Way to Success - Waytosuccess

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

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mailto:[email protected]


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Name of the Book

Way to Success, A Complete Guide for +2 Chemistry

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Chemistry Way To Success Guide through VPP,

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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

+2 CHEMISTRY Atomic Structure II way to success

[email protected] -7- www.waytosuccess.org

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



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



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.



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.



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.



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.



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)



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



∆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



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



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.



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



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

+2 CHEMISTRY Periodic Classification II way to success


2. PERIODIC CLASSIFICATION – II


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)



= 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


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.



10. Compare ionisation energy of C and B. [Jun-2012, Mar-17]

B C


EC: 1s2 2s

22p

1 EC: 1s

2 2s

22p

2

Nuclear charge of C is greater than B.



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


EC: 1s2 2s

22p

4 EC: 1s

2 2s

22p

5

Nuclear charge of F is greater than O.



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


EC: 1s2 2s

2 2p

5 EC: 1s

2 2s

2 2p

6

Nuclear charge of Ne is greater than F.



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


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


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.



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



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.



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.



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


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




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



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


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


E.AI.E

= 2.82

413

= 3.03

+2 CHEMISTRY p – Block Elements way to success


3. p -BLOCK ELEMENTS



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



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



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



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



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.



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.....



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



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



I

F

F

FF

F

Hybridisation sp3d

2



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.



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.



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



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



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


a) linear b) tetrahedral c) squareplanar d) none


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



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



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.



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).



3s 3p 3d

sp3d Hybridisation

Atom in Ground state

Atom in Excited state

Only one unpaired


covalent bond

Three unpaired

electrons permit

three covalent bonds

Formation of ClF3

molecule by sp3d

hybridisation

Cl

F

F

F

Trigonal Bipyramidal

Hybridisation sp3d



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


covalent bond

Seven unpaired

electrons permit

seven covalent bonds

Formation of IF7

molecule by sp3d3

hybridisation

Iodine Atom in Excited state



I

F

F

F

F

F

F

F

Pentagonalbipyramidal

Hybridisation sp3d

3



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



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

+2 CHEMISTRY d – Block Elements way to success


4. d-BLOCK ELEMENTS

Blue print 1 Mark = 2 3 Mark = 2 5 Mark = 1+1* Total marks = 18


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



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



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



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



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



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.



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 ↓


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.


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 ↑



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


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



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



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.



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.

+2 CHEMISTRY f – Block Elements way to success


5. f - BLOCK ELEMENTS



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



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.



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.



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




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



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.

+2 CHEMISTRY Coordination and Bio-coordination Compounds way to success


6. COORDINATION COMPOUNDS AND BIO-COORDINATION COMPOUNDS



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



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


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


Oxidation Number III or +3


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


Oxidation Number +3






5. [Cr(en)3]Cl3 [Mar-2014]

IUPAC name tris(ethylenediamine) chromium (III) chloride

Central metal ion Cr3+

Ligand en (ethylenediamine)


Oxidation Number +3




6. [Co(NH3)3 (NO2)3] [June-2014]

IUPAC name triamminetrinitrocobalt(III)


Ligand NH3 ammine & NO2– nitro


Oxidation Number +3


Charge on the complex ion 0

Nature of the complex Neutral complex

7. [Co(NH3)4Cl2] NO2 [Mar-2015]

IUPAC name tetraamminedichlorocobalt(III)nitrite


Ligand NH3 ammine & Cl– chloro


Oxidation Number +3


Charge on the coordination sphere +1


8. [Co(NH3)5Cl]2+

[Jun-2016]

IUPAC name pentaamminechlorocobalt(III)ion


Ligand NH3 ammine & Cl– chloro


Oxidation Number +3


Charge on the coordination sphere +2


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.



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



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.



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 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.



The geometry of the molecule is square planar.

Hybridisation dsp2

Geometry or shape square planar


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


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



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


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 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



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.



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.


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



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


38. The function of ferredoxin and rubredoxin is


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



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



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



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

–

+2 CHEMISTRY Nuclear Chemistry way to success


7. NUCLEAR CHEMISTRY



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]


17. In the nuclear reaction 92U238

82Pb206

the number of αand βparticles emitted are

[June-2014, Mar-2015, Sep-2015]




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



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]


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


238 = 206 + 4a + b × 0

4a = 238 – 206

4a = 32

a = 8




92 = 82 + 2 ×a + (–1) b

= 82 + 2a – b

2a–b = 92 – 82 = 10

2(8) – b = 10

b = 16 –10

= 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


232 = 220 + 4a + b × 0

4a = 232 – 220

4a = 12

a = 3


92 = 89 + 2 ×a + (–1)b

= 89 + 2a – b

2a–b = 92 - 89 = 3

2(3) – b = 3

b = 6–3

= 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



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


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


232 = 208 + 4a + b 0)

4a = 232 – 208

= 24

a = 6


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


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



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]




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.



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


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



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


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?


18 Hydrogen bomb is based on


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?


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



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


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

+2 CHEMISTRY Solid State - II way to success


8. SOLID STATE - II



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



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.



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]


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 ]


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



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 -



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.



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



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.


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



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.



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.

+2 CHEMISTRY Thermodynamics - II way to success


9. THERMODYNAMICS - II



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



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



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]


9. Give Entropy statement of second law of thermodynamics [Sep-2008]




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



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”.



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)



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


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



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



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.

+2 CHEMISTRY Chemical Equilibrium - II way to success


10. CHEMICAL EQUILIBRIUM–II



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



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



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



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



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



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



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.



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



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%



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



reaction.

N2O4 (g) 2NO2(g) H = +59 kJ mol-1endothermic

exothermic


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.



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


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



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



reaction.

N2(g) + O2(g) 2NO(g) H0f = +43.2 kCal mol-1

endothermic

exothermic


forward reaction.

+2 CHEMISTRY Chemical Kinetics - II way to success


11. CHEMICAL KINETICS – II



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



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



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.



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]


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



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]


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]


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



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+]



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



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)



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%




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

+2 CHEMISTRY Surface Chemistry way to success


12. SURFACE CHEMISTRY



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



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.



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.



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



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



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]




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.



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.



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

+

+-

-



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

+++




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.



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



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.



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

+2 CHEMISTRY Electro Chemistry - I way to success


13. ELECTRO CHEMISTRY –I



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



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



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.



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



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



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.



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



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]


[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.



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




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]



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



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



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).



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

+2 CHEMISTRY Electro Chemistry - II way to success


14. ELECTRO CHEMISTRY – II



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



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


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)


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.



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)



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



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



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+


–

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


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


E = Eo

– n

0591.0 log

][Zn

12



– 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


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



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


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


n

0591.0 log K


0 = Eo

– n

0591.0 log K

Eo

= n

0591.0 log K

– 0.0281 = 1

0591.0 log K

log K = – 0.4751



∴ 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



18. Calculate the equilibrium constant for the reaction,

2Ag+ + Zn ⇌ Zn

2+ + 2Ag V56.1cell

0E [Sep-2015]


n

0591.0 log K


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


–

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




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+]



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


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

+2 CHEMISTRY Isomerimin Organic Chemistry way to success


15. ISOMERIM IN ORGANIC CHEMISTRY



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



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.



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



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



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



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’



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



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



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


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



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.

+2 CHEMISTRY Hydroxy Derivatives way to success


16. HYDROXY DERIVATIVES

Blue print 1 Mark = 1 3 Mark = 2 5 Mark = 1* Total marks = 12


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)



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



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.



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



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



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



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)



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



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


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



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


16. n-propyl alcohol and ethylmethyl ether exhibit ------- isomerism


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



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



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


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


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

+



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.



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)



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



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



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



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



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



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

+2 CHEMISTRY Ethers way to success


17. ETHERS



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]


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



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



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



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



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



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.



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


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



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.

+2 CHEMISTRY Carbonyl Compounds way to success


18. CARBONYL COMPOUNDS



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



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



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



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



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.


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.


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



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.



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



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



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




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


11. Pentanal and 2,2-dimethylpropanal exhibits ------- isomerism


12. Acetone, propanaldehyde and 2-propen-1-ol exhibit ------- isomerism


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


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



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



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



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



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



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)



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



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



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



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



5. How will you prepare the following by dry distillation of calcium salt of fatty acids?

a] CH3COCH3 b] C6H5COCH3 c] C6H5COC6H5

a] CH3COCH3


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


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

+2 CHEMISTRY Carboxylic Acids way to success


19. CARBOXYLIC ACIDS



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



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



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



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+



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



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



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



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



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)


16. How are the following conversions carried out? [June-2010]



c) Lactic acid Lactide (Refer-5 mark-Q-12)





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)


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


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


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



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



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



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



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


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



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?


25. Which of the following is optically active?




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


a) CH3CH2COOH < CH3COOH < HCOOH b) CH3CH2COOH >CH3COOH >HCOOH

c) CH3COOH < CH3CH2COOH < HCOOH d) CH3CH2COOH < HCOOH < CH3COOH


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



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.



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



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

+2 CHEMISTRY Organic Nitrogen Compounds way to success


20. ORGANIC NITROGEN COMPOUNDS



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



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



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



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+



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)



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



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)



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 )



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



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


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



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



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)



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




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

+



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


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


4. 1-nitro propane and 2-nitro propane exhibit --------- isomerism


5. Nitro methane (CH3NO2) and Methyl nitrite (CH3– O – N = O) exhibit ------- isomerism


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



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



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



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.

+2 CHEMISTRY Biomolecules way to success


21. BIOMOLECULES



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



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



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



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



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



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.


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



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



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.

+2 CHEMISTRY Chemistry in Action way to success


22. CHEMISTRY IN ACTION



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]


3. What are anaesthetics ? Give example [June-2007, June-2008, Sep-2010, Sep-2012]


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]


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.



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]


14. Write the preparation and use of Buna-N rubber [Sep-2008, Sep-2013]


15. How is nylon-66 prepared? Give its use [Mar-2009, June-2015]


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.



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



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.



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




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.

+2 CHEMISTRY (Q-70) Hydroxy Derivatives way to success


(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



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)



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)



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)



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)



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)



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



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+


Benzaldehyde (C)

(B) (D)



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+


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)



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


400 K , 4-7 atm

dil. HCl

OH

COOH

salicylic acid

+ NaCl

(D)(C)(B)



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


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




(A)(D)



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)



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]


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



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]


3

iii)

C6H

6O

C6H

5OH

ii)

ONa

H

OH

COONaCO

2


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]


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)



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


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)

+2 CHEMISTRY (Q-70) d-Block Elements way to success


(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



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




A Cu Copper

B CuO Cupric oxide

C Cu2O Cuprous oxide


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


A Cu Copper


C CuSO4 Copper sulphate


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


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



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



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]



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



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



c) Zn +2NaOH Na2ZnO2 + H2

Compound (C) is Na2ZnO2

A Zn Zinc


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]


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



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]




c) ZnSO4 + 2NaHCO3 ZnCO3 + Na2SO4 + H2O + CO2

Compound (C) is ZnCO3 which is used as ointment for curing skin diseases.

A Zn Zinc


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



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)

+2 CHEMISTRY (Q-70) Corbonyl Compounds way to success


(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



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)



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


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)



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 +


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 +


Acetaldehyde

(B)

(A)

i)

3 CH3-CHO

Con. H2SO

4O O

CH-CH3

CH3-CH

CH3

CH

O

Paraldehyde

Acetaldehyde (A)

(C)

ii)



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



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



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



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


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)



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


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


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)



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


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)



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)



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)



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


(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



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]


[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]


[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]


[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



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



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



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



- Notes –



- Notes -

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