Chemical Bonding and Molecular Structure - SelfStudys

64 CHEMISTRY

The attraction force which holds various constitutents (atoms,ions, etc) together in different chemical species is called a chemicalbond.

KOSSEL - LEWIS APPROACH TO CHEMICALBONDING(i) Lewis postulated that atoms achieve stable octet when they

are linked by chemical bonds.(ii) The outer shell electrons are called valence electrons and it

is these electrons which take part in chemical bonding.(iii) The valence electrons are represented by putting dots

around the symbol of the element. These are called Lewis

symbols. For ex: Li , Be, Bg g

g g gg , etc.(iv) Kossel provided the basis for ion formation by electron

transfer and formation of ionic crystalline compounds.

IONIC AND COVALENT BOND

Ionic or Electrovalent BondThe bond formed as a result of electrostatic attraction betweenpositive and negative ions is called ionic bond. Ionic bonds areformed readily between elements with low IE and elements withhigh –ve value of electron gain enthalpy.Factors favouring the formation of ionic bond :(a) Ionisation energy (IE)(b) Electron affinity(c) Lattice energyNote : A qualitative measure of the stability of an ionic compoundis provided by its lattice enthalpy which is defined as the energyrequired to completely separate one mole of a solid ionic compoundinto gaseous constituent ions

Covalent Bond and Lewis StructuresThe bond formed as a result of sharing of electrons betweenelements is called covalent bond. When electrons shared bybonded atoms are contributed entirely by one of the bonded atomsbond formed is known as a coordinate bond. Lewis dot structuresprovide a picture of bonding in molecules and ions in terms ofshared pair of electrons and the octet rule.Covalency : It is defined as the number of electrons contributedby an atom of the element for sharing with other atoms so as toachieve noble gas configuration.

FORMAL CHARGE AND RESONANCE

Formal ChargeIt is the difference between the number of valence electrons in anisolated atom and no of electrons assigned to that atom in Lewisstructure.

( )

( )

Formal change F.C Total no. of valenceon an atom in electrons in the free

atomLewis structureTotal no. ofTotal no. of 1non bonding bonding shared

2electrons (lone pairs) electrons

æ öç ÷=ç ÷è ø

æ öæ öç ÷ç ÷- - -

ç ÷ ç ÷è ø è ø

Note : Among the valence electrons, those pair of electrons whichare involved in bonding are called bond pair of electrons whilethose which are not involved in bonding are called lone pair ofelectrons.

ResonanceIt has been found that the observed properties of certain moleculescannot be satisfactorily explained by one structure. The moleculeis then supposed to have many structures, each of which canexplain most of the properties but not all. This phenomenon iscalled resonance.

BOND PARAMETERS(i) Bond length : It is defined as the equilibrium distance between

the nuclei of two bonded atoms in a molecule. The covalentradius is half of the distance from the centre of nucleus ofone atom to the centre of nucleus of the other atom providedbonded atoms will be of same element in a molecule.The van der Waal’s radius represents the overall size of theatom which includes its valence shell in a non-bondedsituation.Van der Waal’s radius > Covalent radius

(ii) Bond angle : It is defined as the angle between the orbitalscontaining bonding electron pairs around the central atom ina molecule / complex ion.

(iii) Bond Enthalpy : It is defined as the amount of energy requiredto break one mole of bonds of a particular type between twoatoms in a gaseous state. Its unit is kJ mol–1. Larger the bonddissociation enthalpy, stronger is the bond.

4Chemical Bonding

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65Chemical Bonding and Molecular Structure

(iv) Bond order : It is given by the number of bonds betweentwo atoms in a molecule. Isoelectronic molecules and ions

have identical bond order. For ex: F2 and 22O - have B.O.= 1.

In general,

1Bond order Bond Enthalpy

Bond lengthµ µ

POLARITY OF BONDS AND DIPOLE MOMENT

When covalent bond is formed between two similar atoms like H2,O2, Cl2, N2 etc, the shared pair of electrons is equally attracted bytwo atoms. As a result, electrons pair is exactly situated betwenthe two atoms. Such a bond is called non-polar covalent bond.In heteronuclear molecules like HF, the shared pair ofelectrons gets shifted towards the more electronegative element(here F), resulting in polarity in bond. Such a bond is called polarcovalent bond. This polarisation of the bond gives rise to a dipolemoment given by:Dipole moment (m) = charge (q) × distance of separation (d). It is

represented as H F-uuuuuur

. Dipole moment is a vector quantity

Covalent Character in Ionic Compounds

Just as covalent bonds have partial ionic character, ionicbonds have partial covalent character. The partial covalentcharacter of ionic bonds was discussed by Fajan in terms offollowing rules:

(i) Smaller the cation and larger the anion, greater is the covalentcharacter of the ionic bond.

(ii) Greater the charge on cation, greater is the covalent characterof the ionic bond.

(iii) For cations of same size and charge, the one with electronicconfiguration ns2np6nd10 is more polarising than the onewith noble gas configuration ns2 np6.The covalent character of an ionic bond is explained on thebasis of polarisation. The cation polarises the anion, pullingthe electronic charge towards itself and thereby increasingelectronic charge between the two. This is what happens ina covalent bond, i.e. build up of electron charge densitybetween the nuclei. The polarising power of the cation, thepolarisability of the anion and the extent of polarisation ofthe anion determine the percent covalent character of theionic bond.

Applications of Dipole Moment

(i) In determining the polarity of bonds :As m = q × d,

(ii) In the calculation of ionic character :i.e., % ionic character

= Observed dipole moment

Dipole moment considering 100% ionic character × 100

(iii) Determination of shape or symmetry of molecules:Note : In A— B moleculeIonic nature µ XA – XB% Ionic character = 0 when XA = XB (100% covalent bond)% Ionic character = 0 – 15% Non polar covalent bond% Ionic character = 15 – 50% Polar covalent bond% Ionic character = > 50% Ionic bond

THE VALENCE SHELL ELECTRON PAIR REPULSION(VSEPR) THEORY

The main postulates of this theory are:(i) The geometry of a molecule depends upon the number of

valence shell electron pairs (bonded or non-bonded) aroundthe central atom

(ii) Pairs of electrons in valence shell repel one another sincetheir electron clouds are negatively charged.

(iii) These pairs of electrons tend to occupy such positions inspace that minimise repulsion and thus maximise distancebetween them.

(iv) The valence shell is taken as a sphere with electron pairslocalising on the spherical surface at maximum distance fromone another.

(v) A multiple bond is treated as if it is a single electron pair(vi) Where two or more resonance structures can represent a

molecule, VSEPR model is applicable to any such structure.The decreasing order of repulsive interaction of electron pairsislp – lp > lp – bp > bp – bpWhere lp ® lone pair and bp ® bond pairFor prediction of geometrical shape of molecules, we dividedmolecules in two categories :(i) molecules in which central atom has no lone pair(s)(ii) molecules in which central atom has one or more lone

pair(s).

66 CHEMISTRY

Geometry of molecules when central atom has no lone pair of electrons.

No. of electron

pairs

Arrangement of electron pairs

Molecular Geometry Examples

2 180°

LinearA

B — A — B Linear

BeF , HgCl2 2

3

A120°

Trigonal planar

A

Trigonal planar

B

BB

BF , AlCl3 3

4

A

Tetrahedral

109.5°

A

Tetrahedral

B

BB

B

CH ,SiF4 4

5

A

Trigonalbipyramidal

A

Trigonalbipyramidal

B

B

B

B

B

PCl , AsF5 5

6

Octahedral

A

Octahedral

A

BB

BB B

B

SF6

120°90°

90°

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Geometry of molecules when central atom has one or more lone pair of electrons

Molecule No. of No. of Total Arrangement of Shape Example type bonding pairs lone pairs electron pairs electron pairs (geometry)

AB2L 2 1 3A

B BTrigonal planar

Bent or Vshaped SO2, O3, SnCl2

AB3L 3 1 4A

B BB

Tetrahedral

Pyramidal NH3

AB2L2 2 2 4A

B BTetrahedral

V shaped H2O

AB4L 4 1 5A

BB

BB

Trigonal bipyramidal

See saw SF4

AB3L2 3 2 5A

B

B

B


T shaped CIF3

AB2L3 2 3 5A

B

B


Linear XeF2

AB5L 5 1 6

Octahedral

A

B

BB

B

B

Square pyramidal BrF5

AB4L2 4 2 6

Octahedral

A

B

B

B

B

Square planar XeF4

68 CHEMISTRY

VALENCE BOND THEORY (VBT)

(i) This theory is based on the knowledge of atomic orbitals,electronic configurations of elements, the overlap criteria ofatomic orbitals, the hybridisation of atomic orbitals and theprinciples of variation and superposition.

(ii) According to orbital overlap concept, the formation of acovalent bond between two atoms results by pairing of e–1 spresent in the valence shell having opposite spins greater isthe overlap, stronger is the bond formed between the atoms.

(iii) Co–axial overlapping ® extent of overlapping more® high bond strength

Collateral overlapping ® extent of overlapping less® low bond strength

In Co–axial overlapping, extent of overlapping isp – p < s – p < s – s

(iv) VBT explains the formation and directional properties ofbonds in polyatomic molecules like CH4, NH3, etc. in termsof overlap and hybridisation of atomic orbitals.

(v) Two types of bonds are formed on account of overlapping.(a) Sigma (s) bond (b) Pi (p) bond

Sigma ( )sbond Pi bond ( )p

(i) It is formed by sideways overlapping, perpendicular to internuclear axis.

(ii) This bond is formed by overlap of p-p orbitals above and below the plane of participating atoms.

(iii) Less strong and more reactive.(iv) Always exist along with a bond (v) Due to resistance to rotation around the bond

the groups attached to it are not free to rotate.

sp

(i) It is formed by end to end (head on) overlap of bonding orbitals along internuclear axis.

(ii) This bond is formed by overlapping of s with s, s with p(along axis) and p with p orbitals.

(iii) Stronger and less reactive.(iv) Can be independently exist(v) The groups or atoms can undergo bond rotation about

single sigma ( ) bondss

HYBRIDISATIONAccording to Pauling, the atomic orbitals combine to form a newset of equivalent orbitals known as hybrid orbitals. It is thesehybrid orbitals which are used in bond formation. The phenomenonis known as hybridisation.Features of Hybridisation(i) The number of hybrid orbitals is equal to the number of atomic

orbitals that get hybridised.(ii) The hybridised orbitals are always equal in energy and shape.(iii) The hybrid orbitals are more effective in forming stable bonds

than pure atomic orbitals.(iv) The type of hybridisation indicates the geometry of

molecules.Conditions of Hybridisation(i) The orbitals present in valence shell of the atom are

hybridised.(ii) The orbitals undergoing hybridisation should have almost

equal energy.(iii) Promotion of electron is not essential condition prior to

hybridisation.

(iv) It is not necessary that only half filled orbitals participate inhybridisation. In some cases, even filled orbitals of valenceshell take part in hybridisation.

Types of HybridisationDepending upon the type and number of orbitals involved inintermixing, the hybridization can be of various types namely sp,sp2, sp3, sp3d, dsp2, sp3d2, sp3d3.Method for Finding the Type of HybridisationThe structure of any molecule can be predicted on the basis ofhybridization which in turn can be known by the following generalformulation :

1H (V M C A)2

= + - +

where H = Number of orbitals involved in hybridization viz. 2, 3, 4,5, 6 and 7, hence nature of hybridization will be sp, sp2, sp3, sp3d,sp3d2, sp3d3 respectively.

V = Number of electrons in valence shell of the central atom,M = Number of monovalent atoms,C = Charge on cation,A = Charge on anion.

( )( )

( ) ( ) [ ]

2 2 2 2 22 2

2 33

4 2 32 22

24

Linear BeCl , C H , CO , HgCl ,

Trigonal planar 2 SnCl , CO

Tetrahedral 3 CH , H O, NH

Square planar 2 Ni CN , PtCl

Trigonal b

sp s p

sp s p

sp s p

dsp d s p

-

- -

+

+

+

é ù+ + ë û

Shape of Hybridisation AtomicExamples

molecule ion Type orbitals

( )( ) ( )

( ) ( ) ( ) ( )( ) ( )

[ ]

35 5

3 25

3 2 2 3 36 6

3 37

ipyramidal 3 PF , PCl

Square pyramidal 3 2 BrF

3 2 , 2 3Octahedral , SF CrFPentagonal bipyramidal 3 3 IF

sp d s p d

sp d s p d

s p d d s psp d d sps p dsp d

-

+ +

+ +

+ + + +

+ +

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MOLECULAR ORBITAL THEORY

Important Features of MOT(i) The electrons in a molecule are present in various molecular

orbitals (MO’ s) as the electrons of atoms are present invarious atomic orbitals (AO’s).

(ii) The atomic orbitals of comparable energies and propersymmetry combine to form MO’s.

(iii) An electron in a MO is influenced by two or more nucleidepending upon no. of atoms in molecule. Thus, a MO ispolycentric.

(iv) The no. of MO’s formed is equal to the number of combiningAO’s. When two AO’s combine, two MO’s are formed. Oneis bonding MO (BMO) and the other is antibonding MO(ABMO).

(v) BMO has lower energy and higher stability thancorresponding ABMO.

(vi) The MO’s are filled in accordance with Aufbau’s principle,obeying the Pauli’s exclusion principle and Hund’s rule.

Conditions for Atomic Orbitals to Form M.O.(i) The combining AO’s must have same or nearly same energy.(ii) The combining AO’s must have same symmetry about the

molecular axis.(iii) The combining AO’s must overlap to the maximum extent.

Types of MOs - MO’s of diatomic molecules are designated as s (sigma), p (pi),d (delta) etc. s - MO’s are symmetrical around bond-axis while p-MO’s are not symmetrical.

Filling of Electrons in MO’s(i) For O2 and F2, the increasing order of energies of various

MO’s is :s 1s < s* 1s < s 2s < s * 2s < s 2pz < (p 2px = p 2py)< (p* 2px = p* 2py) < s* 2pz.

(ii) For Li2, Be2, B2, C2, N2 the increasing order of energies ofvarious MO’s is :s 1s < s* 1s < s 2s < s* 2s < (p2px = p2py) < s 2pz< (p* 2px = p* 2py) < s* 2pz.The main difference between the two types of sequences inenergy level is that for molecules O2, F2, the s2pz M.O.islower in energy than p2px and p2py.

Electronic Configuration and Molecular BehaviourThe distribution of electrons among various MO’s is calledelectronic configuration of the molecule. It gives importantinformation about the molecule as discussed below:(i) Stability of molecules :

If Nb > Na, molecule is stableIf Nb < Na, molecule is unstableWhere Nb ® No. of electrons in BMOWhere Na ® No. of electrons in ABMO

(ii) Bond order (B. O.)

( )b a1B. O. N N2

= -

\ + ve B.O. (i.e. Nb > Na) means stable molecule whereas– ve B.O. (i.e. Na > Nb) means unstable molecule molecule.

(iii) Nature of bond : B. O. values of 1, 2 or 3 correspond to single,double and triple bonds. According to MOT even a fractionalbond order is possible.

(iv) Bond length (BL) : BL µ 1/B.O.Bonding in Few Molecules/Ions(i) He2 : EC is s 1 s2 s* 1s2.

\ ( )1B.O. 2 2 02

= - =

He2 does not exist.(ii) C2 : EC is s 1s2 s* 1s2 s 2 s2 s* 2s2.

(p2px2 = p2py

2)

\ ( )1B. O. 8 4 22

= - =

It is diamagnetic due to absence of unpaired electron(s)(iii) O2 : EC is s 1 s2 s* 1s2. s 2 s2 s* 2s2 s 2pz

2

2 1x x

2 1y y

2p *2p

2p * 2p

æ ö æ öp pç ÷ ç ÷ç ÷ ç ÷p pè ø è ø

\ ( )1B. O. 10 6 22

= - =

It is paramagnetic due to unpaired electrons in p*2px andp*2py.

(iv) H2– :

M.O. configuration – (s 1s)2 (s*1s)1

ParamagneticBond order = ½[2 – 1] = ½

(v) N2+ :

M.O. configuration= (s1s)2(s*1s)2(s2s)2 (s*2s)2(p2px)

2(p2py)2(s2pz)

1

Paramagnetic

Bond order = [ ]1 9 4 2.52

- =

(vi) O2+:

M.O. configuration

= KK (s 2s)2(s* 2s)2(s 2pz)2(p 2px)

2(p 2py)2(p* 2px)

1

Bond order = [ ]18 3 2.5

2- =

(vii) O2– :

M.O. configuration

= KK (s2s)2(s*2s)2(s2pz)2(p2px)

2(p2py)2(p*2px)

2 (p*2py)1

Paramagnetic, Bond order = 12

[8 – 5] = 1.5

70 CHEMISTRY

HYDROGEN BONDING

It is defined as an attractive force which binds H-atom of onemolecule with the electronegative atom (F, O or N) of anothermolecule. The magnitude of H-bonding depends on the physicalstate of the compound, it is maximum in solid state and minimum ingaseous state.Types of H–bonding :(i) Intermolecular H-bond : H-bond formed between two

different molecules of same or different compounds.These are of two types :(a) Homo intermolecular : H–bond between molecules of

same compoundsExample : H—F........H—F........H—F

(b) Hetero intermolecular : H–bond between molecules ofdifferent compoundsExample : Solution of alcohol and water.

(ii) Intramolecular H-bond : Formed when H-atom is in betweentwo highly electronegative (F, O, N) atoms present within thesame molecule. For example in o-nitrophenol i.e.

O

N

H

O

O

Effect of H–bond on Physical Properties:(i) Melting point and boiling point :– Due to intermolecular

H-bonding, M.P. & B.P. of compounds increases.(ii) Molecular weight : Increases due to H–bonding(iii) Solubility :

(a) Intermolecular H–bonding –Few organic compounds(Non–polar) are soluble in water (Polar solvent) due toH–bonding.

(b) Intramolecular H–bonding : It decreases solubility as itforms chelate by H–bonding, so H–atom is not free forother molecule.

(iv) Viscosity and surface tension – H–bond associatesmolecules together so viscosity and surface tension increases.

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CO

NC

EPT

MA

P

72 CHEMISTRY

1. The following group does not contain a dative bond(a) – NO2 (b) – N2Cl(c) – NC (d) all of these

2. Which of the following pairs will form the most stable ionicbond ?(a) Na and Cl (b) Mg and F(c) Li and F (d) Na and F

3. Which of the following conceivable structures for CCl4 willhave a zero dipole moment ?(a) Irregular tetrahedron(b) Square planar(c) Square pyramid (carbon at apex)(d) None of these

4. Among the following which compound will show the highestlattice energy ?(a) KF (b) NaF (c) CsF (d) RbF

5. The compound which contains both ionic and covalentbond :(a) KCl (b) KCN (c) CH4 (d) H2

6. Which one of the following contains a co-ordinate covalentbond ?

(a) OH2 (b) HCl (c) 2BaCl (d) 2 5N H+

7. The strength of bonds formed by s–s and p–p, s–p overlapin the order of(a) s–p > s–s > p–p (b) p–p > s–s > s–p(c) s–s > p–p > s–p (d) s–s > s–p > p–p

8. The types of bonds present in CuSO4. 5H2O are only(a) electrovalent, covalent and co-ordinate covalent bonds(b) electrovalent, covalent and hydrogen bond(c) electrovalent and covalent bonds(d) electrovalent and co-ordinate covalent bonds

9. Which of the following combination will form an electrovalentbond ?(a) P and Cl (b) NH3 and BF3(c) H and Ca (d) H and S

10. Which one of the following will dissolve in water mostreadily?(a) I2 (b) BaCO3 (c) KF (d) PbI2

11. When a metal atom combines with non-metal atom, thenon-metal atom will(a) lose electrons and decrease in size(b) lose electrons and increase in size(c) gain electrons and decrease in size(d) gain electrons and increase in size

12. Indicate the nature of bonding in CCl4 and CaH2(a) Covalent in CCl4 and electrovalent in CaH2(b) Electrovalent in both CCl4 and CaH2(c) Covalent in both CCl4 and CaH2(d) Electrovalent in CCl4 and covalent in CaH2

13. What is the correct mode of hybridisation of the centralatom in the following compounds?

NO–2 SF4 PF6

–

(a) sp sp2 sp3

(b) sp2 sp3d sp3d2

(c) sp2 sp3 d2sp3

(d) sp3 sp3 sp3d2

14. In PO43– ion, the formal charge on each oxygen atom and

P—O bond order respectively are(a) –0.75, 0.6 (b) – 0.75, 1.0(c) – 0.75, 1.25 (d) –3, 1.25

15. Which of the following hydrogen bonds are strongest invapour phase?(a) HF---HF (b) HF---HCl(c) HCl---HCl (d) HF---HI

16. As the s-character of hybridised orbital increases, the bondangle(a) increase (b) decrease(c) becomes zero (d) does not change

17. Among the following the electron deficient compound is(a) BCl3 (b) CCl4 (c) PCl5 (d) BeCl2

18. How many sigma bonds are there in P4O10?(a) 4 (b) 8(c) 12 (d) 16

19. The number of possible resonance structures for 23CO - is

(a) 2 (b) 3 (c) 6 (d) 920. The strongest hydrogen bond is :

(a) O – H...........S (b) S – H.............O(c) F – H............F (d) F – H.............O

21. Which of the following has the highest dipole moment?

(a)H

HC = O (b)

H|

H|C

CH|

CH|C

3

3

=

(c)

3

3

CH|

H|C

CH|

H|C = (d)

3

3

CH|

Cl|C

CH|

Cl|C =

22. In the cyanide ion, the formal negative charge is on(a) C (b) N(c) Both C and N (d) Resonate between C and N

23. Of the following hydrides which one has the lowest boilingpoint ?(a) 3AsH (b) 3SbH (c) 3PH (d) 3NH

24. Resonance structures can be written for(a) O3 (b) NH3 (c) CH4 (d) H2O

25. Which one of the following is the correct order ofinteractions?(a) covalent < hydrogen bonding < van der Waals < dipole-

dipole(b) van der Waals < hydrogen bonding < dipole-dipole

< covalent(c) van der Waals < dipole-dipole < hydrogen bonding

< covalent(d) dipole-dipole < van der Waals < hydrogen bonding

< covalent26. Which one of the following molecules will form a linear

polymeric structure due to hydrogen bonding?(a) NH3 (b) H2O (c) HCl (d) HF

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27. Among the following, the species having square planargeometry for central atom are(i) XeF4 (ii) SF4(iii) [NiCl4]

2– (iv) [PtCl4]2–

(a) (i) and (iv) (b) (i) and (ii) .(c) (ii) and (iii) (d) (iii) and (iv)

28. In [Ag (CN2)]–, the number of p bonds is :(a) 2 (b) 3(c) 4 (d) 6

29. Which of the following does not contain any coordinatebond?

(a) H3O+ (b) 4BF-

(c) 2HF- (d) 2SO -

30. Hydrogen bonding could not affect the boiling point of :(a) CH3OH (b) HI (c) H2O (d) NH3

31. The low density of ice compared to water is due to :(a) induced dipole-induced dipole interactions(b) dipole-induced dipole interactions(c) hydrogen bonding interactions(d) dipole-dipole interactions

32. Intramolecular hydrogen bonding is found in :(a) p-nitrophenol (b) m-nitrophenol(c) o-nitrophenol (d) phenol

33. The vapour pressure of

2NO

OH is higher than

2NO

OH due to

(a) dipole moment (b) dipole-dipole interaction(c) H-bonding (d) lattice structure

34. The bond length between C–C bond in sp2 hybridisedmolecule is :(a) 1.2 Å (b) 1.39 Å(c) 1.33 Å (d) 1.54 Å

35. Which one of the following pairs of molecules will havepermanent dipole moments for both members ?(a) NO2 and CO2 (b) NO2 and O3(c) SiF4 and CO2 (d) SiF4 and NO2

36. The correct sequence of increasing covalent character isrepresented by(a) LiCl < NaCl < BeCl2 (b) BeCl2 < LiCl < NaCl(c) NaCl < LiCl < BeCl2 (d) BeCl2 < NaCl < LiCl

37. Which of the following would have a permanent dipolemoment?(a) SiF4 (b) SF4 (c) XeF4 (d) BF3

38. Dipole moment is shown by :(a) cis-1, 2-dichloroethene(b) trans-1, 2-dichloroethene(c) trans-2, 3-dichloro-2 pentene(d) Both (a) and (c)

39. The correct order of hybridisation of the central atom in thefollowing species :NH3, [PtCl4]

2–, PCl5 and BCl3 is(a) dsp2, dsp3, sp2, sp3 (b) sp3, dsp2, dsp3, sp2

(c) dsp2, sp2, sp3, dsp3 (d) dsp2, sp3, sp2, sp3

40. Which of the following salt shows maximum covalentcharacter?(a) 3AlCl (b) 2MgCl (c) CsCl (d) 3LaCl

41. Pauling’s electronegativity values for elements are useful inpredicting :(a) polarity of bonds in molecules(b) ionic and covalent nature of bonds(c) coordination number(d) Both (a) and (b)

42. Identify the non polar molecule in the following compounds:(a) H2 (b) HCl(c) HF and HBr (d) HBr

43. Which of the following substances has the greatest ioniccharacter ?(a) Cl2O (b) NCl3 (c) PbCl2 (d) BaCl2

44. Methanol and ethanol are miscible in water due to(a) Covalent character(b) Hydrogen bonding character(c) Oxygen bonding character(d) None of the above

45. As compared to covalent compounds, electrovalentcompounds generally have(a) Low melting points and low boiling points(b) Low melting points and high boiling points(c) High melting points and low boiling points(d) High melting points and high boiling points

46. Which bond angle q would result in the maximum dipolemoment for the triatomic molecule YXY(a) q = 90° (b) q = 120°(c) q = 150° (d) q = 180°

47. Which does not show resonance :(a) Benzene (b) Aniline(c) Ethylamine (d) Toluene

48. Polarisibility of halide ions increases in the order(a) F –, I – , Br–, Cl– (b) Cl –, Br – , I–, F–

(c) I –, Br – , Cl–, F– (d) F –, Cl –, Br–, l–49. For AB bond if percent ionic character is plotted against

electronegativity difference (XA – XB), the shape of the curvewould look like

100

50

01 2 3

A

BC

D

Perc

ent i

onic

cha

ract

er

(X – X )A B

The correct curve is(a) (A) (b) (B) (c) (C) (d) (D)

50. In which of the following compound electrovalent, covalentand co-ordinate bonds are present?(a) 4NH Cl (b) 4CCl (c) 2CaCl (d) 2H O

51. Following are the molecules or ions of the type AH3 where Ais the central atom. In which of them does the central atom Ause its sp2 hybrid orbitals to form A-H bonds ?(a) PH3 (b) NH3 (c) CH3

+ (d) CH3–

74 CHEMISTRY

52. Equilateral shape has(a) sp hybridisation (b) sp2 hybridisation(c) sp3 hybridisation (d) None of these

53. Which one of the following has the shortest carbon carbonbond length ?(a) Benzene (b) Ethene (c) Ethyne (d) Ethane

54. Which of the following does not have a tetrahedral structure?

(a) BH–4 (b) BH3 (c) +

4NH (d) CH455. The bond length in LiF will be

(a) less than that of NaF (b) more than that of KF(c) equal to that of KF (d) equal to that of NaF

56. Which of the following molecules does not have a lineararrangement of atoms ?(a) H2S (b) C2H2 (c) BeH2 (d) CO2

57. In which one of the following molecules the central atomsaid to adopt sp2 hybridization?(a) BeF2 (b) BF3 (c) C2H2 (d) NH3

58. In an octahedral structure, the pair of d orbitals involved in2 3d sp hybridization is

(a) 2 2 2,x y zd d

- (b) 2 2,xz x yd d

-

(c) 2 , xzzd d (d) ,xy yzd d

59. Which of the following shows linear structure?(a) Ethane (b) Ethene (c) Acetylene(d) 4CCl

60. The element X [1s2, 2s22p6, 3s23p5] reacts with Y (atomicno. = 1) to form(a) XY2 (b) X2Y (c) XY (d) None

61. Which one of the following does not follow octate rule?(a) PF3 (b) BF3 (c) CO2 (d) CCl4

62. In BrF3 molecule, the lone pairs occupy equatorial positionsto minimize(a) lone pair - bond pair repulsion only(b) bond pair - bond pair repulsion only(c) lone pair - lone pair repulsion and lone pair - bond pair

repulsion(d) lone pair - lone pair repulsion only

63. Match List I and List II and pick out correct matching codesfrom the given choices :

List I List IICompound Structure

A. ClF3 1. Square planarB: PCl5 2. TetrahedralC. IF5 3. Trigonal bipyramidalD. CCl4 4. Square pyramidalE. XeF4 5. T-shapedCodes(a) A-5, B-4, C-3, D-2, E-1 (b) A-5, B-3, C-4, D-2, E-1(c) A-5, B-3, C-4, D-1, E-2 (d) A-4, B-3, C-5, D-2, E-1

64. Bond order in benzene is :(a) 1 (b) 2(c) 1.5 (d) none of these

65. The decreasing values of bond angles from NH3 (106º) toSbH3 (101º) down group-15 of the periodic table is due to(a) decreasing lp-bp repulsion(b) decreasing electronegativity(c) increasing bp-bp repulsion(d) increasing lp-bp repulsion

66. Hybridization present in CIF3 is :(a) sp2 (b) sp3 (c) dsp2 (d) sp3d

67. Which one of the following has the pyramidal shape?(a) CO3

2– (b) SO3 (c) BF3 (d) PF368. Fluorine molecule is formed by

(a) the axial p - p orbital overlap(b) the sideways p - p orbital overlap(c) the s - s orbital overlap(d) the s - p orbital overlap

69. In NO3– ion number of bond pair and lone pair of electrons

on nitrogen atom respectively are(a) 2, 2 (b) 3, 1 (c) 1, 3 (d) 4, 0

70. Among the following the pair in which the two species arenot isostructural is(a) SiF4 and SF4 (b) 33 XeOandIO-

(c) +-44 NHandBH (d) 66 SFandPF-

71. 2N and 2O are converted into monoanions, –2N and -

2Orespectively. Which of the following statements is wrong ?(a) In –N2 , N – N bond weakens

(b) In –O2 , O - O bond order increases

(c) In –O2 , O - O bond order decreases

(d) –N2 becomes paramagnetic72. In a regular octahedral molecule, MX6 the number of

X - M - X bonds at 180° is(a) three (b) two (c) six (d) four

73. Which of the following molecules has trigonal planargeometry?(a) BF3 (b) NH3 (c) PCl3 (d) IF3

74. The pair of species having identical shapes for molecules ofboth species is(a) XeF2, CO2 (b) BF3, PCl3(c) PF5, IF5 (d) CF4, SF4

75. Which contains both polar and non - polar bonds?(a) NH4Cl (b) HCN(c) H2O2 (d) CH4

76. The values of electronegativity of atoms A and B are 1.20and 4.0 respectively. The percentage of ionic character of A- B bond is(a) 50% (b) 72.24% (c) 55.3% (d) 43%

77. The electronegativities of F, Cl, Br and l are 4.0, 3.0, 2.8, 2.5respectively. The hydrogen halide with a high percentage ofionic character is(a) HF (b) HCl (c) HBr (d) HI

78. Which one of the following has the regular tetrahedralstructure ?

(a) -4BF (b) SF4

(c) XeF4 (d) -24 ])CN(Ni[

(Atomic nos. : B = 5, S = 16, Ni =28, Xe = 54)79. The species in which the central atom uses sp2 hybrid orbitals

in its bonding is

(a) PH3 (b) NH3 (c) +3CH (d) SbH3

80. Which molecule is planar ?(a) 4SF (b) 4XeF (c) 3NF (d) 4SiF

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81. Amongst the following, the molecule/ion that is linear is :

(a) SO2 (b) CO2 (c) 2ClO- (d) 2NO-

82. The trigonal bipyramidal geometry is obtained from thehybridisation :(a) dsp3 or sp3d (b) dsp2 or sp2d(c) d 2sp3 or sp3d 2 (d) None of these

83. The true statements from the following are1. PH5 and BiCl5 do not exist2. pp - dp bond is present in SO23. Electrons travel at the speed of light4. SeF4 and CH4 have same shape

5. +3I has bent geometry

(a) 1, 3 (b) 1, 2, 5(c) 1, 3, 5 (d) 1, 2, 4

84. The hybrid state of S in SO3 is similar to that of(a) C in C2H2 (b) C in C2H4(c) C in CH4 (d) C in CO2

85. Which of the following set contains species having sameangle around the central atom?(a) SF4, CH4, NH3 (b) NF3, BCl3, NH3(c) BF3, NF3, AlCl3 (d) BF3, BCl3, BBr3

86. The compound MX4 is tetrahedral. The number of Ð XMXformed in the compound are(a) Three (b) Four(c) Five (d) Six

87. The bond angle between two hybrid orbitals is 105°. Thepercentage of s-character of hybrid orbital is between(a) 50 - 55% (b) 9 - 12%(c) 22 - 23% (d) 11 - 12%

88. Which of the following has the shortest C – C bond length?(a) C2H5OH (b) C2H6(c) C2H2 (d) C2H4

89. In piperidine NH

, the hybrid state assumed by N is

(a) sp (b) sp2 (c) sp3 (d) dsp2

90. The compound 1, 2 - butadiene has(a) only sp hybridized carbon atoms(b) only sp2 hybridized carbon atoms(c) both sp and sp2 hybridized carbon atoms(d) sp, sp2 and sp3 hybridized carbon atoms

91. In the following molecule, the two carbon atoms marked byasterisk (*) possess the following type of hybridized orbitals

* *3 3H C C C CH- º -

(a) sp3 orbital (b) sp2 orbital(c) sp orbital (d) sp3 and sp respectively

92. Which has the least bond angle(a) NH3 (b) BeF2 (c) H2O (d) CH4

93. The shape of 24SO - ion is

(a) square planar (b) tetrahedral(c) trigonal bipyramidal (d) hexagonal

94. The largest bond angle is in(a) AsH3 (b) NH3 (c) H2O (d) PH3

95. Which one has a pyramidal structure(a) CH4 (b) NH3 (c) H2O (d) CO2

96. Each carbon in carbon suboxide (C3O2) is(a) sp2 - hybridized(b) sp3 - hybridized

(c) sp - hybridized(d) sp2 - hybridized but linked with one co-ordinate bond

97. Which statement is NOT correct ?(a) A sigma bond is weaker than a p -bond.(b) A sigma bond is stronger than a p -bond.(c) A double bond is stronger than a single bond.(d) A double bond is shorter than a single bond.

98. Linear combination of two hybridized orbitals belonging totwo atoms and each having one electron leads to a(a) sigma bond(b) double bond(c) co-ordinate covalent bond(d) Pi bond.

99. Main axis of a diatomic molecule is z, molecular orbital pxand py overlap to form which of the following orbital?(a) p - molecular orbital (b) s - molecular orbital(c) d - molecular orbital (d) No bond will be formed

100. The number of anti-bonding electron pairs in 22O- molecular

ion on the basis of molecular orbital theory is, (Atomicnumber of O is 8)(a) 5 (b) 2 (c) 3 (d) 4

101. In which of the following species is the underlined carbonhaving sp3 - hybridisation ?(a) OOHC–CH3 (b) OHHCCH 23

(c) 33 OCHCCH (d) 32 -= CHHCCH102. Which of the following statements is not correct for sigma

and pi-bonds formed between two carbon atoms?(a) Sigma-bond determines the direction between carbon

atoms but a pi-bond has no primary effect in this regard(b) Sigma-bond is stronger than a pi-bond(c) Bond energies of sigma- and pi-bonds are of the order

of 264 kJ/mol and 347 kJ/mol, respectively(d) Free rotation of atoms about a sigma-bond is allowed

but not in case of a pi-bond103. The bond order in N2

+ is(a) 1.5 (b) 3.0 (c) 2.5 (d) 2.0

104. How many s- and p-bonds are there in?

(a) 14s, 8p (b) 18s, 8p (c) 19s, 4p (d) 14s, 2p105. Which of the following has p π – d π bonding ?

(a) –NO3 (b) -23SO (c) -3

3BO (d) -23CO

106. How many s and p bonds are present in toluene?(a) 3p + 8s (b) 3p + 10s(c) 3p + 15s (d) 6p + 3s

107. The calculated bond order in O2– ion is

(a) 1 (b) 1.5 (c) 2 (d) 2.5

108. The molecular electronic configuration of 2H- ion is?

(a) ( )21ss (b) ( ) ( )22 *1 1s ss s

(c) ( ) ( )12 *1 1s ss s (d) ( )31ss109. In pyrophosphoric acid, H2P2O7, number of s and dp – pp

bonds are respectively(a) 8 and 2 (b) 6 and 2(c) 12 and zero (d) 12 and 2

76 CHEMISTRY

110. Arrange the following ions in the order of decreasing X – Obond length, where X is the central atom in

(a) 2 34 4 4 4ClO ,SO , PO ,SiO- - - -

(b) 4 3 24 4 4 4SiO ,PO ,SO ,ClO- - - -

(c) 4 3 24 4 4 4SiO ,PO ,ClO ,SO- - - -

(d) 4 2 34 4 4 4SiO ,SO ,PO ,ClO- - - -

111. In the change of NO+ to NO, the electron is added to(a) s - orbital (b) p - orbital(c) s* - orbital (d) p* - orbital

112. Which of the following represents the given mode ofhybridisation sp2 –sp2 – sp - sp from left to right ?(a) NCCHCH º-=2 (b) CHC–CHC ºº

(c) 22 === CHCCCH (d) H C2

CH2

113. In 2O- , O2 and 22O- molecular species, the total number of

antibonding electrons respectively are(a) 7, 6, 8 (b) 1, 0, 2 (c) 6, 6, 6 (d) 8, 6, 8

114. If Nx is the number of bonding orbitals of an atom and Ny isthe number of antibonding orbitals, then the molecule/atomwill be stable if(a) Nx > Ny (b) Nx = Ny (c) Nx < Ny (d) Nx £ Ny

115. Which of the following is the correct electron dot structureof N2O molecule?

(a) :ONN:..

== (b) -+

== :ONN: ....

(c) :ONN......

.. == (d) :ONN: ..

..==

116. The correct statement with regard to 2H+ and 2H- is

(a) Both 2H+ and 2H- do not exist

(b) 2H- is more stable than 2H+

(c) 2H+ is more stable than 2H-

(d) Both 2H+ and 2H- are equally stable117. In which of the following pairs hydrogen-bonding is not

possible ?

(a) NH3, NH3 (b) NH3,CH4(c) H2O, CH3OCH3 (d) CH3OH,CH3OCH3

118. Dipole moment of CO2 is zero which implies that(a) Carbon and oxygen have equal electronegativities(b) Carbon has no polar bond(c) CO2 is a linear molecule(d) Carbon has bond moments of zero value

119. In which of the following pairs, the two species are iso-structure?(a) SO3

2– and NO3– (b) BF3 an NF3

(c) BrO3– and XeO3 (d) SF4 and XeF4

120. Four diatomic species are listed below in different sequences.Which of these presents the correct order of their increasingbond order ?(a) 2

2 2 2O NO C He- - +< < < (b) 22 2 2NO C O He- - +< < <

(c) 22 2 2C He NO O- + -< < < (d) 2

2 2 2He O NO C+ - -< < <121. What is the dominant intermolecular force or bond that must

be overcome in converting liquid CH3OH to a gas?(a) Dipole-dipole interactions(b) Covalent bonds(c) London dispersion forces(d) Hydrogen bonding

122. In which one of the following species the central atom hasthe type of hybridization which is not the same as that presentin the other three?(a) 4SF (b) –

3I (c) 2–5SbCl (d) 5PCl

123. Some of the properties of the two species, 3NO- and H3O+

are described below. Which one of them is correct?(a) Similar in hybridization for the central atom with different

structures.(b) Dissimilar in hybridization for the central atom with

different structures.(c) Isostructural with same hybridization for the central

atom.(d) Isostructural with different hybridization for the central

atom.124. In which of the following molecules the central atom does

not have sp3 hybridization?(a) +

4NH (b) CH4 (c) SF4 (d) –4BF

125. Which one of the following species does not exist undernormal conditions?(a) 2Be+ (b) 2Be (c) 2B (d) 2Li

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1. In the anion HCOO– the two carbon - oxygen bonds are foundto be of equal length. What is the reason for it ?(a) Electronic orbitals of carbon atom are hybridised(b) The C = O bond is weaker than the C –O bond(c) The anion HCOO– has two resonating structures(d) The anion is obtained by removal of a proton from theacid molecule

2. In compounds of type ECl3, where E = B, P, As or Bi, theangles Cl - E- Cl for different E are in the order.(a) B > P = As = Bi (b) B > P > As > Bi(c) B < P = As = Bi (d) B < P < As < Bi

3. Molecular shapes of SF4, CF4 and XeF4 are(a) the same, with 2, 0 and 1 lone pairs of electrons

respectively(b) the same, with 1, 1 and 1 lone pairs of electrons

respectively(c) different, with 0, 1 and 2 lone pairs of electrons

respectively(d) different, with 1, 0 and 2 lone pairs of electrons

respectively4. The molecules 3BF and 3NF are both covalent compounds,

but 3BF is non polar whereas 3NF is polar. The reason forthis is(a) atomic size of Boron is larger than nitrogen(b) Boron is metal while nitrogen is gas(c) B – F bonds are non-polar while N – F bonds are polar(d) 3BF is planar but 3NF is pyramidal

5. The common features among the species CN–, CO and NO+

are(a) bond order three and isoelectronic(b) bond order three and weak field ligands(c) isoelectronic and weak field ligands(d) None of these

6. The correct order of bond energies in NO, NO+ and NO– is:(a) +- >> NONONO (b) +- >> NONONO(c) -+ >> NONONO (d) NONONO >> -+

7. A s bonded molecule 3MX is T-shaped. The number ofnon bonded pair of electrons is(a) 0(b) 2(c) 1(d) can be predicted only if atomic number is known

8. Experiment shows that H2O has a dipole moment while CO2has not. Point out the structures which best illustrate thesefacts

(a)O

HO = C = O ; H (b) O = C = O ; H – O – H

(c) OC

O ;H – H –O (d)

O HC = O; O–H

9. The enolic form of a acetone contains(a) 9 sigma bonds, 1 pi bond and 2 lone pairs(b) 8 sigma bonds, 2 pi bonds and 2 lone pairs(c) 10 sigma bonds, 1 pi bond and 1 lone pair(d) 9 sigma bonds, 2 pi bonds and 1 lone pair

10. In which case hydrogen bond will not be observed(a) H3O2

– (b) H2O(c) H5O2

+ (d) H3O+

11. Which one of the following formulae does not correctlyrepresent the bonding capacities of the two atoms involved?

(a)

+

úúú

û

ù

êêê

ë

é

H—

H

H

P—H|

| (b) F F

O

(c) NO ¬O

O – H(d)

O

O – HH – C = C

|H

12. Trimethylamine is a pyramidal molecule N

3CH3CHCH3

and formamide is a planar molecule

O||C

N|

H

HH

, The

hybridisation of Nitrogen in both is

(a) 22 sp,sp (b) 23 sp,sp

(c) 33 sp,sp (d) sp,sp2

13. If climbing of water droplets is made to occur on a coatedmicroscope slide, the slide would have to be coated in whichof the following way

A B C D(a) A (b) B(c) C (d) D

14. Mark the incorrect statement in the following(a) The bond order in the species O2, O2

+ and O2–

decreases as -+ >> 222 OOO(b) The bond energy in a diatomic molecule always

increases when an electron is lost(c) Electrons in antibonding M.O. contribute to repulsion

between two atoms.(d) With increase in bond order, bond length decreases

and bond strength increases.

78 CHEMISTRY

15. Which of the following is not expected to be resonancestructure ?

(a) :CH –N = O:2+–

. .CH3

(b) CH =N – O:2+ –

. .

. .

CH3

(c) CH =N – O:2+ –

. .

. .

CH3

(d) CH =N = O2CH3

16. The boiling point of p-nitrophenol is higher than that ofo-nitrophenol because(a) NO2 group at p-position behave in a different way from

that at o-position.(b) intramolecular hydrogen bonding exists in p-

nitrophenol(c) there is intermolecular hydrogen bonding in

p-nitrophenol(d) p-nitrophenol has a higher molecular weight than

o-nitrophenol.17. Which is the correct order of dipole moments of the

compounds :

(I)NO2

OH(II)

Cl

Cl

(III)

CH3

CH3

(a) I > II > III (b) II > I > III(c) III > I > II (d) III > II > I

18. The dipole moment of

X

is 1.5 D. The dipole moment of

X

XX

X is :

(a) 1 D (b) 1.5 D(c) 2.25 D (d) 3 D

19. The BCl3 is a planar molecule whereas NCl3 is pyramidalbecause(a) B-Cl bond is more polar than N-Cl bond(b) N-Cl bond is more covalent than B-Cl bond(c) nitrogen atom is smaller than boron atom(d) BCl3 has no lone pair but NCl3 has a lone pair of

electrons20. Dipole moment of p-nitroaniline, when compared to

nitrobenzene (X) and aniline (Y) will be :(a) smaller than both (X) and (Y)(b) greater than both (X) and (Y)(c) greater than (Y) but smaller than (X)(d) equal to zero

21. The cylindrical shape of an alkyne is due to the fact that ithas(a) three sigma C – C bonds

(b) two sigma C – C and one 'p' C – C bonds(c) three 'p' C – C bonds(d) one sigma C– C and two 'p' C – C bonds

22. N2 and O2 are converted into monocations, N2+ and O2

+

respectively. Which of the following statements is wrong ?(a) In N2

+, the N—N bond weakens(b) In O2

+, the O—O bond order increases(c) In O2

+, paramagnetism decreases(d) N2

+ becomes diamagnetic23. The AsF5 molecule is trigonal bipyramidal. The hybrid orbitals

used by the As atom for bonding are(a) 2 2

2, , , ,z x yx yd d s p p

-(b) dxy, s, px, py, pz

(c) s, px, py, pz, dz2 (d) 2 2x yd

-, s, px, py, pz

24. The dipole moments of diatomic molecules AB and CD are10.41D and 10.27 D, respectively while their bond distancesare 2.82 and 2.67 Å, respectively. This indicates that(a) bonding is 100% ionic in both the molecules(b) AB has more ionic bond character than CD(c) AB has lesser ionic bond character than CD(d) bonding is nearly covalent in both the molecules

25. Consider the two molecules :

(I) H3C

CH3C

OH

O

CCH

(II) H3C CH3

C

OH

O

CC

H H

H.

Which of the following statements about them is true ?(a) II is capable of forming intramolecular hydrogen

bonding whereas I is not(b) Both are capable of forming equally stable

intramolecular hydrogen bonding(c) I is likely to form more stable intramolecular hydrogen

bonding than II(d) Both of them cannot form intramolecular hydrogen

bonding

26. The hybridization of P is -34PO is the same as of

(a) S in SO3 (b) N in NO–3

(c) I in ICl+2 (d) I in ICl4–

27. The relationship between the dissociation energy of N2 andN2

+ is :(a) Dissociation energy of N2

+ > dissociation energy ofN2

(b) Dissociation energy of N2 = dissociation energy ofN2

+

(c) Dissociation energy of N2 > dissociation energy ofN2

+

(d) Dissociation energy of N2 can either be lower or higherthan the dissociation energy of N2

+

28. In X — H --- Y, X and Y both are electronegative elements(a) thus electron density on X will increase and on H will

decrease(b) in both electron density will decrease(c) in both electron density will increase and on H will

decrease(d) thus electron density will decrease on X and will

increase on H

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29. The electronegativity difference between N and F is greaterthan that between N and H yet the dipole moment ofNH3 (1.5 D) is larger than that of NF3 (0.2D). This is because(a) in NH3 the atomic dipole and bond dipole are in the

same direction whereas in NF3 these are in oppositedirections

(b) in NH3 as well as NF3 the atomic dipole and bond dipoleare in opposite directions

(c) in NH3 the atomic dipole and bond dipole are in theopposite directions whereas in NF3 these are in thesame direction

(d) in NH3 as well as in NF3 the atomic dipole and bonddipole are in the same direction

30. In the process : ++ ¾®¾+ 43 XHHXH , (X being N, P, Asand Sb), the maximum increase in the bond angle will be incase of(a) NH3 (b) PH3(c) AsH3 (d) SbH3

31. On changing N2 to N2+, the dissociation energy of N–N bond

..... and on changing O2 to O+

2 the dissociation energy of O–O bond....(a) increases, decreases (b) decreases, increases(c) decreases in both cases(d) increases in both cases

32. H2O is dipolar, whereas BeF2 is not. It is because(a) the electronegativity of F is greater than that of O(b) H2O involves hydrogen bonding whereas BeF2 is a

discrete molecule(c) H2O is linear and BeF2 is angular(d) H2O is angular and BeF2 is linear

33. The bond order in NO is 2.5 while that in NO+ is 3. Which ofthe following statements is true for these two species ?(a) Bond length in NO+ is equal to that in NO(b) Bond length in NO is greater than in NO+

(c) Bond length in NO+ is greater than in NO(d) Bond length is unpredictable

34. Which is correct about the directional orientation of theorbital?(a) p > sp (b) p < sp(c) p = sp (d) none of these

35. Pb4+ is less stable than Sn4+. It is because of(a) Higher value of IE1 + IE2 + IE3 + IE4 for Pb than Sn(b) Lesser polarising power of Pb4+ than Sn4+

(c) Inert pair effect in Pb(d) None of these

36. According to molecular orbital theory which of the followingstatement about the magnetic character and bond order iscorrect regarding +

2O(a) Paramagnetic and Bond order < O2(b) Paramagnetic and Bond order > O2(c) Diamagnetic and Bond order < O2(d) Diamagnetic and Bond order > O2

37. Select the most stable cation

(a) 4NH+ (b) 5CH+

(c) 3OH+ (d) 2FH+

38. Which of the following represents the correct order of Cl–O

bond lengths in – – – –2 3 4ClO ,ClO ,ClO ,ClO ?

(a) – – – –4 3 2ClO ClO ClO ClO= = =

(b) – – – –2 3 4ClO ClO ClO ClO< < <

(c) – – – –4 3 2ClO ClO ClO ClO< < <

(d) – – – –3 4 2ClO ClO ClO ClO< < <

39. The correct order of increasing bond angles in the followingtriatomic species is :(a) 2 2 2NO NO NO- +< < (b) 2 2 2NO NO NO- +< <(c) 2 2 2NO NO NO+ -< < (d) 2 2 2NO NO NO+ -< <

40. In which of the following molecules / ions BF3, 2NO ,- 2NH-

and H2O , the central atom is sp2 hybridized ?(a) 2NH- and H2O (b) 2NO- and H2O(c) BF3 and 2NO- (d) 2NO- and 2NH-

41. According to MO theory which of the following lists ranksthe nitrogen species in terms of increasing bond order?(a) 2– –

2 2 2N N N< < (b) 2– –2 2 2N N N< <

(c) – 2–2 2 2N N N< < (d) – 2–

2 2 2N N N< <42. Which of the two ions from the list given below have the

geometry that is explained by the same hybridization oforbitals, NO2

–, NO3–, NH2

–, NH4+, SCN– ?

(a) NO2– and NO3

– (b) NH2– and NO3

–

(c) SCN– and NH2– (d) NO2

– and NH2–

43. Which of the following has the minimum bond length ?(a) O2

+ (b) O2– (c) O2

2– (d) O244. The pairs of species of oxygen and their magnetic behaviours

are noted below. Which of the following presents the correctdescription ?(a) 2

2 2O ,O- - – Both diamagnetic(b) 2

2 2O ,O+ - – Both paramagnetic(c) 2 2O ,O+ – Both paramagnetic(d) None of these

45. Which one of the following pairs is isostructural (i.e., havingthe same shape and hybridization) ?

(a) 3 3BCl and BrCl-é ùë û (b) 3 3NH and NO-é ù

ë û

(c) [ ]3 3NF and BF (d) 4 4BF and NH- +é ùë û

46. Bond order of 1.5 is shown by :(a) 2O+ (b) 2O-

(c) 22O - (d) O2

47. Which of the following hydrogen bonds is the strongest?(a) O – H - - - F (b) O – H - - - H(c) F – H - - - F (d) O – H - - - O

48. The bond dissociation energy of B – F in BF3 is 646 kJ mol–1

whereas that of C – F in CF4 is 515 kJ mol–1. The correctreason for higher B – F bond dissociation energy as comparedto that of C – F is(a) stronger s bond between B and F in BF3 as compared

to that between C and F in CF4.(b) significant pp – pp interaction between B and F in BF3

whereas there is no possibility of such interactionbetween C and F in CF4.

(c) lower degree of pp – pp interaction between B and F inBF3 than that between C and F in CF4.

(d) smaller size of B– atom as compared to that of C– atom.

80 CHEMISTRY

49. The number of types of bonds between two carbon atoms incalcium carbide is(a) One sigma, One pi (b) Two sigma, one pi(c) Two sigma, two pi (d) One sigma, two pi

50. In which of the following pairs the two species are notisostructural ?(a) 2

3CO - and 3NO- (b) 4PCl+ and SiCl4(c) PF5 and BrF5 (d) 3

6AlF - and SF651. Among the following, the paramagnetic compound is

(a) Na2O2 (b) O3(c) N2O (d) KO2

52. The species having bond order different from that in CO is(a) NO– (b) NO+

(c) CN– (d) N253. Assuming that Hund’s rule is violated, the bond order and

magnetic nature of the diatomic molecule B2 is :(a) 1 and diamagnetic (b) 0 and dimagnetic(c) 1 and paramagnetic (d) 0 and paramagnetic

54. The species having pyramidal shape is :(a) SO3 (b) BrF3

(c) 2–3SiO (d) OsF2

55. Geometrical shapes of the complexes formed by the reactionof Ni2+ with Cl– , CN– and H2O, respectively, are(a) octahedral, tetrahedral and square planar(b) tetrahedral, square planar and octahedral(c) square planar, tetrahedral and octahedral(d) octahedral, square planar and octahedral

56. In allene (C3H4), the type(s) of hybridisation of the carbonatoms is (are) :(a) sp and sp3 (b) sp and sp2

(c) only sp3 (d) sp2 and sp3

57. Which one of the following molecules is expected to exhibitdiamagnetic behaviour ?(a) C2 (b) N2(c) O2 (d) S2

58. Which of the following is the wrong statement ?(a) ONCl and ONO– are not isoelectronic.(b) O3 molecule is bent(c) Ozone is violet-black in solid state(d) Ozone is diamagnetic gas.

59. In which of the following pairs of molecules/ions, both thespecies are not likely to exist ?(a) 2

2 2H ,He+ - (b) 22 2H ,He- -

(c) 22 2H ,He+ (d) 2

2 2H ,He- +

60. Stability of the species Li2, 2Li- and 2Li+ increases in theorder of :

(a) 2 2 2Li Li Li+ -< < (b) 2 2 2Li Li Li- +< <

(c) 2 2 2Li Li Li- +< < (d) 2 2 2Li Li Li- +< <61. Considering the state of hybridization of carbon atoms, find

out the molecule among the following which is linear ?(a) CH3– CH = CH–CH3(b) CH3 – C º C – CH3(c) CH2 = CH – CH2 – C º CH(d) CH3 – CH2 – CH2 – CH3

62. Which of the following species contains three bond pairsand one lone pair around the central atom ?

(a) H2O (b) BF3 (c) 2NH- (d) PCl363. The pair of species with the same bond order is :

(a) 2–2O , B2 (b) +

2O , NO+

(c) NO, CO (d) N2, O264. Which of the following species exhibits the diamagnetic

behaviour ?(a) NO (b) O2

2– (c) O2+ (d) O2

65. The charge/size ratio of a cation determines its polarizingpower. Which one of the following sequences representsthe increasing order of the polarizing power of the cationicspecies, K+, Ca2+, Mg2+, Be2+ ?(a) Ca2+ < Mg2+ < Be2+ < K+

(b) Mg2+ < Be2+ < K+ < Ca2+

(c) Be2+ < K+ < Ca2+ < Mg2+

(d) K+ < Ca2+ < Mg2+ < Be2+

66. In which of the following ionization processes, the bondorder has increased and the magnetic behaviour haschanged?(a) 2 2N N +® (b) 2 2C C +®(c) NO NO+® (d) 2 2O O +®

67. Which one of the following pairs of species have the samebond order?(a) CN– and NO+ (b) CN– and CN+

(c) 2O- and CN– (d) NO+ and CN+

68. The molecule having smallest bond angle is :(a) NCl3 (b) AsCl3(c) SbCl3 (d) PCl3

69. The electronegativities of four atoms labeled as D, E, F andG are as follows. D = 3.8, E = 3.3, F = 2.8 and G = 1.3. If theatoms form the molecules DE, DG, EG and DF, the order ofarrangement of these molecules in the increasing order ofcovalent bond character is(a) DG < EG < DF < DE (b) DF < DG < DE < EG(c) DG < DF < EG < DE (d) DE < EG < DG < DF

70. Which one of the following is the correct statement?(a) O2 molecule has bond order 2 and is diamagnetic(b) N2 molecule has bond order 3 and is paramagnetic(c) H2 molecule has bond order zero and is diamagnetic(d) C2 molecule has bond order 2 and is diamagnetic

71. Among the following molecules : SO2, SF4, CIF3, BrF5 andXeF4, which of the following shapes does not describe anyof the molecules mentioned?(a) Bent (b) Trigonal bipyramidal(c) See-saw (d) T-shape

72. The type of hybridization in xenon atom and the number oflone pairs present on xenon atom in xenon hexafluoridemolecule are respectively(a) sp3d3 , one (b) sp3d3 , two(c) sp3d3 , two (d) sp3d2 , zero

DIRECTIONS for Qs. 73 to 75: These are Assertion-Reasontype questions. Each of these question contains two statements:Statement-1 (Assertion) and Statement-2 (Reason). Answer thesequestions from the following four options.(a) Statement- 1 is True, Statement-2 is True, Statement-2 is a

correct explanation for Statement -1

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(b) Statement -1 is True, Statement -2 is True ; Statement-2 isNOT a correct explanation for Statement - 1

(c) Statement - 1 is True, Statement- 2 is False(d) Statement -1 is False, Statement -2 is True73. Statement-1 : Ice is less dense than liquid water.

Statement-2 : There are vacant spaces between hydrogenbonded water molecules in ice.

74. Statement-1 : A resonance hybrid is always more stable thanany of its canonical structures.Statement-2 : This stability is due to delocalization ofelectrons.

75. Statement-1 : o and p-nitrophenols can be separated by steamdistillation.Statement-2 : o-nitrophenol have intramolecular hydrogenbonding while p-nitrophenol exists as associated molecules.

Exemplar Questions1. Isostructural species are those which have the same shape

and hybridisation. Among the given species identify theisostructural pairs.

(a) [NF3 and BF3] (b) 4 4[BF and NH ]- +

(c) [BCl3 and BrCl3] (d) [NH3 and 3NO- ]2. Polarity in a molecule and hence the dipole moment depends

primarily on electronegativity of the constituent atoms andshape of a molecule. Which of the following has the highestdipole moment?(a) CO2 (b) HI(c) H2O (d) SO2

3. The types of hybrid orbitals of nitrogen in 2 3NO , NO+ - and4NH+ respectively are expected to be

(a) sp, sp3 and sp2 (b) sp, sp2 and sp3

(c) sp2, sp and sp3 (d) sp2, sp3 and sp4. Hydrogen bonds are formed in many compounds e.g., H2O,

HF, NH3. The boiling point of such compounds depends toan extent on the strength of hydrogen bond and the numberof hydrogen bonds. The correct decreasing order of theboiling points above compounds is(a) HF > H2O > NH3 (b) H2O > HF > NH3(c) NH3 > HF > H2O (d) NH3 > H2O > HF

5. In 43PO - ion the formal charge on the oxygen atom of

P –O bond is(a) + 1 (b) – 1(c) – 0.75 (d) + 0.75

6. In 3NO- ion, the number of bond pairs and lone pairs ofelectrons on nitrogen atom are(a) 2, 2 (b) 3, 1(c) 1, 3 (d) 4, 0

7. Which of the following species has tetrahedral geometry?(a) 4BH- (b) 2NH-

(c) 23CO - (d) 3H O+

8. Number of p bonds and s bonds in the following structure isH

H

H

H

H

H

H

H

(a) 6, 19 (b) 4, 20(c) 5, 19 (d) 5, 20

9. Which molecule/ion out of the following does not containunpaired electrons?(a) 2N+ (b) O2

(c) 22O - (d) B2

10. In which of the following molecule/ion all the bonds are notequal?(a) XeF4 (b) 4BF-

(c) C2H4 (d) SiF411. In which of the following substances will hydrogen bond be

strongest?(a) HCl (b) H2O(c) HI (d) H2S

12. If the electronic configuration of an element is1s2 2s2 2p6 3s2 3p6 3d2 4s2, the four electrons involved inchemical bond formation will be(a) 3p6 (b) 3p6, 4s2

(c) 3p6, 3d2 (d) 3d2, 4s2

13. Which of the following angle corresponds to sp2

hybridisation?(a) 90° (b) 120°(c) 180° (d) 109°

Direction (Q. no. 14-17) The electronic configurations of theelements A, B and C are given below. Answer the questions from14 to 17 on the basis of these configurations.

A 1s2 2s2 2p6

B 1s2 2s2 2p6 3s2 3p3

C 1s2 2s2 2p6 3s2 3p5

14. Stable form of A may be represented by the formula(a) A (b) A2(c) A3 (d) A4

15. Stable form of C may be represented by the formula(a) C (b) C2(c) C3 (d) C4

16. The molecular formula of the compound formed from B andC will be(a) BC (b) B2C(c) BC2 (d) BC3

17. The bond between B and C will be(a) ionic (b) covalent(c) hydrogen (d) coordinate

82 CHEMISTRY

18. Which of the following order of energies of molecular orbitalsof N2 is correct?(a) (p2py ) < (s2pz ) < (p*2px ) » (p*2py )(b) (p2py ) > (s2pz ) > (p*2px ) » (p*2py )(c) (p2py ) < (s2pz) < (p*2px) » (p*2py )(d) (p2py ) > (s2pz) < (p*2px) » (p*2py )

19. Which of the following statement is not correct from theview point of molecular orbital theory?(a) Be2 is not a stable molecule(b) He2 is not stable but 2He+ is expected to exist(c) Bond strength of N2 is maximum amongst the

homonuclear diatomic molecules belonging to thesecond period

(d) The order of energies of molecular orbitals in N2molecule iss2s < s*2s < s2pz < (p2px ; p2py) < (p*2px ; p*2py )< s*2pz

20. Which of the following options represents the correct bondorder?(a) 2 2 2O O O- +> > (b) 2 2 2O O O- +< <

(c) 2 2 2O O O- +> < (d) 2 2 2O O O- +< >21. The electronic configuration of the outer most shell of the

most electronegative element is(a) 2s22p5 (b) 3s23p5

(c) 4s24p5 (d) 5s25p5

22. Amongst the following elements whose electronicconfiguration are given below, the one having the highestionisation enthalpy is(a) [Ne]3s23p1 (b) [Ne]3s23p3

(c) [Ne]3s23p2 (d) [Ar]3d104s24p3

NEET/AIPMT (2013-2017) Questions23. Which of the following is a polar molecule ? [2013]

(a) SF4 (b) SiF4(c) XeF4 (d) BF3

24. The outer orbitals of C in ethene molecule can be consideredto be hybridized to give three equivalent sp2 orbitals. Thetotal number of sigma (s) and pi (p) bonds in ethene moleculeis [NEET Kar. 2013](a) 1 sigma (s) and 2 pi (p) bonds(b) 3 sigma (s) and 2 pi (p) bonds(c) 4 sigma (s) and 1 pi (p) bonds(d) 5 sigma (s) and 1 pi (p) bonds

25. In which of the following ionisation processes the bondenergy increases and the magnetic behaviour changes fromparamagnetic to diamagnetic? [NEET Kar. 2013](a) N2 ® N2

+ (b) O2 ® O2+

(c) C2 ® C2+ (d) NO ® NO+

26. The pair of species that has the same bond order in thefollowing is: [NEET Kar. 2013](a) O2, B2 (b) CO, NO+

(c) NO–, CN– (d) O2, N2

27. In which of the following pair both the species have sp3

hybridization? [NEET Kar. 2013](a) H2S, BF3 (b) SiF4, BeH2

(c) NF3, H2O (d) NF3, BF328. XeF2 is isostructural with [2013]

(a) ICl2– (b) SbCl3

(c) BaCl2 (d) TeF229. Which of the following is paramagnetic ? [2013]

(a) 2O- (b) CN–

(c) NO+ (d) CO30. Which one of the following molecules contains no p bond?

[2013](a) H2O (b) SO2

(c) NO2 (d) CO2

31. Which of the following is electron - deficient ? [2013](a) (SiH3)2 (b) (BH3)2

(c) PH3 (d) (CH3)232. Be2+ is isoelectronic with which of the following ions?

[2014](a) H+ (b) Li+(c) Na+ (d) Mg2+

33. Which of the following molecules has the maximum dipolemoment ? [2014](a) CO2 (b) CH4(c) NH3 (d) NF3

34. Which one of the following species has plane triangularshape ? [2014](a) N3

– (b) NO3–

(c) NO2– (d) CO2

35. The correct bond order in the following species is: [2015](a) 2 –

2 2 2O O O+ +< < (b) – 22 2 2O O O+ +< <

(c) – 22 2 2O O O+ +< < (d) 2 –

2 2 2O O O+ +< <36. Which of the following pairs of ions are isoelectronic and

isostructural ? [2015]

(a) – 2–3 3ClO , CO (b) 2– –

3 3SO , NO

(c) – 2–3 3ClO , SO (d) 2– 2–

3 3CO , SO37. Which of the following options represents the correct bond

order ? [2015](a) –

2 2 2O O O+< < (b) –2 2 2O O O+> <

(c) –2 2 2O O O+< > (d) –

2 2 2O O O+> >38. Maximum bond angle at nitrogen is present in which of the

following ? [2015](a) –

2NO (b) 2NO+

(c) –3NO (d) NO2

39. Which of the following species contains equal number of s-and p-bonds : [2015](a) XeO4 (b) (CN)2

(c) CH2(CN)2 (d) –3HCO

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40. Decreasing order of stability of O2, 2 2O ,O, ∗ and 22O , is :

(a) 22 2 2 2O O O O∗ , ,= = = [2015 RS]

(b) 22 2 2 2O O O O, , ∗= = =

(c) 22 2 22O O O O∗ , ,= = =

(d) 22 2 22O O O O, , ∗= = =

41. In which of the following pairs, both the species are notisostructural ? [2015 RS]

(a) 4 4SiCl ,PCl∗

(b) diamond, silicon carbide(c) NH3, PH3

(d) XeF4, XeO4

42. The hybridization involved in complex [Ni(CN)4]2– is(At. No. Ni = 28) [2015 RS](a) dsp2 (b) sp3

(c) d2sp2 (d) d2sp3

43. Consider the molecules CH4, NH3 and H2O. Which of thegiven statements is false? [2016](a) The H–C–H bond angle in CH4, the H–N–H bond angle

in NH3, and the H–O–H bond angle in H2O are allgreater than 90°

(b) The H–O–H bond angle in H2O is larger than the H–C–H bond angle in CH4.

(c) The H–O–H bond angle in H2O is smaller than the H–N–H bond angle in NH3.

(d) The H–C–H bond angle in CH4 is larger than the H–N–H bond angle in NH3.

44. Predict the correct order among the following : [2016](a) lone pair- lone pair > lone pair - bond pair > bond pair -

bond pair(b) lone pair - lone pair > bond pair - bond pair > lone pair

- bond pair(c) bond pair - bond pair > lone pair - bond pair > lone pair

- lone pair(d) lone pair - bond pair > bond pair - bond pair > lone pair

- lone pair45. Which of the following pairs of compounds is isoelectronic

and isostructural ? [2017](a) TeI2,XeF2 (b) 2 2IBr , XeF-

(c) IF3, XeF2 (d) BeCl2,XeF2

46. The species, having bond angles of 120° is :- [2017](a) CIF3 (b) NCl3(c) BCl3 (d) PH3

47. Which of the following pairs of species have the same bondorder ? [2017](a) O2, NO+ (b) CN–, CO(c) 2 2N ,O- (d) CO, NO

84 CHEMISTRY

EXERCISE - 11. (b) Dative bond is formed between two atoms when the

electrons of shared pair are contributed solely by oneof the two atoms and it is directed toward that atomwhich do not contribute the electrons. It is representedby an arrow on the bond which is directed towards theelectron recipient atom.

So, –N NCl+

- º is the only group among the givenoptions which does not contain a dative bond.

(a) NO

O

; (b) N N Cl+

-- º - ;

(c) –N C2. (b) The stability of the ionic bond depends upon the lattice

energy which is expected to be more between Mg andF due to +2 charge on Mg atom.

3. (d) In CCl4, each C – Cl bond is polar but due to the very

symmetrical tetrahedral arrangement they exactly canceleach other and hence CCl4 have zero dipole moment.

CCl

Cl

Cl

Cl

So, none on these i.e. option (d) is correct choice.

4. (b) For compounds containing cations of same charge,lattice energy increases as the size of the cationdecrease. Thus, NaF has highest lattice energy. Thesize of cations is in the order Na+ < K+ < Rb+ < Cs+

5. (b) In KCN, K+ and CN– have ionic bondand C º N has covalent bond.

6. (d) H – O H – Cl Cl – Ba – Cl|H

2 2H N– NH

H+¯

. .

(a) (b) (c) (d)7. (d) The strength of a bond depends upon the extent of

overlapping. s-s and s-p overlapping results in theformation of s bond but extent of overlapping alonginternuclear axis is more in case of s-s overlapping thanin s-p. p-p overlapping may result in s bond ifoverlapping takes place along internuclear axis or mayresult in p–bond if sideways overlapping takes place.In any case the extent of overlapping is lesser in p - pthan that of the other two, s - s and s-p. Hence thecorrect order is

s - s > s - p > p - p.8. (d) CuSO4.5H2O can be represented as

H O2

H O2

OH2

OH2

OH2

Cu

2+

SO42–

So, in CuSO4.5H2O one electrovalent bond, 5coordinate bonds and covalent bonds are present.

9. (c) Higher the difference in electronegativity between thetwo atoms, more will be electrovalent character of thebond. Among given choices, calcium and hydrogenhave maximum difference in their electronegativities.

10. (c) KF being highly ionic compound will dissolve mostreadily in water. (Like disolves like).

11. (d) When a metal for example Na combines with a nonmetal e.g., Cl2. Following reaction occurs

22Na Cl 2NaCl+ ¾¾®In this process Na loses one electron to form Na+ andCl accepts one electron to form Cl–

Na Na e+ -¾¾® +Cl e Cl- -+ ¾¾®

Therefore, in this process Cl gain electrons and henceits size increases.

12. (a)13. (b) Hybridisation of the central atom in compound is given

by

H = 1 [V M C A]2

+ - +

where V = no. of valency electrons in central metalatom,M = no. of monovalent atoms surrounding the centralatom,C = charge on cation and A = charge on anion

· For ,NO2- H =

1 [5 0 0 1] 32

+ - + =

sp2 hybridisation

· For SF4, H = 1 [6 4 0 0] 52

+ - + =

sp3d hybridisation

· For PF6–, H =

1[5 6 0 1] 6

2+ - + =

sp3d2 hybridisation.So, option (a) is correct choice.

14. (c) Bond order between P – O

structuresresonatingof.nototaldirectionpossibleallinbondsof.no

= 25.145

==

–O P O–

O–

O

O P O–

O–

O–

–O P O–

O–

O

–O P O

O–

O–

Formal charge on oxygen = 3

0.754

- = -

Hints & Solutions

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15. (a) A compound having element with highestelectronegativity will form strongest hydrogen bond.

16. (a) Bond angle increases with increase in s-character ofhybridised orbital. The table given below shows thehybridised orbitals, their % s-chatracter and bondangles.Hybridised % s-character Bondorbitals anglesp3 25 109.5°sp2 33 120°sp 50 180°

17. (a) Boron in BCl3 has 6 electrons in outermost shell. HenceBCl3 is a electron deficient compound.

18. (d) The structure of P4O10 is

P||O

O||P

OO

O

OO O

O = P P = O

The number of s bonds in it = 16[Note: Single bonds are s-bonds. Double bond consistsof 1s and 1p bond].

19. (b) There are three resonance structures of 23CO - ion.

O

C–O O–

O–

C–O O

O–

CO O–

(I) (II) (III)20. (c) Higher the electronegativity of the other atom, greater

is the strength of hydrogen bond. Strongest hydrogenbond is between H and F.

F – H ------- F.21. (a) Greater the difference in electronegativity between the

two atoms, larger will be polarity and hence dipolemoment. Thus (a) has maximum dipole moment.

H – C = OH

(C–O bond is more polar)

C = CH3C

H3C

HH

(a) (b)(very less polar)

CC

H

H3C H

CH3

(c)

CC

Cl

H3C

CH3

(d)Cl

Symmetrical molecules (µ = 0)

22. (b) In CN– ion formal negative charge is on nitrogen atomdue to lone pair of electrons.

23. (c) NH3 undergoes H-bonding and hence has the highestb.p. Among the remaining hydrides i.e. PH3, AsH3 andSbH3 as we move from PH3 to BiH3, the molecular massincreases. As a result the van der waal’s forces ofattraction increases and the boiling point increasesregularly from PH3 to BiH3.

24. (a) Resonating structures can be written only for suchmolecules in which multiple bonds are present, eg, O3

25. (c)26. (d) HF form linear polymeric structure due to hydrogen

bonding.27. (a) Square Plannar Structure : XeF4 and [PtCl4]

2–

28. (c) In one – C º N, No. of p bonds = 2So in [Ag (CN2)]– , No. of p bonds= 2 ×2 = 4

29. (c) HF2 is the only compound among the given optionswhich does not contain any coordinate bond becauseit has hydrogen bonding.

[H – F•••H]–

30. (b) The compounds having O or F or N along withhydrogen show hydrogen bonding and have highboiling points. So HI does not form hydrogen bondsand its boiling point is not affected.

31. (c) Ice has many hydrogen bonds which give rise to cagelike structure of water molecules. This, structure possesslarger volume and thus makes the density of ice low.

32. (c) Intramolecular hydrogen bonding is formed (withinmolecule) when the two groups capable of formingH-bonding are very near to each other.

OH

N

O

O

o-Nitrophenol33. (c) ortho-Nitrophenol has intramolecular H-bonding

N

||

O

O OH.

.

..

.

and para-nitrophenol has

intermolecular H-bonding.

|| |O– HO– H O– H

2NO 2NO 2NO

,

hence o-Nitrophenol is more volatile thanp-Nitrophenol intermolecular H-bonds are stronger thenintramolecular H-bonds.

34. (c) The bond length between carbon & carbon in sp2

hybridisation (double bond) is 1.33 Å.35. (b) Both NO2 and O3 have angular shape and hence will

have net dipole moment.

86 CHEMISTRY

36. (c) As difference of electronegativity increases % ioniccharacter increases and covalent character decreasesi.e., electronegativity difference decreases covalentcharacter increases.Further greater the charge on the cation and smallerthe size more will be its polarising power. Hencecovalent character increases.

37. (b) SF4 has permanent dipole moment.SF4 has sp3d hybridization and see saw shape (irregulargeometry).

F

S

F

F

F

m ¹ 0Whereas XeF4 shows squre planar geometry SiF4 hastetrahedral shape and BF3 has triangular planar shape.All these are symmetric molecules. Hence m = 0.

38. (d) Dipole moment is a vector quantity, hence dipolemoment of a symmetrical molecule is zero.

C = C

Cl

Hcis-1, 2-dichloroethene

(a)

Cl

H

C = C

H

Cltrans-1, 2-dichloroethene

(b)

Cl

H

Symetrical hence dipole moment = 0

C = C

Cl

trans-2, 3-dichloro-2-pentene(c)

C H2 5

CH3 Cl

\ (a) and (c), both show dipole moment.39. (b) In NH3 sp3 hybridisation is found.40. (a) According to Fajan’s rule, as the charge on the cation

increases, and size decreases, its tendency to polarisethe anion increases. This brings more and morecovalent nature to electrovalent compounds. HenceAlCl3 shows maximum covalent character.

41. (d)42. (a) In H2, both atoms are identical, so the molecule is non

polar.43. (d) According to Fajan's rule :

1Covalent character size of cation

µ

size of anionµAmong the given species order of size of cationsN3+ < O2+ < Pb2+ < Ba2+

order of size of anions O2– > Cl–.Hence the order of covalent character is

3 2 2 2NCl Cl O PbCl BaCl> > >BaCl2 is most ionic in nature.

44. (b) A substance is said to be soluble in water if it is capableof forming hydrogen bonds with water molecules as inmethanol and ethanol.

O H O H O H

R H RHydrogen bonding between alcohol and water.However, this hydrogen bonding is restricted whenalkyl group of alcohol has four or more carbon atoms.

45. (d) Ionic compounds generally have high melting andboiling points because of the strong electrostatic forceof attraction between oppositely charged ions.Consequently, a considerable amount of energy isrequired to overcome strong attractive inter-ionic forcesand to break down the crystal lattice.

46. (a) The dipole moment of two dipoles inclined at an angle

q is given by the equation 2 2X Y 2XY cosm = + + qcos 90° = 0. Since the angle increases from 90° – 180°,the value of cos q becomes more and more – ve andhence resultant decreases. Thus, dipole moment ismaximum when q = 90°.

47. (c) Benzene and the compounds having benzene ring showresonance. Hence, ethylamine does not showresonance.

48. (d) In case of anions having same charge as the size ofanion increases, polarisibility of anion also increases.

49. (c) Percent ionic character is given by following equation.% of ionic character = 16(XA – XB) + 3.5(XA – XB)2

From the above relation, it is clear that as soon as(XA – XB) increases, % ionic character will also increase.Therefore, curve C show a correct path.

50. (a) Covalent bonds are formed by sharing of electrons. Nforms 3 covalent bonds with 3 hydrogens.

H••

••H N• •••H

This leaves one lone pair of electron free which can bedonated to H+ to form ammonium ion NH4

+. This iscalled coordinate bond.

H••

••H N• •••H

+ H+ ¾¾®

|

|

H

H – N H

H

+é ùê ú

®ê úê úë û

Covalent bonds are shown by straight line & coordinateby an arrow.This NH4

+ cation can gain electrons from Cl– to form aelectrovalent bond.

|

|

H

H – N H

H

+é ùê ú

®ê úê úë û

–• •

• ••Cl••

é ù¾¾®ê ú

ê úë ûNH4Cl

51. (c) +3CH has 2sp hybridization because S orbital has one

electron and p orbitals of central atom C have only twoelectrons. Therefore s and 2p orbitals participate inhybridization.

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52. (b) Equilateral or triangular planar shape involves sp2

hybridization.53. (c) The bond length decreases in the order sp3 > sp2 > sp.

Because of the triple bond, the carbon-carbon bonddistance in ethyne is shortest.

54. (b) BH3 has sp2 hybridization and hence does not havetetrahedral structure while all others have tetrahedralstructures.

55. (a) We know that size of Na+ ion is greater than Li+ ion.Since bond length is directly proportional to atomicradius. Therefore, the bond length in LiF is less thanthat of NaF.Option (a) is correct.

56. (a) For linear arrangement of atoms the hybridisationshould be sp (linear shape, 180° angle). Only H2S hassp3-hybridization and two lone pair of electrons onsulphur atom hence has angular shape while C2H2,BeH2 and CO2 all involve sp-hybridization and hencehave linear arrangement of atoms.

57. (b) BF3 involves sp2-hybridization.58. (a) Only those d orbitals whose lobes are directed along

X, Y and Z directions hybridise with s and p orbitals.In other three d orbitals namely dxy, dyz and dxz, thelobes are at an angle of 45° from both axis, hence theextent of their overlap with s and p orbitals is muchlesser than 22 yx -

d and 2zd orbitals.

59. (c) Acetylene has sp hybridization and linear structure CH º CHHence, option (c) is correct.

60. (c) X is deficient of e–. Y has 1e– only. Therefore, Y ishydrogen. It forms covalent bond with X (halogen) bysharing 2 valence electrons

X Y61. (b) BF3 does not follow octate rule because central atom,

boron lacks an electron pair. Thus, it also acts as Lewisacid.

×F

××× ××

× B ×

×

×F××× ××

×F

××

× × ×

62. (c)

Br

F

F

F

:

:

In BrF3, both bond pairs as well as lone pairs ofelectrons are present. Due to the presence of lone pairsof electrons (lp) in the valence shell, the bond angle iscontracted and the molecule takes the T-shape. This isdue to greater repulsion between two lone pairs orbetween a lone pair and a bond pair than between thetwo bond pairs.

63. (b)List I List II

Compound Structure(A) ClF3 T-shaped(B) PCl5 Trigonal bipyramidal(C) IF5 Square pyramidal(D) CCl4 Tetrahedral(E) XeF4 Square planar

64. (c) Benzene has the following resonance structures–

¾®¬

Hence, its bond order isno of possible resonating structures

2= 1.5.

65. (b) The bond angle decreases on moving down the groupdue to decrease in bond pair-bond pair repulsion.NH3 PH3 AsH3 SbH3 BiH3107º 94º 92º 91º 90ºThis can also be explained by the fact that as the size ofcentral atom increases and its electronegativitydecreases. Thus distance between bond pairs ofelectron increases and bp-bp repulsion decreases. Asa result bond angle decrease from NH3 to BiH3.

66. (d) Hybridisation present in a molecule can be findout bythe following formula.

1H (V M C A)2

= + - +

Where V = no. of electrons in valence shell of centralatomM = no. of singly charged atomsC = charge on cationA = charge on anion

So, H(in ClF3) = 1

[7 3 0 0]2

+ - + = 5

Þ sp3d Hybridisation67. (d) PF3 has pyramidal shape. Phosphorus exist in sp3

hybridization state hence it exist in tetrahedral shape.But due to presence of lone pair its shape is pyramidal.

68. (a)

69. (d) N : 1s2 2s2 2p3

To form -3NO , nitrogen uses one p-electron for

p-bond formation two p-electrons for s-bond formation.2s electrons are used for coordinate bond formation.Thus there is no lone pair on nitrogen but four bondpairs

N

O

O O

88 CHEMISTRY

70. (a) Shape (structure ) of a species can be predicted on thebasis of hybridisation of its central atom, which in turncan be known by determining the number of hybrid

bonds (H). Thus 1H (V X C A)2

< ∗ , ∗

For 3sp;4)0044(21H:SiF4 =+-+=

(Tetrahedral structure)

For 41

: (6 4 0 0) 5;2

3SF H sp d= + - + =

(See saw structure due to presence of one lp of electronson central atom)Hence the two compounds have different structures

71. (c) Species N2 O2–N2

–O2

Bond order 3 2 2.5 1.5

The O-O bond is –O2 decreases

72. (a)X4

X3 X2

X1

X6

X5

M

Thus here bond angles between°=-- 180XMX 24

°=-- 180XMX 31

°=-- 180XMX 6573. (a) BF3 is sp2 hybridised. So, it is trigonal planer. NH3,

PCl3 has sp3 hybridisation hence has trigonal pyramidalshape. IF3, has sp3d hydridisation and has T-shape.

74. (a) Both XeF2 and CO2 have a linear structure.F — Xe — F O = C = O

75. (c)76. (b) Electronegativity difference is 4.0 - 1.20 = 2.8 percentage

ionic character is 72.24% when the electronegativitydifference is 1.7, the % ionic character is approx 51%.

77. (a) Ionic character follows the order HF > HCl > HBr > HI

78. (a)

-24 ])CN(Ni[ (dsp2 square planar ),

-4BF (sp3 tetrahedral),

SF4 (sp3d see saw shaped)

F F

FF

Xe

3 24XeF (sp d square planar),

79. (c) From amongst given species PH3, NH3 and SbH3 areall sp3 hybridised. Their central atom has both bondpair as well as lone pair of electrons. The lone pairoccupy the fourth orbital. CH3

+ has only three pairs ofelectrons so it is sp2 hybridised.

80. (b) Geometry of compounds can be determined byhybridisation, which in turn can be known by thefollowing relation.

Hybridisation (H) = 12

[No. of electrons in valence shell

of central atom + No. of monovalent atoms –charge oncation + charge on anion]

(i) SF4 : ( )1H 6 4 0 0 5

2= + - + =

sp3d (trigonal bipyramidal but actual geometry is seesaw due to presence of one lone pair of electrons.

(ii) XeF4 : ( )1H 8 4 0 0 62

= + - + =

sp3d2 (Octahedral but actually square planar due topresence of two lone pairs of electrons)

(iii) NF3 : ( )1H 5 3 0 0 4

2= + - + =

sp3 (tetrahedral but distorted due to one lone pair ofelectrons)

(iv) SiF4 : ( )1H 4 4 0 0 42

= + - + =

sp3 (tetrahedral)81. (b) Molecule or ion having sp hybridisation and no lone

pair of electrons is linear.

Hybridisation 21

= [No. of e– in valence shell of central

atom + No. monovalent atoms + Charge on anion –Charge on cation](a) SO2

H 21

= (6 + 0 + 0 – 0)

= 3 sp2 (trigonal planar shape)(b) CO2

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H 21

= (4 + 0 + 0 – 0) = 2 sp (linear shape)

(c) ClO2–

H 21

= (7 + 0 + 1 – 0) = 4 sp3 (tetrahedral)

(d) NO2–

H 21

= (5 + 0 + 1 – 0) = 3 sp2 (trigonal planar)

\ CO2 has linear shape.82. (a) According to VSEPR theory, trigonal bipyramidal

geometry results from sp3d or dsp3 hybridisation.dsp2 hybridisation results in square planar geometry,while d2sp3 leads to octahedral shape.

83. (b) SeF4 has distorted tetrahedral geometry while, CH4 hastetrahedral geometry.

84. (b) Molecule HybridizationSO3 sp 2

C2H2 spC2H4 sp 2

CH4 sp 3

CO2 sp

85. (d)

86. (d)X

X

XX

M three angle below M and three above M hence

= 687. (c) s-character µ bond angle

For 25% s character (as in sp3 hybrid orbital), bondangle is 109.5°, for 33.3% s character (as in sp2 hybridorbital), bond angle is 120° and for 50% s character (asin sp hybrid orbital), bond angle is 180°.Similarly, when the bond angle decreases below 109.5°,the s-character will decrease accordingly.Decreasing in angle = 120° – 109.5° = 10.5°Decrease in s-character = 33.3 – 25 = 8.3Actual decrease in bond angle = 109.5° – 105° = 4.5°Expected decrease in s-character

8.3 4.5 3.56%10.5

< ´ <

Thus, the s-character should decrease by about 3.56%i.e., s-character = 25 – 3.56 = 21.44%

88. (c) Because of the triple bond the carbon - carbon bonddistance in ethyne is shortest.

89. (c) Hybridisation of N = ½ [5 + 3 + 0 – 0] = 4 hence sp3

90. (d)2 2 3

32CH C CH CHsp sp sp sp

= = - 1, 2 -butadiene.

91. (c)92. (c) H2O, NH3 and CH4 all are sp3 hybridized due to lp – lp

repulsions, bond angle in H2O (104.5°) is lower than inNH3 (107°) which have one lp and CH4 (109° 28’) whichhave no lp BeF2 on the other hand, has sp hybridizationand hence has a bond angle of 180°.

93. (b) 24SO - ion is tetrahedral since hybridization of S is sp3.

94. (b) O105° HH

. . N

107° HH

. .

HP

HH. .

107°H Less then

AsHH

. .

H

95. (b) NH3 has pyramidal structure, as nitrogen is sp3

hybridised. This is due to the presence of lone pair ofelectrons on N atom.

96. (c) O = C = C = C = O is sp hybridised97. (a) A -s bond is stronger than a p -bond hence option

(a) is not correct. Sigma (s) bonds are formed by headon overlap of unhybridised s–s, p–p or s–p orbitalsand hybridised orbitals (sp, sp2, sp3, sp3d and sp3d2)hence s bonds are strong bonds. Whereas Pi (p)-bondsare formed by sideways overlap of unhybridisedp- and d-orbitals hence p bonds are weak bonds.

98. (a) Linear combination of two hybridized orbitals leads tothe formation of sigma bond.

99. (a) For p-overlap the lobes of the atomic orbitals areperpendicular to the line joining the nuclei.

100. (d) Total no. of electrons in -22O = 16 + 2 = 18

Distribution of electrons in molecular orbital2 * 2 2 * 2 21 , 1 , 2 , 2 , 2 ,s s s s s zs s s s p

2 22 2 ,x yP Pp = p * 2 * 22 2x yp pp = pAntibonding electrons = 8 (4 pairs)

101. (b) In CH3 CH2OH underlined C is forming 4 σ bonds,hence sp3 hybridisation. In others it is sp2 hybridised(due to 3 σ bonds).

102. (c) As sigma bond is stronger than the p (pi) bond, so itmust be having higher bond energy than p (pi) bond.

103. (c) 2N + = 7 + 7 – 1 = 13 electronsConfiguration is

2 2 2 2 21 , *1 , 2 , *2 , 2 xs s s s ps s s s p = 2 12 , 2y zp pp sBond order =

1 No. of in bonding No. of in antibonding2 molecular orbital molecular orbital

e s e s- -æ ö-ç ÷è ø

= 1 1(9 4) 5 2.52 2

- = ´ =

104. (b) Only d of S and p of O can be involved in π bonding.105. (c) The first bond between any two C atoms having

multiple bonds is s and rest are p bonds. Draw detailedstructure of compound and find the total bonds in it

C

H

CC

C

C

C

H

ps s

s

sss

H s

H s

s

p

p

s

CC

HHss

s C

Hs

H

Hs

Hs

sp

ss

\ 19 s bonds and 4p bonds.

90 CHEMISTRY

106. (c) Hs CH Hs

s

s

C

H

CC

C

C

Cp

s s

s

sss

H s

H s

s

p

p

s

Hs

Hs

15s + 3p

107. (b) 2O 16 1 17- = + = Molecular orbital configuration ofO2

– (superoxide ion).2 2 2 2 2 2 12 *2 2 2 2 , *2 * 2z x y x yKK s s p P p P Ps s s p = p p = p

Bond order 8 5 3 1.52 2-

= = =

108. (c)109. (d) Bond structure of H2P2O7 is following

H – O – P – O – P – O – Hs s sss

s

Os p

Os p

O – Hs

s

O – Hs

s

12s, 2dp – pp bonds.110. (b) More will be the electronegativity of X lesser will be

the bond length of X-O bond.111. (d) M.O. configuration of NO+ is :

(s1s)2 (s * 1s)2 (s2s)2 (s * 2s)2 (s2pz)2 (p2px)

2 (p2py)2

and M.O. configuration of NO is :(s1s)2(s * 1s)2 (s2s)2 (s * 2s)2 (s2pz)

2 (p2px)2

(p2py)2 (p * 2px)

1.112. (a)

43212 NCHCCH º-=

3 σ bonds (sp2 hybridisation); 2 σ bonds(sp - hybridisation)

C1 = 3 σ bonds, C2 = 3 σ bonds,C3 = 2 σ bonds

113. (a) Molecular orbital electronic configuration of thesespecies are :

2 2 2 2 22O (17 ) 1 , *1 , 2 , *2 , 2 ,ze s s s s p- - = s s s s s

2 2 2 12 2 , *2 2*x y x yp p p pp = p p = p

2 2 2 2 22 (16 ) 1 , *1 , 2 , *2 , 2 ,zO e s s s s p- = s s s s s

2 2 1 12 2 , *2 *2p = p p = px y x yp p p p2 2 2 2 2 22 (18 ) 1 , *1 , 2 , *2 , 2 ,zO e s s s s p- - = s s s s s

2 2 2 22 2 , *2 *2p = p p = px y x yp p p pHence number of antibonding electrons are 7, 6 and 8respectively.

114. (a)

115. (b)F HF HF H

N N. .

N N– O––– octet of each atom is complete.

116. (c) 12H :( 1 )s+ s

1 1B.O. (1 0)2 2

\ = - =

2 12H :( 1 )( *1 )s s- s s

1 1B.O. (2 1)2 2

\ = - =

Even though the bond order of 2H+ and 2H- are equalbut 2H+ is more stable than 2H- as in the latter, anelectron is present in the antibonding (s * 1s) orbital ofhigher energy.

117. (b) In H - bonding atoms involved are F, O and N.

118. (c) –– OCO d+d=d = linear molecule hence 0m = .

119. (c) Both 3BrO- and XeO3 have sp3 hybridization butdue to presence of one lp of electrons they havetrigonal pyramidal geometry.

120. (d) Calculating the bond order of various species.2 2 2

2O : KK 2 , 2 , 2 ,zs s p- *s s s

2 2 2 12 2 , 2 2x y x yp p p p* *p = p p = p

1B.O. (N N )2 b a= -

= 8 52- or 1.5

NO = 2 2 2 22 , 2 ,62 , 2z xK K s s p p*s s p 2 12 , 2y xp p*= p p

B.O. = N N 8 32 2

b a- -= or 2.5

= 2 2 2 22C : KK 2 , 2 , 2 xs s p- *s s p 2 22 , 2y zp p= p s

B.O. = N N 8 22 2

b a- -= or 3

2 2 12He 1 1s s+ *= s s

B. O. = N N 2 1

2 2b a- -= or 0.5

From these values we conclude that the correct orderof increasing bond order is

2 22 2 2He O NO C+ - -< < <

121. (d) Due to intermolecular hydrogen bonding in methanol,it exist as assosiated molecule.

122. (c) For 2–5SbCl , H =

5+5+22

= 6 means 3 2sp d hybridization

–3I , 4SF , and 5PCl ; all have 3sp d hybridization.

123. (b) In 3NO - , nitrogen have sp2 hybridisation, thus planar

in shape. In 3H O+ , oxygen is in sp3 hybridisation,thus tetrahedral geometry is expected but due topresence of one lp of electrons on central oxygen atomit is pyramidal in shape.

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124. (c) +4NH : sp3 hybridisation

CH4: sp3 hybridisationSF4: sp3d hybridisation

–4BF : sp3 hybridisation

125. (b) Bond order of 2Be = 0, hence 2Be cannot exist.

EXERCISE - 2

1. (c) H – C H – CO O

O O

s

s

or Resonance

hybrid H – CO

O due to resonance C – O bond

length is the same.

2. (b) BiCl3: BiClCl

Cl ; sp2 - Hybridisation (Trigonal

geometry); Bond angle = 120º

P

As

Bi

Cl Cl Cl(Pyramidalgeometry)

In P lC 3

In AsCl3

In BiCl3

(Pyramidalgeometry)

Bond angle =below 109 28’and decreases fromPCl to BiCl

o

3 3

sp(Pyramidalgeometry)

Cl Cl Cl

Cl Cl Cl

3

sp3

sp3

In these, order of bond angle :BCl3 > PCl3 > AsCl3 > BiCl3

3. (d)No. of electrons No. of mono- charge charge

1H in valenceshell valent atoms on on2

of centralatom anion cation

é ùê ú= + + -ê úê úë û

For SF4 :1 (6 4 0 0) 52

= + + - =

S is sp3d hybridised in SF4. Thus SF4 has 5 hybridorbitals out of which only four are used to form bondswith four F atoms leaving one lone pair of electrons onsulphur.

For CF4 : 4]0044[21H =-++= \ sp3 hybridisation

Since all the four orbitals of carbon are involved inbond formation, no lone pair is present onc atom

For XeF4 : 6)0048(21H =-++= , \ sp3d2

hybridization of the six hybrid orbitals, out of whichonly four form bonds with F atoms, leaving behind twolone pairs of electrons on Xe.

4. (d) The shape of 3BF is trigonal planar FF

-d+d-d

–BF

-d and

m = 0 hence it is non polar. The shape of 3NF is

pyramidal +d

F-dN

:

F -dF -d

and m ¹ 0 hence it is polar..

5. (a) Number of electrons in each species areCN 6 7 1 14- = + + =CO 6 8 14= + =

NO 7 8 1 14+ = + - =Each of the species has 14 electrons which aredistributed in MOs as belowFor CO and CN–

2 2 2 2 2 2 21 , *1 , 2 , *2 ,{ 2 2 , 2x y zs s s s p p ps s s s p = p s

Bond order = 32

410=

-

For NO+

2 2 2 2 2 2 21 , *1 , 2 , *2 , 2 , 2 2z x ys s s s p p ps s s s s p = p

Bond order10 4 3

2-

= =

6. (c) Bond energy a Bond orderBond order can be determined by MO configuration.NO : No. of electrons = 7 + 8 = 15

s1s2, s*1s2, s2s2, s*2s2, 2 2 2 12 , 2 2 , *2z x y xP p p ps p = p p

\ Bond order = 10 5 2.5

2-

=

NO+ : No. of electrons = 15 – 1 = 14

Delete * 12p xp from NO configuration

\ Bond order = 10 4

32-

=

NO– : No. of electrons = 15 + 1 = 162 2 2 2 2 2 21 , *1 , 2 , *2 , 2 , 2 2 ,z x ys s s s p p ps s s s s p = p

1*2 xpp 1*2 yp= p

\ Bond order = 10 6

22-

=

\ Bond order; hence bond strengthNO+ > NO > NO–

7. (b) For T-shape geometry the molecule must have 3 bondedpair and 2 lone pair of electrons.

8. (a) Structure of CO2 is linear O = C = O while that of

H2O is H H O

i.e ., bent structure so in CO2 resultantdipole moment is zero while that of H2O has significantvalue.

9. (a) O – H

3 2CH C CH- = has 9 ,1 and 2lone pairs.s p

92 CHEMISTRY

10. (d) H3O2– species H bonding in H2O

(i) O H.....O H|H

-- -é ùê úê úë û

(ii) H H| |

O H.....O H- -H5O2

+ (H-bonding) H3O+ (no H-bonding)(iii)

HO

HH.....H – O

H

+(iv)

HO

HH

+

11. (d) *

OH|

H C C O H, < , ,P

The star marked carbon has a valency of 5 and hencethis formula is not correct.

12. (b) In amine the nitrogen is sp3 hybridised and in amide thenitrogen is sp2 hybridised.

13. (c) Since water is polar in nature and like dissolves like,the coating must be nonpolar to polar manner.

14. (b) The removal of an electron from a diatomic moleculemay increase the bond order as in the conversionO2 (2) )5.2(O 2

+¾®¾ or decrease the bond order as inthe conversion, N2(3.0) ),5.2(N2

+¾®¾ As a result,the bond energy may increase or decrease. Thus,statement (b) is incorrect.

15. (d) Nitrogen can not be pentavalent as it has no d-orbitals.16. (c) The b.p. of p-nitrophenol is higher than that of

o-nitrophenol because in p-nitrophenol there isintermolecular H-bonding but in o-nitrophenol it isintramolecular H-bonding.

17. (a) The group moments of –NO2, –OH, –Cl and –CH3respectively are 3.98, 1.6, 1.55 and 0.4D.

18. (b) Resultant of two C–X dipoles in 1, 4 positions is zero.The resultant of other two C–X dipoles in 3, 5 positions

2 21.5 1.5 2 1.5 1.5cos120º 1.5= + + ´ ´ = D19. (d) As there is no lone pair on boron in BCl3 therefore no

repulsion takes place. But there is a lone pair onnitrogen in NCl3. Therefore repulsion takes place. ThusBCl3 is planar molecule but NCl3 is a pyramidalmolecule.

20. (b) Group moment of –NO2 is –3.98 D (acting towards thebenzene ring) and that of –NH2 is + 1.5D (acting awayfrom the group). Hence moments of both the groups inp-positions will act in the same direction.

21. (d) In alkynes the hybridisation is sp i.e, each carbon atomundergoes sp hybridisation of one s and one p orbitalto form two sp-hybrid orbitals. Out of which oneoverlaps to form C – C s bond and other overlaps withhalf filled orbital of H forming s-bond. The other two2p-orbitals remain unhybridised and form two p-bondsvia sidewise overlaping.Hence two p bond and one sigma bond between C —C lead to cylindrical shape.

22. (d) N2+ is paramagnetic

s1s2, s*1s2,s*2s2,p2p2x = p2p2

y ,s2p1z

23. (c) The electronic configuration of As is

As = 3

10 1 1 1 1 1

hybridisation

[ ]3 4 4 4 4 4x y z

sp d

Ar d s p p p d

¯1444442444443

Thus the hybridisation involved in the AsF5 moleculeis trigonal bipyramidal. So, the hybrid orbitals usedare s, px, py, pz, dz

2.24. (c) As dipole moment = electric charge × bond length

D. M. of AB molecule

D53.131082.2108.4 810 =´´´= --

D.M. of CD molecule10 84.8 10 2.67 10 12.81D, ,< ´ ´ ´ <

Now % ionic character

= Actual dipole moment of the bond 100

Dipole moment of pure ionic compound´

then % ionic character in

AB %94.7610053.1341.10

=´=

% ionic character in

CD 10.27 100 80.17%12.81

< ´ <

25. (c) Structure I has two resonance forms, which make thehydrogen bonding more stable. The II molecule hasno such resonance stabilization.

26. (c) Number of hybrid orbitals of P in -34PO = ½ [5 + 0 + 3]

= 4 (sp3)No. of hybrid orbitals of N in NO–

3 = ½ [5 + 0 + 1] = 3(sp2)No. of hybrid orbitals of I+ (5s2 5p2

x 5p1y 5p1

z) : (sp3)2

(sp3)2 (sp3)1 (sp3)1

I-atom in excited state : 5s2 5p4 2 2 21 05 5

-z x yd d

I-atom is sp3d2 hybridised state in ICl–4 : (sp3d2) (sp3d2)

(sp3d2)1 (sp3d2)1 (sp3d2)1 (sp3d2)0 – Two lps occupythe axial positions of octahedron and the shape ofICl–4 becomes square planar.

27. (c) Dissociation energy of any molecule depends uponits bond order. Bond order in N2 molecule is 3 whilebond order in +

2N is 2.5. Further we know that morethe Bond order, more is the stability and more is thebond dissociation energy.

28. (c) In X — H - - - Y, X and Y both are electronegativeelements (i.e attracts the electron pair) then electrondensity on X and Y will increase and on H will decrease.

29. (a) In NH3 the atomic dipole (orbital dipole due to lonepair) and bond dipole are in the same direction whereasin NF3 these are in opposite direction so in the formercase they are added up whereas in the latter case netresult is reduction of dipole moment. It has been shownin the following figure :

. . N

. . N

H FH F

HF

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30. (d) The bond angles HXH vary in the order : NH3 > PH3AsH3 > SbH3. In the formation of XH+

4, the bond angleHXH increases to 109º 28'. Hence, maximum increasein the bond angle will be for SbH3.

31. (b) On changing N2 to N+2, B.O. decreases from 3 to 2.5

whereas on changing O2 to O+2, B.O. increases from 2

to 2.5. In former case, the bond dissociation energydecreases and in the latter case, it increases.

32. (d) In a linear symmetrical molecule like BeF2, the bondangle between three atoms is 180°, hence the polaritydue to one bond is cancelled by the equal polarity dueto other bond, while it is not so in angular molecules,like H2O.

33. (b) Now since bond order of NO+ (3) is higher than that ofNO (2.5). Thus bond length of NO+ will be shorter.

34. (b) The p-orbital has equal sized lobes whereas an sphybrid orbital has more probability density on one side.Hence, sp orbital is more directional in character.

35. (c) ns2 electrons in the havier elements, like lead, remainpaired and do not take part in bonding. Thus, Pb2+ ismore stable than Pb4+.

36. (b) 2 * 2 2 * 2 22O : 1 , 1 , 2 , 2 , 2 ,s s s s s zs s s s p

2 * 1

2 * 1

2 2,

2 2

ì ìp pï ïí í

p pï ïî î

x x

y y

p p

p p

Bond order 22

610=

-=

(two unpaired electrons in antibonding molecular orbital)2 * 1

2 * 2 2 * 2 22 2 * 0

2 , 2O : 1 , 1 , 2 , 2 , 2 ,

2 , 2+

ì ìp pï ïs s s s s í íp pï ïî î

x xz

y y

p ps s s s p

p p

Bond order 5.22

510=

-=

(One unpaired electron in antibonding molecular orbital)

Hence O2 as well as +2O both are paramagnetic, and

bond order of +2O is greater than that of O2.

37. (a) Its conjugate base (i.e. NH3) is most stable.38. (c) The b.o in Cl – O– is 1

The b.o in O = Cl – O– is 1.5

The b.o. in O = Cl = O

O–

is 5 1.663

=

The b.o. in O = Cl – OO

O

– is 7 1.754

=

The bond length increases as b.o. decreases.39. (b) From the structure of three species we can determine

the number of lone pair electron(s) on central atom (i.e.N atom) and thus the bond angle.

O....| ||N

O....(NO )2

O....| ||N

O.. ..

. .

NO2-

NO2+

O....| ||O.. ..

+.N.. .. ..

We know that higher the number of lone pair electron(s)on central atom, greater is the lp – lp repulsion. Thussmaller is bond angle.The correct order of bond angle is

2 2 2NO NO NO+

- < < i.e., option (b) is correct.40. (c) On determining hybridisation from H = 1/2 (V + M – C + A).

The hybridisation of BF3, NO2–, NH2

– and H2O aresp2, sp2, sp3 and sp3 respectively.

41. (a) Molecular orbital configuration of2– 2 2 2 22N 1 *1 2 *2s s s s= s s s s –

2 12

2 1

2 * 22

2 * 2x x

zy y

p pp

p p

ì ìp pï ïsí íp pï ïî î

Bond order = 10 – 6 2

2=

– 2 2 2 22N 1 *1 2 *2s s s s= s s s s

2

2

2

2x

y

p

p

ìpïí

pïî

1

20

*22

*2x

zy

pp

p

ìpïs ípïî

Bond order = 10 5 2.5

2-

=

2 2 2 22N 1 *1 2 *2s s s s= s s s s

22

2

2, 2

2x

zy

pp

p

ìpï sípïî

Bond order = 10 – 4

32

=

\ The correct order is = 2– –2 2 2N N N< <

42. (a) Hybridisation = 12

[No. of valence electrons of central

atom + No. of monovalent atoms attached to it +Negative charge if any – Positive charge if any]

NO2– 21H [5 0 1 0] 3

2sp= + + - = =

NO3– 21H [5 0 1 0] 3

2sp= + + - = =

NH2– 31H [5 2 1 0] 4

2sp= + + + = =

NH4+, 31H [5 4 0 1] 4

2sp= + + - = =

SCN– = spi.e., NO2

– and NO3– have same hybridisation.

43. (a) 2 2 2 2 22 (16) 1 , *1 , 2 , *2 , 2 ,zO s s s s p= s s s s s

2 2 1 12 2 , *2 *2x y x yp p p pp = p p = pB.O. = ½ (Nb – Na) = ½ (10 – 6) = 2

2 2 2 2 22 (15) 1 , *1 , 2 , *2 , 2 ,zO s s s s p+ = s s s s s

2 2 1 02 2 , *2 *2x y x yp p p pp = p p = p

94 CHEMISTRY

B.O. = ½ (Nb – Na) = ½ (10 – 5) = 2.52 2 2 2 2

2 (17) 1 , *1 , 2 , *2 , 2 ,zO s s s s p- = s s s s s

2 2 2 12 2 , *2 *2x y x yp p p pp = p p = p

B.O. = ½ (Nb – Na) = ½ (10 – 7) = 1.52 2 2 2 2 2

2 (18) 1 , *1 , 2 , *2 , 2 ,zO s s s s p- = s s s s s

2 2 2 22 2 , *2 *2x y x yp p p pp = p p = p

B.O. = ½ (Nb – Na) = ½ (10 – 8) = 1Since, the bond length decreases as the bond orderincreases, hence, O2

+ have least bond length.44. (c) MOT configurations of O2 and O2

+ :O2

+ : (s1s)2 (s*1s)2 (s2s)2 (s*2s)2 (s2pz)2

(p2p2x = p2p2

y) (p*2px1 = p*2p0

y )Number of unpaired electrons = 1, so paramagnetic.O2 : (s1s)2 (s*1s)2 (s2s)2 (s*2s)2 (s2pz)

2

(p2p2x = p2p2

y) (p*2px1 = p*2p1

y)Number of unpaired electrons = 2, so paramagnetic.

45. (d) 4BF- hybridisation sp3, tetrahedral structure.

4NH+ hybridisation sp3, tetrahedral structure.

46. (b) (O2) = 2 2 2 2 21 , 1 , 2 , 2 , 2 ,zs s s s p* *s s s s s2 2 1 12 2 , 2 * 2x y x yp p p P*p = p p = p

Bond order = 10 6 4 2

2 2 2b aN N- -

= = =

( )2O ion+ =2 2 2 2 21 , 1 , 2 , 2 , 2 ,zs s s s p* *s s s s s

22 xpp 2 12 , 2y xp p*= p p

Bond order 10 5 5 12

2 2 2 2b aN N- -= = =

( )2O- = 2 2 2 2 21 , 1 , 2 , 2 , 2 ,zs s s s P* *s s s s s

2 2 2 12 2 , 2 2x y x yp p p p* *p = p p = p

Bond order ( ) 10 7 3 112 2 2 2

b aN N- -= = = =

( )22O - = 2 2 2 2 21 , 1 , 2 , 2 , 2 ,zs s s s p* *s s s s s

2 2 2 22 2 , 2 2x y x yp p p p* *p = p p = p

Bond order 10 8 2 1

2 2 2b aN N- -= = =

47. (c) Greater the difference between electro-negativity ofbonded atoms, stronger will be the bond. Since F ismost electronegative hence F – H ...... F is the strongestbond.

48. (b) The delocalised pp – pp bonding between filledp-orbital of F and vacant p-orbital of B leads toshortening of B–F bond length which results in higherbond dissociation energy of the B–F bond.

B FF

F

Vacant2 -orbitalp

Filled2 -orbitalp

B = FF

FB – F

F

F

B – FF

F

F

FFB

+

+

+ +1/3

+1/3

+1/3

49. (d) Calcium carbide exists as Ca2+ and C22–. According to

the molecular orbital model, C22– should have molecular

orbital configuration :2 2 2 21 , *1 , 2 , *2 ,s s s ss s s s

2 2 2{ 2 2 }, 2x y zp p pp = p s

Thus M.O. configuration suggests that it contains ones & 2p bonds.

50. (c) PF5 trigonal bipyramidal

P

F

F

FF

F

BrF5 square pyramidal (distorted due to presence ofone lp of electrons on central atom)

Br

FF

F

F

F

51. (d) (i) In Na2O2, we have 22O - ion. Number of valence

electrons of the two oxygen in 22O - ion = 8 × 2 + 2

=18 which are present as follows

s1s2, s*1s2, s2s2, s*2s2, 22s zp , { 22p xp = 22p yp ,

{ 2*2p xp = 2*2p yp

\ Number of unpaired electrons = 0, hence, 22O - is

diamagnetic.(ii) No. of valence electrons of all atoms in

O3 = 6 × 3 = 18.Thus, it also, does not have any unpaired electron,hence it is diamagnetic.

(iii) No. of valence electrons of all atoms in N2O= 2 × 5 + 6 = 16. Hence, here also all electrons arepaired. So it is diamagnetic.

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(iv) In KO2, we have 2O- . No. of valence electrons ofall atoms in 2O- = 2 × 8 + 1 = 17,

2 2 2 2 21 , *1 , 2 , *2 , 2 zs s s s ps s s s s

2 2 2 12 2 , *2 *2x y x yp p p pp = p p = p

Thus it has one unpaired electron, hence it isparamagnetic.

52. (a) Molecular electronic configuration of2 2 2 2 2 2 2CO: 1 , *1 , 2 , *2 ,{ 2 2 , 2s s s s p = p sx y zs s s s p p p

Therefore, bond order 10 4 3

2 2b aN N- -

= = =

2 2 2 2 2 2 2NO : 1 , *1 , 2 , *2 , 2 ,{ 2 2+ s s s s s p = pz x ys s s s p p p

Bond order 10 4 3

2-

= =

2 2 2 2CN 1 , *1 , 2 , *2 ,s s s s- = s s s s2 2 2{ 2 2 , 2p = p sx y zp p p

Bond order 10 4 3

2-

= =

2 2 2 2 2 2 22N : 1 , *1 , 2 , *2 ,{ 2 2 , 2s s s s p = p sx y zs s s s p p p

Bond order 10 4

32-

= =

– 2 2 2 2 2NO : 1 , *1 , 2 , * 2 , 2 ,s s s s s zs s s s p2 2{ 2 2 ,p = px yp p 1 1{ * 2 * 2p = px yp p

Bond order 10 6 2

2-

= =

\ NO– has different bond order from that in CO.53. (a) Molecular orbital configuration of B2(10) as per the

condition given in question will be

s1s2, s* 1s2, s2s2, s* 2s2, 22 xpp

Bond order of B2 = 6 4

12-

= .

B2 will be diamagnetic.

54. (d) OSF2 : 6 2H 4

2+

= = .

It has 1 lone pair.

S

O F F

:

(Shape is trigonal pyramidal)

The shapes of SO3, BrF3 and 23SiO - are triangular planar

respectively.

55. (b)3

2 24Ni 4Cl [NiCl ]+ - -+ ¾¾®

sp

[NiCl4]2–. = 3d8 configuration with nickel in + 2oxidation state, Cl– being weak field ligand does notcompel for pairing of electrons.So,

24[NiCl ] -

3d 4s 4p

sp3 hybridisationHence, complex has tetrahedral geometry

Cl

ClClCl

Ni

2–

2 24Ni 4CN [Ni(CN) ]+ - -+ ¾¾®

[Ni(CN)4]2– = 3d 8 configuration with nickel in + 2oxidation state, CN– being strong field ligand compelsfor pairing of electrons.So,

[NiCN4]–2

dsp2 hybridisation

3d4s 4p

Hence, complex has square planar geometry.

NC CN

CNNC

2–

Ni

2 22 2 6Ni 6H O [Ni(H O) ]+ ++ ¾¾®

[Ni(H2O)6]2+ = 3d 8 configuration with nickel in + 2oxidation state. As with 3d 8 configuration twod-orbitals are not available for d2sp3 hybridisation. So,hybridisation of Ni (II) is sp3d 2 and Ni (II) with sixco-ordination will have octahedral geometry.

[Ni(H2O)6]2+

sp d3 hybridisation2

3d4s 4p 4d

H O2

H O2

OH2

OH2H O2

H O2

Ni

2+

96 CHEMISTRY

56. (b) Allene (C3H4) is = =2 2

2 2H C C CHsp sp sp

57. (a) & (b) The molecular orbital structures of C2 and N2 are2 2 2 2 2 2 2

2 x y zN 1 *1 2 *2 2 2 2= s s s s s p ps s s s p p p2 2 2 2 2 2

2 zC 1 *1 2 *2 2 2= s s s s p ps s s s py PBoth N2 and C2 have paired electrons, hence they arediamagnetic.

58. All options are correct,

(a) ONCl 8 7 17 32e not isoelectronicONO 8 7 8 1 24e

-

-ü= + + = ý- = + + + = þ

(b)O. .

1.278Å 1.278A°

116.8°OThe central atom is hybridized with one lone pair.

sp2

O(c) It is a pale blue gas. At – 249.7°, it forms violet black

crystals.(d) It is diamagnetic in nature due to absence of unpaired

electrons.59. (c) H2

2+ = s1s0 s*1s0

Bond order for H22+ =

1 (0 0) 02

- =

He2 = s1s2s*1s2

Bond order for He2 = 1 (2 2) 02

- =

so both H22+ and He2 does not exist.

60. (b) Li2 = s1s2 s*1s2 s2s2

\ Bond order = 1 (4 2) 12

, <

Li2+ = s1s2 s*1s2 s2s1

B.O. 1 (3 2) 0.52

< , <

Li2– = s1s2 s*1s2s2s2s*2s1

B.O. = 1 (4 3) 0.52

, <

The bond order of Li2+ and Li2

– is same but Li2+ is more

stable than Li2– because Li2

+ is smaller in size and has 2electrons in antibonding orbitals whereas Li2

– has 3electrons in antibonding orbitals. Hence Li2

+ is morestable than Li2

–.

61. (b) H C — C C — CH3 3ºsp 3 sp 3sp sp

linear due to hybridized C atom.sp62. (d) PCl3

ClP

ClCl

63. (a) Both 2–2O and B2 has bond order equal to 1.

( ) 2 * 2 2 * 22

1 110 1 1 2 2

2 2x y

B s s s sp p

= s s s sp = p

Bond order = 2

b aN N-

= 6 4 2 1

2 2- = =

B2 is known in the gas phase

( )22 2 2 2 22O ( 1 ) ( *1 ) ( 2 ) ( *2 ) 2 zs s s s p- = s s s s s

( ) ( ) ( ) ( )2 222 * *2 2 2 2x y x yp p p pp = p p = p

Bond order ( )1 10 8 12

= - =

64. (b) Diamagnetic species have no unpaired electrons2

2O - Þ s1s2, s*1s2, s2s2, s*2s2, 22s zp ,{p2px

2 = p2py2, {p*2px

2 = p*2py2

Whereas paramagnetic species has one or moreunpaired electrons as in

2 2 2 2 22O 1 , 1 , 2 , 2 , 2* *® s s s s s zs s s s p , 2 2{ 2 2 ,p = px yp p

1{ 2*p =xp 12 2 unpaired electrons*p =yp

2 2 2 2 22O 1 , 1 , 2 , 2 , 2 zs s s s p+ * *® s s s s s ,

22 xpp = 2 1 02 2 *2 1 unpaired electrony x yp p p*p p = p = unpaired electron2 2 2 2 2NO 1 , 1 , 2 , 2 , 2* *® s s s s s

zs s s s p ,

2 2 1 02 2 , 2 2 1 unpaired electron* *p = p p = p =x y y zp p p p65. (d) Smaller the size and higher the charge more will be the

polarising power of cation. Since the order of the sizeof cation is K Ca Mg Be+ ++ ++ ++> > > . So thecorrect order of polarising power isK+ < Ca2+ < Mg2+ < Be2+

66. (c) (i) N2 : bond order = 3, diamagneticN2

– : bond order = 2.5, paramagnetic(ii) C2 : bond order = 2, diamagnetic

C2+ : bond order = 1.5, paramagnetic

(iii) NO : bond order = 2.5, paramagneticNO+ : bond order = 3, diamagnetic

(iv) O2 : bond order = 2, paramagneticO2

+ : bond order = 2.5, paramagnetic67. (a) For any species to have same bond order we can expect

them to have same number of electrons. Calculatingthe number of electrons in various species.

–2O (8 8 1 17)+ + = ; CN (6 7 1 14)- + + =

NO+(7 + 8 – 1= 14); CN+ (6 + 7 –1 = 12)We find CN– and NO+ both have 14 electrons and bondorder 3.

68. (c) All the members form volatile halides of the typeAX3. All halides are pyramidal in shape. The bondangle decreases on moving down the group due todecrease in bond pair-bond pair repulsion.

69. (a) Compound having maximum electronegativitydifference is more ionic. Hence the correct order isDG < EG < DF < DE

70. (d)71. (b) SO2 – bent

SF4 – see-sawClF3 – T-shapeBrF5 – square pyramidalXeF4 – square planar.

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72. (a) XeF4 having one lone pair of electron show distortedpentagonal bipyramidal shape and sp3d3 hybridisation

73. (a) It is correct that during formation of ice from waterthere are vacant spaces between hydrogen bondedmolecules of ice. Ice has a cage like structure. Due tothis reason ice is less dense than liquid water. Henceboth statement-1 and statement-2 are true andstatement-2 is correct explanation of statement-1.

74. (a)75. (a) o and p-nitrophenols can be separated by steam

distillation because o-nitrophenol is steam volatile dueto presence of weak intramolecular hydrogen bondingwhereas p-introphenol contains intermolecular hydrogenbonding. Here, both statement-1 and statement-2 arecorrect and statement-2 is correct explanation ofstatement-1.

EXERCISE - 3Exemplar Questions

1. (b) 4BF- and 4NH+ both the species are tetrahedral andsp3 hybridised.

2. (c) CO2 being symmetrical has zero dipole moment.Amongst HI, SO2 and H2O. H2O exhibit the highestdipole moment as the central atom in H2O contains 2lone pairs.

= 0.38DH—I

m

m = 1.84 DH

O ... .

H m = 1.62 D

S

..

O O

3. (b) Applying the formula to find the hybridisation ofcentral atom (nitrogen) :

12

[No. of valence e – of central atom +

no. of monovalent atoms attached to it +(–ve) charge if any – (+ve) charge if any]

For, NO2+ =

12

[5 + 0 + 0 – 1] = 2 Þ sp-hybridised

NO3– =

12

[5 + 0 + 1 – 0] = 3 Þ sp2-hybridised

NH4+ =

12

[5 + 4 + 0 – 1] = 4 Þ sp3-hybridised

4. (b) Strength of H-bonding is in the order :HF > H2O > NH3

Also, each H2O molecule is linked to four other H2Omolecules through H-bonds while each HF molecule islinked only to two other HF molecules.Hence, decreasing order of b.p. will be : H2O > HF >NH3

5. (c) In 34PO - ion, formal charge on each O-atom of P – O

bond Total charge 3 0.75

Number of O atom 4= = - = -

6. (d) In NO3– ion,

number of bond pairs (or shared pairs) = 4number of lone pairs = 0

7. (a) 4BH- Þ 4 bond pairs and 0 lone pair ® sp3 hybridised.\ Tetrahedral geometry

8. (c) The correct structure of the given compound will be asfollows :

H

H

H

H

H

H

H

HNow, there are 5 p-bonds and 19 s-bonds are present inthe above molecule.

9. (c) The electronic configuration of the given moleculesare :

2 2 2 2 2 2 12N 1 , *1 , 2 , *2 , 2 , 2+ = s s s s p = p sx y zs s s s p p p ;

1 unpaired e–

2 2 2 2 2 2 22O 1 , *1 , 2 , * 2 , 2 , 2 2 ,= s s s s s p » pz x ys s s s p p p

1 1* 2 * 2p » px yp p ; 2 unpaired e–s

2 2 2 2 2 2 2 22O 1 , *1 , 2 , * 2 , 2 , 2 2 ,- = s s s s s p » pz x ys s s s p p p

2 2*2 *2p » px yp p ; no unpaired e–s

2 2 2 2 1 12 2B 1 , *1 , 2 , * 2 , 2= s s s s p » px ys s s s p p ;

2 unpaired e–s10. (c) XeF4 Þ square planar, all bonds are equal.

4BF- Þ tetrahedral, all bonds are equal.

C2H4 Þ C = CHH

HH, C = C bond is not equal to

C – H bond.SiF4 Þ tetrahedral, all bonds are equal.

11. (b) HCl, HI and H2S do not from H-bonds. Only H2O formshydrogen bonds and each H2O molecule is linked withfour H2O molecules.

HO

H H

HO

HHO

HHO

HHO

12. (d) The given electronic configuration shows that theelement belongs to d-block of the periodic table andknown to be a transition element. In transition elements,electrons of ns and (n – 1)d subshell take part in bondformation.

98 CHEMISTRY

13. (b) For sp2 hybridisation, generally the geometry is usuallytaken to be triangular planar.

X120°

Y

Y Y

Thus, bond angle should be 120°.14. (a) The given electronic configuration of A shows that it

is a noble gas because the octet is complete and henceit will be the stable form.

15. (b) The electronic configuration of C represent chlorine.Its stable form is Cl2 i.e., C2.

16. (d) The electronic configuration show that B representsphosphorus and C represents chlorine. The stablecompound formed will be PCl3 i.e., BC3.

17. (b) The bond between B and C will be covalent as both Band C are non-metal atoms.

18. (a) The correct increasing order of energies of molecularorbitals of N2 is :s1s < s*1s < s2s < s*2s < (p2px » p2py ) < s2pz < (p*2px; p*2py ) < s*2pz

19. (d) (a) Be2 = s1s2, s*1s2, s1s2, s* 2s2

(B.O) = 12

[Number of bonding electrons (Nb ) –

Number of anti-bonding electrons Na ]

4 4 02-

= =

Bond order of Be2 is zero thus, it does not exist.(b) He2 = s1s2, s*1s2

2 2B.O 02-

= =

Now, 2 12He 1 , *1+ = s ss s

2 1B.O 0.52-

= =

As the bond order is not zero, He2+ molecule is

expected to exist.(c) N2 (7 + 7 = 14) = s1s2, s*1s2, s2s2, s*2s2,

2 2 22 2 , 2p » p sx y zp p p

10 4B.O 3

2-

= =

Thus, dinitrogen (N2 ) molecule contain triple bondand as bond order = 3 while no molecule of secondperiod have more than double bond. Thus, bondstrength of N2 its maximum amongst thehomonuclear diatomic molecules belonging to thesecond period.

(d) The correct order of energies of molecular orbitalsin N2 molecule is

s1s < s*1s < s2s < s*2s < (p2px ; p2py ) < s2pz <p*2px » p*2py < s*2pz

20. (b) E.C of O2 :

s1s2, s*1s2, s2s2, s*2s2, 2 2 22 , 2 2 ,s p » pz x yp p p

1 1*2 *2p » px yp p

B.O = ( ) ( )b a1 1N N 10 6 22 2

- = - =

E.C of 2O+ :2 2 2 2 2 2 21 , *1 , 2 , *2 , 2 , 2 2 ,s s s s s p » pz x ys s s s p p p

1 0*2 *2p » px yp p

B.O = ( )1 10 5 2.52

- =

E.C of 2O- :2 2 2 2 2 2 21 , *1 , 2 , *2 , 2 , 2 2 ,s s s s s p » pz x ys s s s p p p

2 1*2 *2p » px yp p

Bond order = ( )110 7 1.5

2- =

Hence, the correct order of bond order will be :

2 2 2O O O- +< < .21. (a) Out of the given electronic configuration 2s22p5

represents fluorine which is the most electronegativeelement while 3s23p5, 4s24p5 and 5s25p5 representschlorine, bromine and iodine respectively.

22. (b) The electronic configuration of (b) and (d) have exactlyhalf-filled p-orbitals but (b) being smaller in size than(d) will have the highest ionisation enthalpy.

NEET/AIPMT (2013-2017) Questions

23. (a) SF4 has 4 bond pairs and 1 lone pair of electrons, sp3d

hybridisation leads to irregular shape S|

|

F

F

F

F

and

resultant m ¹ 0.

24. (d) H— C = C —H| |

H H

s s s

s sp

25. (d) (a) 2 2N NB.O. 3 2.5

+¾¾®

Bond energy decreasesMagnetic behaviour changes from diamagnetic toparamagnetic

(b) 2 2O OB.O. 2 2.5

+¾¾®

Bond energy increasesMagnetic behaviour does not change.

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(c)2 2C C

B.O. 2 2.5

+¾¾®

Bond energy decreasesMagnetic behaviour changes from diamagnetic toparamagnetic

(d)NO NO

B.O. 2 2.5

+¾¾®

bond energy increasesMagnetic behaviour changes from paramagnetic todiamagnetic

26. (b) No. of electrons in CO = 6 + 8 = 14No. of electrons in NO+ = 7 + 8 – 1 = 14\ CO and NO+ are isoelectronic species.Isoelectronic species have identical bond order.

27. (c) Applying VSEPR theory, both NF3 and H2O are sp3

hybridized.

28. (a) F Xe F¾ ¾ sp3d and Linear

Cl I Cl¾ ¾– sp3d and Linear

29. (a) Molecular orbital configuration of 2O- is

2O- (17) = s1s2, s*1s2, s2s2, s*2s2,s2pz

2, p2px2 = p2py

2, p*2px2 = p*2py

1

30. (a) OH H

ss O S O

s

p¬ = O N O

s

p¬ = O C O

s s

p p= =

31. (b) (BH3)2 or (B2H6)

H H HB B

H H H

It contains two 3 centre - 2 electron bonds and presentabove and below the plane of molecules compoundswhich do not have sufficient number of electrons toform normal covalent bonds are called electron defi-cient molecules.

32. (b) Be2+ = (4 – 2) = 2is isoelectronic with Li+ (3 – 1 = 2)Since both have same number of electrons in theiroutermost shell.

33. (c) Dipole moment of NH3 > NF3

Nd–. .

H H Hd+

m = 1.4D

N

F F F

d+

m = 0.23D

( )d –( )d– ( )d–

(F is more electronegative than N)

34. (b) Hybridization of NO3– =

1(5 0 1 0)

2+ + -

= 263

2= = sp hence geometry is trigonal planner..

NO2– (nitrite ion) also has sp2 hybridization and gives

a trigonal planner geometry but because there are only

two outer atoms, the molecular geometry is bent withÐ120º bond angles.

35. (c) +2O ion - Total number of electrons (16 – 1) = 15.

Electronic configuration2 2 2 2 2σ1 σ 1 σ2 σ 2 σ2 xs s s s p* *< < < <

2π2 yp< 2 1π2 π 2z yp p*= <

Bond order = b aN N 10 5 5 122 2 2 2- -

= = =

–2O (Super oxide ion): Total number of electrons

(16 + 1) = 17.Electronic configuration

2 2 2 2 2σ1 σ 1 σ2 σ 2 σ2 xs s s s p* *< < < <2 2 2 1π2 π2 π 2 π 2y z y zp p p p* *< = < =

Bond order b a(N N ) 10 7 3 112 2 2 2- -

= = = =

22O+ ion: Total number of electrons

= (16 – 2) = 14 Electronic configurations1s2 < s*1s2 < s2s2 < s*2s2 < s2px

2 < p2py2 = p2pz

2

Bond order = b a(N – N ) 10 – 4 6 32 2 2

= = =

So bond order: O2– < O2

+ < O22+

36. (c) ClO3– and SO3

–2 both have same number of electrons(42) and central atom in each being sp3 hybridised. Bothare having one lone pair on central atom hence they arepyramidal.

37. (a) Oxygen molecule (O2) – Total number of electrons = 16and electronic configuration is

2 2 2 2 2σ1 σ 1 σ2 σ 2 σ2* *< < < < xs s s s p2π2< yp 2 1 1π2 π 2 π 2* *= < =z y zp p p

Bond order = b aN N 10 6 4 22 2 2- -

= = =

+2O ion - Total number of electrons (16 – 1) = 15.

Electronic configuration2 2 2 2 2σ1 σ 1 σ2 σ 2 σ2* *< < < < xs s s s p

2π2< yp 2 1π2 π 2*= <z yp p

Bond order = b aN N 10 5 5 122 2 2 2- -

= = =

–2O (Super oxide ion) Total number of electrons

17)116( =+ . Electronic configuration2 2 2 2 2σ1 σ 1 σ2 σ 2 σ2* *< < < < xs s s s p

2 2 2 1π2 π2 π 2 π 2* *< = < =y z y zp p p p

Bond order b a(N N ) 10 7 3 112 2 2 2- -

= = = =

38. (b) NO2+ has sp hybridisation so it is linear with bond angle

= 180°.

100 CHEMISTRY

39. (a) Xe

O

O O ONumber of s bonds = 4Number of p bonds = 4

40. (a) According to molecular orbital theory as bond orderdecreases stability of the molecule decreases

1Bond order (N – N )b a2=

Bond order for 21O (10 5) 2.52

+ = - =

Bond order for 21O (10 6) 22

= - =

Bond order for 21O (10 7) 1.52

- = - =

Bond order for 22

1O (10 8) 1.02

- = - =

hence the correct order is

– 2–2 2 2 2O O O O+ > > >

41. (d) XeF4, XeO4

XeF F

F F(Square planar)

Xe

O

OO O

[Tetrahedral]42. (a) Ni2Å = [Ar]18 4s0 3d8

Valence bond theory can be used to predict shape.

3d 4s 4p

dsp2 hybridization

(In presence of ligand, pairing of electron occurs)\ Square planar.

43. (b) CH4 NH3H2O

H

HH

HC 109°28'

HH

HN

107°

H HO

104.5°Tetrahedral; Trigonal Bent

pyramidalNote: The geometry of H2O should have beentetrahedral if there are all bond pairs. But due to presenceof two lone pairs the shape is distorted tetrahedral.Hence bond angle reduced to 104.5° from 109.5°.

44. (a) According to VSEPR theory order of repulsion inbetween lp – lp, lp – bp and bp – bp is as underlp – lp > lp – bp > bp – bp

45. (b) 2 2IBr , XeF-

Total number of valence electrons are equal in both thespecies and both the species exhibit linear shape.

46. (c) BCl3 is trigonal planar and hence the bond angle is120°.

Cl

Cl ClB 120°

47. (b) CN– and CO have same no. of electrons and have samebond order equal to 3.

EB

D_7

753

Chemical Bonding and Molecular Structure - SelfStudys

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