cr2O7
A FeO.Cr2O3 Chromite or chrome ore
B Na2CrO4 Sodium chromate
C Na2Cr2O7 Sodium dichromate
D K2Cr2O7 Potaasium dichromate
25. Compound 'A' is a powerful oxidising agent and also it is a red orange crystal which melts at
396°C. 'A' reacts with chloride salt and conc.H2SO4 to give 'B' which is reddish orange in colour.
'A' also reacts with KOH to give 'C' which is yellow in colour. Find out (A), (B) and (C). Explain
the reaction. [Sep-2014, Mar-17]
a) Compound (A) is K2Cr2O7
b) K2Cr2O7 + 4KCl + 6H2SO4 2CrO2Cl2 + 6 KHSO4 + 3H2O Compound (B) is CrO2Cl2
c) K2Cr2O7 + 2KOH 2K2CrO4 + H2O
Red orange Yellow
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Compound (C) is K2CrO4
A K2Cr2O7 Potassium dichromate
B CrO2Cl2 Chromyl chloride
C K2CrO4 Potassium chromate
26. 'A' is a yellow coloured metal soluble in aqua regia. The roasted ore of this metal reacts with dil.
KCN to form a soluble complex ‘B’. This complex ‘B’ reacts with zinc dust to form another
complex ‘C’ along with the metal ‘A’. Identify ‘A’, ‘B’ and ‘C’. Explain the reaction. [Jun-2016]
a) 4Au + 8KCN + 2H2O + O2 4K[Au(CN)2] + 4KOH
(A) Potassium Aurocyanide (B)
b) 2K[Au(CN)2] + Zn K2[Zn(CN)4] + 2 Au
Potassiumtetracyanozincate(II) (C)
A Au Gold
B K[Au(CN)2] Potassium Aurocyanide
C K2[Zn(CN)4] Potassiumtetracyanozincate(II)
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(Q-70) CARBONYL COMPOUNDS
Hints:
1. General formula of aliphatic aldehydes and ketones: CnH2n O
HCHO, CH3CHO, CH3COCH3
2. Aldehydes:Formaldehyde HCHO,AcetaldehydeCH3CHO,Benzaldehyde C6H5CHO
3. Ketones: Acetone CH3COCH3,Acetophenone C6H5COCH3, Benzophenone C6H5COC6H5
4. Aldehydes reduce Tollen’s reagent and Fehling’s solution and restore original red colour of Schiff’s
reagent
5. Benzaldehyde reduces only Tollen’s reagent but not Fehling’s solution
6. Benzaldehyde oil of bitter almonds
7. Aldehydes and ketones undergo addition reactions with NaHSO4 and HCN
8. Aldehydes and ketones form oximes with hydroxylamine
9. Ketones do not reduce Tollen’s reagent and Fehling’s solution
10. Aldehydes and ketones containing CH3CO- group answer haloform test (form iodoform with I2 /
NaOH) {CH3CHO, CH3COCH3, C6H5COCH3}
11. Aldehydes with no α-hydrogen undergo cannizzaro reaction with 50% NaOH (HCHO,C6H5CHO)
12. Paraldehyde is used in hypnotic
13. In Clemmenson reduction (Zn/Hg + HCl) and in Wolff-Kishner reduction (N2H4 /C2H5ONa) –CO–
group is converted into –CH2– group
14. Cinnamic acid is unsaturated acid and Cinnamaldehyde is unsaturated aldehyde
15. Lactic acid is an optically active compound
CH3 C
O
H COOH
H
16. Carboxylic acids give brisk effervescence with NaHCO3
17. Aromatic hydrocarbons: Benzene C6H6, Toluene C6H5CH3
18. Sodalime (NaOH + CaO) is used for decarboxylation reaction
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1. An organic compound (A) of molecular formula C2H4O reduces Tollen’s reagent. A reacts with
HCN followed by hydrolysis in acid medium gives B (C3H6O3) which is optically active.
Compound B on oxidation by Fenton’s reagent gives the compound C (C3H4O3). This answers
iodoform reaction and gives effervescence with NaHCO3. Find A, B and C. Explain the reactions. [June-2008, Sep-2008, Mar-2010, Sep-2015]
+CH3 C
H
O
H-CN CH3 C
H
O
CN
H
H2O/H+
CH3 C
H
O
COOH
H
i)
Acetaldehyde Acetaldehyde cyanohydrin Lactic acid
ii) +CH3 C
H
O
COOH
H
H2O2/Fe2+
(O) +CH3 C
O
COOH H2O
(A) (B)
(C)Lactic acid (B) Pyruvic acid
2. An organic compound (A) C2H3OCl on treatment with Pd and BaSO4 gives (B) C2H4O which
answers iodoform test. (B) when treated with con.H2SO4 undergoes polymerisation to give (C) a
cyclic compound. Identify (A), (B), and (C) and explain the reactions. [Sep-2009]
i) (A) (B) Pd / BaSO
4
C2H
3OCl C
2H
4O
answers iodoform test (B) is acetaldehyde
H+Pd/BaSO4
CCH3
O
Cl H2 CCH3
O
+ HCl
Acetyl chloride Acetaldehyde(A) (B)
ii)
CH3
CH
3 CH3-CHO OO
CH3-CH CH-CH3
O
con.H2SO4
Paraldehyde
Acetaldehyde (B)
(C)
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3. An organic compound (A) C2H6O liberates hydrogen on treatment with metallic sodium. (A) on
mild oxidation gives (B) C2H4O which answers iodoform test. (B) when treated with conc. H2SO4
undergoes polymerization to give (C), a cyclic compound. Identify (A), (B) and (C) and explain the
reactions. [June-2006]
i) (A) H2
Na
C2H
6O
ii)
iii)
3 CH3CHO
con.H2SO4
CH3
CH
OO
CH CH CH3CH3
O
Paraldehyde
Acetaldehyde (B)
(C)
(A) Alcohol Ethyl alcohol
CH3CH
2OH + (O) CH
3CHO + H
2O
K2Cr
2O
7 / H+
Ethyl alcohol Acetaldehyde(A) (B)
4. Compound A (C2H4O) reduces Tollen’s reagent. A on treatment with zinc amalgam and conc.
HCl gives compound B. In presence of conc. H2SO4 A forms a cyclic compound C which is used as
hypnotic. Identify A, B & C and explain the reactions. [June-2011]
i) (A) reduces Tollen's reagent Acetaldehyde
C2H
4O
ii) CH3 C
O
CH3
CH3
Acetaldehyde Ethane
HCon. HCl
Zn / Hg
(A) (B)
iii)
3 CH
3-CHO
Con. H2SO
4O O
CH-CH3
CH3-CH
CH3
CH
O
Paraldehyde
Acetaldehyde (A)
(C)
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5. An organic compound (A) of molecular formula C2H4O reacts with HCN to give (B) (C3H5ON).
(B) on hydrolysis gives (C) (C3H6O3) which is an optically active compound. (C) also undergoes
iodoform test. Find A, B and C. Explain the reactions. [Sep-2011]
O
H
CH3 C + HCN
O
H
CH3 C
H
CNH
2O / H+
O
H
CH3 C
H
COOH
Acetaldehyde Acetaldehyde cyanohydrin Lactic acid(A) (B) (C)
6. An organic compound A (C2H3N) on reduction with SnCl2 / HCl gives B (C2H4O) which reduces
Tollen’s reagent. Compound B on reduction with N2H4 / C2H5ONa gives C (C2H6). Identify
(A),(B),(C) and explain the reactions. [Sep-2012]
i)
CH3 C N
HH
SnCl2
HClCH
3CH NH HCl
CH3
CH NH HCl
O H2
HydrolysisCH
3CH NH
4ClO +
Methyl cyanide Iminimum hydrochloride
Acetaldehyde
(A)
(B)
ii)
CH
3 C
ON
2H
4
C2H
5ONa
CH3
CH3
Acetaldehyde Ethane
H
(B) (C)
7. An organic compound (A) of molecular formula C2H4O is prepared by the reduction of compound
(B) of molecular formula C2H3N dissolved in ether, with SnCl2 and HCl. Compound (A) reduces
Tollen’s reagent. When a drop of Conc.H2SO4 is added to compound (A), it polymerises to give a
cyclic compound (C). Identify (A), (B) and (C). Explain the reactions. [Mar-2017]
CH3 C N
HH
SnCl2
HClCH
3CH NH HCl
CH3
CH NH HCl
O H2
HydrolysisCH
3CH NH
4ClO +
Methyl cyanide Iminimum hydrochloride
Acetaldehyde
(B)
(A)
i)
3 CH3-CHO
Con. H2SO
4O O
CH-CH3
CH3-CH
CH3
CH
O
Paraldehyde
Acetaldehyde (A)
(C)
ii)
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8. Compound (A) with molecular formula C2H4O reduces Tollen’s reagent. (A) on treatment with
HCN gives compound (B). Compound (B) on hydrolysis with an acid gives compound (C) with
molecular formula C3H6O3 which is an optically active compound. Compound (A) on reduction
with N2H4 / C2H5ONa gives a hydrocarbon (D) of molecular formula C2H6. Identify (A),(B),(C),
(D) and explain the reactions. [Mar-2013]
i) (A) Reduces Tollen's reagent Aldehyde Acetaldehyde
C2H
4O
ii)
O
H
CH3 C + HCN
O
H
CH3 C
H
CNH
2O / H+
O
H
CH3 C
H
COOH
Acetaldehyde Acetaldehyde cyanohydrin Lactic acid(A) (B) (C)
iii) CH3 C
ON
2H
4
C2H
5ONa
CH3
CH3
Acetaldehyde Ethane
H
(A) (D)
9. An organic compound (A) of molecular formula C7H6O is not reduced by Fehling’s solution but
will undergo Cannizaro reaction. Compound (A) reacts with aniline to give compound (B).
Compound (A) also reacts with Cl2 in the presence of catalyst to give compound (C). Identify (A),
(B) and (C) and explain the reactions. [Mar-2006]
i) Aromatic aldehydes reduce Tollen's reagent but not Fehling's solution
ii) Since compound (A) undergoes Cannizzaro reaction, it is benzaldehyde
iii)
NC6H5 C
H
O
+
H
H C6H5
H
NC6H5C C6H5
- H2O
O HH
H
NC6H5C C6H5
Schiffs base(A) (B)Benzaldehyde
iv)
CHO CHO
Cl
Cl2
FeCl3
Benzaldehyde m - chloro benzaldehyde(A) (C)
Aniline
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10. An organic compound (A) of molecular formula C7H6O is not reduced by Fehling’s solution but
will undergo Cannizaro reaction. Compound (A) reacts with aniline to give compound (B).
Compound (A) also reacts with Cl2 in the absence of catalyst to give compound (C). Identify (A),
(B) and (C) and explain the reactions. [Mar-2014]
i) Aromatic aldehydes reduce Tollen's reagent but not Fehling's solution
ii) Since compound (A) undergoes Cannizzaro reaction, it is benzaldehyde
iii)
NC6H5 C
H
O
+
H
H C6H5
H
NC6H5C C6H5
- H2O
O HH
H
NC6H5C C6H5
Schiffs base(A) (B)Benzaldehyde
iv)
(A)(C)
Aniline
CC6H
5
O
+H ClCl CC6H
5
O
Cl + HCl
BenzaldehydeBenzoyl chloride
Absence of catalyst
11. An organic compound A (C7H6O) reduces Tollen’s reagent. On treating with an alkali compound
A forms B and C. B on treating with sodalime forms benzene and C (C7H8O) is an antiseptic.
Identify (A), (B) and (C) and explain the reactions. [Sep-2006]
i) (A)
C7H
6O
Reduces Tollen's reagent Aldehyde Benzaldehyde
ii)
C6H5CHO
H2O50% NaOH
C6H5CHO
+ C6H5CH2OHC6H5COOH +
Benzyl alcoholBenzoic acid
iii) C6H
5COOH C
6H
6 +CO
2
NaOH / CaO
(A) (B) (C)
(B)Benzoic acid Benzene
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12. An organic compound A (C7H6O) forms a bisulphite. A when treated with alcoholic KCN forms B
(C14H12O2) and A on refluxing with sodium acetate and acetic anhydride forms an acid C
(C9H8O2). Identify A, B, and C. Explain the conversion of A to B and C. [June-2007]
i) (A) forms a bisulphite
C7H
6O
It is a carbonyl compound
(Benzaldehyde)
ii)
H
+
O
C6H
5 C
H
O
alcoholic KCN
H
O
C6H
5 C
H
O
Benzoin
CC6H
5 C
6H
5 C
Benzaldehyde (A) (B)
iii)
C6H
5-C = O
H
+ H2 CH-CO
O
CH3-CO H
OH
CH3COONa
C6H
5-C =
H
CH-COOH + CH3COOH
Cinnamic acidBenzaldehyde
(A) (C)
13. An aromatic aldehyde (A) of molecular formula C7H6O which has the smell of bitter almonds on
treatment with (CH3CO)2O and CH3COONa to give compound (B) which is an aromatic
unsaturated acid. (A) also reacts with (A) in the presence of alc.KCN to give dimer (C). Identify
(A), (B) and (C) and explain the reactions. [Sep-2007]
i) (A) has the smell of bitter almonds Benzaldehyde
C7H
6O
ii) C6H
5-C = O
H
+ H2 CH-CO
O
CH3-CO H
OH
CH3COONa
C6H
5-C =
H
CH-COOH + CH3COOH
Acetic anhydride
Cinnamic acid
Benzaldehyde
(A)
(B)
iii) C6H
5-C
H
+
O
C6H
5 C
H
O
Alcoholic KCNC
6H
5-C
H
O
C6H
5 C
H
O
Benzoin (C)(A)
Benzaldehyde
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14. Compound A with molecular formula C7H6O reduces Tollen’s reagent and also gives Cannizaro
reaction. A on oxidation gives the compound B with molecular formula C7H6O2. Calcium salt of B
on dry distillation gives the compound C with molecular formula C13H10O. Identify (A), (B) and
(C) and explain the reactions. [Mar-2008]
i) (A) Reduces Tollen's reagent undergoes Cannizzaro reaction
C7H
6O
&
So, (A) is benzaldehyde
ii) C6H
5CHO + (O) C
6H
5COOH
alkaline KMnO4
Benzaldehyde Benzoic acid(A) (B)
iii)
Calcium benzoate
Benzophenone
Dry distillation+ CaCO
3
C
O
O
C
O
OCa C
OC
6H
5
C6H
5
C6H
5 C6H
5
(C)
15. An aromatic compound (A) with molecular formula C7H6O has the smell of bitter almonds. (A)
reacts with Cl2 in the absence of catalyst to give (B) and in the presence of catalyst compound (A)
reacts with chlorine to give (C). Find A, B and C. Explain the reactions. [Mar-2009, Sep-2014]
i) (A) has the smell of bitter almonds Benzaldehyde
C7H
6O
ii)
iii)
C6H5C
O
H ClCl2+ HClC6H5C
O
+
Benzaldehyde Benzoyl chloride
in the absence of catalyst
(A) (B)
(C)
CHO
Cl
CHO
+ Cl2
FeCl3
Benzaldehyde m-chloro benzaldehyde(A)
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16. Compound (A) of molecular formula C7H8 when treated with air in presence of V2O5 at 773 K
gives a compound (B) of molecular formula C7H6O which has the smell of bitter almonds.
Alkaline KMnO4 oxidises compound (B) to (C) of molecular formula C7H6O2. Compound (B) on
treatment with N2H4 and KOH gives back compound (A). Identify (A), (B) & (C) and explain the
reactions. [June-2010]
i) (A) (B) smell of bitter almonds benzaldehydeair / V
2O
5
773 K
C7H
8 C7H
6O
C6H
5CH
3 +
2 (O) C
6H
5CHO + H
2O
air / V2O
5
773 K
Toluene Benzaldehyde(A) (B)
ii) C6H
5CHO + (O) C
6H
5COOH
alkaline KMnO4
Benzaldehyde Benzoic acid(B) (C)
iii) C6H
5 C H
ON
2H
4 / KOH
C6H
5CH
3
Benzaldehyde Toluene (A)(B)
17. An organic compound A (C7H8) on oxidation by air in the presence of V2O5 at 773 K gives B
(C7H6O), which reduces Tollen’s reagent. B when heated with acetic anhydride and sodium
acetate gives C (C9H8O2). Identify (A), (B) & (C) and explain the reactions.
[Mar-2011, June-2013, Sep-2013]
i) (A) (B) {(B) reduces Tollen's reagent Benzaldehyde}air / V
2O
5
C7H
8C
7H
6O
C6H
5CH
3 + 2(O) C
6H
5CHO + H
2O
air / V2O
5
Toluene Benzaldehyde(A) (B)
ii)
C6H
5-C = O
H
+ H2 CH-CO
O
CH3-CO H
OH
CH3COONa
C6H
5-C =
H
CH-COOH + CH3COOH
Cinnamic acid
(C)
Benzaldehyde
(B)
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18. Compound (A) of molecular formula C7H8 when treated with air in presence of V2O5 at 773 K
gives a compound (B) of molecular formula C7H6O which has the smell of bitter almonds. (B) on
reduction with lithium aluminium hydride to form (C) of molecular formula C7H8O. Identify (A),
(B) & (C) and explain the reactions. [Mar-2015]
C6H
5CH
3 + 2(O) C
6H
5CHO + H
2O
air / V2O
5
Toluene Benzaldehyde(A) (B)
ii) C6H
5CHO + 2(H) C
6H
5CH
2OH
LiAlH4
Benzyl alcohol (C)
i)
19. An organic compound (A) of molecular formu1a C7H6O is called as oil of bitter almonds. (A) on
oxidation gives (B) of molecular formula C7H5O2 which gives brisk effervescence with NaHCO3
solution. When (A) is refluxed with alcoholic KCN, compound (C) is formed. Identify A, B and C
and write the equations. [June-2014]
i) (A) Oil of bitter almonds Benzaldehyde
C7H
6O
ii)
iii)
C6H
5CHO + (O) C
6H
5COOH
alkaline KMnO4
Benzaldehyde Benzoic acid(A) (B)
Benzaldehyde
C6H
5-C
H
+
O
C6H
5 C
H
O
alcoholic KCNC
6H
5-C
H
O
C6H
5 C
H
O
Benzoin (C)(A)
20. An organic compound (A) C7H6O reduces Tollen’s reagent. Compound (A) reacts with acetic
anhydride in the presence of anhydrous sodium acetate and gives an unsaturated acid (B).
compound (A) reacts with acetone in the presence of alkali and gives (C). What are (A), (B) and
(C)? Explain the reactions [June-2015]
i) (A) reduces Tollen's reagent Benzaldehyde C7H
6O
C6H
5-C = O
H
+ H2
CH-CO
O
CH3-CO H
OH
CH3COONa
C6H
5-C =
H
CH-COOH + CH3COOH
Cinnamic acid
(B)
Benzaldehyde
(A)
ii)
C6H
5-C = O
H
+ H2 CH-CO-CH
3
Benzaldehyde (A)
iii)
acetone
NaOHC
6H
5-C =
H
CH-CO-CH3
Benzalacetone (C)
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21. An organic compound (A) of molecular formu1a C3H6O answers iodoform test. Another organic
compound (B) of molecular formu1a C7H6O is known as oil of bitter almonds. ‘A’ reacts with ‘B’
to form an unsaturated compound ‘C’ of molecular formu1a C10H10O. Compound ‘B’ reacts with
malonic acid in the presence of pyridine to form an unsaturated acid ‘D’ of molecular formula
C9H8O2. Identify A, B, C and D. Explain the reactions. [Jun-2016]
i) Compound (A) answers iodoform test, so it is acetone
ii) Compound (B) is known as oil of bitter almonds, so it is benzaldehyde
C6H
5-C = O
H
+ H2 CH-CO-CH
3
Benzaldehyde (A)
iii)
acetone
NaOHC
6H
5-C =
H
CH-CO-CH3
Benzalacetone (C)(B)
C6H
5-C = O
H
+ H2 C Pyridine
COOH
COOH
C6H
5-C =
H
COO H
COOH
C C6H
5-C =
H
CH-COOH
Cinnamic acid
- CO2
Malonic acidBenzaldehyde (B) (D)
iv)
22. An organic compound (A) with molecular formula C3H6O undergoes iodoform reaction. Two
molecules of compound (A) react with dry HCl to give compound (B) of molecular formula
C6H10O. Compound (B) reacts with one more molecule of compound (A) to give compound (C) of
molecular formula C9H14O. Identify (A), (B) & (C) and explain the reactions. [Sep-2010]
i) (A) undergoes iodoform reaction acetone
C3H
6O
ii) CH3 C O
CH3
+ H2 CH C
O
CH3
Dry. HClCH
3 C
CH3
CH C
O
CH3
Mesityl oxideAcetone (A) (B)
iii) CH3 C
CH3
CH C
O
CH H2
CH3
C CH3O+ CH
3 C
CH3
CH C
O
CH
CH3
C CH3
PhoroneMesityl oxide (B) Acetone (A) (C)
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23. Compound (A) of molecular formula C3H6O does not reduce Tollen’s reagent and Fehling’s
solution. Compound (A) undergoes Clemmenson reduction to give compound (B) of molecular
formula C3H8. Compound (A) in the presence of conc. H2SO4 condences to give an aromatic
compound (C) of molecular formula C9H12. Identify (A), (B), (C) and explain the reactions.
[Mar-2012, Mar-2016]
i) (A) does not reduce Tollen's reagent & Fehling's solution
So, (A) is a ketone Acetone
ii)
C3H
6O
CH3 C
O
CH3
Zn / Hg Con. HClCH
3 C CH3H
2
AcetonePropane
(A) (B)
iii)
3 CH
3-CO-CH
3
Con. H2SO
4
CH3
CH3CH
3
+ 3 H2O
Mesitylene (C)
+
Clemmenson reduction
24. Compound (A) of molecular formula C3H6O does not reduce Tollen’s reagent but undergoes
haloform reaction. Compound (A) undergoes dehydration reaction in the presence of dry HCl to
give compound (B) of molecular formula C6H10O. Three molecules of compound (A) undergoes
condensation reaction in the presence of conc. H2SO4 to give a cyclic hydrocarbon (C). Identify
(A),(B),(C) and explain the reactions. [June-2012]
i) (A) does not reduce Tollen's reagent ketone So, (A) is acetone C3H
6O
ii)
CH
3 C O
CH3
+ H2 CH C
O
CH3
Dry. HClCH
3 C
CH3
CH C
O
CH3
Mesityl oxideAcetone (A) (B)
iii)
3 CH
3-CO-CH
3
Con. H2SO
4
CH3
CH3CH
3
+ 3 H2O
Mesitylene
Acetone (A)
(C)
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25. An organic compound A of molecular formula C3H6O on reduction with LiAlH4 gives B.
Compound B gives blue colour in Victor Meyer’s test and also forms a chloride C with SOCl2. The
chloride on treatment with alcoholic KOH gives D. Identify (A), (B), (C) and (D) and explain the
reactions. [Mar-2007]
i) (A)
C3H
6O C
nH
2nO (A) is aliphatic aldehyde or ketone
ii) (A) (B)LiAlH
4
Reduction
Victor Meyer's testBlue colour
(B) Secondary alcohol & (A) Ketone
(Isopropyl alcohol) (Acetone)
CH3 C
O
CH3 2[H]+LiAlH4
CH3 C
O
CH3
H
H
(A) (B)
Isopropyl alcoholAcetone
iii) (B) (C)SOCl
2
CH3 C
O
CH3
H
HSOCl2 CH3 C CH3H
Cl
Isopropyl alcohol
(B)
Isopropyl chloride
(C)
Alcoholic KOH(D)
Alcoholic KOHCH3 C CH2
H + KCl + H2O
Propene
(D)
26. An organic compound ‘A’ (C8H8O) undergoes iodoform test. When reduced with zinc amalgam
and HCl it gives ‘B’ (C8H10). ‘A’ with Br2 in ether at 273 K gives ‘C’ (C8H7OBr). Identify (A), (B),
(C) and explain the reactions. [Sep-2016]
i) Compound ‘A’ (C8H8O) undergoes iodoform test. So, it is acetophenone
ii) C6H5COCH3 )H(4,HClHg/Zn C6H5CH2CH3
Acetophenone (A) Ethyl benzene (B)
iii) C6H5COCH3 + Br –Br C6H5COCH2Br + HBr
Phenacyl bromide (C)
+2 CHEMISTRY (Q-70) Electro Chemistry - I way to success
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(Q-70) ELECTRO CHEMISTRY-I
Hints:
1. pH = ]H[log10
, pOH = ]OH[log10
& pH + pOH = 14
2. pH of acidic buffer, pH = pKa + log [acid]
[salt]
3. pOH of basic buffer, pOH = pKb + log [base]
[salt]
4. Ionic product of water Kw = [H+] [OH
–] = 1 10
–14 mol
2 dm
–6
5. Dissociation constant of the weak acid, Ka =
1
C 2
6. For weak acids, [H+] = Cα = CKa
7. Specific conductance, R
1
a
l
8. Specific conductance Resistance
constantCell
9. Equivalent conductance, C =N
103
mho cm2 (gm equiv)
–1
C =N
10 3
mho m2 (gm equiv)
–1
10. Molar conductance, C =
M
10 3 mho m
2 mol
–1
11. Cell constant = a
l
12. Degree of dissociation, α =
λ
cλ
13. m = Z I t
14. Electrochemical equivalent, Z = 96495
massEquivalent
& Equivalent mass =
Valency
massAtomic
15. CH3COOH = CH3COONa + HCl – NaCl & NH4OH = NH4Cl + NaOH – NaCl
16. Equivalent mass of Cu = 31.77, Ag = 108 , I = 127, Al = 9
1. Calculate the pH of 0.02M aqueous Ba(OH)2 solution assuming Ba(OH)2 as a strong electrolyte.
[Sep-2016]
Ba(OH)2 Ba2+
+ 2OH–
0.02M 2 × 0.02M = 0.04M
pOH = – log [OH
–] = – log 0.04 = – log 4 × 10
–2 = 2 – log 4 = 2 – 0.6021
= 1.3979
pH = 14 – pOH = 14 – 1.3979 = 12.6021
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2. What is the pH of a solution containing 0.5 M propionic acid and 0.5 M sodium
propionate? The Ka of propionic acid is 1.34 × 10–5
. What happens to the pH of the solution when
volume is doubled by adding water? [Mar-2006, June-2010, Mar-2014]
pH = pKa + log [acid]
[salt]
= – log Ka + log [acid]
[salt]
= – log (1.34 × 10–5
) + log 0.5
0.5
= 5 – log 1.34 + log 1
= 5 – 0.1271 + 0
= 4.8729
When volume is doubled by adding water, the concentrations of salt and acid are reduced to the same
extent and therefore the ratio [acid]
[salt] remains the same. So, there will be no change in pH of the solution.
3. Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.15 mole per
litre acetic acid. Ka for acetic acid is 1.8 × 10–5
.
[June-2006, Sep-2006, Sep-2007, June-2011, Sep-2011]
pH = pKa + log [acid]
[salt]
= –log Ka + log [acid]
[salt]
= – log 1.8 × 10–5
+ log 0.15
0.20
= 5 – log 1.8 + log 20 – log 15
= 5 – 0.2553 + 1.3010 – 1.1761
= 4.8696
4. Find the pH of a buffer solution containing 0.30 mole per litre of CH3COONa and 0.15 mole per
litre of CH3COOH. Ka for acetic acid is 1.8 × 10–5
. [Sep-2008]
pH = pKa + log [acid]
[salt]
= –log Ka + log [acid]
[salt]
= – log (1.8 × 10–5
) + log 0.15
0.30
= 5 – log 1.8 + log 2
= 5 – 0.2553 + 0.3010
= 5.0457
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5. Calculate the pH of a buffer solution containing 0.04 M NH4Cl and 0.02 M NH4OH. For NH4OH
Kb is 1.8 × 10–5
. [Mar-2007]
pOH = pKb + log [base]
[salt]
= –log Kb + log [base]
[salt]
= – log (1.8 × 10–5
) + log 0.02
0.04
= 5 – log 1.8 + log 2
= 5 – 0.2553 + 0.3010
= 5.0457
pH + pOH = 14
pH = 14 – pOH
= 14 – 5.0457
= 8.9543
6. Calculate the pH of 0.1 M CH3COOH solution. Dissociation constant of acetic acid is 1.8 × 10–5
[June-2009, Mar-2011, June-2012, June-2014]
For weak acid, [H+] = CKa
[H+] = 1.0108.1 5
= 1.34 × 10–3
M
pH = – log [H+]
= – log 1.34 × 10–3
= 3 – log1.34
= 3 – 0.1227
= 2.8773
7. The equivalent conductance of HCl, CH3COONa and NaCl at infinite dilution are 426.16, 91.0
and 126.45 ohm–1
cm2 (gram equivalent)
–1 respectively. Calculate the equivalent conductance of
acetic acid. [June-2008, Sep-2010, Sep-2013]
λ∞ CH3COOH = λ∞ CH3COONa + λ∞ HCl - λ∞ NaCl
= 91 + 426.16 – 126.45
= 390.71 ohm–1
cm2 gm.equivalent
–1
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8. Ionic conductance at infinite dilution of Al3+
and SO42–
are 189 ohm–1
cm2 gm.equiv
–1 and 160
ohm–1
cm2 gm.equiv.
–1. Calculate equivalent and molar conductance of the electrolyte at infinite
dilution. [Mar-2010, Mar-2013, Sep-2015, Mar-17]
Al2(SO4)3 2Al3+
+ 3SO42–
Equivalent conductance of Al2(SO4)3 at infinite dilution,
λ∞ Al2(SO4)3 = 3
1 λ∞ Al
3+ +
2
1 λ∞ SO4
2–
= 3
1 (189) +
2
1 (160)
= 143 mho cm2 gm.equiv
–1
Molar conductance of Al2(SO4)3 at infinite dilution,
μ∞ Al2(SO4)3 = 2 μ∞ Al3+
+ 3 μ∞ SO42–
= 2 (189) + 3 (160)
= 858 mho cm2 mol
–1
9. Equivalent conductivity of acetic acid at infinite dilution is 390.7 and for 0.1 M acetic acid the
equivalent conductance is 5.2 mho cm2 gm.equiv
–1. Calculate the degree of dissociation, hydrogen
ion (H+) concentration and dissociation constant of the acid. [June-2007, Sep-2014, Mar-2015]
a) Degree of dissociation, α =
C =
7.390
2.5= 0.0133 = 1.33%
b) Hydrogen ion concentration, [H+] = C α = 0.1 × 0.0133 = 0.00133 M
c) Dissociation constant of the acid, Ka =
1
C 2
= 0133.01
)0133.0(1.0 2
= 2.38 × 10
–5 M
10. What current strength in amperes will be required to liberate 10 g of iodine from potassium
iodide solution in one hour? (Equivalent mass of iodine is 127) [Mar-2012]
m = z I t
∴ I = tz
m
=
t96495
massEquivalent
m
=
606096495
127
10
= 6060127
1096495
= 2.11 ampere
11. 0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the
electrochemical equivalent of copper? [Mar-2009, June-2015]
According to Faraday’s First law of electrolysis,
m = z I t
z = tI
m =
60502.0
0.1978
= 3.296 × 10–4
g coulomb–1
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12. If 50 milli ampere of current is passed through copper coulometer for 60 min, calculate the
amount of copper deposited. [Jun-2016]
I = 50 milli ampere = 50 × 10–3
ampere
t = 60 min = 60 × 60 sec
m = z I t
= 96495
massEquivalent × I × t
= 96495Valency
massAtomic
× I × t
= 964952
5.63
× 50 × 10
-3 × 60 × 60
= 0.0592 g
13. An electric current is passed through three cells in series containing respectively the solutions of
copper sulphate, silver nitrate and potassium iodide. What weights of silver and iodine will be
liberated while 1.25 g of copper is being deposited? [Mar-2008, Sep-2012, June-2013, Mar-2016]
iodineofWeight
copperofWeight =
iodineofweightEquivalent
copperofweightEquivalent
iodineofWeight
25.1 =
127
7.31
∴ Weight of iodine = 5 g
silverofWeight
copperofWeight =
silverofweightEquivalent
copperofweightEquivalent
silverofWeight
25.1
= 108
7.31
∴ Weight of silver = 4.25 g
14. 0.04 N solution of a weak acid has a specific conductance of 4 × 10–4
mho cm-1
. The degree of
dissociation of the weak acid at this dilution is 0.05. Calculate the equivalent conductance of the
weak acid at infinite dilution. [Sep-2009]
λC = N
103 =
04.0
10410 43 = 10 mho cm
2 (gram equiv.)
–1
Degree of dissociation α =
C
∴ Equivalent conductance of the weak acid at infinite dilution,
λ∞ =
C
= 05.0
10 = 200 mho cm
2 (gram equiv.)
–1
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- Notes –
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- Notes -
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