1
2
UNIT
CHEMICAL BOND
2.1 CHEMICAL BOND
Chemical bond is a force which binds two or more atoms together.
Eg: H2, Cl2, HCl, NaCl etc.
2.2 WHY CHEMICAL BONDS ARE FORMED?
(1) Octet Rule:
Inert gases are very stable. They have eight electrons in the valence shell. They do not
form bond. Other elements also have tendency to attain stable octet configuration
similar to nearest inert gas. For this, atoms lose, gain or share electrons. This is why
bonds are formed.
(2) Energy consideration of chemical bond formation
Each and every system of this universe has natural tendency to attain lowest energy state.
Atoms also have tendency to attain lowest energy state. Therefore, they form bonds to minimise
its energy.
A bond is formed between two atoms A and B when EA + EB > EAB where, EA = Energy of
A, EB = Energy of B, EAB = Energy of AB
A + B  AB
EA EB EAB
57
58 A NEW APPROACH TO INORGANIC CHEMISTRY
2.3 TYPES OF CHEMICAL BONDS
Chemical bonds can be classified into the following class
2.4 ION IC BOND OR ELECTROVALENT BOND


Ionic bond is electrostatic force of attraction which binds two oppositely charged ions together.
An ionic bond is formed between metals and non metals.
Na +  Cl  Na+ + Cl–  NaCl
2, 8, 1 2, 8, 7
2, 8
2, 8, 8
Mg  +  O   Mg++ + O– –  Mg++ O– –  MgO
2, 8, 2 2, 8, 6
2, 8 2, 8


(ii)






Ex- (i)
Cl 


Cl 
 
 


 Mg++ (Cl–)2  MgCl2


(iii) Mg 




(iv) Ca  +  O   Ca++ O 2–
 CaO

Ionic bond is formed between metals having 1 or 2 electrons and non-metals having 6 or 7
electrons in the Valence Shell. Ionic bond is formed by complete transfer of electrons from metals
to non metals.
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CHEMICAL BOND 59
2.5 FACTORS WHICH FAVOUR IONIC BOND FORMATION
Following factors favour formation of ionic bond.
(i) Large size of metal.
(ii) Low ionization enthalpy value of metals.
(iii) Small size of non-metals
(iv) High electron gain enthalpy value of non-metals
(v) Electronegativity difference between metal and non-metals must be greater than 1.8.
(vi) Lattice enthalpy of ionic compound must be high
Ionic bond is formed when IE < Eg + Uo
where, IE = ionisation enthalpy of metal, Eg = electron gain enthalpy of non-metals,
U0 = Lattice enthalpy of ionic compounds.
2.6 PROPERTIES OF IONIC COMPOUNDS
Compounds having ionic bonds are called ionic compounds.
(i) It is composed of oppositely charged ions i.e. cations and anions.
(ii) It is formed between metals and non-metals by complete transfer of electrons.
(iii) Ions are held together by strong electrostatic force of attraction
(iv) All ionic compounds are hard solids.
(v) They are very brittle.
(vi) lonic compounds do not form molecules but they form crystal lattice i.e. cations are
surrounded by anions and anions are surrounded by cations.
(vii) lonic compounds are souble in water or polar solvents and insoluble in non-polar
solvents.
(viii) lonic compounds undergo ionization in solution (Break into ions).
+
–
+
–
– Na
NaCl + water 
–
–
–
–
+
+
+
–
+
+
+
+
Cl–
–
+
–
(ix) Reactions of ionic compounds are very fast.
(x) They are insulators in solid state but become conductors in fused or solution state. In
solid state ions can’t move and in liquid state they become mobile and carry charge.
(xi) lonic compounds undergo electrolysis in fused or solution state.
(xii) lonic compounds do not show space isomerism because ionic bond is spherically
symmetrical.
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60 A NEW APPROACH TO INORGANIC CHEMISTRY
2.7 LATTICE ENTHALPY
Amount of energy released when one mole of ionic solid is formed from its constituent gaseous
ions is called lattice enthalpy
It is denoted by U0. It is calculated with the help of Born-lande equation. Born-Lande equation
is given as
 NAZ Z e 2
U0 =
r0
 1
1  n 
Where,
Uo = Lattice enthalpy of ionic solid
N = Avogadro no. (6.022 × 1023 particles)
A = Madelung’s constant. Value of ‘A’ depends on electronic configurations of ions.
Z+ = Charge of cation
Z– = Charge of anion
e = Charge of electron : (1.60 × 10–19C)
r0 = Inter nuclear distance [distance between nuclei of cation and anion.]
r0 = r c + ra
where, rc = radius of cation
ra = radius of anion
n = Born exponent. Value of n depeneds on structure of solids
Thus
U0 
Z Z
r0
i.e. Lattice enthalpy value increases with increasing charge and decreasing size of ions.
Ex.(i) U0 of NaF > NaCl > NaBr
Ex. (ii) U0 of NaCl < Mg SO4
2.7.1 Born Haber Cycle
Lattice enthalpy value is determined with the help of Born – Haber Cycle. It is based on Hess’s
Law of Constant Heat summation.
Hess’s Law : Energy change of a chemical reaction is path and step independent
Ex. NaCl can be prepared from Na, and Cl2 in two ways
(i) One step process
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CHEMICAL BOND 61
1
f H 
Na(s)  Cl 2 (g) 
 NaCl(s)
2
(ii) Multi step process
 H
iH
S
Na(s) 
 Na(g) 
 Na  (g)
dH
Eg
Cl 2 (g) 
 Cl(g)  Cl (g)
U
0
Na  (g)  Cl  (g) 
NaCl(s)
Total change in energy by path II = sH(Na) + iH(Na) +
1
dH(Cl 2 ) + Eg(Cl2) + U0
2
Hence from Hess law of constant heat summation
fH = sH + iH +
Where,
fH = Enthalpy of formation of NaCl,
sH = Enthalpy of sublimation of Na,
dH = Enthalpy of dissociation of Cl2 ,
1
 H + Eg + U0
2 d
iH = Ionisation energy of Na
Eg = Electron gain enthalpy of Cl
U0 = Lattice enthalpy of NaCl
If all the other values are known U0 can be calculated.
2.8 POLARISATION OF IONIC COMPOUND
Polarisation: The process of deformation of shape of ions by the action of oppositely charged
ions is called polarization.
An ionic compound consist of cations and anions. Net positive charge of cation attracts electron
cloud of anion towards itself and repels the positive nucleus. As a result spherical shape of anion
is deformed. On the contrary, net negative charge of anion attracts the positive nucleus of cation
towards itself and repels the electron cloud.
–
+
+
+
+
+
+
+
+
–
–
–
–
+
–
–
+
–
–
+
+
+
+
+
+
+
+
+
+
+
+
– –
–
–– – –
– –
–
– –
– ––
–
–
–
–
But, electrons of cations are tightly held by the nucleus. Hence, it cannot be polarized, However
electron of anions are loosely held and get polarizaed easily.
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62 A NEW APPROACH TO INORGANIC CHEMISTRY
i.e. cation have polarizing power and anions have polarisability.
Fajan’s rule :1. Polarizing power of cation to polarize the anion increases with
(a) increasing charge of cation
e.g. Polarizing power of Na+ < Mg+ + < AI + + +
(b) decreasing size of cation
e.g. Polarizing power of Li+ > Na+ > K+ > Rb+ > Cs+
(c) presence of d–electron s
e.g. Polarizing power Ca+ +< Fe+ +
2, 8, 8 2, 8, 14
It is because screening effect of d–electrons is very poor, hence effective nuclear
charge is higher
2. Polarisability of anion to get polarized by cation increases with
(a) increasing negative charge of anion
e.g. Polarisability of F– < O2– < N3–
(b) increasing size of anion
e.g. Polarisability of F– < Cl– < Br– <1–
Effect of Polarisation :Polarisation in an ionic compound produces covalent character. Increasing polarization increases
covalent character in ionic compound.
Increasing covalent character decreases melting point as well as solubility of ionic compound
in water.
(i) Ex.melting point and boiling point of LiCI < Na Cl < KCl
(ii) Ex.covalent character of NaF < NaCl < NaBr < Nal
(iii) Ex.covalent character of Na NO3 < Na2 SO4 < Na3 PO4
2.9 COVALENT BOND
The force of attraction arising due to mutual sharing of electrons between two atoms which bind
them together is called covalent bond
It is denoted by a dash (–) between two bonded atoms

 Cl +
 

++
+ Cl +
+ ++
+ ++

 Cl

  + +Cl++


 Cl


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Cl
CHEMICAL BOND 63
Covalent bond is formed between two non-metal atoms or a less metallic metal and a nonmetal atom.
Type of Covalent bond
(A) on the basis of number of electrons shared it is of three types
Covalent bond
Single bond
Double bond
Triple bond
Single bond: A covalent bond formed by sharing of one pair of electrons between two atoms
is called single bond. All single bonds are sigma () bonds.
e.g. (i)

F


F

O

O
(ii)


H
H
H
H
Double bond : A Covalent bond formed by sharing of two pair of electrons between two
atoms is called a double covalent bond.
O
Ex.
+
+
++
O+
+
or,
O

—
—

+++
O+
Double bond consist of one sigma () and one pi() bonds.
Triple bond : A covalent bond formed by sharing of three pair of electrons between two
atoms is called triple bond.
N
N

N
N

N
N
A triple bond has one sigma () and two pi() bonds.
(B) Classification of Covalent bond on the basis of mode of formation of bond
Covalent bond
Sigma bond ()
pi bond ()
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64 A NEW APPROACH TO INORGANIC CHEMISTRY
Sigma bond : A covalent bond formed by head-on (axial or linear) overlap of atomic orbitals
is called sigma bond. It is denoted by 
Pictorial representation of formation of  bond
(i) S-S overlap

+
H
 bond
H2
H
(ii) S-P overlap
Z
Y
 bond
Z
Y
X 
+
X
H
HF
F
(iii) P-P overlap
Z
Z
Y
X
+
X
 bond
Cl
Cl
Cl2
Pi () bond : A covalent bond formed by parallel (lateral or side wise) overlap of atomic
orbitals is called  bond.
(i)  bonds are stronger than  bond because  bond has longer area of overlap.
(ii) Only smaller atoms can from  bond e.g. B,C,N,O,F. Smaller atoms complete its octet
by forming  bond and larger atoms undergo polymerization.
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CHEMICAL BOND 65
Example of -bond

P
H
P
H
H
(a)
C
C
H
C
or
sp2 – sp 2
H
H

C

Ethene
H
H


H
or
(b)
C

H
C
H–CC–H
Ethyne
2nd and 3rd p-p overlap between two atoms forms pi () bond.
Some Important Questions
Q1. BCl 3 is monomer but AlCl3 is dimer.
Q2. CO2 is monomer but SiO2 is polymer.
Q3. Nitrogen forms N2 but Phosphorus forms P4.
Q4. Oxygen forms O2 but Sulphur forms S8.
Ans. (1):
In BCl3 , size of B atom is small, it can form  bond, it completes its octet by accepting a pair
of electrons from any of the chlorine atom.
Size of aluminium is large. It cannot form -bond. Hence, two AlCl3 combine together in
which Cl- atom of one molecule donates a pair of electron to the Al atom of 2nd molecule and viceversa.
Cl
Cl
B
— Cl
Cl
Cl
Al
Cl
Cl
Al
Cl
Cl
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66 A NEW APPROACH TO INORGANIC CHEMISTRY
Ans. (2): Carbon atom is small in size.it can form –bond. It complete its octet by forming
two  and two  bonds. But Sin cannot form bond. Hence silicon atoms combines with four Oatoms and each O-atom combines with two Si–atoms and complete their octet. In this way, a large
polymer is formed.
O

C


O

|
Si
|
O
|
Si — O — Si — O — Si —
|
O
|
Si
|
Ans. (3): Nitrogen atom has three unpaired electrons in the valence shell with which it can
form three bonds. Size of nitrogen atom is small, it can form  bond. Thus nitrogen forms N2 with
the help of one - bond and two -bonds.
 N  N


N - 1S2 2S2 2P3
Size of phosphorus atom is large, it cannot form -bonds. Phosphorus atom satisfies its valency.
by forming three -bonds with three other phosphorus atoms. This is why nitrogen forms N2 but
Phosphorus forms P4
P
|
P—P—P
|
P
Ans.(4):
Oxygen atom has two unpaired electrons in the valence shell with which it can form two
bonds Thus Oxygen atom is small in size, it an form  bonds. Oxygen forms O2 with one  and
one  bonds.
2
2
4 :O
8O :- 1 S 2S 2P

O:

Sulphur atom has large size. It can’t form -bond. So, it satisfies its valency by forming two
 bonds with other sulphur atoms. But, a triplet is not stable. Thus, a ring of 8 sulphur atoms is
formed to attain stability.
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CHEMICAL BOND 67
S
S
S
S
S
S
S
S
(C) CLASSIFICATION OF COVALENT BOND BASED ON NATURE OF BONDS
Covalent bond
Non-Polar covalent bond
Polar covalent bond
Non-polar covalent bond : A covalent bond in which shared electron pair is symmetrically
distributed between two bonded atoms is called non-popar covalent bond.
e.g. Cl
Cl , F2, H2, Br2 l2, O2, N2
It is formed between atoms of same elements or atoms of same electronegativity.
S C S
P
EN of P = 2.1 and
EN 2.5 2.5 2.5
|
H
= 2.1 are same, hence
e.g. (i) EN of C and S are same, hence (ii) H H H
P – H bond is non-polar covalent
C = S bond is non polar
Polar covalent bond : A covalent bond in which shared electron pair is unsymmetrically
distributed between two bonded atoms is called polar covalent bond. It is formed between atoms
of different electronegativity.
+
Ex. (a) H
–
Cl ,
(b)
+/2
H
O–
–/2
H
Polarity of a covalent bond is expressed in terms of (i) Percent ionic character and
(ii) Dipole Moment.
(i) Percent ionic character is given by P  µ  100
de
or,
P=
Where,
µobs
 100
µth
P = Percent ionic character,
d = Bond length,
µobs = Observed dipole moment
µ = Dipole moment
e = Charge of electron
µth = Theoretical dipole moment
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68 A NEW APPROACH TO INORGANIC CHEMISTRY
(ii) Dipole moment
Product of charge and bond length is called dipole moment. It is denoted by µ
µ=qxd
Where µ = Dipole moment.
q = Charge on the atom
d = Bond length
Unit of dopole moment = Coulumb – metre or Debye
Bond length: Distance between nuclei of two bonded atoms is called bond length.
Dipole moment of molecule: Dipole moment is a vector quantity. A molecule may have
many bonds. Hence dipole moment of a molecule will be resultant of dipole moments of all the
bonds present in a molecule.
In some molecules, the bond dipoles cancel each other and resultant dipole moment is zero. In
such case, molecule is non- polar even if bonds are polar.
Ex. (i) BF3
F
B—F
F
Net µ = 0
(ii) CO2


O=C=O
Net  = O
Example of molecule in which net dipole moment is Zero.
1. Symmetrical Linear:
6. Symmetrical Pentagonal bipyramidal eg. IF7
e.g. (i) CO2,
F
F |
l
F |
F F
O=C=O
(ii) BeCl2 Cl––––Be –––– Cl
2. Symmetrical Triangular :
F
F
7. Symmetrical Hexagonal Planar e.g. C6H6
Cl
e.g. BX3
B — Cl
Cl
3. Symmetrical Tetrahedral :
e.g. CCl4, Ni(CO)4
Cl
|
C
Cl |
Cl
Cl
Benzene
8. Symmetrical Trans Isomer :
e.g.
Cl
H
Cl = C
H
Cl
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Trans dichloro
ethylene
CHEMICAL BOND 69
4. Symmetrical Trigonal bipyramidal
9. Symmetrical Para Isomer:
Cl
e.g. PCl5
Cl
Cl |
P — Cl
Cl |
Cl
p-dichloro benzene
e.g.
Cl
5. Symmetrical Octahedral
SF6
e.g.
F
F |
F
S
F |
F
F
Q.5 Which one has minimum dipole moment?
(a) Butene-1
(b) Cis butene-2
(c) Trans-2 butene (d) 2-methyl propene.
Hints: Structures of the given molecule are as follows.
(a) CH3CH2CH = CH2
Butene-1
(b) CH
3
CH3
C=C
H
H
Cis-butene-2
H
(c) CH3
C=C
H
CH3
(d)
CH3 — C = CH2
|
CH3
2-methyl Propene
Trans-2-butene
Trans-2-butene has minimum (µ = 0) dipole moment
Determination of Structure from Dipole Moment Value
(i) AB2 type Molecule: (i) µ = 0 (Nonpolar) - Symmetrical linear
(ii) µ > 0 - Angular (Polar)
Q.6 (i) CO2 is non-polar but SO2 is polar why ?
Explanation: CO2 is symmetrical linear, hence net dipole moment is zero but SO2 is angular
hence net dipole moment is greater than Zero.
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70 A NEW APPROACH TO INORGANIC CHEMISTRY
S
O
C
linear
O
O
O
Angular
AB3 type Molecule: (i) µ = 0 (Nonpolar) - Symmetrical Triangular
(ii) µ > 0 - Pyramidal
Q.7 (ii) BF3 is non-polar but NF3 is polar. Why?
Explanation: BF3 is symmetrical triangular, hence net dipole moment is zero, but NF3 is
pyramidal net dipole moment > 0 (Finite dipole moment)
N
|
F F F
Pyramidal
F
B—F
F
Triangular Planar
Problems for Practice
Q.8 Which one of the following molecule has µ = 0?
(a) SO2
(b) HCl
(c) NH3
(d) BF3
Q.9 Which one of the following molecule is polar?
(a) C6H6
(b) CO2
(c) CH4
(d) H2O
Q.10 Which one of the following is most polar?
(a) CCl4
(b) CHCl3
(c) CH3OH (d) CH3Cl
2.10 THEORIES OF COVALENT BOND
There are two theories of covalent bond formation
1. Valence Bond Theory: Proposed by Heitlor and London and modified by Pauling and
Slater
2. Molecular Orbital Theory: Proposed by Hund and Mulliken
Valence Bond Theory
This theory was proposed by Linus Pauling. He was awarded nobel prize in chemistry in 1954.
1. A covalent bond is formed by the overlap of atomic orbitals of combining atoms.
2. Axial overlap of atomic orbitals give  bond whereas parallel overlap of atomic orbitals
give -bond.
3. Atomic orbitals containing unpaired electron only undergo overlap.
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CHEMICAL BOND 71
4. Atomic orbitals having paired electron may also take part in overlap by unpairing the
paired electron provided unpairing is energetically favourable i.e. amount of energy
required for promotion of electron must be less than the amount of energy released in
two extra bond formation.
5. Strength of bond increases with increasing extent of orbital overlap.
Some Solved Problems
Q.11. Nitrogen shows valency 3 only but phosphorus shows a valency of 3 and 5 both. Or
Nitrogen forms NCl3 only but phosphorus forms PCl3 and PCl5 both. Why?
Ans. Nitrogen has three unpaired electron in the valence shell with which it can form three
covalent bonds. Hence, nitrogen shows trivalency and forms NCl3.
1 1 1
2
7 N - 1S
2S2
2P3
Nitrogen has no vacant orbitals in the valence shell, hence promotin of electron is energetically
unfavourable.
In ground state phosphorus has three unpaired electron in the valence shell with which it can
form three covalent bonds and forms PCl3. 15P – 1S22S2 2P6 3S2 3P3 3d0
1 1 1
3S2
3P3
3d0
Phosphorus has five empty d-orbitals in the valence shell. Hence in excited state promotion of
one electron from 3 S to 3 d –orbital takes place. This promotion is energetically fovourable.
Thus, in excited state phosphorus shows pentavalency and forms PCl5
P*
excited state
3S1 3P3
3d1
Q.12 Oxygen shows valency two only but sulphur shows valency 2,4 and 6 why?
Ans: 8O 1S2 2S2 2p4
 
2S2
2p4
Oxygen has two unpaired electrons in the valence shell with which it can form two covalent
bonds. Oxygen has no vacant orbitals in the valence shell. Hence promotion of electron is
energetically unfavourable hence it shows valency two only.
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72 A NEW APPROACH TO INORGANIC CHEMISTRY
In ground state. Sulphur has two unpaired electrons in the valence shell with which it can
form two bonds.
16S – 1S2 2S2 2P6 3S2 3P4 3d0
 
3s2
3p4
3d0
Sulphur has five empty d-orbitals in the valence shell. Hence in excited state promotion of
one electron from 3p to 3d – orbital takes place and sulphur shows tetravalency. Further transfer
of one electron from 3 S to 3d orbital may also take place, due to which sulphur shows a valancy
of six. These two transfers are energetically favourable.
S*
first excited state
  
3p3

3s2
 
3s1
  
3p3
S*
2nd excited state
3d1
3d2
2.11 MOLECULAR ORBITAL THEORY (MO THEORY)
1. A covalent bond is formed by linear combination (addition and subtraction) of atomic
orbitals of combining atoms.
2. Orbitals having same symmetry and similar energy undergo linear combination.
3. Symmetric combination of atomic orbitals give molecular orbitals of lower energy
called bonding molecular orbitals.
4. Antisysmmetric combination of atomic orbitals give molecular orbitals of higher energy
called anti-bonding molecular orbitals.
Antibonding molecular orbital
Energy
Atomic
orbital
Atomic orbital
Bonding
Molecular orbital
5. Electrons enter into molecular orbitals according to aufbau principle, Hund’s rule and
Pauli’s exclusion principle.
6. Molecular orbitals are polycentric
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CHEMICAL BOND 73
Bond order: Number of bonds between two atoms is called bond order and given as
Nb  Na where, N b  No. of electrons in bonding molecular orbital
N a = No. of electrons in anti bonding molecular orbital
2
Magnetic properties: If molecular orbital contains all paired electron then molecule is
diamagneticie repelled by the magnet but if molecular orbital contain any unpaired electron, then
the molecule will be paramagenticie attracted by the magnet.
Comparision of valence bond theory and molecular orbital theory
(i) Valence Bond Theory
Bond order =
V.B.T
+
AO
partial overlap of atomic
orbitals, retain their identity
AO
(ii) Molecular Orbital Theory
Anti Symmetric
–
+
Antibonding molecular
orbital ()
MOT
+
AO
+
+
AO
+
+
symmetric
Bonding molecular orbital ()
Pictures of Molecular orbitals
(i) mixing of two s-orbital:
–
+
+
s (Anti-bonding
MO)
+
+
s(AO)
s(AO)
s
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(Bonding MO)
74 A NEW APPROACH TO INORGANIC CHEMISTRY
(ii) mixing of two p-orbitals
–
Antibonding
MO

p
p
+


 (Bonding
MO)
–

(iii)
+
+
+
–
(MO)
Antibonding
+
p 
p 
+
+
(MO)
Bonding
Molecular orbital energy level diagram of homonuclear diatomic molecules
1. H2
1S*
1S1
H
1S1
1S
H
BO = 2 – 0 = 1
2
H2 = 1s2
H2 is diamagnetic
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CHEMICAL BOND 75
2. H2+
1S*
1S1
H
1S0
1S
H+
BO = 1 – 0 = 1/2
2
H2+ = 1S1
Paramagnetic
3. He2
1S*
1S2
4.
1S2
1S
BO = 2 – 2 = 0
2
He2 = 1S2 1S*2
He2 will not exist.
He2+
1S*
1S1
1S
He
BO = 2 – 1 = 1/2
2
He2+ = 1S21S*1
He2+ is Paramagnetic
He+
5. He2++
1S*
He+
1S1
1S
He+
BO = 2 – 0 = 1
2
He2++ = 1S2
Diamagnetic
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76 A NEW APPROACH TO INORGANIC CHEMISTRY
6. Li2
2S*
2S1
2S1
2S
BO = 4 – 2 = 1
2
Li = 1S2 1S*2 2S2
1S*
2
diamagnetic
1S2
Li
1S2
1S
Li
7. Be2
1S*
2S2
2S2
2S
BO = 4 – 4 = 0
2
Be2 = 1S2 1S*2 2S2 2S*2
1S*
1S2
Be
1S2
Be2 will not exist
1S
Be
Remember:
(i) In boron, Carbon and nitrogen energy of pz is higher than px and py. In all other
molecules energy of pz is lower than px and py.
(ii) Energy of both  MO, px and py are equal. Similarly energy of both * MO, *px and
*py are also equal.
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CHEMICAL BOND 77
8. B2
Pz*
Px*
Py*
2p1
2p1
Pz
Px
Py
2S*
2S2
2S2
BO = 2 – 0 = 1
2
2S
B2 = 1S2 1S*2 2S2 2S*2 px1 = py1
1S*
1S2
Paramagnetic
1S2
1S
B
B
9. C2
Pz*
2p2
Px* Py*
2p2
Px Py
2S*
2S2
BO = 4 – 0 = 2
2
C = 1S2 1S*2 2S2S*px2 = py2
2
2S2
diamagnetic
2S
1S*
1S2
1S2
1S
C
C
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78 A NEW APPROACH TO INORGANIC CHEMISTRY
10. C2+
Pz*
2p2
Px* Py*
2p1
BO = 3 – 0 = 3/2
2
+
C2 = 1S2 1S*2 2S2 2S*2, Px2 = Py1
Paramagnetic
Px Px
2S*
2S2
2S 2
2S
1S*
1S2
C
1S 2
1S
C
11. N2
Pz*
2p3
Px* Py*
Px Py
2S*
2S2
2S2
2S
2p3
BO = 6 – 0 = 3
2
N2 = 1S2, 1S*2, 2S2, 2S*2, Px2 = Py2, Pz2
diamagnetic
1S*
1S2
N
1S2
1S
N
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CHEMICAL BOND 79
12. N 2+
Pz*
Px* Py*
2p3
Px
2p2
Py
2S*
2S2
BO = 5 – 0 = 5/2
2
N + =  2,  *2, 
2S2
2
2S
1S
1S
2S2
,
2S*2
, Px2 = Py2, Pz1
Paramagnetic
1S*
1S2
1S2
1S
N
N+
13. O2
Pz*
2p4
2p4
Px Py
Pz
BO = 6 – 2 = 2
2
O2= , 1S2 1S*2 , 2S2 , 2S*2 , Pz2
Px2 = Py2, *Px1 = p*Py 1
Paramagnetic
2S*
2S2
2S2
2S
1S*
1S2
O
1S2
1S
O
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80 A NEW APPROACH TO INORGANIC CHEMISTRY
14. O+2
Pz*
2p3
2p4
Px Py
Pz
BO = 6 – 1 = 5/2
2
O2* = 1S2, 1S*2, 2S2, 2S*2, Pz2,
2S*
2S2
2S2
2
Px2 = py,  *1
px
2S
Paramagnetic
1S*
1S2
O
1S2
1S
O+
15. O2++
Pz*
*
 px
*
 py
2p3
2p3
Px Py
Pz
2S*
`
2S2
BO = 6 – 0 = 3
2
O2++ = 1S2, 1S*2, 2S2, 2S*2, Pz2,
2S2
2S
Px2 = Py2
diamagnetic
1S*
1S2
O+
1S2
1S
O+
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CHEMICAL BOND 81
16. O2–
Pz*
*
*
 px
py
2p5
2p4
Px Py
Pz
BO = 6 – 3 = 3/2
2
O – = 1S, 1S*, 2S, 2S*2, Pz2
2S*
2
2S2
2S2
2S
Px2 = Py2, *Px2 = *Py1
Paramagnetic
1S*
1S2
O
1S2
1S
O–
17. O2– –
Pz*
*
*
 px
py
2p5
2p5
Px Py
2S*
2S2
2S2
2S
Pz
BO = 6 – 4 = 1
2
2–   
O = 1S , 1S*, 2S, 2S*2, Pz2
Px2 = Py2, *Px2 = *Py2
diamagnetic
1S*
1S2
O–
1S2
1S
O–
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82 A NEW APPROACH TO INORGANIC CHEMISTRY
18. F2
Pz*
*
*
px
py
2p5
2p5
Px Py
Pz
2S*
2S2
BO = 6 – 4 = 1
2
2S2
F2 = 1S2, 1S*2, 2S2, 2S*2, Pz2
* 2
Px2 = Py2, * 2 = 
2S
Px
1S*
1S2
Py
diamagnetic
1S2
1S
F
F
2.10 CO-ORDINATE BOND
Co-ordinate bond is a special case of co-valent bond in which both the electrons for sharing is
donated by only one atom. A co-ordinate bond is formed by the overlap of an empty orbital and a
filled orbital.
It is also known as dative bond or Semi-polar bond.
filled orbital
empty orbital
co-ordinate bond
A co-ordinate bond is formed between two atoms out of which one has eight electrons and
other lacks two electrons.
Ex.1 In NH3 N atom has 8 electrons in the valence shell whereas H+ ion has no electron.
Hence a co-ordinate bond will be formed.
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CHEMICAL BOND 83
H+
H
°°
N
H
N
H
H
H
H
H
O–
S
Ex. 3 SO42–
Ex. 2 SO2
O
O
S
O–
O
O
Co-ordinate bond is denoted by an arrow ()
An atom which donate electron pair for sharing is called donor atom
An atom which accepts electron pair for sharing is called acceptor atom
Properties of co-ordinate bond is like polar covalent bond. This is why it is called semipolar
covalent bond.
Bond is directed towards an atom. Hence it is called dative bond.
2.12 METALLIC BOND
The force which binds metal atoms together in a metallic crystal is called metallic bond.
Ex. Force between iron atoms in an iron rod
Theories of Metallic Bond
There are three theories to explain the formation of metallic bond.
I. Free-electron Theory (Electrostatic Theory)
II. Valence Bond Theory (Resonance Theory)
III. Molecular Orbital Theory (Band Theory)
I. Free Electron Theory
This theory was proposed by Drude in 1900 and refined by Lorentz in 1923. According to this
theory all the metal atoms lose their valence electron and form a sea of electrons in which cations
are doped. i.e Cations, and electrons surround each other.
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84 A NEW APPROACH TO INORGANIC CHEMISTRY
e– Na+ e– Na+ e– Na+ e– Na+ e– Na+
Na+ e– Na+ e– Na+ e– Na+ e– Na+ e–
e– Na+ e– Na+ e– Na+ e– Na+ Na+ e–
The electrostatic force of attraction among ions and electrons hold them together and this
force is known as metallic bond. This theory explains many properties of metals.
Q.13 Transition metals are hard. Why?
Transition metals have incompletely filled d-orbital which undergo d-d overlap. As a result,
transition metals are held together by two forces :
1. lonic bonding due to ion-electron attraction. and
2. covalent bonding due to d-d overlap.
Eg. Fe2+ –18[Ar] 3d6
Fe++ 2e– Fe++ 2e– Fe++ 2e–
2e– Fe++ 2e– Fe++ 2e– Fe++
Fe++ 2e– Fe++ 2e– Fe++ 2e–
Ion-electron attraction
It has four unpaired electrons in 3d orbital which undergo d-d overlap
3d6
d-d overlap
This is why iron or transition metals are hard. They have high melting point and boiling point
and very high enthalpy of fusion.
II. Valence Bond model or resonance model
According to this model metal-metal bond is covalent bond formed by mutual sharing of
electron between atoms.
These bonds are delocalized and resonate among all the neighbouring atoms. Thus a large
number of atoms remain associated together.
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CHEMICAL BOND 85
Example: In a lithium crystal each Li atom is surrounded by 8 Li atoms. And its structure is
resonance hybrid of all the possible structures formed by exchange of bonds
Li —– Li
Li —– Li
(i)
Li
|
Li
Li
|
Li
(ii)
Li
|
–
Li —– Li
Li –—– Li–
|
+
Li
Li
(iii)
(iv)
Li+
Li+
Li
|
Li––—– Li
(v)
Li––—– Li
|
Li
Li+
(vi)
The crystal has three dimensional network of atoms
III. Molecular Orbital model (Band model)
According to this model atomic orbitals of all the metal atoms
having corresponding energy undergo linear combination to give
same number of molecular orbitals.
2p
A metal crystal has very large number of atoms thus a large 2p
number of molecular orbitals are formed. These molecular orbitals
are very close together and looks like continuous energy bands
Example: let crystal of Li containing ‘n’ atoms.
Lithium atom has 1s, 2s and 2p orbitals.1s orbitals of n Li 2s
2s
atom undergo LCAO to give M.O. All are fully filled.
2s orbitals of n Li atoms undergo LCAO to give M.O half of
which is filled
2P orbitals of n Li atoms undergo LCAO to give 3n molecular
1s
1s
orbitals which are completely empty.
Completely filled atomic orbitals are called Valence band.
Incomplete or empty molecular orbitals are called conduction band.
The energy gap between Valence band and conduction band is called forbidden band or brullion
Zones.
This theory explains electrical conductivity, thermal conductivity and metallic lusture.
Explanation of conduction of electricity from M.O. Theory
(i) Conductors: When forbidden band is very small or the valence band and conduction
band are overlapping. The substance is conductor.
Ex.
Grey Tin F.B. = 0, White Tin F.B. = 0 are conductors
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86 A NEW APPROACH TO INORGANIC CHEMISTRY
C.B.
C.B.
C.B.
F.B.
F.B.
V.B.
V.B.
(ii)
Ex.
(iii)
Ex.
V.B.
Conductors
Semi Conductors
Insulators
Semi conductors: When forbidden band is smaller than the thermal energy available
to the electrons then electron can excite from VB to CB and a small current can flow.
These are called semi conductors.
Si (F.B = 63 kJ Mol–1), Ge (F.B = 111 kJ Mol–1) are semi conductors
Insulators: Forbidden band is very large. It is larger than thermal energy available.
Electrons remain confined in valence band. These substance are insulators.
Diamond is an insulator F.B. = 511 kJ mol–1
2.13 HYDROGEN BOND
The idea of H-bond was introduced by Latimer and Rodebush in 1920 :
Electrostatic force of attraction which binds positive pole of hydrogen and negative pole of
electronegative atom is called hydrogen bond.
Ex. (i)
H–
F–
H+
F–
Hydrogen bond
(ii)
H
O
|
H
H
O
|
H
Hydrogen bond
Hydrogen bond is formed when hydrogen atom is linked with highly electronegative atom
such as F, O, N.
Ex. HF,
H2O,
NH3,
ROH,
RCOOH,
RNH2,
R 2 NH etc
form H – bonds
It is denoted by ……………………. (dotted lines)
H—bond is of two types :
1. Intermolecular hydrogen bond and 2. Intramolecular hydrogen bond.
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CHEMICAL BOND 87
(A) Intermolecular H-bond : Hydrogen bond formed between two different molecules
of same or different compounds is called intermolecular H-bond.
Ex. (i) – – H+
F–............... H+F–...............
(ii) H
O ............... H– O............... H+O–
H
H
Different molecule of same compound
H
H
H
(iii)
O ............... H– N–............... H+O ...............
H
H
H
Different molecule of different compound
Consequences of hydrogen bond : Compounds which can form intermolecular hydrogen
bond remain associated with each other and form clusters. Such compounds have high melting
and boiling points and they are highly soluble in water. Such compounds are either liquids or can
be liquefied easily.
Q. 14 H2O is liquid but H2S is gas:
Explanation: H2O molecules are held together by Intermolecular hydrogen bond. Molecules
are associated. H2S molecules are held by weak Vanderwaals force.
------ H – O ------ H – O ------ H – O -----|
|
|
H
H
H
Q. 15 Ice floats on water (or water expands on freezing.)
Explanation: Water when freezes becomes ice. In this state, water molecules occupy fixed
position. Each water molecule is surrounded by four water molecules along the corners of a
tetrahedron.
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88 A NEW APPROACH TO INORGANIC CHEMISTRY
O
H
H
O
H
O
H
H
O
H
H
H
O
H
H
Air is filled in the voids. Hence, volume increases and density decreases. Hence ice floats on
water.
Q. 16 Density of water is maximum at 4°C.
Maximum number of hydrogen bonds are broken upto 4°C. Above 4°C, water molecules
start expanding due to increase in kinetic energy. In this way volume increases and density falls.
Thus, density is maximum at 4°C.
Q.17 NH3 can be liquefied easily but PH3 cannot be?
NH3 can form intermolecular hydrogen bond and remain associated with each other
H
H
|
|
------- H – N ----------- H – N ----------|
|
H
H
PH3 molecules are held together by weak Vanderwalls force. Hence boiling point of NH3 is
high but that of PH3 is low
Q.18 NH3 is highly soluble in water.
NH3 can form intermolecular hydrogen bond with water and dissolves.
H
H
|
|
.....
.....
H—N
H—O
H — N ..... H — O .....
|
|
|
|
H
H
H
H
Q.19 HF can be liquefied easily but HCl can’t be liquefied.
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CHEMICAL BOND 89
HF molecules are held by intermolecular Hydrogen bond and HCl molecules are held by
weak Vanderwall forces. Hence HF can be liquified easily, but HCl cannot be liquified.
..... H — F ..... H — F .....
Q.20 CuSO4·5H2O. When heated loses only four water molecules at a time? Why
Ans. (b) because fifth water molecule is linked with sulphate ion through hydrogen bond.
Q.21 Alcohols and sugars are soluble in water
Ans. Alcohol and sugar can form intermolecular H-bond with water and dissolves.
(B) Intramolecular Hydrogen Bond :
Hydrogen bonds formed between different atoms within the molecule is called intramolecular
H – bond. Intramolecular H-bond form 5 or 6 membered rings.
Ex. 1: In salicylaldehyde H atom of –OH group is bonded with O atom of CHO group.
(i)
O—H
H
(ii)
C=O
H
Salicyladehyde
O
N=O
|
O
H atom of –OH group is linked
with 'O' atom of –NO2group
Ortho nitrophenol
Effect of Intramolecular H-bond
Intramolecular H-bond decreases surface area of molecules. As a result intermolecular force
decreases.
It decreases melting point and boiling point of the compounds. Solubility of compound also
decreases.
Q. 22 Orthonitrophenol is steam volatile but para nitro phenol is not.
Para nitrophenol forms intermolecular hydrogen bond, its boiling point is high. So, it is not
steam volatile.
O
O


........ H — O — O — N = O ........ H — O — O — N = O ........
Para Nitrophenol
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90 A NEW APPROACH TO INORGANIC CHEMISTRY
Orthonitrophenol forms intramolecular H-bond. Its B.P. is low, hence it is steam volatile.
No. of Hydrogen bonds : (i) water (H2O) = 4 (ii) alchohol (ROH) = 3 (iii) ether (R2O) = 2
(iv) Ammonia (NH3) = 4 (v) p-amine (RNH2) = 3 (vi) s-amine(R2NH) = 2 (vii) t-amine(R3N)
=1
H-bond is stronger than vanderwaals force but weaker than covalent bond. Its bond energy
(10 – 40 kJ mol-1) strength of H-bond increases with increasing electronegativity and decreasing
size of atom to which H-atom is bonded.
2.14 MULTI-CENTERED BOND
A covalent bond which binds more than two atoms together is called multicentred bond. It is
formed in electron deficient compounds such as Boranes (Boron hydrides)
e.g. Diborane (B2H6)
B— 1S2 2S2
In ground state
B*
In excited state
2s1
2P1
2p2
one e–from 2S is promoted to 2p orbital
Boron atom has three electron but four orbitals in the valence shell. Two half filled orbitals of
each boron overlaps with half-filled 1s orbital of hydrogen to give normal B-H bond. Half filled
orbital of one Boron and empty orbital of other Boron overlaps with 1s orbital of H-atom to give
two multicentred bond which is known as two-electron three-centred bond (2e-3 c bond)-or banana
bond or hydrogen bridge.
B
B
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CHEMICAL BOND 91
H
H
H
B
H
B
H
H
Hydrogen Bridge
2.15 VANDER WAAL’S BOND
The force of attraction among the non-bonded atoms and molecules is called Vander waal’s bond.
e.g. (i) Component gases of air are held together by week Vander waals forces
(ii) Non-metal molecules. Such as P4, S8, l2, etc. are held by vander waals forces.
Reasons of arising Vander Waal’s bond
(i) Dipole-dipole interaction (or Kessam forces)
Ex. HCl, HBr, HI, H2S
(ii) Ion-dipole interaction
Ex. Cl– ......... H — Cl
(iii) Dipole–induced dipole interaction (or Debye forces)
Ex. H — Cl ......... H — H
(iv) Induced dipole-instantaneous induced dipole interaction (London Dispersion forces)
Ex. inert gases (He, Ne, Ar, Kr, Xe, Rn etc.)
Strength of Van-der Waal’s force increase with increasing size and molecular mass of molecules.
Increasing strength of Vander waals force increases melting and boiling point of substance.
Q.23 O2 is gas but sulphur is solid. Why?
Molecular mass of oxygen is 32u and that of sulphur is 256u. Hence strength of Vander
waales force and melting and boiling point of oxygen is low and that of sulphur is high. This is
why Oxygen is gas but sulphur is solid at room temperature.
Q.24 N2 is gas and P4 is solid.Why?
Molecular mass of N2 is 28 u where as of P4 is 124u. Vander waal’s force are stronger in P4
than in N2. Hence, strength of Vander Waal’s force and melting and boiling point of N2 is low and
P4 is high. This is why Oxygen is gas but sulphur is solid.
Q.25 F2 and Cl2 are gases. Br2 is liquid but I2 is solid.
Molecular mass of F2 and Cl2 are low being 38u and 71u respectively. Thus, Vander waal’s
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92 A NEW APPROACH TO INORGANIC CHEMISTRY
forces are weakest and they are gases. Br2, having a moderate molecular mass of 160u and
moderate melting and boiling point is a liquid. Where as I2 has a molecular mass of 254u. Vander
waal’s force is strong, melting and boiling points are high and thus I2 is solid.
1.
2.
3.
4.
5.
6.
SOME IMPORTANT TERMS
Bond length
7. Isosters
Bond angle
8. Odd electron bond, molecule
Bond energy
9. Singlet Linkage
Bond dissociation energy 10. Resonance
Isomorphous
11. Hybridisation
Iso electronic
12. VSEPR theory
13. Valency
1. Bond Length : Distance between nuclei of two bonded atoms is called bond length.
It is also called inter nuclear distance.
(i) Bond length increases with increasing size of atom
Ex. Bond length of
(a) F2 < Cl2 <Br2 < I2
(b) HF < HCI < H Br < HI
Bond length = d
(c) O–H < S–H
(ii) Bond length increases with decreasing bond order (number of bonds)
Bond length of Single bond
> Double bond > Triple bond
Ex.
(a) C—C
> C == C
> C  C
(b) F—F
> O == O
> N == N
(iii) Bond length increases with increasing p-character of hybrid orbitals
S
<
Sp
<
Sp2
<
Sp3
<
p
0%p
50%p
66.6%p
75%p
100%p
H H
|
|
Ex. H — C — C — H
|
|
H H
H
H
H — C  C — H
C=C
H
H
sp2
sp
sp3 |
H—C—> H—C=> H—C=>
|
3
2
Bond length of Sp-S >
Sp-S
>
Sp-S
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CHEMICAL BOND 93
2. Bond Angle : Angle between two bonds of an atom is called bond angle.
O
In H2O bond angle is 104.5o
Ex.
H 104.5° H
Bond angle is fixed in a given hybridisation
Example
Hybridisation
Sp
Sp2
Sp3
dsp2
Sp3 d
Sp3 d2
Sp3 d3
Bond angle
180o
120o
109o 28'
90o
120o and 90o
90o
72o and 90o
But in a given hybridization, bond angle decreases in three conditions
(i) Bond angle decreases with increasing no. of lone pairs on central atom.
Ex
CH4
NH3
H 2O
Hybridisation
Sp3
Sp3
Sp3
No. of lone pairs
0
1
2
o
o
Bond angle
109 28'
107
104.5°
Because number of lone pairs increases
(ii) Bond angle decreases with increasing size of central atom.
Ex
H 2O
H 2S
3
Hybridisation
SP
Sp3
No. of lone pairs
2
2
o
Bond angle
104.5
90o
Because size of S > O
(iii) Bond angle decreases with increasing electronegativity of terminal atom.
Ex
NH3
NF3
3
Hybridisation
SP
Sp3
No. of lone pairs
1
1
o
Bond angle
107
101o
Because electronegativity of F > H
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94 A NEW APPROACH TO INORGANIC CHEMISTRY
3. Bond Energy : Amount of energy released when one mole of a particular bond is formed
between two atoms is called bond energy.
Bond energy increases with
(i) Increasing bond order and
(ii) Decreasing bond length
e.g. (i) B. E of
F2
< O2
< N2
BO = 1
BO = 2
BO = 3
4. Bond Dissociation Energy: amount of energy required to break one mole of a particular
bond between two atoms is called bond dissociation energy. Magnitude of bond energy
and bond dissociation energy are same.
5. Isomorphous : Molecules or ions having same shape are called isomorphous.
N
|
e.g. (i)
H H H
(ii) MgSO 4 . 7H 2 O
Epsom salt
6.
7.
8.
9.
P
|
H H H
ZnSO 4 . 7H 2 O
White vitreol
Corresponding atoms of isomorphous compounds must have same valency
Isoelectronic : Atoms, molecules or ions having same number of electrons are called
isoelectronic species.
e.g. (i) CO (6 + 8)
NO+ (7 + 8 – 1)
CN– (6 + 7 + 1) = (14)
(ii) N2O (7 × 2 + 8)
CO2 (6 + 8 × 2)
(22)
(iii) CH4 (6 + 1 × 4)
NH3 (7 + 1 × 3)
H2O (8 + 1 × 2) (10)
Isosters : Molecules or ions having same number of atoms and electrons both are
called isosters.
e.g. (i) CO
NO+
CN–
No. of atoms =2
No. of electrons = 14
(ii) N2O
CO2
No. of atoms =3
No. of electrons = 22
Odd electron bond : A bond formed by odd number of electrons are called odd
electron bonds
Ex. H2+
(1e–)
He2+ (3e–)
One electron bond and three electron bonds are of same strength and equal to 1/2
bond.
Odd electron Molecule : Molecules having odd number of electrons are called odd
electron molecule.
Ex. NO
NO2
–
No. of e = 7 + 8 = 15
7 + 16 = 23
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CHEMICAL BOND 95
• Odd electron molecules are unstable and paramagnetic.
• They dimerise. eg.NO forms N 2O2 and NO2 forms N2O4
10. Singlet linkage : A bond formed by only one electron is called singlet linkage. It is
denoted by half arrow.
Cl
Cl
P
Cl
Cl
Cl
11. Resonance: A molecule or ion is said to have resonance if all its properties cannot be
explained by a single structure. Such molecules or ions have many probable structures.
These probable structures are known as resonating structure or contributing structure,
or canonical form. True structure of such molecules or ions lie in between all the
probable structure and known as resonance hybrid.

e.g. (i)
Resonating structure
Resonance Hybrid
(Hypothetical)
(ii) CO2–
3
–
O
O
O C
C
O–
|
–
O  C
–2/3
O
 –2/3 C
O
–2/3
O
O–
O–
O
A molecule or ion is said to have resonance if it has fractional bond order and its bond length
lie in between the bond length of single, double and triple bonds.
Conditions for Resonance Structure:
(i) Atomic arrangement must be same, electronic arrangement may change.
(ii) No. of paired and unpaired electrons remain same
(iii) Like charges should be far apart.
(iv) Unlike charges should be close together.
(v) Negative charge should be on electronegative atom and positive charge on electropositive
atom.
–
O
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96 A NEW APPROACH TO INORGANIC CHEMISTRY
12. Hybridisation: The process of mixing of two or more different orbitals of an atom
to give same number of equivalent orbitals is called hybridization. It is actually
redistribution of energy amoung the orbitals of an atom. The orbitals formed are called
hybrid orbitals.
Example: Be
–1S2
2S2
Ground state
No. of unpaired electrons = 0
Covalency of Be = 0
But, Berylium shows valency
2
Be
Excited state
2S
2p
In excited state promotion of one electron from 2s to 2p orbital takes place. Now it
has two unpaired electron and hence its valency is two.
Two electrons of Berylium are in different orbitals. Hence, two bonds of beryllium
should be different. But the two bonds of beryllium are similar. To explain the equivalent
bonds of Be, the concept of hybridization was introduced.
Be
SP
Hybridised state
SP
In hybridised state one S- and one p- orbital mix together to give two equivalent Sp
hybrid orbitals.
+
2s
2P
SP
SP
linear (180°)
Bond angle = 180o
Structure = Linear. It has 50% S Character
Characteristics of hybridization:
(i) No. of hybrid orbitals are equal to the number of orbitals taking part in hybridization.
(ii) Hybrid orbitals form stronger bonds than the pure orbitals.
(iii) Hybrid orbitals form sigma bonds only.
(iv) Hybrid orbitals are directed in space along fixed direction. Hence, in a given
hybridization geometry of molecule is fixed.
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CHEMICAL BOND 97
Types of Hybridistion : Some important type of hybridizations are
1. Sp
2. Sp2
3. Sp3
4. dsp2
5. Sp3d
6. Sp3d2
7. SP3d3
1. Sp hybridisation: Mixing of one s- and one p-orbital of an atom to give two equivalent
orbitals is called sp-hybridization
+
2Pz
2s
SP
linear (180°)
SP
2. Sp2 : Mixing of one s and two p-orbitals of an atom to give three equivalent orbitals
is called sp2 hybridisation.
SP2
+
SP2
Px
SP2
S
Py
Bond angle = 120o
Structure = Trigonal planar (Triangular) It has 33.33% s-character
3
3. Sp : Mixing of one s and three p-orbitals of an atom to give four equivalent orbitals
is called sp3 hybridisation.
SP3
+
SP3
Px
S
Py
SP3
SP3
Bond angle = 109o28
Structure = tetrahedral
It has 25% s-Character
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98 A NEW APPROACH TO INORGANIC CHEMISTRY
4. ds p2 : Mixing of one d- one S- and two P- orbitals of an atom to give four equivalent
orbitals is called dsp2 hybridisation.
+
+
px
py
dx2 – y2
px + py
S
dsp2
Bond angle = 90o
Structure = Square planar
It is found in transition metals only
5. Sp3d : Mixing of one s, three p- and one d-orbitals of an atom to give five equivalent
orbitals is called, sp3d hybridisation.
+
+
s
Px
Py
Pz

dz2
SP3d
Trigonal bipyramidal
Structure = Trigonal bipyramidal
Bond angle equatorial – 120o, axral = 90o
6. Sp3d2: Mixing of one s three p and two d-orbitals of an atom to give six hybrid orbitals
is called sp3d2 hybridisation.
+
S
+
Px
+
Py
+
Pz

+
dx2 – y2
dz2
Octahedral, sp3d2
Bond angle = 90o
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CHEMICAL BOND 99
7. Sp3d3 : Mixing of one S, three p- and three d- orbitals of an atom to give seven
equivalent orbitals is called sp3d3 hybridisation.
Pz
z
y
x
x
s
Py
Px
dxy
dxz
z
y 
sp3d3 hybridisation
Pentagonal bipyramidal structure
Angle = equatorial (72°), axial = 90°
dyz
2.14 VALENCE SHELL ELECTRON PAIR REPULSION THEORY
VSEPER Theory
This theory was first proposed by sidgwick and powell in 1940 and improved by Gillespie and
Nyholm in 1957.
1. Atomic orbitals present in the valence shell of an atom repel each other, as a result
these orbitals tend to remain at maximum distance of minimum repulsion and mininum
energy.
For a given number of orbitals. Only one orientation of maximum distance is possible
Hence, in a given number of orbitals geometry is fixed.
e.g.
No. of orbitals
Geometry
2
Linear
3
Triangular
4
Tetrahedral
5
Trigonal bipyramidal
6
Octahedral
7
Pentagonal bipyramidal
2. Lone Pair – lone pair repulsion is greater than lone pair – bond pair repulsion which
in turn is greater than bond pair –bond pair repulsion.
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100 A NEW APPROACH TO INORGANIC CHEMISTRY
L.P. – L.P. > L.P. – B.P. > B.P. – B.P.
As a result, presence of lone pair changes shape from structures as well as decreases
the bond angle.
Shape. Arrangement of atoms within the molecule is called shape.
Structure. Arrangement of orbitals within the atom is called structure. Structure and shape
are same when number of lone pair is zero. Shape becomes different from structure when ione
pair is present on central atom.
Table-2
No. of orbitals
Hybridisation
Structure
Bond angle
No. of Lone pairs
Shape
2
sp
Linear
B–A–B
180°
—
—
3
Sp2
Triangular
B
A—B
B
120°
1
Angular
Tetrahedral
109° 28
4
Sp3
A
B
1
B
|
A
B
B
B
Pyramidal
A
B
|
B
B
Angular
A
B
2
B
B
3
linear
B—A—B
5
Sp3d
Trigonal
bipyramidal
B
B
120°/90°
1
K-Shaped
B
B
B
|
A— B
|
B
A
B
B
2
T-Shaped
B
|
A— B
|
B
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CHEMICAL BOND 101
6
Sp3d2
Octahedral
B
|
A
|
B
B
B
90°
3
Linear
B—A—B
1
Square pyramid
B
B
B
B
2
B
|
A
B
Square planar
B
B
7
Sp3d3
Pentagonal
bipyramidal
B
B
B
B
|
A
|
B
72° and 90°
1
B
B
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B
B
A
B
Octahedral
B
|
B
B
A
|
B
B
B
102 A NEW APPROACH TO INORGANIC CHEMISTRY
Determination of shape and structure of simple inorganic molecules or ions:
(a) No. of  bond = No. of atoms –1
(b) No. of lone pairs =
where
vce
2
V = No. of Valence electrons of central atom
C = No. of electrons consumed = No. of terminal atoms × Valency of terminal
atom
e = charge of ion.
(c) No. of Orbitals = No. of sigma bond + No. of lone pairs
= (No. of atoms – 1) +
V–C–e
2
Example
1. SO– –4
O–
|
S
||
O
6  4  2  ( 2)
No. of orbitals = 5  1 
Hybridisation = Sp3
2
= 4
6  8 2
 4 0 4
2
Structure = Tetrahedral
O
O–
Shape = Tetrahedral
2. NH3 : No. of orbitals = 4 – 1 +
= 3+
2
 31 4
2
5 – 3 ×1
2
Hybridisation = Sp3
Structure = Tetrahedral
Shape = Pyramidal
LP = 1
3. CO2
4  2 2
2
=2+0=2
Hybridisation = SP
Structure = linear
Shape = Linear
No. of orbitals = 3  1 
N
H
H
H
O=C=O
4. H2O
No. of orbitals = 3  1  6  2  1
2
Hybridisation = SP3
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O
H
H
CHEMICAL BOND 103
= 2
4
4
2
Structure = Tetrahedral
Lp = 2
Shape = Angular
5. BCl3
3  31
2
=3+0=3
Cl
No. of orbitals = 4  1 

6. BO3–
3 : No. of orbitals = 4 – 1
7. AlCl3 :
Cl
3 3 2 3
2
66
2
=3+0=3
= 3
B
Cl
Hybridisation = sp2
Structure and Shape = Triangular
lone pair = 0
Hybridisation = SP2
O–
Structure = Triangular
Lone pair = 0
Shape = Triangular
O–
3  31
303
2
Hybridisation = SP2
Structure = Triangular
Lp = 0
Shape = Triangular
O–
B
No. of orbitals 4  1 
Cl
Cl
Al
Cl
8. CO
42
=1+1=2
2
Hybridisation = sp, Structure = Linear, Lp = 1,
9. CO2–
3
No. of orbitals = 2  1 
4  2  3  ( 2)
2
462
= 3
2
=3+0=3
No. of orbitals = 4  1 
C=O
Shape = Linear
Hybridisation = Sp2
O–
Structure = Triangular
No. of Lp = 0
Shape = Triangular
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C=O
O–
104 A NEW APPROACH TO INORGANIC CHEMISTRY
10. NCl3
N
5 1 3
No. of orbitals = 4  1 
 3 1 4
2
Hybridisation –SP3
Structure = Tetrahedral
No. of Lp = 1
Shape = Pyramidal
11. NO2–
5  2  2  ( 1)
 2 1 3
No. of orbitals = 3  1 
2
Hybridisation = Sp2
Structure = Triangular
No. of Lone pair = 1
Shape = Angular
12. NO3–
No. of orbitals = 4  1 
N
O
O
O
N
O–
O
5  4 1 1
=4+0=4
2
Hybridisation = Sp3
Structure = Tetrahedral
No. of Lp = 0
Shape = Tetrahedral
14. PH3 : No. of orbitals = 4  1 
Hybridisation = Sp3
Structure = Tetrahedral
No. of Lp = 1
Shape = Pyramidal
Cl
5  3  2  (1)
5  6 1
3
303
2
2
Hybridisation = Sp2
Structure = Triangular
No. of Lp = 0
Shape = Triangular
13. NH4+
No. of Orbitals = 5  1 
Cl
Cl
H+
|
N
H | H
H
5  3 1
 31 4
2
P
H
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H
H
CHEMICAL BOND 105
15. PCl3
No. of orbitals = 4  1 
5  3 1
 3 1 4
2
Hybridisation = Sp3 d
Structure = Tetrahedral
No. of Lp = 1
shape = Pyramidal
16. PCl5
P
Cl
Cl
Cl
No. of orbitals = 6  1 
5  5 1
505
2
Hybridisation = Sp3d
Structure = Trigonal bipyramidal
No. of Lp = 0
shape = Trigonal bipyramidal
17. PO43–
Cl
|
P Cl
Cl |
Cl
Cl
583
5  4  2  (3)
404
= 4
2
2
O–
Hybridisation = Sp3
|
Structure = Tetrahedral
P
No. of Lp = 0, Shape = Tetrahedral
O | O–
O–
18. H2S
No.of orbitals = 5  1 
6  2 1
4
2 224
2
2
Hybridisation = Sp3
Structure = Tetrahedral
No. of Lp = 2
Shape = Angular
19. SO2
No. of orbitals = 3  1 
Hybridisation = Sp2
Structure = Triangular
6  2 2
2
 2   2 1 3
2
2

No. of orbitals = 3  1 

S
H
H
S
O
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O
106 A NEW APPROACH TO INORGANIC CHEMISTRY
Lp = 1
Shape = Angular
20. SO3
No. of orbitals = 4  1 
6  23
303
2
O
S
O
2
Hybridisation = Sp
Structure = Triangular
Lp = 0
Shape = Triangular
21. SO32–
No. of orbitals = 4  1 
Hybridisation = Sp3
Structure = Tetrahedral
No. of Lp = 1
Shape = Pyramidal
22. ClO–
No. of orbitals = 2  1 
Hybridisation = Sp3
Structure = Tetrahedral
No. of Lp = 3
Shape = Linear
23. ClO2–
No. of orbitals = 3  1 
Hybridisation = Sp3
Structure = Tetrahedral
No. of Lp = 2
Shape = Angular
24. ClO3–
No. of orbitals = 4  1 
O
662
6  2  3  (2)
 31 4
= 3
2
2
S
O
O–
O–
7  2  1  (1)
7 2 1
1
1 3  4
2
2
Cl—O–
7  2  2  (1)
7  4 1
2
224
2
2
Cl
O
O–
7  3  2  ( 1)
7  6 1
3
 31 4
2
2
Hybridisation = Sp3
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CHEMICAL BOND 107
Structure = Tetrahedral
No. of Lp = 1
Shape = Pyramidal
25. ClO4–
No. of orbitals = 5  1 
Cl
O
Hybridisation = Sp3
Structure = Tetrahedral
Shape = Tetrahedral
O–
7  4  2  (1)
7 8 1
4
404
2
2
Hybridisation = sp3
Structure = Tetrahedral
No. of Lp = 0
shape = Tetrahedral
26. MnO–4
No. of orbitals = 5  1 
O
O–
Cl
O
7  4  2  (1)
=4+0=4
2
O
O
O–
Mn
O
O
O
2.15 VALENCY
Combining capacity of an element is called valency. It is of three types :
1. Electrovalency
2. Co-valency
3. Co-Ordinate Valency
1. Electrovalency: Valency shown by transfer of electrons. It is equal to
number of electrons lost/gained per atom.
Maximum electro valency = 3
2. Co-valency: Valency shown by sharing of electrons.
Co-valency = No. of electrons shared per atom
Maximum co-valency = 9 [Plutonium]
8-Xenon and Osmium
3. Co-Ordinate Valency: Valency shown by donation of electron pairs
Co-ordinate valency = Number of electron pairs donated or accepted per atom.
Maximum coordinate valency = 6
Variable Valency
Some element show different valency in different compounds. This is called variable valency.
eg.
Fe
2,3
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108 A NEW APPROACH TO INORGANIC CHEMISTRY
Cu
S
Hg
1,2
2,4,6
1,2
Causes of Variable Valency:
1. Inert Pair Effect: In heavier elements of group 13, 14 and 15 of the periodic table
a pair of electrons become inert towards bonding and valency decreases by 2.
Ex.
Group-13
Element Valency
B
3
AI
3
Ga
3,1
In
3,1
Tl
1
Group-14
Element
Valency
C
4
Si
4
Ge
4,2
Sn
4,2
Pb
2
Group-15
Element
Valency
N
3
P
3,5
As
3,5
Sb
3,5
Bi
3
Valency changes by 2 units
2. Expansion of octet : Some elements show variable valency due to expansion of octet
Elements having paired electrons as well as vacant orbitals in the valence shell can
expand its octet and shows variable valency.
Ex. (i) Nitrogen forms NCI3 only where as phosphorus forms PCI3 and PCI5 both.
(ii) Oxygen shows valency 2 only, Sulphur shows valency 2 , 4 , 6
3. Presence of d-orbitals: Elements having incompletely filled d- sub orbit will show
variable valency due to successive use of d-electrons. In this case valency changes by
one unit
Ex.
Fe
2,3
Cu
1,2
CU
2,3
NCERT TEXT BOOK EXERCISES WITH ANSWERS
Q.1. Explain the formation of a chemical bond
Ans. See text page no. .....57
Q.2. Write Lewis dot symbols for atoms of the following elements:
Mg, Na, B, O, N, Br
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CHEMICAL BOND 109

Ans.
12
Mg = 2, 8, 2 = Mg

Na = 2, 8, 1 = Na

5B = 2, 3 =  B 
11




O = 2, 6 = O


7N = 2, 5 = N

8


Br = 2, 8, 18, 7 = Br


35
Q.3. Write Lewis symbols for the following atoms and ions:
S and S2– : Al and Al3+; H and H–
Al3+ ion=
2, 8 = [Al]3+
S



2, 8, 8 =


2–

S2– ions=


Ans. 16S = 2, 8, 6 = S ,


Al = 2, 8, 3 = Al ,

13


H=1= H ,
H– ion=
2= H
Q.4. Draw the Lewis structures for the following molecules and ions:
H2S, SiCl4, BeF2, CO2–
3, HCOOH
Ans.
Cl



Cl


Cl

(iii) BeF2 = F
— Be — F







Cl

H
H

Si

 
 
(ii) SiCl4 =

S

(i) H2S =




1
2



HO



O

O


C= O

(v) HCOOH =

(iv) CO =
H

2–
3


O
||
C
Q.5. Define octet rule. Write its significance and limitations.
Ans. Octet Rule. See text page no. .....57
Significance of octet rule. It helps to explain why different atoms combine with each other
to form compounds.
Limitations: Very few elements obey this rule.
Q.6. Write the favourable factors for the formation of ionic bond.
Ans. See text page No....59
Q.7. Discuss the shapes of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
Ans. See text page No-102-107
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110 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.8. Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in
water is less than that of ammonia. Discuss.



Ans. H N H H O
H
H
In NH3, there is only one lone pair on N-atom to repel the bond pairs whereas in H2O,
there are two lone pairs on O-atom to repel the bond pairs. Hence, the repulsions on bond
pairs are greater in H2O than in NH3 and hence the bond angle is less.
Q.9. How do you express the bond length in terms of bond order?
Ans. Greater the bond order, shorter is the bond length.
Bond length 
Q.10.
Ans.
Q.11.
Ans.
1
Bond order
Ex. Bond length of C—C > C = C > C  C
Define the bond length.
See text page No. .....92
Explain the important aspects of resonance with reference to the CO2–3 ion.
Resonance explains the extra stability and observed bond length and bond angle.
O–
O=C



O

S=O
 
 
 


O

SO
Ans. (i) SO3
 
 
O–
It is expected that CO32– has one C==O and two C—O. But observed bond length lies
between the length of C—O and C== O for all the bonds. It is explained by resonance.
Q.12. H3PO3 can be represented by the structures 1 and 2 shown below. Can these two structures
be taken as the canonical forms of the resonance hybrid of H3PO3? If not, give reason for
the same
H
    
  

H
P O H H O
P O
H O
  

    
O
O

 
H
(1)
(2)
Ans. No, these structures cannot be taken as canoncial forms because the position of atoms
have been changed.
Q.13. Write the resonance structures for SO3, NO2 and NO–3.
O
O
O
SO
O
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CHEMICAL BOND 111
O–
O
N
N
O–
O
O
O
N
N
||
|
O
O
O
O
O
O
O–
Q.14. Use Lewis symbols to show electron transfer between the following atoms to form cations
and anions:
(a) K and S
(b) Ca and O
(c) Al and N.
(iii) NO–3 =
+ O

 
N
2–
or (K+)2S2–

 

Ca2+
O


S



(b)
K+
 
Ca
2, 8, 8, 2

Ans. (a)
K+
K

2, 8, 8, 1 S
K
2, 8, 8, 1 2, 8, 6

(ii) NO2 =
= K 2S
2–
or Ca2+O2–
= CaO
Al3+

N



N

2, 5

Al
2, 8, 3

(c)

2, 6
or Al3+N3– = AlN
Q.15. Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent
while that of CO2 is linear. Explain these on the basis of dipole moment.
Ans. Net dipole moment of CO2 molecule is zero. This is possible only if CO2 is a linear molecule
(O = C = O). On the other hand, H2O molecule is found to have a net dipole moment
O
(1.84D). This shows that it is a bent molecule
H H
Q.16. Write the significance/application of dipole moment.
Ans. See text page No-69 (Determination of structure from dipole moment)
Q.17. Define electronegativity. How does it differ from electron gain enthalpy?
Ans. See unit-1 Periodicity Page No-17
Q.18. Explain with the help of a suitable example polar covalent bond.
Ans. See text page No-67
Q.19. Arrange the following molecules in order of increasing ionic character of their bonds
LiF, K2O, N2, SO2, CIF3
Ans. N2 < SO2 < ClF3 < K2O < LiF
Q.20. The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are
shown incorrectly. Write the correct Lewis structure for acetic acid.
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112 A NEW APPROACH TO INORGANIC CHEMISTRY


H O
H= C C
H
O


H


H O

Acetic acid
Ans. H C C O H

H
Q.21. Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with
four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is
not square planar.
Ans. Carbon undergoes sp3 hybridisation which gives a tetrahedral shape. For squar planar, the
hybridisation required is dsp2 which is not possible for C-atom because carbon atom has
no d-orbital.
Q.22. Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are
polar.
Ans. This is because BeH2 molecule is linear (H–Be–H) so that the two Be–H bond moments
are equal and opposite and hence cancel out.
Q.23. Which out of NH3 and NF3 has higher dipole moment and why?
Ans. NH3 has higher dipole moment than NF3.
H
N
H
Q.24.
Ans.
Q.25.
Ans.
Q.26.
Ans.
H
F
N
F
F
In NH3 the bond moment and lone pair moment work in same direction hence the two
reinforce each other. In NF3 the bond moment and lone pair moment are in opposite direction
hence they oppose each other.
What is meant by hybridization of atomic orbitals? Describe the shapes of sp, sp2 and sp3
hybrid orbitals.
See text pageNo-96-97
Describe the change in hybridization (if any) of the Al atom in the following reaction:
AlCl3 + Cl–  AlCl–4
In AlCl3 Al undergoes sp2 hybridisation to give it triangular planar structure. In AlCl–4, the
hybridization is sp3 and the shape is tetrahedral.
Is there any change in the hybridisation of B and N atoms as a result of the reaction,
BF3 + NH3  F3B.NH3?
In BF3, B is sp2 hybridised and in NH3, N is sp3 hybridised. After the reaction, hybridisation
of B changes to sp3 but that of N remains unchanged.
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CHEMICAL BOND 113
Ans.
Q.28.
Ans.
Q.29.
Ans.
Q.30.


Q.27.
B
F 
N
H
H
H
H
sp2
sp3
sp3
sp3
Draw diagrams showing the formation of a double bond and a triple bond between carbon
atoms in C2H4 and C2H2 molecules.
See text page No-65
What is total number of sigma and pi bonds in the following molecules?
(a) C2H2
(b) C2H4

H    H


(a) H C 
(b)
(3

,
2

)
(5, 1)
=
C
H
C=

H   C H
Considering x-axis as the internuclear axis, which out of the following will not form a
sigma bond and why? (a) 1s and 1s (b) 1s and 2px (c) 2py and 2py (d) 1s and 2s.
Only (c) will not form a -bond because taking x-axis as the intermolecular axis, there
will be lateral (sideway) overlap between the two 2py orbitals forming a  bond.
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH3 – CH3
(b) CH3 – CH = CH2
(c) CH3 – CH2 – OH
(d) CH3 – CHO (e) CH3COOH
Ans. (a)
(b)
(d)
H
H
H
H
H Both C– atoms use sp3 hybrid orbitals.
H
H
H
3 2
H C C = 1C
H
H
H
H C C O

H
H
H
H
H
(e)
C C

(c)
H
F
H

N
B  F+
F
F


F
H
H
, C = sp3, C = sp2, C = sp2
3
2
1
H
, Both C–atoms use sp3 hybrid orbitals.
2 1 H
C C
: C2 = sp3, C1 = sp2
O
O
H
2 1
H C C O H. C2 = sp3, C 1 = sp
H
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114 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one
example of each type.
Ans. The shared pair of electrons present between the bonded atoms are called bond pairs.
The electron pairs present on the atoms which do not take part in bonding are called lone

pair of electrons. For example, In H2O there are two bond pairs and two lone pairs.
H


O
H
Q.32. Distinguish between a sigma bond and a pi bond.
Ans. See text page no-64
Q.33. Explain the formation of H2 molecule on the basis of valence bond theory.
Ans. See text page no-64
Q.34. Write the important conditions required for the linear combination of atomic orbitals to
form molecular orbitals.
Ans. Orbitals must have same symmetry and similar energy.
Q.35. Use molecular orbital theory to explain why the Be2 molecule does not exist.
Ans. E.C. of 4Be = 1s22s2
Molecular orbital electronic configuration of Be2 = 1s2, *21s 2s2 s
1
(4  4) = 0. Hence, Be2 does not exist.
2
Q.36. Compare the relative stability of the following species and indicate their magnetic properties:
 Bond order =
O2, O2+, O2– (superoxide), O22– (peroxide)
Ans. Bond order of O2 = 2, O2+ = 2.5, O2– = 1.5, O2–
2 = 1.0.
+
–
Hence order of stability O2+ > O2 > O2– > O2–
2 , O2, O2 and O2 have unpaired electrons, hence
2–
these are paramagnetic while O2 has no unpaired electron, hence it is diamagnetic. For
details. see page no. 79-81
Q.37. Write the significance of a plus and a minus sign shown in representing the orbitals.
Ans. As orbitals are represented by wave functions, a plus sign in an orbital represents a +ve
wave function and a minus sign represents a –ve wave function.
Q.38. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to
the equatorial bonds?
Ans. Hybridisation of P in PCl5 is sp3d. Axial bonds are longer because it experiences repulsion
of three axial Cl atoms.
Q.39. Define hydrogen bond. Is it weaker or stronger than the vanderwaals forces?
Ans. Hydrogen bond – For definition, see text page No-86
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CHEMICAL BOND 115
Hydrogen bond is stronger than the vanderwaals forces.
Q.40. What is meant by the term bond order? Calculate the bond order of:
N2, O2, O+2 and O–2
Ans. See text page no-78-80
SECTION A: QUESTIONS FROM PREVIOUS IITJEE PAPERS
Type-I Subjective Questions




1. State four major physical properties that can be used to distinguish between covalent and
ionic compounds. Mention the distinguishing features in each case.
(1978, 2M)
2. Write the Lewis dot structual formula for each of the following. Give also, the formula of
a neutral molecule, which has the same geometry and the same arrangement of the bonding
electrons as in each of the following. An example is given below in the case of H3O+ and
NH3.
+
H
H


N
H O
H
H
H


ion
Neutral molecule
(i) O22–
(ii) CO32–
(iii) CN–
(iv) NCS–
(1983, 4M)
3. What effect should the following resonance of vinyl chloride have on its dipole moment?
(1983, 4M)
CH2
CH
Cl
CH–2
CHCl+
4. Arrange the following as stated.
“Increasing strength of hydrogen bonding (X — H — X).”
O, S, F, Cl, N
(1981, 1M)
5. Give reasons in two or three sentences only for the following:
“Hydrogen peroxide acts as an oxidizing as well as a reducing agent”.
(1992, 1M)
–29
6. The dipole moment of KCl is 3.336 × 10 C-m which indicates that it is a highly polar
molecule. The inter atomic distance between K+ and Cl– in this molecule is 2.6 × 10–10 m.
Calculate the dipole moment of KCl molecule if there were opposite charges of one
fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.
(1993, 2M)
7. Explain the difference in the nature of bonding in LiF and LiI.
(1996, 2M)
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116 A NEW APPROACH TO INORGANIC CHEMISTRY
8. Using the VSEPR theory, identify the type of hybridisation and draw the structure of OF2.
What are the oxidation states of O and F?
(1997, 3M)
–
9. In the reaction,  + I2   3, which is the Lewis acid?
(1997, 1M)
+
+
10. Between Na and Ag , which is stronger Lewis acid and why?
(1997, 3M)
+
2+
11. (a) Arrange the following ions in order of their increasing radii: Li , Mg , K+, Al3+.
(1997, M)
(b) Arrange the following sulphates of alkaline earth metals in order of decreasing thermal
stability: BeSO4, MgSO4, CaSO4, SrSO4.
12. Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence
shell electron pair repulsion (VSEPR) theory. (Atomic number: H = 1, P = 15, S = 16, Cl
= 17)
(1998, 4M)
13. (a) Write the MO electron distribution of O2. Specify its bond order and magnetic property.
(2000, 3M)
(b) Draw the molecular structures of XeF2, XeF4 and XeO2F2, indicating the location of
lone pair(s) of electrons.
(2000, 3M)
14. Using VSEPR theory, draw the shape of PCl5 and BrF5.
(2003, 2M)
15. Draw the shape of XeF4 and OsF4 according to VSEPR theory. Show the lone pair of
electrons on the central atom.
(2004, main, 2M)
16. On the basis of ground state electronic configuration arrange the following molecules in
increasing O–O bond length and bond order. KO2, O2, O2
(2004, 2M)
17. Predict whether the following molecules are iso-structural or not. Justify your answer.
(i) NMe3
(ii) N(SiMe3)3
(2005, 2M)
Answers of Subiective Questions
1. Districition between ionic and covalent compounds:
(i) Melting points. Ionic compounds have higher melting points than covalent
compounds.
(ii) Boiling points: Ionic compounds have higher boiling points than covalent compounds.
(iii) Solutbility: Ionic compounds have greater solutbility in water than a covalent
compounds.
(iv) Conductivity in aqueous solution: Ionic compounds have greater electrical
conducitivity in aqueous solution while covalent compounds are usually nonconducting.
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CHEMICAL BOND 117





  
  
C N

–
and

C O (CO)

  
  


 
 
and


F
–
N
  




F   B   F (BF)3
 C
Cl
N (ClCN)

 
(iv) NCS– : S
(Cl)
Cl

2


(iii) CN– : C
and
O




O



Cl

2–


O


C
 
and


(ii) CO32– :
2–

O



2. (i) O2–2 : O

3. Resonance in vinyl chloride decreases dipole moment of the molecule by decreasing the
bond length.
4. X with higher electronegativity and smaller size forms stronger H-bond. Hence, increasing
order of strength of H-bond is
S < Cl < N < O < F
5. In hydrogen peroxide (H2O2), oxygen is in –1 oxidation state. It can be oxidized to O2
(zero oxidation state) or can be reduced to H2O (–2 oxidation state of oxygen). Hence,
H2O2 can act as both oxidising agent and reducing agent.
6. Dipole moment calculated theoretically is
µ = q.d
Here, q = 1.6 × 10–19 C and d = 2.6 × 10–10 m
 µTheo = 1.6 × 10–19 × 2.6 × 10–10 = 4.6 × 10–29 cm
% Ionic characer =
=
µobs
 100
µTheo
3.336  1029
4.16  1029
 100  80.2%
7. LiF is an ionic compound. However, LiI is predominantly covalent because of small size
of Li+ and large size of iodide ion. A smaller cation and a larger anion introduces covalency
in ionic compound.
O
8.
No. of  bond = 2
oxidation state of O = +2
F
F
No. of Ione pairs = 2
V-shaped
No. of orbitals = 4
oxidation state of F = –1
3
Hybridisation = sp
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118 A NEW APPROACH TO INORGANIC CHEMISTRY
9. I2 is Lewis acid because I– coordinate its one lone pair to I2.
10. Ag+ is stronger Lewis acid because it can easily accommodate lone pair of electrons from
Lewis base. On the other hand Na+ has noble gas configuration. It cannot accept lone pair
of electrons.
11. (a) Ionic radii of Li+ < Al3+ < Mg2+ < K+
(b) Thermal stability of BeSO4 < MgSO4 < CaSO4 < SrSO4
12. In H2S, S is sp3 hybridised with two lone pair of electrons giving V-shaped (water like)
shape. In PCl3, P is sp3 hybridised with one lone pair of electrons on it.
Therefore, PCl3 is pyramidal in shape.
*
13. (a) O 2 : 1s 2 , * 1s 2 , 2s 2 ,  2s 2 , 2 p z2 2p2x = 2p2y, px1 = py1
Bond order =
10  6
 2, paramagnetic.
2
F
(b)
Xe
Xe
14.
F
F
Cl
F
F
Br
Cl
Cl
Trigonal bipyramidal
(P is sp3d hybridised)
F
F
Square pyramidal
(Br is sp3d2 hybridised)
F
O
Xe
F
F
F
F
15.
O
F
Seesaw shaped
Square planar
P
Cl
O
Xe
F
F
Linear
Cl
F
F
F
F
Square planar
S
F
F
Trigonal bipyramidal
16. Bond orders: O–2 = 1.5, O2 = 2, O2+ = 2.5
Bond lengths: O2+ < O2 < O2–
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CHEMICAL BOND 119
17. No, (i) NMe3 is pyramidal while (ii) N (SiMe3)3 is planar. In the later case, p – d back
bonding between N and Si makes N sp2 hybridised.
Type II: Objective questions with only one correct option
1. The compound which contains both ionic and covalent bonds is.
(1979, 1M)
(a) CH4
(b) H2
(c) KCN
(d) KCl
2. The total number of electrons that take part in forming the bonds in N2 is
(1980, 1M)
(a) 2
(b) 4
(c) 6
(d) 10
3. Which of the following molecule is covalent?
(1980, 1M)
(a) H2
(b) CaO
(b) KCl
(d) Na2S
4. Element X is strongly electropositive and element Y is strongly electronegative. Both are
univalent. The compound formed would be
(1980, 1M)
+ –
– +
(a) X Y
(b) X Y
(c) X – Y
(d) X  Y
5. If a molecule MX3 has zero dipole moment, the sigma bonding orbitals used by M (atomic
number < 21) are
(1981, 1M)
(a) pure p
(b) sp hybridised
2
(c) sp hybridised
(d) sp3 hybridised
6. Among the following, the linear molecule is
(1982, 1M)
(a) CO2
(b) NO2
(c) SO2
(d) ClO2
7. The ion that is isoelectronic with CO is
(1982, 1M)
–
+
(a) CN
(b) O 2
–
(c) O2
(d) N+2
8. Carbon tetrachloride has no net dipole moment because of
(1983, 1M)
(a) its planar structure
(b) its regular tetrahedral structure
(c) similar sizes of carbon and chlorine atoms
(d) similar electron affinities of carbon and chlorine
9. Which one among the following does not have the hydrogen bond?
(1983, 1M)
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120 A NEW APPROACH TO INORGANIC CHEMISTRY
(a) Phenol
(b) Liquid NH3
(c) Water
(d) HCl
10. On hybridisation of one s and one p-orbital we get
(a) two mutually perpendicular orbitals
(b) two orbitals at 180°
(c) four orbitals directed tetrahedrally
(d) three orbitals in a plane
11. The bond between two identical non-metal atoms has a pair of electrons
(a) unequally shared between the two
(1984, 1M)
(1986, 1M)
(b) transferred fully from one atom to another
(c) with identical spins
(d) equally shared between them
12. The hybridisation of sulphur in sulphur dioxide is
(a) sp
(c) sp2
(1986, 1M)
3
(b) sp
(d) dsp2
13. Of the following compounds, which will have a zero dipole moment?
(1987, 1M)
(a) 1, 1-dichloroethylene
(b) cis-1, 2-dichloroethylene
(c) trans-1, 2-dichloroethylene
(d) None of the above
14. The species in which the central atom uses sp2-hybrid orbitals in its bonding is
1M)
(a) PH3
(b) NH3
(1988,
(c) CH+3
(d) SbH3
15. The Cl—C—Cl angle in 1, 1, 2, 2–tetrachloroethene and tetrachloromethane respectively
will be about
(1989, 1M)
(a) 120° and 109.5°
(b) 90° and 109.5°
(c) 109° and 90°
(d) 109.5° and 120°
16. The molecule which has zero dipole moment is
(1989, 1M)
(a) CH2Cl2
(b) BF3
(c) NF3
(d) ClO2
17. Which of the following is paramagnetic?
(1989, 1M)
(a) O–2
(b) CN–
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CHEMICAL BOND 121
(c) CO
(d) NO+
18. The molecule which has pyramidal shape is
(1989, 1M)
(a) PCl3
(b) SO3
2–
(c) CO3
(d) NO–3
19. The type of hybrid oritals used by the chlorine atom in ClO–2 is
(1992, 1M)
3
2
(a) sp
(b) sp
(c) sp
(d) None of these
20. The maximum possible number of hydrogen bonds a water molecule can form is (1992,
1M)
(a) 2
(b) 4
(c) 3
(d) 1
21. Which one of the following molecules is planar?
(a) NF3
(b) NCl3
(c) PH3
(d) BF3
22. Among the following species, identify the isostructural pairs.
NF3, NO–3, BF3, H3O+, N3H
(a) [NF3, NO–3] and [BF3, H3O+]
(b) [NF3, N3H] and [NO–3, BF3]
(c) [NF3, H3O+] and [NO–3, BF3]
23. The number and type of bonds between two carbon atoms in CaC2 are
(a) one sigma () and one pi ( bonds)
(b) one sigma () and two pi () bonds
(c) one sigma () and one half pi () bonds
(d) one sigma () bond
(1996, 1M)
(1996, 1M)
(1996, 1M)
24. Arrange the following compounds in order of increasing dipole moment, toluene (I), mdichlorobenzene (II), o-dichlorobenzene (III), p-dichlorobenzene (IV)
(1996, 1M)
(a) I < IV < II < III
(b) IV < I < II < III
(c) IV < I < III < II
(d) IV < II < I < III
–
25. The cyanide ion, CN and N2 are isoelectronic, but in contrast to CN–, N2 is chemically
inert, because of
(1997 C, 1M)
(a) low bond energy
(b) absence of bond polarity
(c) unsymmetrical electron distribution
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122 A NEW APPROACH TO INORGANIC CHEMISTRY
(d) presence of more number of electrons in bonding orbitals
26. Among KO2, AlO–2, BaO2 and NO+2, unpaired electron is present in
+
2
(a) NO and BaO2
(c) KO2 only
(b) KO2 and AlO
(d) BaO2 only
27. Which contains both polar and non-polar bonds?
(a) NH4Cl
(b) HCN
(c) H2O2
(1997 C, 1M)
–
2
(1997, 1M)
(d) CH4
28. Which one of the following compounds has sp2-hybridisation?
(a) CO2
(c) N2O
(1997, 1M)
(b) SO2
(d) CO
29. The geometry and the type of hybrid orbital present about the central atom in BF3 is
(1998, 2M)
(a) linear, sp
(b) trigonal planar, sp2
(c) tetrahedral, sp3
(d) pyramidal, sp3
30. The correct order of increasing C—O bond length of CO, CO2–
3, CO2 is
2–
2–
(a) CO 3 < CO2 < CO
(b) CO2 < CO3 < CO
(c) CO < CO2–
3 < CO2
(d) CO < CO2 < CO2–
3
31. The geometry of H2S and its dipole moment are
(a) angular and non-zero
(1999, 2M)
(1999, 2M)
(b) angular and zero
(c) linear and non-zero
(d) linear and zero
32. In compounds of type ECl3, where E = B, P, As or Bi, the angles Cl—E—Cl(1999, 2M)
(a) B > P = As = Bi
(c) B < P = As = Bi
(b) B > P > As > Bi
(d) B < P < As < Bi
33. In the compound CH2 = CH — CH2 — CH2 — C  C H, the hybridisation of C2 – C3
carbon are
(1999, 2M)
2
3
3
(a) sp – sp
(b) sp – sp
(c) sp – sp3
(d) sp2 – sp3
34. The hybridisation of atomic orbitals of nitrogen in NO+2, NO–3 and NH+4 are
3
2
(b)
sp, sp and sp respectively
3
(d)
sp2, sp3 and sp respectively
(a) sp, sp and sp respectively
2
(c) sp , sp and sp respectively
2
(2000, 1M)
3
35. Amongst H2O, H2S, H2Se and H2Te, the one with the highest boiling point is(2000, 1M)
(a) H2O because of hydrogen bonding
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CHEMICAL BOND 123
(b) H2Te because of higher molecular weight
(c) H2S because of hydrogen bonding
(d) H2Se because of lower molecular weight
36. Molecular shape of SF4, CF4 and XeF4 are
(2000, 1M)
(a) the same, with 2, 0 and 1 lone pair of electrons respectively
(b) the same, with 1, 1 and 1 lone pair of electrons respectively
(c) different, with 0, 1, and 2 lone pair of electrons respectively
(d) different, with 1, 0 and 2 lone pair of electrons respectively
37. The common features among the species CN–, CO and NO+ are
(a) bond order three and isoelectronic
(2001, 1M)
(b) bond order three and weak field ligands
(c) bond order two and electron acceptors
(d) isoelectronic and weak field ligands
38. The correct order of hybridisation of the central atom in the following species NH3, [PtCl4]2–,
PCl5 and BCl3 is
(2001, 1M)
(a) dsp2, dsp3, sp2 and sp3
(c) dsp2, sp2, sp3, dsp3
(b) sp3, dsp2, sp3d and sp2
(d) dsp2, sp3, sp2, dsp3
39. Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1
complex of BF3 and NH3
(2002, 3M)
(a) N : tetrahedral, sp3; B: tetrahedral, sp3
(b) N : pyramidal, sp3; B; pyramidal, sp3
(c) N: pyramidal, sp3; B: planar, sp2
(d) N: pyramidal, sp3, B: tetrahedral, sp3
40. The nodal plane in the -bond of ethene is located in
(2002, 3M)
(a) the molecular plane
(b) a plane parallel to the molecular plane
(c) a plane perpendicular to the molecular plane which bisects the carbon-carbon -bond
at right angle.
(d) a plane perpendicular to the molecular plane which contains the carbon-carbon -bond
41. Which of the following molecular species has unpaired electron (s)?
(2002, 3M)
(a) N2
(b) F2
–
(c) O 2
(d) O2–
2
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124 A NEW APPROACH TO INORGANIC CHEMISTRY
42. Which of the following molecular species has highest dipole moment?
(2002, 3M)
(a) CH3Cl
(b) CH2Cl2
(c) CHCl3
(d) CCl4
43. Which of the following are isoelectronic and isostructural?
NO–3, CO2–3, ClO2–3, ClO–3, SO3
(2003, 1M)
–
2–
–
(a) NO3, CO 3,
(b) SO3, NO 3
–
2–
(c) ClO 3, CO 3
(d) CO2–
3, SO3
44. Number of lone pair(s) in XeOF4 is/are
(2004, 1M)
(a) 0
(b) 1
(c) 2
(d) 3
45. According to MO theory,
(2004, 1M)
+
(a) O2 is paramagnetic and bond order greater than O2
(b) O+2 is paramagnetic and bond order less than O2
(c) O+2 is diamagnetic and bond order is less than O2
(d) O+2 is diamagnetic and bond order is more than O2
46. Which of the following contains maximum number of lone pairs on the central atom?
(2005, 1M)
–
(a) ClO 3
(b) XeF4
(c) SF4
(d) I3–
47. Among the following the paramagnetic compound is
(2007, 3M)
(a) Na2O2
(b) O3
(c) N2O
(d) KO2
48. The species having bond order different from that in CO is
(2007, 3M)
–
+
(a) NO
(b) NO
–
(c) CN
(d) N2
49. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic
molecule B2 is
(2010)
(a) 1 and diamagnetic
(b) 0 and diamagnetic
(c) 1 and paramagnetic
(d) 0 and paramagnetic
50. The species having pyramidal shape is
(2010)
(a) SO3
(b) BrF3
2–
(c) SiO 3
(d) SOF2
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CHEMICAL BOND 125
Type-III Objective Questions II [One or more than one correct option]
1. Dipole moment is shown by
(a) 1, 4-dichlorobenzene
(b) cis-1,2-dichloroethene
(c) trans-1,2-dichloroethene
(d) trans-1, 2-dichloro-2-pentene
2. CO2 is isostructural with
(a) HgCl2
(b) C2H2
(c) SnCl2
(d) NO2
3. The linear structure assumed by
(a) SnCl2
(b) CS2
+
(c) NO2
(d) NCO–
(e) SO2
4. Which of the following have identical bond order?
(a) CN–
(b) O2–
(c) NO+
(d) CN+
5. The molecules that will have dipole moment are
(a) 2, 2-dimethyl propane
(b) trans-2-pentene
(c) cis-3-hexene
(d) 2, 2, 3, 3-tetramethyl butane
(1986, 1M)
(1986, 1M)
(1991, 1M)
(1992, 1M)
(1992, 1M)
Type-IV : Assertion and Reason Type Questions
Read the following questions and answer as per the direction given below:
(a) Statement I is true; Statement II is true; Statement II is the correct explanation of Statement.
(b) Statement I is true; Statement II is true; Statement II is not the correct explanation of
Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
1. Statement I: The electronic structure of O3 is
–



O




||
O

+
O
O

||
O structure is not allowed because octet around O cannot be
(1998, 2M)





expanded.

||
Statement II: O
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126 A NEW APPROACH TO INORGANIC CHEMISTRY
2. Statement I: LiCl is predominantly a covalent compound.
Statement II: Electronegativity difference between Li and Cl is too small. (1998, 2M)
Type-V Match the Columns Type Questions
1. Match the reactions in Column-I with nature of the reactions/type of the products in
column-II.
(2007, 6M)
Column-I
(A)
(B)
(C)
(D)
O2–  O2 + O2–
2
+
CrO2–
+
H

4
MnO–4 + NO2– + H+ 
NO–3 + H2SO4 + Fe2+ 
Column-II
(p)
(q)
(r)
(s)
Redox reaction
One of the products has trigonal planar structure
Dimeric bridged tetrahedral metal ion
Disproportionation
2. Match each of the diatomic molecules in Column-I with its property/properties in Column-II.
Column-I
(A)
(B)
(C)
(D)
B2
N2
O2–
O2
Column-II
(p)
(q)
(r)
(s)
(t)
Paramagnetic
Undergoes oxidation
Undergoes reduction
Bond order  2
Mixing of ‘s’ and ‘p’ orbitals
Type-VI Fill in the Blanks
1. The angle between two covalent bonds is maximum in ............ (CH4, H2O, CO2)
(1981; 1M)
2. Pair of molecules which forms strongest intermolecular hydrogen bonds is ............ .(SiH4
and SiF4, acetone and CHCl3, formic acid and acetic acid)
(1981, 1M)
3. There are ............  bonds in a nitrogen molecule.
(1982, 1M)
4. ............ hybrid orbitals of nitrogen atom are involved in the formation of ammonium ion.
(1982, 1M)
+
5. The shape of [CH3] is ............
(1990, 1M)
6. The valence atomic orbitals of C in silver acetylide is ............ hybridized. (1990, 1M)
7. The kind of delocalization involving sigma bond orbitals is called ............ (1994, 1M)
8. The two types of bonds present in B2H6 are covalent and ............
(1994, 1M)
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CHEMICAL BOND 127
9. When N2 goes to N+2, the N—N bond distance ............, and when O2 goes to O+2 the O—
O bond distance ............
(1996, 1M)
+
–
10. Among N2O, SO2, I 3 and I 3, the linear species are ............ and ............ . (1997 C, 1M)
Type-VII : True/False type questions
1. Linear overlap of two atomic p-orbitals leads to a sigma bond.
(1983, 1M)
(1985,
2. All molecules with polar bonds have dipole moment.
1
M)
2
In benzene, carbon uses all the three p-orbitals for hybridisation.
(1987, 1M)
3
sp hybrid orbitals have equal s and p character.
(1987, 1M)
The presence of polar bonds in a polyatomic molecule suggests that the molecule has nonzero dipole-moment.
(1990, 1m)
H2O molecule is linear.
(1993, 1M)
The dipole moment of CH3F is greater than that of CH3Cl.
(1993, 1M)
(1985,
3. SnCl2 is a non-linear molecule.
4.
5.
6.
7.
8.
Type-VIII : Integer Type Questions
1. Based on VSEPR theory, the number of 90° F—Br—F angles in BrF5 is
Answers
Objectives Questions I
1.
8.
15.
22.
29.
36.
43.
50.
1
M)
2
(c)
(b)
(a)
(c)
(b)
(d)
(a)
(d)
2.
9.
16.
23.
30.
37.
44.
(c)
(d)
(b)
(b)
(d)
(a)
(b)
3.
10.
17.
24.
31.
38.
45.
(a)
(b)
(a)
(b)
(a)
(b)
(a)
4.
11.
18.
25.
32.
39.
46.
(a)
(d)
(a)
(b)
(b)
(a)
(d)
5.
12.
19.
26.
33.
40.
47.
(c)
(c)
(a)
(c)
(d)
(a)
(d)
6.
13.
20.
27.
34.
41.
48.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-2\IIIRD PROOF DT. 27/3/12
(a)
(c)
(b)
(c)
(b)
(c)
(a)
7.
14.
21.
28.
35.
42.
49.
(a)
(c)
(d)
(b)
(a)
(b)
(a)
128 A NEW APPROACH TO INORGANIC CHEMISTRY
Objective Questions II
1. (b, d)
2. (a, b)
3. (b, c, d)
4. (a, c)
5. (b, c)
Assertion and Reason
1. (a)
2. (c)
Match the Columns
1. A  p, s; B  r; C  p, q; D  p.
2. A  p, q, r, t; B  q, r, s, t; C  p, q, r, t; D  p, q, r, s, t.
Fill in the Blanks
1.
4.
7.
10.
CO2
sp3
hyperconjugation
N2O, I –3
2. HCOOH and CH3COOH
5. planar
8. three centre two electron bond
3. 2
6. sp
9. increase, decreases,
True/False
1. T
2.
F
3.
T
4.
F
5.
F
6.
F
7.
F
8.
T
Integer Type Questions
1. (0)
HINTS AND SOLUTIONS
Type II: Objective Question I

1. In KCN, the bonding between potassium ion and cyanide ion is ionic while carbon and
nitrogen are covalently bonded
Covalent bonds

–
+
[K] [ C  N]
Ionic bond
2. N2 has triple bond and each covalent bond is associated with two electrons, therefore, six
electrons are involved in forming bonds in N2.
3. H2 is a covalent molecule
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CHEMICAL BOND 129
4. Strongly electropositive, univalent, X will form X+ and strongly electronegative, univalent
Y will form Y–,
X + Y  X+Y–
5. Non-polar MX3, must have triangular planar arrangement, ie, there should be sp 2
hybridisation around M.
6. CO2 is a linear molecule becasue of sp hybridisation around carbon atom.
OCO
7. CO has a total of 14 electrons and CN– also has 14 electrons:
C(6e–) + N(7e–) + e–  CN– (14e–)
8. CCl4 has a regular tetrahedral shape.
Cl
µ
C µ
Cl
Net dipole moment = 0
Cl
Cl


9. HCl does not form hydrogen bond. For formation of hydrogen bond hydrogen atom must
be bonded to most electronegative atom O, N and F.
10. Hybridisation of one ‘s’ and one ‘p’ orbitals gives two sp hybrid orbitals at 180°
s + p  sp
11. Bonds between identical, non-metal, is purely covalent due to same electronegativities of
the bonded atoms. The shared pair of electrons are symmetrically distributed.
12. In SO2, the Lewis-dot structure is:
H
Cl
C C
Cl
||
||
S
O
O
Bond pairs at S = 2 (-bonds) + 1 (lone-pair) = 3 = sp2 hybridised.
13. In trans-1,2-dichlorethylene dipole vectors are directed in opposite direction, cancelling
each other.
Net dipole moment = 0
H
+
3
14. In CH , there are only three electron pairs around carbon atom giving sp2 hybridisation
state.
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130 A NEW APPROACH TO INORGANIC CHEMISTRY
H
+ H
C
H
sp2 hybridised
Cl
Cl
Cl
C  C 120°
15. Cl 
Cl Cl
sp2 hybridised
109°
C
Cl
Cl
sp3 hybridised
16. BF3 has triangular planar arrangement.
F
F
B 120° Net dipole moment = 0
F
sp2
Three identical vectors, acting in outward direction, at equal angles, in a plane, cancel
each other giving zero resultant, hence non-polar.
17. O2– has one upaired odd electron in * MO hence it is paramagnetic.
18. PCl3 has sp3-hybridised phosphorus, with one lone pair. Therefore, molecule has pyramidal
shape

Cl
P
Cl


Cl
19. Bond pairs at Cl = 2 (-bonds) + 2(lone-pairs) = 4
 hybridisation at Cl = sp3
Cl
O
O–
20. A water molecule can form four H-bonds:

H




-----H
H
O ----- H
O
H 
O
H
H
-----H
O----- H
O
H 
Four sites of H-bonding
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CHEMICAL BOND 131












N
N
P
F
F 
F, Cl  Cl , H 
H,
B F
21.
F
Cl
H
F
sp3
sp3
sp3
sp2 (Planar)
F
N
F
22. (i) NF3 : F
(ii) BF3 : F
B
F
Triangular planar
(B—sp2)
F
Pyramidal
(N—sp3)

(iii) NO : O  N
(iv) H3O : H
O
Triangular planar
(N—sp2)
H
Pyramidal
(O—sp3)


–
+
N  N N H
Central nitrogen is sp-hybridised

(v) N3H:
H
+


–
3
+
O
O–


Therefore, NF3, H3O+ and BF3 NO–3 pairs have same shape.
23. The carbide (C2–2) ion has the following bonding pattern:
C  C one sigma and two pi bonds.
24. Resultant Dlipole moment decreases with increasing angle between two dipoles. pdichlorobenzene is non-polar
Cl
Cl
p-dichlorobenzene
The two dipole vectors cancel each other giving zero resultant dipole moment.
o-dichlorobenzene has greater dipole moment than meta isomer.
Cl
µ1
Cl
(o-dichlorobenzene)
dipole vectors are at 60° angle
Cl
µ1 > µ2
µ2
Cl
(m-dichlorobenzene)
dipole vectors are at 120° angle
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132 A NEW APPROACH TO INORGANIC CHEMISTRY
Toluene is less polar than both ortho and para dichloro benzene. Therefore, the overall
order of dipole moments is:
p-dichlorobenzene < toluene < m-dichlorobenzene < o-dichlorobenzene
25. N  N is non polar neutral molecule hence cannot be attacked by any reagent whereas
C  N has charge, hence it is easily attacked by any electrophile.
26. KO2 has O2– ion with one unpaired electron in 2p orbital.
AlO–2 has both oxygen in O2– state, therefore, no unpaired electron is present.
BaO2 has O2–
2 ion with no unpaired electron.
+
NO+2 has [O  N  O] bonding, hence no unpaired electron.
27. H2O2 :
H
OO 
H non-polar polar bond
bond
28. Sulphur in SO2 is sp2 hybridised

S

O
O
EP = 2 (-bonds) + 1(lone pair) = 3
Hybridisation = sp2
Carbon in CO2 is sp hybridised, N in N2O is sp hydridised, carbon in CO is sp hybridised.
F
F  B
29.
F
sp2 hydridisation
(Trigonal planar)
30. Bond-length 
1
Bond order
Bond-orders: CO2 = 2, CO = 3, CO2–3 = 1.33
Therefore, order of bond-length is
CO2–
3 > CO2 > CO
31. H2S has sp3 hybridised sulphur, therefore, angular in shape with non-zero dipole moment:
H
S Non-linear, polar molecule
H
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CHEMICAL BOND 133
32. When E = B, in BCl3, bond angle is 120°. When E = P, As or Bi, in ECl3, hybridisation at
E will be sp3. Also, if central atoms are from same group, bond angle decreases down the
group provided all other things are similar. Hence, the order of bond angles is:
BCl3 > PCl3 > AsCl3 > BiCl3
5
1
2
3
4
6
33. CH2  CH  CH2  CH2  C  CH
Hybridisation at C2 = sp2 and at C3 = sp3.
H
O
+
34. O  N  O ,
sp
–
O  N  O,
sp2
N+
F
sp3
H
H
35. H2O moecules are held together by intermolecular H-bond

H
H
O


H
O

H

O

H
H
H

F
F
F
S

36.
C
F
F
F
Seesaw shape
p = 1
F
F
Xe
Cl
F
F
F

Tetrahedral
p = 0
Square planar
p = 2
All are different with 1, 0 and 2 lone pairs of electrons at central atom.
37. All three have 14 electrons (iso-electronic) with bond order of three.
38. NH3 = sp3, [PtCl4]2– = dsp2, PCl5 = sp3d, BCl3 = sp2
F
H




F
H
39. F  B–
N+ H . Both ‘B’ and ‘N’ have sp3 hybridisation and tetrahedral geometry..
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134 A NEW APPROACH TO INORGANIC CHEMISTRY


H
H
CC
40.
H
H
Pi bond is formed by the p-orbitals whose lobes have minima in the plane of molecule,
hence molecular plane is the nodal plane of pi-bond.
41. O–2 has odd number (17) of electrons, therefore it must contain at least one unpaired electron.

H
C
Cl Cl
C

Cl

C


Cl H
Cl

C


H

43.
Cl


Cl

H
Cl
Cl
42. NO–3 and CO2–3 both have 32 electrons, central atom sp2 hybridised, triangular planar.
H
H
Cl
O
F

F
Xe At central atom (Xe), there is one lone pair.
F
F


43.
44. O+2 (15e–):
*
*
*
1s 2 , 1s 2 , 2 s 2 ,  2s 2 , 2 p z2
Bond-order =
O2(16e–):
2 p 2y  2 p1y
2 px2
*
*
 2 pz0
 2 px0
10  5
 2.5; paramagnetic.
2
*
2 *
2 *
2
1s 1s 2 s  2 s
2
2 px2
2 p 2y  2 p1y
2 pz2 * 2 p1
z
*
 2 pz
10  6
2
2
Hence, (a) is the correct answer.


Bond-order =

46. ClO–3: –O  Cl  O one lone pair at Cl.
O
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

CHEMICAL BOND 135
XeF4 : F  Xe 

F
F


F

F
two lone pairs at Xe.
F
 one lone pair at S.
S
F
F





I3– : I I
I three lone pairs at central iodine.
47. O–2 in KO2 has 17 electrons, species with odd electrons are always paramagnetic.
48. The bond order of CO = 3. NO+ , CN– and N2 are isoselectronic with CO, have the bond
orders 3 But NO– (16e–) has bond order of 2.
49. For molecules lighter than O2, the increasing order of energies of moleculer orbitals is:
 *
*
 2 p y 
 2 p y...  2 p
1s,  1s,  2s,  2s 

2
p
z *
x

 2 p x 
  2 pz
If Hund’s rule is violated in B2, electronic arrangement would be :
*
*
*
*
1s 2 , 1s 2 , 2s 2 ,  2 s 2 ,   2 px2
No unpaired electron=0-diamagnetic.
Bond-order =
Nb  N a
64
1
=
2
2

50.
S
F F O
Pyramidal
S is sp3 hybridised
SO3 is planar (S is sp2 hybridised), BrF3 is T-shaped and SiO32 – is planar (Si is sp2
hybridised).
Type III: Objective Questions II
1. 1, 4-dichlorobenzene is non-polar, individual dipole vectors cancel each other.
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136 A NEW APPROACH TO INORGANIC CHEMISTRY
 Cl
H
CC
Cl
C2H5
µ=0
Non-polar

Polar
H



Cl
H
ClCH2




H
CC
Cl


CC
H
Cl

µ0
Cl

Cl 
Polar
2. CO2, HgCl2, C2H2 are all linear. SnCl2 and NO2 are angular
3. S = C = S
Linear
O=N =O
Linear
Sn
N
Cl O
Cl
O—C  N
Linear
O


Sn
Cl
–

+

O = C = O, Cl – Hg – Cl, H – C  C – H
S
Cl
O
Bent
O
Bent
4. CN –and NO+ are isoelectronic, have the same bond orders of 3.

CH3


(b)
CH3

CH3
C

5. (a)
H3C
C=C
H
H


CH3
CH2CH3
Polar
Symmetric, non-polar


C
C


(d) H3C



H
 
C=C
H
CH3 CH3
CH2H3C

(c)

CH3H2C
CH3
CH3 CH3
Symmetric, non-polar
Polar
Type IV: Assertion and Reason Type Questions
1. Assertion is correct, given structure is one of the resonance structure of ozone.


O–
–O
+
O




O
+
O
O
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CHEMICAL BOND 137
Reason is also correct because oxygen cannot expand its octet. It is also the explanation
for the given structure of ozone.
2. Assertion is correct but reason is wrong. The covalency in LiCl is due to small size
of Li+ ion which brings about large amount of polarization in bond.
Type V: Match the Columns
1. (A) In the reaction : O2–  O2 + O2–
2
– 1/2
oxygen (O
) is simultaneously oxidized (O2) and reduced (O2–
2 ). This is
disproportion reaction.
(B) In acid medium, CrO2–4 is converted into Cr2O2–7 which is a dimeric, bridged
tetrahedral:






O
Cr
O
O

O–

O–
O
Cr

O
(C) MnO–4 + NO–2 + H+  Mn2+ + NO–3
It is a redox reaction and a product NO–3 has Trigonal planar structure.
(D) NO–3 + H2SO4 + Fe2+  Fe3+ + NO
It is a redox reaction
*
*
2. (A) B2 : 1s 2 1s 2  2s 2  2s 2  2 p1x  2 p1y
64
 1 , Paramagnetic
2
Bond is formed by mixing of s and p orbitals.
B2 undergoes both oxidation and reduction as
Bond order =
Heat
B2 + O2 
 B 2O3 (Oxidation)
B2 + 3H2  B 2H6 (Reduction)
*
(B)
*
N 2 : 1s 2 , 1s 2 , 2s 2 ,  2s 2 , 2 Pz2 2 px2  2 p 2y Diamagnetic
10  4
3 2
2
N2 undergoes both oxidation and reduction as
Bond order =
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138 A NEW APPROACH TO INORGANIC CHEMISTRY

N 2  O 2 
 NO
Catalyst
N 2  3H 2 
 NH 3
In N2 , bonds are formed by mixing of s and p orbitals.
*
*
(C) O 2– : 1s 2 , 1s 2 , 2s 2 ,  2s 2 , 2 pz2
2px2 = 2py2, 2*px2 = *2py1
Paramagnetic with bond-order = 1.5. O2– undergoes both oxidation and reduction
and bond involves mixing of s and p-orbitals.
*
*
(D) O 2 : 1s 2 , 1s 2 , 2 s 2 ,  2s 2 , 2 p z2
2px2 = 2p2y, *2px1 = *2py1
Paramagnetic with bond order = 2
O2 undergoes reduction and the bond involves mixing of s and p- orbitals.
Type VI: Fill in the Blanks
1. CO2 180o. ,CH4(109o 28'), H2O (104.5o)
O
 
C
OH and CH3
 
2. H




O
C
OH Both are capable of forming H-bonds.
3. In nitrogen molecule, there are two pi bonds and one sigma bond.

N  N

3
4. sp hybrid orbital is involved in formation of ammonium ion.
 
H
N
H
H
H
5. Triangular planar. Carbon in CH+3 is sp2 hybridised.

H
+
CH

H
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CHEMICAL BOND 139


6. sp hybridized.
Ag C  C Ag
7. Hyperconjugation involves delocalization of -electrons.
8. Three centred 2 electron. (multicentred bond)
H
H
H
H
B

B
H
H
9. Bond order in N2 is 3 while in N+2 is 2.5, hence bond distance increases as N2 goes
to N+2.
Bond order in O2 is 2 while in O+2 it 2.5, hence bond distance decreases as O2 goes
to O+2.
10. N2O and I3– are linear species.
+
I – I – I–, O
N = N = O,
sp
S
O
sp2
I+
I sp3 I
Type VII: True/False


1. Linear overlapping of p-orbitals form sigma bond.
2. The resultant of individual bond dipoles may not be non-zero. Some molecules with
polar bonds are non-polar.
3. Sn in SnCl2 has sp2-hybridised hence it is angular.
4. Carbon in benzene is sp2 hybridised, ie, uses only two of its p- orbitals in hybrisation.
5. In sp3 hybrid, there is 25% s-character and 75% p- character.
6. Presence of polar bond in polyatomic molecule may have zero or non-zero dipole
moment.
O



7. H2O is V-shaped molecule : H
H
sp3 (V-shaped)
8. F is more electronegative than Cl, hence C—F bond is more polar than C—Cl.

Type VIII: Integer Type Questions
F
1.
F
F
Br
F

F
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140 A NEW APPROACH TO INORGANIC CHEMISTRY
Lone pair will push the Br—F bond pairs in upward directions and all Br—F bond
angles will decrease and becomes less that 90°.
SECTION B: AIEEE SECTION
Q.1 In which of the following species the interatomic bond angle is 109o 28’ ?
(a) NH3, (BF4)–1
(b) (NH4)+ , BF3
(c) NH3, BF3
(d) (NH2)–1, BF3
Ans. (a) : Both under go sp3 hybridization. In BF4– bond angles is 109o28’ but in NH3 actual
bond angle is less than 109o28’ because of the repulsion between lone pair and bond pairs due to
which bond angle decreases.
Q.2 In which of the following species is the underlined carbon having sp3 hybridisation?
(a) CH3COOH
(b) CH3CH2 OH
(c) CH3COCH3
(d) CH2 = CH –CH3
O


||

CH
—C
— OH
Ans. (b) : In molecules (a)
3
2

sp

CH—CH3).
O




||
 , (c)  CH 3 —C — CH3


sp 2




 and (d) (CH =
2


the underlined carbon atom has a multiple bond, only (b) CH 3CH 2OH has sp3 hybridization
sp 3
Q.3 Number of sigma bonds in P4O10 is
(a) 6
(b) 7
(c) 17
(d) 16
O

P
O | O
| O |
Ans. (d) structure of P4O10 is O = P
O P=O
O—P O

O
No. of  bonds = 16
No. of  bonds = 4
Q.4 A square planar complex is formed by hybridization of which atomic orbitals?
(a) s, px, py, dyz
(b) s, px,py, dx2 – y2
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CHEMICAL BOND 141
(b) s, px, py, dz2
(d) s, py, pz, dxy
2
Ans. (b) : dsp hybridisation gives square planar structure with s, px,py and dx2–_y2 orbitals.
Q.5 Which of the following are arranged in an increasing order of their bond strengths ?
(a) O2– < O2 < O2+ < O22–
(b) O22– < O2– < O2 < O2+
(c) O2– < O22– < O2 < O2+
(d) O2+ < O2 < O2–< O22–
Ans. (b) : From Molecular orbital configuration
O2  BO = 2
O2+  B.O. 2.5
O2–  B.O. = 1.5
Q22–  B.O. = 1.0
Hence bond order and therefore bond strength of
–
+
O 2–
2  O 2  O2  O2
Q.6 Which one of the following compounds has the smallest bond angle in its molecule ?
(a) SO2
(b) OH2
(c) SH2
(d) NH3.
Ans. (c) Molecule :
SO2
OH2
SH2
NH3
o
o
o
Bond angle :
119.5
104.5
92.5
106.5o
Q.7 The pair of species having identical shapes for molecules of both species is
(a) CF4,SF4
(b) XeF2, CO2
(c) BF3, PCl3
(d) PF5, IF5.
Ans. (b): linear shape.
F — Xe — F
O=C=O
Q.8 The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is
(a) H2S < SiH4 < NH3 < BF3
(b) H2S < NH3 < SiH4 < BF3
(c) NH3 < H2S < SiH4 < BF3
(d) H2S < NH3 < BF3 < SiH4
Ans. (b) : The correct order of bond angle (smallest first) is
H2S < NH3 < SiH4 < BF3
92.6° < 107o < 109o28’ < 120°

92.6°
S
H
N
H H
120°
H
|
109°28 Si
107°
H
H
H H
|
B
H
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142 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.9 The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements
is true for these two species?
(a) Bond length in NO+ is greater than in NO
(b) Bond length in NO is greater than in NO+
(c) Bond length in NO+ is equal to that in NO
(d) Bond length is unpredictable
Ans. (b) Small the bond order, longer will be the bond length. Thus NO+ is having higher bond
order than that of NO so NO+ has shorter bond length.
Q.10 Which one of the following has the regular tetrahedral structure?
(a) XeF4
(b) SF4
–
(c) BF4
(d) [Ni(CN)4]2–
[Atomic nos.: B = 5, S =16, Ni = 28, Xe = 54]
see-saw shaped
F
Xe
F
(a)
(b)
–
F
|
F
|
square planar
Ans. (c)
F
|
F
F
|
S—F
|
F
F
|
F
F
regular tetrahedral
NC
CN
NC
CN
2–
square planar
(c)
(d)
o
Q.11 The maximum number of 90 angles between bond pair-bond pair of electrons is
observed in
(a) dsp3 hybridisation
(b) sp3d hybridisation
(c) dsp2 hybridisation
(d) sp3d2 hybridisation
Ans. (d)

dsp3 hybridisation
(six 90° angles between
bond pair and bond pair)
(c)

dsp2 hybridisation
(four 90° angles between
bond pair and bond pair)
(d)
sp3d 2 hybridisation
(twelve 90° angles between bond
pair and bond pair)
Q.12 Which one of the following species is diamagnetic in nature?
(a) He2+
(b) H2
+
(c) H2
(d) H2–
Ans. (b) :
He2+  (1s)2 *(1s)1, one unpaired electron
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CHEMICAL BOND 143
H2  (1s)2 *(1s)0, no unpaired electron
H2+  (1s)1 *(1s)o, one unpaired electron
H2–  (1s)2 *(1s)1, one unpaired electron
Due to absence of unpaired electrons, H2 will be diamagnetic.
Q.13 Of the following sets which one does NOT contain isoelectronic species?
(a) PO43– , SO42– , ClO4–
(b) CN– , N2 , C22–
(c) SO32– , CO32– , NO3–
(d) BO33–, CO32– , NO3–
Ans. (c): Number of electrons in SO32– = 16 + 8 × 3 + 2 = 42
Number of electrons in CO32– = 6 + 8 × 3 + 2 = 32
and
NO3– = 7 + 8 × 3 + 1 = 32
These are not isoelectronic species
Q.14 Which of the following molecules/ions does not contain unpaired electrons?
(a) O22–
(b) B2
+
(c) N2
(d) O2
Ans. (a) See M.O. diagram page No–76-81
(a) The molecular orbital configuration of O22– contains no unpaired electrons
(b) The molecular orbital diagram of B2 molecule. contains 2 unpaired electrons.
(c) The molecular orbital diagram of N2+ contains one unpaired electron.
(d) The molecular orbital diagram of O2 molecule. contains 2 upaired electrons.
Q.15 Among the following mixtures, dipole-dipole as the major interaction, is present in
(a) benzene and ethanol
(b) acetonitrile and acetone
(c) KCl and water
(d) benzene and carbon tetrachloride
Ans. (b): Dipole-dipole interactions occur among the polar molecules.
Ex. CH3–CN and
CH3COCH3
acetone
acetonitrite
CH3
+
–
+ –
CH 3  C  N
C=O
CH3
Q.16 In which of the following molecules/ions are all the bonds not equal?
(a) SF4
(b) SiF4
(c) XeF4
(d) BF4–
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144 A NEW APPROACH TO INORGANIC CHEMISTRY
F
F
S
Ans. (a)
F
F
SF4 molecule has sp3d hybridisation but its expected Trigonal bipyramidal geometry
gets distorted due to presence of a lone pair of electrons and it becomes distored
tetrahedral or see-saw with the bond angles equal to 89o and 177o instead of the
expected angles of 90o and 180o respectively. Equatorial and axial bonds are not equal
SiF4 : sp3 hybridisation and tetrahedral geometry. Bond angle 109°28'. All bonds are
equal.
F
109.5°
F
F
F
XeF4 : sp3d2 hybridisation, square planar shape due to presence of two lone pair of
electron on Xe atom. Bond angle 90o. All bonds are equal
F
F
Xe
F
F
F
–
3
BF4 : sp hybridisation and tetrahedral geometry.
F
B
F
F
Bond angle 109o28'. All bonds are equal
Q.17 Which of the following species exhibits the diamagnetic behaviour?
(a) NO
(b) O22–
(c) O2+
(d) O2
Ans. (b) : From Molecular orbital configuration
NO  paramagnetic (n = 1)
O2  paramagnetic (n = 2)
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CHEMICAL BOND 145
O22–  diamagnetic (n = 0)
O2+  paramagnetic (n = 1)
See MO diagram page no.78-80
Q.18 The charge/size ratio of a cation determines its polarizing power. Which one of the
following sequences represents the decreasing trend of polarising power of cationic
species, K+, Ca2+, Mg2+, Be2+?
(a) Ca2+ < Mg2+ < Be+ < K+ (b) Mg2+ < Be+ < K+ < Ca2+
(c) Be2+ < K+ < Ca2+ < Mg2+ (d) K+ < Ca2+ < Mg2+ < Be2+
Ans. (d) : high charge and small size of the cations increases polarization.
As the size of the given cation decreases as
K+ > Ca2+ > Mg2+ > Be2+
Hence, polarizing power decrease as
K+ < Ca2+ < Mg2+ < Be2+
Q.19 In which of the following ionization processes, the bond order has increased and the
magnetic behavior has changed?
(a) N2  N2+
(b) C2  C2+
(b) NO  NO+
(d) O2  O2+
Ans. (a) N2  1s2 *1s2 2s2 *2s2 2px2 = 2py2 2pz2

Diamagnetic
10  4
3
2
N+2  1s2 *1s2 2s2 *2s2 2p2x = 2py2 2pz1

Paramagnetic
Bond order =
94
 2.5
2
(b) C2  1s2 *1s2 2s2 *2s2 2px2 = 2p2y

diamagnetic
Bond order =
84
2
2
C2+  1s2 *1s22s2 *2s2 2px2 = 2py1

Paramagnetic
Bond order =
Bond order =
74
 1.5
2
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146 A NEW APPROACH TO INORGANIC CHEMISTRY
(c) NO  1s2 *1s22s2*2s2 2p2z 2px2 = 2p2y *2px1

Paramagnetic
10  5
 2.5
2
NO+  1s2 *1s22s2*2s22p2z 2p2x = 2p2y

Diamagnetic
Bond order =
10  4
3
2
(d) molecular orbital configuration of
O2  1s2 *1s2 2s2 *2s2 2p2z 2p2x = 2py2 *2px1 = *2py1
 Paramagnetic
Bond order =
10  6
2
2
O2+  1s2 *1s2 2s2 *2s2 2pz2 2px2 2p2y *2p1x

Paramagnetic
Bond order =
10  5
 2.5
2
Q.20 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
Ans. (c) : Because of highest electronegativity of F, hydrogen bonding in F — H --- F is
strongest.
Q.21 Which one of the following pairs of species have the same bond order?
(a) NO and CN–
(b) CN– and NO+
(c) CN– and NO–
(d) O2– and CN–
Ans. (b) : In the given pair of species
NO+ and CN– are isoelectronic in nature and have same bond order.
Q.22 Which one of the following constitutes a group of the isoelectronic species?
–
(a) N2, O–2, NO+, CO
(b) C2–
2 , O2 , CO, NO
–
(c) NO+, C2–
(d) CN–, N2, O22–, C2–
2 , CN , N2
2
Ans. (c) : Number of electrons in each species are given below
N2 = 14
CN– = 14
Bond order =
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CHEMICAL BOND 147
O2– = 17
C22– = 14
NO+ = 14
O22– = 18
CO = 14
NO = 15
+
2–
–
Thus NO , C 2, CN , N2 and CO are isoelectronic in nature. Hence option (c) is correct.
o————o
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