Chemical bonding by_rohit_raj_ranjan

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CHEMICAL BONDING

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INTRODUCTION
A molecule is formed if it is more stable and has lower energy than the individual atoms. Normally only electrons
in the outermost shell of an atom are involved in bond formation and in this process each atom attains a stable
electronic configuration of inert gas. Atoms may attain stable electronic configuration in three different ways by
loosing or gaining electrons by sharing electrons. The attractive forces which hold various constituents (atoms,
ions etc) together in different chemical species are called chemical bonds. Elements may be divided into three
classes.
 Electropositive elements, whose atoms give up one or more electrons easily, they have low ionization
potentials.
 Electronegative elements, which can gain electrons. They have higher value of electronegativity.
 Elements which have little tendency to loose or gain electrons.
Three different types of bond may be formed depending on electropositive or electronegative character of atoms
involved.
Electropositive element + Electronegative element = Ionic bond (electrovalent bond)
Electronegative element + Electronegative element = Covalent bond
or less electro positive + Electronegative element = Covalent bond
Electropositive + Electropositive element = Metallic bond.
ELECTROVALENCY
This type of valency involves transfer of electrons from one atom to another, whereby each atom may attain octet
in their outermost shell. The resulting ions that are formed by gain or loss of electrons are held together by
electrostatic force of attraction due to opposite nature of their charges. The reaction between potassium and
chlorine to form potassium chloride is an example of this type of valency.
K Cl
x
x
x
x
x
x
x K Cl
x
x
x
x
x
x
x or K Cl
 
Here potassium has one electron excess of it‘s octet and chlorine has one deficit of octet. So potassium donates
it‘s electron to chlorine forming an ionic bond.
Ca++
O2–
(Ionic bond)
x
x
Ca
O
Here the oxygen accepts two electrons from calcium atom. It may be noted that ionic bond is not a true bond as
there is no proper overlap of orbitals.
Criteria for Ionic Bond:
One of the species must have electrons in excess of octet while the other should be deficit of octet. Does this
mean that all substance having surplus electron and species having deficient electron would form ionic bond? The
answer is obviously no. Now you should ask why? The reasoning is that in an ionic bond one of the species is
cation and the other is anion. To form a cation from a neutral atom energy must be supplied to remove the
electron and that energy is called ionization energy. Now it is obvious that lower the ionization energy of the
element the easier it is to remove the electron. To form the anion, an electron adds up to a neutral atom and in
this process energy is released. This process is called electron affinity.
So for an ionic bond one of the species must have low ionization energy and the other should have high electron
affinity. Low ionization energy is mainly exhibited by the alkali and alkaline earth metals and high electron affinity
by the halogen and chalcogens. Therefore this group of elements are predominant in the field of ionic bonding.
Energy Change During the Formation of Ionic Bond
The formation of ionic bond can be consider to proceed in three steps
(a) Formation of gaseous cations
+ -
A(g)+I.E. A (g)+ e

The energy required for this step is called ionization energy (I.E)
(b) Formation of gaseous anions

3
- -
X(g)+ e X (g)+E.A

The energy released from this step is called electron affinity (E.A.)
(c) Packing of ions of opposite charges to form ionic solid
+ -
A (g)+ X (g) AX(s)+ energy

The energy released in this step is called lattice energy.
Now for stable ionic bonding the total energy released should be more than the energy required.
From the above discussion we can develop the factors which favour formation of ionic bond and also determine
its strength. These factors have been discussed below :
(a) Ionization energy: In the formation of ionic bond a metal atom loses electron to form cation. This process
required energy equal to the ionization energy. Lesser the value of ionization energy, greater is the tendency
of the atom to form cation. For example, alkali metals form cations quite easily because of the low values of
ionization energies.
(b) Electron affinity: Electron affinity is the energy released when gaseous atom accepts electron to form a
negative ion. Thus, the value of electron affinity gives the tendency of an atom to form anion. Now greater the
value of electron affinity more is the tendency of an atom to form anion. For example, halogens having
highest electron affinities within their respective periods to form ionic compounds with metals very easily.
(c) Lattice energy: Once the gaseous ions are formed, the ions of opposite charges come close together and
pack up three dimensionally in a definite geometric pattern to form ionic crystal.
Since the packing of ions of opposite charges takes place as a result of attractive force between them, the
process is accompanied with the release of energy referred to as lattice energy. Lattice energy may be
defined as the amount of energy released when one mole of ionic solid is formed by the close packing of
gaseous ion.
In short, the conditions for the stable ionic bonding are:
(a) I.E. of cation forming atom should be low:
(b) E.A. of anion – forming atom should be high;
(c) Lattice energy should be high.
Born Haber Cycle
Determination of lattice energy
The direct calculation of lattice enthalpy is quite difficult because the required data is often not available.
Therefore lattice enthalpy is determined indirectly by the use of the Born – Haber cycle. The cycle uses ionization
enthalpies, electron gain enthalpies and other data for the calculation of lattice enthalpies. The procedure is
based on the Hess‘s law, which states that the enthalpy of a reaction is the same, whether it takes place in a
single step or in more than one step. In order to understand it let us consider the energy changes during the
formation of sodium chloride from metallic sodium and chlorine gas. The net energy change during the process is
represented by Hf.
2 f
1
Na(s) Cl (g) NaCl(s) ;Energychange ( H )
2
  

S
1
D
2
Na(g) Cl(g)
IE
EA
Na (g)

Cl (g)

e

2
S Sublimation of sodium
D Dissociation energyof Cl
IE IE of sodium(IE)
EA EA of chlorine(EA)
U Lattice energy of NaCl





U
Example1. Calculate the lattice enthalpy of 2
MgBr . Given that
Enthalpy of formation of 2
MgBr = -524 kJ -1
mol
Some of first & second ionization enthalpy (IE1 + IE2 ) = 148 kJ mol1
Sublimation energy of Mg = +2187 kJ -1
mol

Vaporization energy of 2
Br (I) = +31kJ -1
mol
Dissociation energy of 2
Br (g) = +193kJ -1
mol
Electron gain enthalpy of Br(g) = -331 kJ -1
mol
Solution: f vap
H S I.E H D 2 E.A. U
        
or  
f vap
U H S I.E H D 2 E.A.
        
or U = 524  [2187  148 + 31 + 193 + 2  (331)]
= 524 1897
  = 1
2421kJ mol

Characteristics of ionic compounds:
The following are some of the general properties shown by these compounds
(i) Crystalline nature: These compounds are usually crystalline in nature with constituent units as ions. Force of
attraction between the ions is non-directional and extends in all directions. Each ion is surrounded by a number of
oppositely charged ions and this number is called co-ordination number. Hence they form three dimensional solid
aggregates. Since electrostatic forces of attraction act in all directions, therefore, the ionic compounds do not
posses directional characteristic and hence do not show stereoisomerism.
(ii) Due to strong electrostatic attraction between these ions, the ionic compounds have high melting and boiling
points.
(iii) In solid state the ions are strongly attracted and hence are not free to move. Therefore, in solid state, ionic
compounds do not conduct electricity. However, in fused state or in aqueous solution, the ions are free to move
and hence conduct electricity.
(iv) Solubility: Ionic compounds are fairly soluble in polar solvents and insoluble in non-polar solvents. This is
because the polar solvents have high values of dielectric constant which defined as the capacity of the solvent to
weaken the force of attraction between the electrical charges immersed in that solvent. This is why water, having
high value of dielectric constant, is one of the best solvents.
The solubility in polar solvents like water can also be explained by the dipole nature of water where the oxygen of
water is the negative and hydrogen being positive, water molecules pull the ions of the ionic compound from the
crystal lattice. These ions are then surrounded by water dipoles with the oppositely charged ends directed
towards them. These solvated ions lead an independent existence and are thus dissolved in water. The
electrovalent compound dissolves in the solvent if the value of the salvation energy is higher than the lattice
energy of that compounds.
AB Lattice energy A B
 
  
These ions are surrounded by solvent molecules. This process is exothermic and is called solvation.
   x
A x solv. A solv. energy


 
  
 
   y
B y solv. B solv. energy

  
  
 
The value of solvation energy depend on the relative size of the ions. Smaller the ions more is the solvation. The
non-polar solvents do not solvate ions and thus do not release energy due to which they do not dissolve ionic
compounds.
(v) Ionic reactions: Ionic compound furnish ions in solutions. Chemical reactions are due to the presence of these
ions. For example
2
2 4 4
Na SO 2Na SO
 

 
2
2
BaCl Ba 2Cl
 

 
COVALENCY
This type of valency involves sharing of electrons between the concerned atoms to attain the octet configuration
with the sharing pair being contributed by both species equally. The atoms are then held by this common pair of
electrons acting as a bond, known as covalent bond. If two atoms share more than one pair then multiple bonds
are formed. Some examples of covalent bonds are

5
Cl Cl – Cl N  N
 


 
Cl
 


 
N
x
x
x
x
x
x
Nx
x
x
x
Sigma and Pi Bonding: When two hydrogen atoms form a bond, their atomic orbitals overlap to produce a
greater density of electron cloud along the line connecting the two nuclei. In the simplified representations of the
formation of H2O and NH3 molecules, the O—H and N—H bonds are also formed in a similar manner, the bonding
electron cloud having its maximum density on the lines connecting the two nuclei. Such bonds are called sigma
bonds (-bond).
A covalent bond established between two atoms having the maximum density of the electron cloud along the line
connecting the centre of the bonded atoms is called a -bond. A -bond is thus said to possess a cylindrical
symmetry along the internuclear axis.
Let us now consider the combination of two nitrogen atoms. Of the three singly occupied p-orbitals in each, only
one p-orbital from each nitrogen (say, the px may undergo ―head –on‖ overlap to form a -bond. The other two p-
orbitals on each can no longer enter into a direct overlap. But each p-orbital may undergo lateral overlap with the
corresponding p-orbital on the neighbour atom. Thus we have two additional overlaps, one by the two py orbitals,
and the other by the two pz orbitals. These overlaps are different from the type of overlap in a -bond. For each
set of p-orbitals, the overlap results in accumulation of charge cloud on two sides of the internuclear axis. The
bonding electron cloud does no more posses an axial symmetry as with the -bond; instead, it possess a plane of
symmetry. For the overlap of the pz atomic orbital, the xy plane provides this plane of symmetry; for the overlap of
the py atomic orbitals, the zx plane serves the purpose. Bonds arising out of such orientation of the bonding
electron cloud are designated as -bonds. The bond formed by lateral overlap of two atomic orbitals having
maximum overlapping on both sides of the line connecting the centres of the atoms is called a -bond. A -bond
possess a plane of symmetry, often referred to as the nodal plane.
-Bond
(a) s-s
overlapping
s s
(b) s-p
overlapping +
s p
(c) p-p
overlapping
+
p p
 - Bond: This type of bond is formed by the sidewise or lateral overlapping of two half filled atomic orbitals.
p p
CO-ORDINATE COVALENCY
A covalent bond results from the sharing of pair of electrons between two atoms where each atom contributes one
electron to the bond. It is also possible to have an electron pair bond where both electrons originate from one
atom and none from the other. Such bonds are called coordinate bond or dative bonds. Since in coordinate
bonds two electrons are shared by two atoms, they differ from normal covalent-bond only in the way they are
formed and once formed they are identical to normal covalent –bond.
It is represented as []
Atom/ion/molecule donating electron pair is called Donor or Lewis base. Atom / ion / molecule accepting electron
pair is called Acceptor or Lewis acid [] points donor to acceptor
NH4
+
, NH3 has three (N – H) bond & one lone pair on N – atom. In NH4
+
formation this lone pair is donated to H
+
(having no electron)

NH3 + H
+
 NH4
+
Lewis base Lewis acid
N
H
H
H
H
+
N
H
H
H
or H
O S
O
O
SO3
P
Cl
Cl
Cl
Cl
Cl Cl
PCl 6
Sb
F
F
F
F
F F
SbF6
Properties of the coordinate compounds are intermediates of ionic and covalent compounds.
Comparison of ionic, covalent & coordinate compounds
Property Ionic Covalent Coordinate
1. binding force Between ions
strong
(coulombic)
Between molecules
smaller (Vander Waal’s)
in between
2. mp/bp High less than ionic in between
3. condition conductor of
electricity in
fused state &
in aqueous
solution
bad conductor Greater than
covalent
4. solubility in polar
solvent (H2O)
High Less in between
5. Solubility in non
polar solvent
(ether)
Low High in between
6. Physical state generally solid liquid & gaseous solid, liquid
gas
Example 2. Which of the following statement is/are not true for -bond.
1. It is formed by the overlapping of s  s or s  p orbitals
2. It is weaker than pi bond
3. It is formed when bond exists already.
4. It is resulted from partial overlapping of orbitals.
(A) 1, 2, 3, 4 (B) 2, 3 and 4
(C) 2 and 4 (D) 1, 2 and 4 Solution: (B)
Example 3. The types of bond present in ZnSO4.7H2O are only
(A) Electrovalent and covalent
(B) Electrovalent and co-ordinate

7
(C) Electrovalent, Covalent and coordinate
(D) None of these Solution: (C)
Example 4. Classify the following bonds as ionic, polar covalent or covalent and give your answer
(a) Si Si bond in Cl3 SiSiCl3 (b) SiCl bond in Cl3SiSiCl3
(c) CaF bond in CaF2 (d) NH bond in NH3
Solution: (a) Covalent due to identical electronegativity
(b) One electron pair is shared between Si & Cl and thus, covalent bond is expected but electron
negativity of Cl is greater than that of Si & some polarity develops giving polar – covalent
nature
(c) Ionic since Ca completes its octet by transfer of two outershell electrons thus, completing
their octets
Ca [Ar]4s
2
, F[He)2s
2
2p
5
(d) Polar covalent, explanation as in (b)
Example 5. Arrange the bonds in order of increasing ionic character in the molecules:
LiF, K2O, N2, SO2 and ClF3
Solution: N2 < ClF3 < SO2 < K2O < LiF
Example 6. Arrange the following in order of increasing ionic character:
CH, FH, BrH, NaI, KF and LiCl
Solution: CH < BrH < FH < NaI < LiCl < KF
HYBRIDIZATION
The tetravalency shown by carbon is actually due to excited state of carbon which is responsible for carbon
bonding capacity.
 
C excited state
1s 2s x
2P y
2P z
2P
If the bond formed is by overlapping then all the bonds will not be equivalent so a new concept known as
hybridization is introduced which can explain the equivalent character of bonds.
s and p orbital belonging to the same atom having slightly different energies mix together to produce same
number of new set of orbital called as hybrid orbital and the phenomenon is called as hybridization.
Important characteristics of hybridization
(i) The number of hybridized orbital is equal to number of orbitals that get hybridized.
(ii) The hybrid orbitals are always equivalent in energy and shape.
(iii) The hybrid orbitals form more stable bond than the pure atom orbital.
(iv) The hybrid orbitals are directed in space in same preferred direction to have some stable arrangement and
giving suitable geometry to the molecule.
Depending upon the different combination of s and p orbitals, these types of hybridization are known.
(i) sp
3
hybridization: In this case, one s and three p orbitals hybridise to form four sp
3
hybrid orbitals. These
four sp
3
hybrid orbitals are oriented in a tetrahedral arrangement.
For example in methane CH4
1090
28'
2s
2px
2py
2pz
(ii) sp
2
hybridization: In this case one s and two p orbitals mix together to form three sp
2
hybrid orbitals and are
oriented in a trigonal planar geometry.

1200
2s
2px 2py
The remaining p orbital if required form side ways overlapping with the other unhybridized p orbital of other C
atom and leads to formation of  bond as in H2C = CH2
(iii) sp hybridization: In this case, one s and one p orbital mix together to form two sp hybrid orbitals and are
oriented in a linear shape.
2s 2ps
1800
The remaining two unhybridised p orbitals overlap with another unhybridised p orbital leading to the formation
of triple bond as in HC  CH.
Shape Hybridisation
Linear sp
Trigonal planar sp
2
Tetrahedral sp
3
Trigonal bipyramidal sp
3
d
Octahedral sp
3
d
2
Pentagonal bipyramidal sp
3
d
3
Example 7. Which of the following molecule has trigonal planer geometry?
(A) CO2 (B) PCl5
(C) BF3 (D) H2O
Solution. BF3 has trigonal planer geometry (sp
2
- hybridized Boron).
Hence (A) is correct.
Rule for determination of total number of hybrid orbitals
 Detect the central atom along with the peripheral atoms.
 Count the valence electrons of the central atom and the peripheral atoms.
 Divide the above value by 8. Then the quotient gives the number of  bonds and the remainder gives the
non-bonded electrons. So number of lone pair
=
non bonded electrons
2
.
 The number of  bonds and the lone pair gives the total number of hybrid orbitals.
An example will make this method clear
SF4 Central atom S, Peripheral atom F
 total number of valence electrons = 6+(4 7) = 34
Now 34/8= 4
2
8
 Number of hybrid orbitals = 4 bonds + 1 lone pair)
So, five hybrid orbitals are necessary and hybridization mode is sp
3
d and it is trigonal bipyramidal (TBP).
Note:
Whenever there are lone pairs in TBP geometry they should be placed in equatorial position so that
repulsion is minimum.

9
F
F
F
S
F
S
F
F
F
F
(A) (B)
1. NCl3 Total valence electrons = 26
Requirement = 3  bonds + 1 lone pair
Hybridization = sp
3
Shape = pyramidal
N
Cl
Cl
Cl
2. BBr3 Total valence electron = 24
Requirement = 3 bonds
Hybridization = sp
2
Shape = planar trigonal
B
Br Br
Br
3. SiCl4 Total valence electrons = 32
Requirement = 4 bonds
Hybridization = sp
3
Shape = Tetrahedral
Si
Cl Cl Cl
Cl
4. CI4 Total valence electron = 32
Requirements = 4  bonds
Hybridization = sp
3
Shape = Tetrahedral
C
l
l l l
5. SF6 Total valence electrons = 48
Requirement = 6  bonds
hybridization = sp
3
d
2
shape = octahedral / square
bipyramidal
S
F
F
F F
F
F
6. BeF2 Total valence electrons : 16
Requirement : 2  bonds
Hybridization : sp
Shape : Linear
F – Be – F
7. ClF3 Total valence electrons : 28
Requirement : 3  bonds + 2 lone pairs
Hybridization : sp
3
d
Shape : T – shaped
Cl F
F
F
We have already discussed that whenever there are lone pairs they should be placed in equatorial positions. Now
a question that may come to your mind that though the hybridization is sp
3
d, so the shape should be T.B.P. But
when all the bonds are present the actual shape is TBP. But when instead of bond there are lone pairs in TBP the
actual geometry is determined by the bonds not by the lone pairs. Here in ClF3 the bond present (2 in axial and 1
in equatorial) gives the impression of T shape.
8. PF5 Total valence electrons : 40
Hybridization : sp
3
d
Shape : Trigonal bipyramidal (TBP)
P
F
F
F
F
F
9. XeF4 Total valence electrons : 36
Requirement:4 bonds+ 2 lone pairs
Hybridisation : sp
3
d
Shape : Square planar

Xe
F
F
F
F
(A)
Xe
F
F
F
F
(B)
Xe
F
F
F
(C)
F
Now three arrangements are possible out of which A and B are same. A and B can be inter converted by simple
rotation of molecule. The basic difference of (B) and (C) is that in (B) the lone pair is present in the anti position
which minimizes the repulsion which is not possible in structure (C) where the lone pairs are adjacent. So in a
octahedral structure the lone pairs must be placed at the anti positions to minimize repulsion. So both structure
(A) and (B) are correct.
10. XeF2 Total valence electrons : 22
Requirements : 2 bonds + 3 lone pairs
Hybridisation: sp
3
d
Shape : Linear
Xe
F
F
[ l.p. are present in equatorial position and ultimate shape is due to the bonds that are formed]
11. PF2Br3 Total valence electrons : 40
Requirements : 5  bonds
Hybridisation: sp
3
d
Shape : trigonal bipyramidal
P
F
Br
Br
F
Br
Here we see that fluorine is placed in axial position whereas bromine is placed in equatorial position. It is the
more electronegative element that is placed in axial position and less electronegative element is placed in
equatorial position. Fluorine, being more electronegative pulls away bonded electron towards itself more than that
is done by bromine atom which results in decrease in bp – bp repulsion and hence it is placed in axial position.
In this context it can also be noted that in T.B.P. shape the bond lengths are not same. The equatorial bonds are
smaller than axial bonds. But in square bipyramidal shape, all bond lengths are same.
12. 3-
4
PO Total valence electrons : 32
Hybridisation: sp
3
Shape: tetrahedral
O
P
O
O O
O–
P
O
O O
O
P
O
O O
etc.
Here all the structures drawn are resonating structures with O
–
resonating with double bonded oxygen.
13. NO2
–
Total valence electron: 18
Requirement : 2 bonds + 1 lone pair
Hybridisation: sp
2
Shape: angular
N
O O
14. CO3
2–
Total valence electrons: 24
Requirement = 3  bonds
Hybrdisation = sp
2
Shape: planar trigonal
But C has 4 valence electrons of these 3 form  bonds  the rest will form a  bond.
C
O O
O
C
O O
O
–
C
O O
O
–
In the structure one bond is a double bond and the other 2 are single. The position of the double bonds keeps
changing in the figure. Since peripheral atoms are isovalent, so contribution of the resonanting structures are
equal. Thus it is seen that none of the bonds are actually single or double. The actual state is
C
O
–2/3
O
–2/3
O
–2/3
Bond order = 3/2 = 1.5
15. CO2 Total valence electrons : 16 O = C = O

11
Requirement: 2 bonds
Hybridisation: sp
Shape: linear
16. -
4
BF Total valence electrons = 32
Requirement= 4  bonds
Hybridisation: sp
3
Shape: Tetrahedral
B
F
F
F
F
–
17. -
3
ClO Total valence electron = 26
Requirement = 3  bond + 1 lone pair
Hybridization: sp
3
Shape: pyramidal
Cl
O
O
O
-
Cl
O
O
- O
Cl
O
-
O
O
18. XeO2F2 Total valence electrons : 34
Requirement: 4  bonds +1 lone pairs
Hybridization : sp
3
d
Shape: Distorted TBP (sea-saw geometry)
O
O
Xe
F
F
19. XeO3 Total valence electrons : 26
Requirement: 3  bonds + 1 lone pair
Hybridization: sp
3
Shape: Pyramidal
Xe
O O
O
20. XeOF4 Total valence electrons : 42
Requirement: 5  bonds + 1 lone pair
Hybridization: sp
3
d
2
Shape: square pyramidal.
Xe
F
F
F
F
O
Example 8. Predict the hybridization for the central atom in 3
POCl , 4
OSF , 5
OIF
Solution: 3
POCl Total No. of V.E. =
5 6 21 32
4
8 8
 
 
So, hybridization = 3
sp
4
OSF =
6 6 28 40
5
8 8
 
 
So, hybridization of s = 3
dsp
5
OIF
6 7 35 48
6
8 8
 
  
So, hybridization of I = 2 3
d sp
Example 9. Out of the three molecules XeF4, SF4 and SiF4 one which has tetrahedral structures is
(A) All of three (B) Only SiF4
(C) Both SF4 and XeF4 (D) Only SF4 and XeF4
Solution: Hybridization of XeF4 = sp
3
d
2
, SF4 = sp
3
d, SiF4 = sp
3
Hence (B) is correct.
Example 10. Among the following compounds in which case central element uses dorbital to make bonds with
attached atom
(A) 2
BeF (B) 2
XeF
(C) 4
SiF (D) 3
BF
Solution: In 2
XeF . Xe atom has 3
sp d hybridisation.
Hence (B) is correct.
Example 11. When NH3 is treated with HCl, state of hybridisation on central nitrogen

(A) Changes from sp
3
to sp
2
(B) Remains unchanged
(C) Changes from sp
3
to sp
3
d (D) Changes from sp
3
to sp
Solution: On NH4
+
state of hybridisation on central nitrogen atom is sp
3
as
in NH3.
N
H
H
H
H
+
Hence (B) is the correct answer.
Exercise 4.
Among the following which has bond angle very near to 90°
(A) NH3 (B) XeF4
(C) BF3 (D) H2O
Exercise 5.
Homolytic fission of C – C bond in hexafluoroethane gives an intermediate in which hybridization
state of carbon is
(A) sp
2
(B) sp
3
(C) sp (D) cannot be determined
Exercise 6.
A molecule XY2 contains two , two  bonds and one lone pair of electrons in valence shell of X. The
arrangement of lone pair as well as bond pairs is
(A) Square pyramidal (B) Linear
(C) Trigonal planar (D) Unpredictable
Exercise 7.
Draw the structure the following indicating the hybridisation of the central atom.
(i) SOF2, (ii) SO2, (iii) POCl3, (iv) I3
–
Exercise 8.
The type of hybridisation of orbitals employed in the formation of SF6 molecule is …………..
Exercise 9.
The angle between two covalent bonds is maximum for……………(CH4, H2O, CO2)
Exercise 10.
The bond angle in 2
4
SO 
ion is ………………..
Maximum Covalency
Elements which have vacant d-orbital can expand
their octet by transferring electrons, which arise
after unpairing, to these vacant d-orbital e.g. in
sulphur.
↿⇂ ↿⇂ ↿ ↿
↿ ↿ ↿ ↿ ↿ ↿
In excited state
In ground state
3s 3p 3d
In excited state sulphur has six unpaired electrons and shows a valency of six e.g. in SF6. Thus an element can
show a maximum covalency equal to its group number e.g. chlorine shows maximum covalency of seven.

13
VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY (SHAPES AND GEOMETRY OF
MOLECULES)
Molecules exist in a variety of shapes. A number of physical and chemical properties of molecules arise from and
are affected by their shapes. For example, the angular shape of the water molecules explains its many
characteristic properties while a linear shape does not.
The determination of the molecular geometry and the development of theories for explaining the preferred
geometrical shapes of molecules is an integral part of chemical bonding. The VSEPR theory (model) is a simple
treatment for understanding the shapes of molecules.
Strictly speaking VSEPR theory is not a model of chemical bonding. It provides a simple recipe for predicting the
shapes of molecules. It is infact an extension of the Lewis interpretation of bonding and is quite successful in
predicting the shapes of simple polyatomic molecules.
The basic assumptions of the VSEPR theory are that:
Pairs of electrons in the valence shell of a central atom repel each other
1. These pairs of electrons tend to occupy position in space that minimize repulsions and thus maximize
distance between them.
2. The valence shell is taken as a sphere with the electron pairs localizing on the spherical surface at maximum
distance from one another.
3. A multiple bonds are treated as a single super pair.
4. Where two or more resonance structures can depict a molecule, the VSEPR model is applicable to any such
structures
For the prediction of geometrical shapes of molecules with the help of VSEPR model, it is convenient to divide
molecules into two categories
Regular Geometry
Molecules in which the central atom has no lone pairs
Irregular Geometry
Molecules in which the central atom has one or more lone pairs, the lone pair of electrons in molecules occupy
more space as compared to the bonding pair electrons. This causes greater repulsion between lone pairs of
electrons as compared to the bond pairs repulsions. The descending order of repulsion
(lp – lp) > (lp – bp) > (bp – bp)
where lp-Lone pair; bp-bond pair
Regular Geometry
Number of
electron pairs
Arrangement of electrons Molecular geometry Examples
2
A
180
Linear
B – A – B
Linear
2 2
BeCl ,HgCl
3
A
Trigonal planar
 = 120
B
B B
A
Trigonal planar
 = 120
3 3
BF ,AlCl
4
A
1090
28'
A
1090
28'
4 4 4
CH ,NH SiF


5
trigonal bipyramidal
900
1200
trigonal bipyramidal
900
1200
5 5
PCl , PF
6
A 90
A
B
B
B
B
B
B
Octahedral
6
SF
Irregular Geometry
Molecule
Type
No. of
Bonding pairs
No. of
lone pair
Arrangement of
electrons pairs
Shape
(Geometry)
Examples
2
AB E 2 1 A
B B
Trigonal planar
Bent 2 3
SO ,O
3
AB E 3 1
A
B B
B
Tetrahedral
Trigonal
pyramidal
3
NH
2 2
AB E 2 2
A
B
B
Tetrahedral
Bent 2
H O
4
AB E 4 1
B
A
B
B
B
Trigonal bi-pyramidal
See saw 4
SF
3 2
AB E 3 2
B
A
B
B
Trigonal bi-pyramidal
T – shaped 3
CIF
5
AB E 5 1
B
A
B
B
B
B
Octahedral
Square
pyramidal
5
BrF
4 2
AB E 4 2
A
B
B
B
B
Octahedral
Square
planar
4
XeF
Example 13. Why the bond angle of H – C – H in methane (CH4) is 109° 28’ while
H – N – H bond angle in NH3 is 107° though both carbon and nitrogen are sp
3
hybridized
Solution: In CH4 there are 4 bond pair of electrons while in NH3 are 3 bond pair of electrons and 1 lone pair
of electrons. Since bond pair bond pair repulsion is less than lone pair bond pair repulsion, in NH3

15
bond angle is reduced from 109°28‘
to 107°.
Example 14. Why bond angle in NH3 is 107° while in H2O it is 104.5°?
Solution: In NH3, central nitrogen atom bears only one lone pair of electrons whereas in H2O central
oxygen atom bears two lone pair of electrons.
Since the repulsion between lone pair-lone pair and lone pair – bond pair is more than that
between bond pair-bond pair, the repulsion in H2O is much greater than that in NH3 which results
in contraction of bond angle from 109°28‖ to 104.5° in water while in NH3 contraction is less i.e.
from 109°28‖ to 107°.
“If the electronegativity of the peripheral atoms is more, then the bond angle will be less”.
For example if we consider NH3 and NF3, F – N – F bond angle will be lower than H – N – H bond angle. This is
because in NF3 the bond pair is displaced more towards F and in NH3 it is displaced more towards N. So
accordingly the b.p. – b.p. interaction is less in NF3 and more in NH3.
Example 15. The bond angle of H2O is 104° while that that of F2O is 102°.
Solution: Both H2O and F2O have a lone pair of electrons. But fluorine being highly electronegative, the
bond pair of electrons are drawn more towards F in F2O, whereas in H2O it is drawn towards O.
So in F2O the bond pairs being displaced away from the central atom, has very little tendency to
open up the angle. But in H2O this opening up is more as the bond pair electrons are closer to
each other. So bond  of F2O is less than H2O.
O
H H
repulsion more O
F F
Repulsion less
RESONANCE
There may be many molecules and ions for which it is not possible to draw a single Lewis structure. For example
we can write two electronic structures of O3.
O
O
O O
O
O
(A) (B)
In (A) the oxygen  oxygen bond on the left is a double bond and the oxygenoxygen bond on the right is a single
bond. In B the situation is just opposite. Experiment shows however, that the two bonds are identical. Therefore
neither structure A nor B can be correct.
One of the bonding pairs in ozone is spread over the region of all the three atom rather than associated with
particular oxygenoxygen bond. This delocalised bonding is a type of bonding in which bonding pair of electrons
is spread over a number of atoms rather than localised between two.
O
O
O
(C)
Structures (A) and (B) are called resonating or canonical structures and C is the resonance hybrid. This
phenomenon is called resonance, a situation in which more than one plausible structure can be written for a
species and in which the true structure cannot be written at all.
Some other examples
(i) CO3
2–
ion
O O
O
O O
O
O O
O
O O
O
(ii) Carbonoxygen bond lengths in carboxylate ion are equal due to resonance.
(+) (+)
()
()


R
O
-
O
R
O
O
-
R
O
O
(iii) Benzene
(iv) Vinyl Chloride
C
H2
Cl
H2C
Cl
+
Difference in the energies of the canonical forms and resonance hybrid is called resonance stabilization energy
and provides stability to species.
Rules for writing resonating structures
 Only electrons (not atoms) may be shifted and they may be shifted only to adjacent atoms or bond
positions.
 The number of unpaired electrons should be same in all the canonical form.
 The positive charge should reside as far as possible on less electronegative atom and negative charge on
more electronegative atom.
 Like charge should not reside on adjacent atom
 The larger the number of the resonating structures greater the stability of species.
 Greater number of covalency add to the stability of the molecule.
Example 17. Out of the following resonating structures for CO2 molecule, which are important for describing
the bonding in the molecule and why?
O C O O C O O C O O C O
(I) (II) (III) (IV)
2
Solution: Out of the structures listed above, the structure (III) is wrong since the number of electron pairs on oxygen
atoms are not permissible. Similarly, the structures (II) has very little contribution towards the hybrid because one of the oxygen
atoms (electronegative) is show to have positive charge. Carbon dioxide is best represented by structures (I) and (IV).
FACTORS GOVERNING POLARIZATION AND POLARISABILITY (FAJAN’S RULE)
 Cation Size: Smaller is the cation more is the value of charge density () and hence more its polarising
power. As a result more covalent character will develop. Let us take the
example of the chlorides of the alkaline earth metals. As we go down from Be to Ba the cation size
increases and the value of  decreases which indicates that BaCl2 is less covalent i.e. more ionic. This is
well reflected in their melting points. Melting points of
BeCl2 = 405°C and BaCl2 = 960°C.
 Cationic Charge: More is the charge on the cation, the higher is the value of  and higher is the
polarising power. This can be well illustrated by the example already given, NaBr and AlBr3. Here the
charge on Na is +1 while that on Al in +3, hence polarising power of Al is higher which in turn means a
higher degree of covalency resulting in a lowering of melting point of AlBr3 as compared to NaBr.
 Noble Gas vs Pseudo Noble Gas Cation: A Pseudo noble gas cation consists of a noble gas core
surrounded by electron cloud due to filled d-subshell. Since
d-electrons provide inadequate shielding from the nuclei charge due to relatively less penetration of
orbitals into the inner electron core, the effective nuclear charge (ENC) is relatively larger than that of a
noble gas cation of the same period. NaCl has got a melting point of 800°C while CuCl has got melting
point of 425°C. The configuration of Cu
+
= [Ar] 3d
10
while that of Na
+
= [Ne]. Due to presence of d
electrons ENC is more and therefore Cl
–
is more polarised in CuCl leading to a higher degree of
covalency and lower melting point.
 Anion Size: Larger is the anion, more is the polarisability and hence more covalent character is
expected. An e.g. of this is CaF2 and CaI2, the former has melting

17
point of 1400°C and latter has 575°C. The larger size of I
–
ion compared to F
–
causes more polarization of
the molecule leading to a lowering of covalency and increasing in melting point.
 Anionic Charge: Larger is the anionic charge, the more is the polarisability. A well illustrated example is
the much higher degree of covalency in magnesium nitride
(3Mg
++
2N
3–
) compared to magnesium fluoride (Mg
++
2F
—
). This is due to higher charge of nitride
compare to fluoride. These five factors are collectively known as Fajan‘s Rule.
Example 18. The melting point of KCl is higher than that of AgCl though the crystal radii of Ag
+
and K
+
ions are
almost same.
Solution: Now whenever any comparison is asked about the melting point of the compounds which are fully
ionic from the electron transfer concept it means that the compound having lower melting point
has got lesser amount of ionic character than the other one. To analyse such a question first find
out the difference between the 2 given compounds. Here in both the compounds the anion is the
same. So the deciding factor would be the cation. Now if the cation is different, then the answer
should be from the variation of the cation. Now in the above example, the difference of the cation
is their electronic configuration. K
+
= [Ar]; Ag
+
= [Kr] 4d
10
. This is now a comparison between a
noble gas core and pseudo noble gas core, the analysis of which we have already done. So try
to finish off this answer.
Example 19. AlF3 is ionic while AlCl3 is covalent.
Solution: Since F
–
is smaller in size, its polarisability is less and therefore it is having more ionic character.
Whereas Cl being larger in size is having more polarisability and hence more covalent character.
Example 20. Which compound from each of the following pairs is more covalent and why?
(a) CuO or CuS (b) AgCl or AgI ‘
(c) PbCl2 or PbCl4 (d) BeCl2 or MgCl2
Solution: (a) CuS (b) AgI
(c) PbCl4 (d) BeCl2
DIPOLE MOMENT
Difference in polarities of bonds is expressed on a numerical scale. The polarity of a molecule is indicated in
terms of dipole moment  
 . To measure dipole moment, a sample of the substance is placed between two
electrically charged plates. Polar molecules orient themselves in the electric field causing the measured voltage
between the plates to change.
The dipole moment is defined as the product of the distance separating charges of equal magnitude and opposite
sign, with the magnitude of the charge. The distance between the positive and negative centres called the bond
length.
Thus,
electric charge bond length
   = q d

As q is in the order of 10
10
esu and d is in the order of 8
10
cm, is the order of 18
10 esu cm

. Dipole moment is
measured in ‗Debye‘ unit (D)
18 30
1D 10 esu cm 3.33 10 coulomb metre
 
  
Note:
(i) Generally as electronegativity difference increase in diatomic molecules, polarity of bond between the
atoms increases therefore value of dipole moment increases.
(ii) Dipole moment is a vector quantity

(iii) A symmetrical molecule is non- polar even though it contains polar
bonds. For example, 2 3 4
CO , BF , CCl etc because summation of all
bond moments present in the molecules cancel each other.
Cl
C
Cl
Cl
Cl
(iv) Unsymmetrical non-linear polyatomic molecules have net value of dipole moment. For example,
2 3 3
H O, CH OH, NH etc.
Calculation of Resultant Bond Moments
Let AB and AC are two polar bonds inclined at an angle  their
dipole moments are 1
 and 2
 .
Resultant dipole moment may be calculated using vectorial method.

A


B


C


1

R

2

2 2
R 1 2 1 2
2 cos
        
when  = 0 the resultant is maximum
R 1 2
    
when,  = 180, the resultant is minimum
R 1 2
   

180o

180o
Example 21. The compound which has zero dipole moment is
(A) CH2Cl2 (B) NF3
(C) PCl3F2 (D) ClO2
Solution: (C)
Example 22. Sketch the bond moments and resultant dipole moment in
(i) 2
SO (ii) 2 2 2
Cis C H Cl
 and (iii) 2 2 2
trans C H Cl

Solution:
S
O
O
(i)
C C
H
Cl Cl
H
(ii)
(iii)
C C
Cl
H Cl
H
Resultant  = 0
Example 23. CO2 has got dipole moment of zero. Why?
Solutions: The structure of CO2 is O = C = O . This is a highly symmetrical structure with a plane of
symmetry passing through the carbon. The bond dipole of C–O is directed towards oxygen as it is
the negative end. Here two equal dipoles acting in opposite direction cancel each other and
therefore the dipole moment is zero.
Example 24. Dipole moment of CCl4 is zero while that of CHCl3 is non zero.
Solution: Both CCl4 & CHCl3 have tetrahedral structure but CCl4 is symmetrical while CHCl3 is non-
symmetrical.

19
Cl
C
Cl
Cl
Cl
Symmetrical
H
C
Cl
Cl
Cl
Non-Symmetrical
Due to the symmetrical structure of CCl4 the resultant of bond dipoles comes out to be zero. But
in case of CHCl3 it is not possible as the presence of hydrogen introduces some dissymmetry.
Example 25. Compare the dipole moment of H2O and F2O.
Solution: Let‘s draw the structure of both the compounds and then analyse it.
O
H H
O
F
F
In both H2O and F2O the structure is quite the same. In H2O as O is more electronegative than
hydrogen so the resultant bond dipole is towards O, which means both the lone pair and bond
pair dipole are acting in the same direction and dipole moment of H2O is high. In case of F2O the
bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are acting in
opposition resulting in a low dipole.
In C-H, carbon being more electronegative the dipole is projected towards C. Now the question
comes whether hybridization has anything to do with the dipole moment. The answer is obviously
yes. If yes, why? Depending on the hybridization state the electronegativity of carbon changes
and therefore the dipole moment of C-H bond will change. As the s character in the hybridized
state increases, the electronegativity of C increases due to which C attracts the electron pair of C-
H bond more towards itself resulting in a high bond dipoles.
Now as we have said about carbon hydrogen bonds, the question that is coming to your mind is
whether we would be dealing with organic compounds or not. Yes we would be dealing with the
organic compounds.
For instance but -2- ene. It exists in two forms cis and Trans.
Cis
H CH3
H CH3
C
C
Trans
H CH3
H
CH3
C
C
The trans isomer is symmetrical with the 2 methyl groups in anti position. So the bond dipoles the
two Me– C bonds acting in opposition, cancel each other results into a zero dipole. Whereas in
cis isomer the dipoles do not cancel each other resulting in a net dipole.
Example 26. The molecule having largest dipole moment among the following is
(A) CH4 (B) CHCl3
(C) CCl4 (D) CHI3
Solution: (B)
Example 27. Compare the dipole moment of cis 1, 2 dichloroethylene and trans 1, 2 dichloroethylene.
Solution: Cl
Cis
Cl
H
H
C C
H
Trans
Cl
Cl
H
C C
In the trans compound the C-Cl bond dipoles are equal and at the same time acting in opposition
cancel each other while in cis compound the dipoles do not cancel each other resulting in a
higher value.

Generally all Trans compounds have a lower dipole moment corresponding to Cis isomer, when
both the substituents attached to carbon atom are either electron releasing or electron
withdrawing.
PERCENTAGE OF IONIC CHARACTER
Every ionic compound having some percentage of covalent character according to Fajan‘s rule. The percentage
of ionic character in a compound having some covalent character can be calculated by the following equation.
The percent ionic character =
Observed dipole moment
100
Calculated dipole moment assuming 100% ionic bond

Example 28. Dipole moment of KCl is 3.336  10
–29
coulomb metre which indicates that it is highly polar
molecule. The interatomic distance between K
+
and Cl
–
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.
Solution: Dipole moment  = e  d coulomb metre
For KCl d = 2.6  10
–10
m
For complete separation of unit charge
e = 1.602  10
–19
C
Hence  = 1.602  10
–19
 2.6  10
–10
= 4.1652  10
–29
Cm
KCl = 3.336  10
–29
Cm
 % ionic character of KCl =
29
29
3.336 10
100
4.165 10





= 80.09%
Example 29. Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as it’s observed
dipole moment.
Solution: To calculate  considering 100% ionic bond
= 4.810
–10
 0.83 10
–8
esu cm
= 4.8  0.83  10
–18
esu cm = 3.984 D
 % ionic character = 68
.
45
100
984
.
3
82
.
1 

The example given above is of a very familiar compound called HF. The % ionic character is nearly 43.25%,
so the % covalent character is (100 – 43.25) = 56.75%. But from the octet rule HF should have been a purely
covalent compound but actually it has some amount of ionic character in it, which is due to the
electronegativity difference of H and F. Similarly knowing the bond length and observed dipole moment of
HCl, the % ionic character can be known. It was found that HCl has 17% ionic character. Thus it can be
clearly seen that although we call HCl and HF as covalent compounds but it has got appreciable amount of
ionic character. So from now onwards we should call a compound having more of ionic less of covalent and
vice versa rather than fully ionic or covalent.
BOND CHARACTERISTICS
1. Bond Length: The distance between the nuclei of two atoms bonded together is termed as bond length or
bond distance. It is expressed in angstrom A
 
 
 
units or picometer (pm).
8 12
1A 10 cm;1pm 10 m
 
 
 
 
 
Bond length in ionic compound = c a
r r
 

Similarly, in a covalent compound, bond length is obtained by adding up the covalent (atomic) radii of two
bonded atoms.
Bond length in covalent compound (AB) = A B
r r


21
The factors such as resonance, electronegativity, hybridization, steric effects, etc., which affect the radii of
atoms, also apply to bond lengths.
Important features
(i) The bond length of the homonuclear diatomic molecules are twice the covalent radii.
(ii) The lengths of double bonds are less than the lengths of single bonds between the same two atoms, and
triple bonds are even shorter than double bonds.
Single bond > Double bond > Triple bond (decreasing bond length)
(iii) Bond length decreases with increase in s-character since s-orbital is smaller than a
p – orbital.
3
sp C H 1.112A ;
  2
sp C H 1.103A ;
  spC H 1.08A ;
 
(25% s-character as in alkanes) (33.3% s-character as in alkenes) (50% s-character as in alkynes)
(iv) Bond length of polar bond is smaller than the theoretical non-polar bond length.
2. Bond Energy or Bond Strength: Bond energy or bond strength is defined as the amount of energy required
to break a bond in molecule.
Important features
(i) The magnitude of the bond energy depends on the type of bonding. Most of the covalent bonds have
energy between 50 to 100 kcal 1
mol
(200-400 kJ 1
mol
). Strength of sigma bond is more than that of a  -
bond.
(ii) A double bond in a diatomic molecules has a higher bond energy than a single bond and a triple bond has a
higher bond energy than a double bond between the same atoms.
C C C C C C
     (decreasing bond length)
(iii) The magnitude of the bond energy depends on the size of the atoms forming the bond, i.e. bond length.
Shorter the bond length, higher is the bond energy.
(iv) Resonance in the molecule affects the bond energy.
(v) The bond energy decreases with increase in number of lone pairs on the bonded atom. This is due to
electrostatic repulsion of lone pairs of electrons of the two bonded atoms.
(vi) Homolytic and heterolytic fission involve different amounts of energies. Generally the values are low for
homolytic fission of the bond in comparison to heterolytic fission.
(vii) Bond energy decreases down the group in case of similar molecules.
(viii) Bond energy increase in the following order:
     
2 3
s p sp sp sp
C C N N O O
No lone pair One lone pair Two lone pair
   
    
3. Bond angles: Angle between two adjacent bonds at an atom in a molecule made up of three or more atoms
is known as the bond angle. Bond angles mainly depend on the following three factors:
(i) Hybridization: Bond angle depends on the state of hybridization of the central atom
3 2
4 3 2
Hybridization sp sp sp
Bond angle 109 28 120 180
Example CH BCl BeCl

Generally s- character increase in the hybrid bond, the bond angle increases.
(ii) Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A
lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due
to this, the bonds are displaced slightly inside resulting in a decrease of bond angle.
(iii) Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.
HYDROGEN BONDING
Introduction:
An atom of hydrogen linked covalently to a strongly electronegative atom can establish an extra weak attachment
to another electronegative atom in the same or different molecules. This attachment is called a hydrogen bond.
To distinguish from a normal covalent bond, a hydrogen bond is represented by a broken line eg X – H
…
Y where
X & Y are two electronegative atoms. The strength of hydrogen bond is quite low about 2-10 kcal mol
–1
or 8.4–42
kJ mol
–1
as compared to a covalent bond strength 50–100 kcal mol
–1
or 209 –419 kJ mol
–1

Conditions for Hydrogen Bonding:
 Hydrogen should be linked to a highly electronegative element.
 The size of the electronegative element must be small.
These two criteria are fulfilled by F, O, and N in the periodic table. Greater the electronegativity and
smaller the size, the stronger is the hydrogen bond which is evident from the relative order of energies of
hydrogen bonds.
Types of Hydrogen Bonding:
 Intermolecular hydrogen bonding: This type of bonding takes place between two molecules of the
same or different types. For example,
O—H----- O — H ------ O — H ------
H H H
Intermolecular hydrogen bonding leads to molecular association in liquids like water etc. Thus in water only a few
percent of the water molecules appear not to be hydrogen bonded even at 90°C. Breaking of those hydrogen
bonds throughout the entire liquid requires appreciable heat energy. This is indicated in the relatively higher
boiling points of hydrogen bonded liquids. Crystalline hydrogen fluoride consists of the polymer (HF)n. This has a
zig-zag chain structure involving
H-bond.
F
F
H
H
H H
F
F
H
F
 Intramolecular hydrogen bonding: This type of bonding occurs between atoms of the same molecule
present on different sites. Intramolecular hydrogen bonding gives rise to a closed ring structure for which
the term chelation is sometimes used. Examples are
o-nitrophenol, salicylaldehyde.
O
O
O
H
N
o-nitrophenol Salicaldehyde
H
O
O
H
C
Effect of Hydrogen Bonding
Hydrogen bonding has got a very pronounced effects on certain properties of the molecules. They have got
effects on
 State of the substance
 Solubility of the substance
 Boiling point
 Acidity of different isomers
These can be evident from the following examples.
Example 30. H2O is a liquid at ordinary temperature while H2S is a gas although both O and S belong to the
same group of the periodic table.
Solution: H2O is capable of forming intermolecular hydrogen bonds. This is possible due to high
electronegativity and small size of oxygen. Due to intermolecular H-bonding, molecular
association takes place. As a result the effective molecular weight increases and hence the
boiling point increases. So H2O is a liquid. But in H2S no hydrogen bonding is possible due to
large size and less electronegativity of S. So it‘s boiling point is equal to that of an isolated H2S
molecule and therefore it is a gas.
Example 31. Ethyl alcohol (C2H5OH) has got a higher boiling point than dimethyl ether
(CH3-O-CH3) although the molecular weight of both are same.

23
Solution: Though ethyl alcohol and dimethyl ether have the same molecular weight but in ethyl alcohol the
hydrogen of the O-H groups forms intermolecular hydrogen bonding with the OH group in another
molecule. But in case of ether the hydrogen is linked to C is not so electronegative to encourage
the hydrogen to from hydrogen bonding.
H O
C2H5
H O H O
C2H5
C2H5
Due to intermolecular H-bonding, ethyl alcohol remains in the associated form and therefore boils
at a higher temperature compared to dimethyl ether.
Importance of Hydrogen Bonding in Biological Systems:
Hydrogen bonding plays a vital role in physiological systems. Proteins contain chains of amino acids. The amino
acid units are arranged in a spiral form somewhat like a stretched coil spring (forming a helix). The N-H group of
each amino acid unit and the fourth C=O group following it along the chain, establishes the N–H---O hydrogen
bonds. These bonds are partly responsible for the stability of the spiral structure. Double helix structure of DNA
also consists of two strands forming a double helix and are joined to each other through hydrogen bond.
INTERMOLECULAR FORCES
We have enough reasons to believe that a net attractive force operates between molecules of a gas. Though
weak in nature, this force is ultimately responsible for liquifaction and solidification of gases. But we cannot
explain the nature of this force from the ideas of ionic and covalent bond developed so far, particularly when we
think of saturated molecules like H2, CH4, He etc. The existence of intermolecular attraction in gases was first
recognised by Vanderwaal‘s and accordingly intermolecular forces have been termed as Vanderwaal‘s forces. It
has been established that such forces are also present in the solid and liquid states of many substances.
Collectively they have also been termed London forces since their nature was first explained by London using
wave mechanics.
Nature of Vanderwaal’s Forces:
The Vanderwaal‘s forces are very weak in comparison to other chemical forces. In solid NH3 it amount to about 39
KJ mol
–1
(bond energy N-H bond = 389 KJ mol
–1
). The forces are non directional. The strength of Vanderwaal‘s
force increases as the size of the units linked increases. When other factors (like H-bonding is absent), this can
be appreciated by comparison of the melting or boiling points of similar compounds in a group.
Origin of Intermolecular Forces:
Intermolecular forces may have a wide variety of origin.
 Dipole-dipole interaction: This force would exist in any molecule having a permanent dipole e.g. HF,
HCl, H2O etc.
 Ion-dipole interaction: These interactions are operative in solvation and dissolution of ionic compounds
in polar solvents.
 Induced dipole interaction: These generate from the polarisation of a neutral molecule by a charge or
ion.
 Instantaneous dipole-induced dipole interaction: In non polar molecules dipoles may generate due to
temporary fluctuations in electron density. These transient dipole can now induce dipole in neighbouring
molecules producing a weak temporary interaction.
METALLIC BONDING
Metals are characterised by bright, lustre, high electrical and thermal conductivity, malleability, ductility and high
tensile strength. A metallic crystal consists of very large number of atoms arranged in a regular pattern.

Different model have been proposed to explain the nature of metallic bonding two most important modules are as
followsThe electron sea Model
+

+

+

+

+

+

+

+

+

+

+

+

Positive
Kernles
Mobile
electrons
In this model a metal is assumed to consist of a lattice of positive ion (or kernels) immersed in a sea of mobile valence
electrons, which move freely within the boundaries of a crystal. A positive kernel consists of the nucleus of the atom
together with its core on a kernel is, therefore, equal in magnitude to the total valence electronic charge per atom. The
free electrons shield the positively charged ion cores from mutual electrostatic repulsive forces which they would
otherwise exert upon one another. In a way these free electrons act as ‗glue‘ to hold the ion cores together.
The forces that hold the atoms together in a metal as a result of the attraction between positive ions and surrounding
freely mobile electrons are known as metallic bonds.
Through the electron sea predated quantum mechanics it still satisfactorily explains certain properties of the metals.
The electrical and thermal conductivity of metals for example, can be explained by the presence of mobile electrons in
metals. On applying an electron field, these mobile electrons conduct electricity throughout the metals from one end
to other. Similarly, if one part of metal is heated, the mobile electrons in the part of the metals acquire a large amount
of kinetic energy. Being free and mobile, these electrons move rapidly throughout the metal and conduct heat to the
other part of the metal.
On the whole this model is not satisfactory.
Example 32. Sodium metal conducts electricity because it
(A) is a soft metal
(B) contains only one valence electron
(C) has mobile electron
(D) reacts with water to form H2 gas
Solution: (C)
MOLECULAR ORBITAL THEORY
In Molecular Orbital Theory (MOT) the atoms in a molecule are supposed to loose their individual control over the
electrons. The nuclei of the bonded atoms are considered to be present at equilibrium inter-nuclear positions. The
orbitals where the probability of finding the electrons is maximum are multicentred orbitals called molecular
orbitals extending over two or more nuclei.
In MOT the atomic orbitals loose their identity and the total number of electrons present are placed in MO‘s
according to increasing energy sequence (Auf Bau Principle) with due reference to Pauli‘s Exclusion Principle and
Hund‘s Rule of Maximum Multiplicity.
When a pair of atomic orbitals combine they give rise to a pair of molecular orbitals, the bonding and the anti-
bonding. The number of molecular orbitals produced must always be equal to the number of atomic orbitals
involved. Electron density is increased for the bonding MO‘s in the inter-nuclear region but decreased for the anti-
bonding MO‘s, Shielding of the nuclei by increased electron density in bonding MO‘s reduces inter nuclei
repulsion and thus stabilizes the molecule whereas lower electron density even as compared to the individual
atom in anti-bonding MO‘s increases the repulsion and destabilizes the system.
In denotation of MO‘s,  indicates head on overlap and  represents side ways overlap of orbitals. In simple
homonuclear diatomic molecules the order of MO‘s based on increasing energy is
y y
x x
z
z
2p 2p
1s 1s 2s 2s 2p 2p
2p
2p
*
* * *
*
 
 
 
   
     
   

   

   

25
This order is true except B2, C2 & N2. If the molecule contains unpaired electrons in MO‘s it will be paramagnetic
but if all the electrons are paired up then the molecule will be diamagnetic.
Bond order =
no. of e s occupying bonding MO's no. of e soccupyingantibonding MO's
2
 

Application of MOT to homonuclear diatomic molecules.
H2 molecule : Total no. of electrons = 2
Arrangement : 2
1s

Bond order : ½ (2 – 0) = 1
2
H
molecule : Total no. of electrons = 1
Arrangement : 1
1s

Bond order : ½ (1 – 0) = 1/2
He2 molecule : Total no. of electrons = 4
Arrangement : 2
1s
 2
1s
*

Bond order : ½ (2 – 2) = 0
He2 molecule does not exist.
2
He
molecule : Total no. of electrons = 3
Arrangement : 2
1s
 1
1s
*

Bond order : ½ (2 – 1) = 1/2
So 2
He
exists and has been detected in discharge tubes.
Li2 molecule : Total no. of electrons = 6
Arrangement : 2
1s
 2
1s
*
 2
2s

Bond order : ½ (4 – 2) = 1
No unpaired e‘s so diamagnetic
Be2 molecule : Total no. of electrons = 8
Arrangement : 2
1s
 2
1s
*
 2
2s
 2
2s
*

Bond order : ½ (4 – 4) = 0
No unpaired e
–
s so diamagnetic
B2 molecule : Total no. of electrons = 10
Arrangement : 2
1s
 2
1s
*
 2
2s
 2
2s
*
 x
2
2p

Bond order : ½ (6 – 4) = 1 diamagnetic
But observed Boron is paramagnetic
C2 molecule : Total no. of electrons = 12
Arrangement : 2
1s
 2
1s
*
 2
2s
 2
2s
*
 x
2
2p
 y
z
1
2p
1
2p
 

 
 

 
 
Bond order : ½ (4 – 0) = 2
It is paramagnetic
But observed C2 is diamagnetic
N2 molecule : Total no. of electrons = 14
Arrangement : 2
1s
 2
1s
*
 2
2s
 2
2s
*
 2
2px
 y
z
2
2p
2
2p
 

 
 

 
 
Bond order : ½ (6 – 0) = 3
 It is diamagnetic
O2 molecule : Total no. of electrons = 16
Arrangement : 2
1s
 2
1s
*
 2
2s
 2
2s
*
 x
2
2p
 y
z
2
2p
2
2p
 

 
 

 
 
y
z
1
2p
1
2p
*
*
 

 
 
 

 
Bond order : ½ (6 – 2) = 2
It is paramagnetic
F2 molecule : Total no. of electrons = 18

Arrangement : 2
1s
 2
1s
*
 2
2s
 2
2s
*
 x
2
2p
 y
z
2
2p
2
2p
 

 
 

 
 
2
2py
2
2pz
*
*
 

 
 
 

 
2px
*

Bond order : ½ (6 – 4) = 1
It has been seen that in case of B2, C2 & N2 the order of filling the e‘s is different from the normal sequence.
B2 : 2
1s
 2
1s
*
 2
2s
 2
2s
*
 y
z
1
2p
1
2p
 

 
 

 
 
It is paramagnetic
C2 : 2
1s
 2
1s
*
 2
2s
 2
2s
*
 y
z
2
2p
2
2p
 

 
 

 
 
It is diamagnetic
N2 : 2
1s
 2
1s
*
 2
2s
 *2
2s

2
2py 2
2px
2
2pz
 

 

 

 
 
It is diamagnetic
Example 33. Compare the bond energies of O2, 2
O
& 2
O
Solution: Higher the bond order greater will be the bond energy.
Now configuration of O2 = 2
1s
 2
1s
*
 2
2s
 2
2s
*
 x
2
2p
 y
z
2
2p
2
2p
 

 
 

 
 
y
z
1
2p
1
2p
*
*
 

 
 
 

 
Now formation of 2
O
means to remove an electron from anti-bonding one, which means increase
in B.O.
B.O. of 2
O
= ½ (6-1) = 2.5
2
O
means introduction of an e
–
in the anti-bonding thereby reducing the bond order.
Bond order of 2
O
= ½ (6 – 3) = 1.5
So bond energy of 2
O
> O2 > 2
O
Example 34. Give MO configuration and bond orders of H2, H2
–
, He2 and He2
–
. Which species among the
above are expected to have same stabilities?
Solution: H2 = 1s
2
Bond order = 1
H2
–
= 1s
2
*1s
1
Bond order = 0.5
He2 = 1s
2
1s
*2
Bond order = 0
He2
–
= 1s
2

*
1s
2
2s
1
Bond order =
(3 2)
2

= 0.5
H2
–
and He2
–
are expected to have same stabilities.
Example 35. Which of the following species have the bond order same as 2
N ?
(A) CN
(B) OH
(C)NO
(D) CO
Solution. In 2
N no. of bonding electrons and antibonding electrons are 10 and 4 respectively. Therefore
the bond order is
10 4
2

= 3. Out of those given only CN
is isoelectronic with 2
N . Therefore
CN
has the same bond order as 2
N .
Hence (A) is correct

27
M.O. of Some Diatomic Heteronuclei Molecules
The molecular orbitals of heteronuclei diatomic molecules should differ from those of homonuclei species
because of unequal contribution from the participating atomic orbitals. Let‘s take the example of CO.
The M.O. energy level diagram for CO should be similar to that of the isoelectronic molecule N2. But C & O differ
much in electronegativity and so will their corresponding atomic orbitals. But the actual MO for this species is very
much complicated since it involves a hybridisation approach between the orbital of oxygen and carbon.
HCl Molecule: Combination between the hydrogen 1s A.O‘s. and the chlorine 1s, 2s, 2p & 3s orbitals can be
ruled out because their energies are too low. The combination of H 1s
1
and 1
x
3p gives both bonding and anti-
bonding orbitals, and the 2 electrons occupy the bonding M.O. leaving the anti-bonding MO empty.
NO Molecule: The M.O. of NO is also quite complicated due to energy difference of the atomic orbitals of N and
O.
As the M.O.‘s of the heteronuclei species are quite complicated, so we should concentrate in knowing the bond
order and the magnetic behaviour.
Molecules/Ions Total No. of electrons Magnetic behaviour
CO 14 Diamagnetic
NO 15 Paramagnetic
NO
+
14 Diamagnetic
NO
–
16 Diamagnetic
CN 13 Paramagnetic
CN
–
14 Diamagnetic
INERT PAIR EFFECT
Heavier p-block and d-block elements show two oxidation states. One is equal to group number and second is
group number minus two. For example Pb(5s
2
5p
2
) shows two OS, +II and +IV. Here +II is more stable than +IV
which arises after loss of all four valence electrons. Reason given for more stability of +II O.S. that 5s
2
electrons
are reluctant to participate in chemical bonding because bond energy released after the bond formation is less
than that required to unpair these electrons (lead forms a weak covalent bond because of greater bond length).
Example 36. Why does PbI4 not exist?
Solution: Pb(+IV) is less table than Pb(+II) due to inert pair effect and therefore Pb(+IV) is reduced to
Pb(+II) by I
–
which changes to I2(I
–
is a good reducing agent)
MISCELLANEOUS EXERCISES
Exercise 1: Compare the bond energies of N2, +
2
N & -
2
N .
Exercise 2: Though the electronegativities of nitrogen and chlorine are same, NH3 exists as liquid whereas
HCl as gas. Why?
Exercise 3: Explain giving reason:
-
2
ClF is linear, but the ion +
2
ClF is bent
Exercise 4: Explain with reason:
Two different bond lengths are observed in PF5 but only one bond length observed in SF6.
Exercise 5: (a) Amongst BBr3 and BF3 which is a stronger acid and why?
Exercise 6: Explain why the measured resultant dipole moment for FNO is 1.81 D, is so much higher than the
value for nitryl fluoride FNO2 (0.47 D).

Exercise7: Why
O
H
OH
(I)
is liquid at room temperature while
O
H
OH
(II)
is a high melting solid?
Exercise 8: (i) Among the compounds CH3COOH, NH3, HF and CH4 in which maximum hydrogen bonding is
present?
(ii) Which one of the following has strongest bond?
HF, HCl, HBr, HI
Exercise 9: Why BeF2 and BF3 are stable though Be and B have less than 8 electrons? Which one is more
stable?
Exercise10: Ether
R
O
R
and water
H
O
H
have same hybridisation at oxygen. What
angle would you expect for them?
ANSWER TO MISCELLANEOUS EXERCISES
Exercise 1: The configuration of N2 is
2
1s
 2
1s
*
 2
2s
 *2
2s
 y
z
2p 2
2px
2p
 

 

 

 
 
y
z
2p
2p
*
*
 

 
 
 

 
Now 2
N
means removal of an electron from a bonding M.O. This will decrease the B.O.
B.O. of 2
N
= ½ (5 – 0) = 2.5
Now again for 2
N
bond order is ½ (6–1) = 2.5
So from the bond order it may seem that both 2
N
& 2
N
may have the same bond energy. But
removal of an electron from a diatomic species tend to decrease the inter electronic repulsion and
thereby shortens the bond length. So the bond energy becomes more than compared to 2
N
N2 > 2
N
> 2
N
Exercise 2: The size of nitrogen is less than the size of chlorine. Therefore, electron density in nitrogen is
more than that of chlorine. So, nitrogen forms hydrogen bonding leading to association of
molecules. Hence, NH3 is liquid. Hydrogen bonding is not possible with chlorine.
Exercise 3: In 2
ClF
central chorine atom involves sp
3
d hybridisation, to have minimum electronic repulsion
three lone pairs should be in equitorial position as follows; giving linear shape to the ion.
Whereas, in case of 2
ClF
ion central atom Cl involves sp
3
hybridisation having two lone pairs,
resulting in bent shape for the ion, (bond angle less than 109°28 due to repulsion of bond pair by
lone pair)
F
F
120°
Linear shape-bond angle-180°
Cl
F F
Bent shape
+
Exercise 4: PCl5 has trigonal pyramidal structure (sp
3
d hybridisation of central atom) in which bond angles
are 90° and 120° respectively and there are two types of bond axial and equatorial. In case of SF6
the structure is octahedral (sp
3
d
2
hybridization of the central atom - S) resulting only one type of
bond, bond angle (90°) and one type of bond length.

29
(PF5)
(Bond angle 90° & 120°)
(SF6)
F
F
F
F
F
F
Bond angle - 90°
P
F
F
F
F
F
Exercise 5: (a) BBr3 since back bonding is present in BF3.
Exercise 6: Molecular symmetry (in terms of bond angles) leads to lesser dipole moment.
Exercise 7: (i) shows intramolecular hydrogen bonding while
(ii) shows intermolecular hydrogen bonding.
Exercise 8: (i) HF due to maximum electronegativity of fluorine.
(ii) HF
Exercise 9: The stability is explained by symmetrical linear structure for BeF2 and triangular planar structure
for BF3. BeF2 is more stable because its greater bond angle (180
o
).
Exercise 10: In H2O bond angle is less than 109
o
.28‘ due to lone pair and bond pair repulsion. But in ether, due
to strong mutual repulsion between two alkyl groups bond angle becomes greater than 109
o
.28‘.
Board Type QuestionS .
Prob 2. Arrange the bonds in order of increasing ionic character in the molecules:
LiF, K2O, N2, SO2 and ClF3
Sol. N2 < ClF3 < SO2 < K2O < LiF
Prob 3. Arrange the following in order of increasing ionic character:
CH, FH, BrH, NaI, KF and LiCl
Sol. CH < BrH < FH < NaI < LiCl < KF
IIT Level Questions
Prob 6. Predict the shapes of the following ions
(a) BeF3
-
(b) BF4
-
(c) IF4
-
(d) IBr2
-
Sol. (a) Triangular
(b) Tetrahedral
(c) Square planar
(d) Linear
Prob 8. Arrange the following in increasing order of stability O2, O2
+
, O2
-
, O2
2-
Sol. O2
2-
< O2
-
< O2 < O2
+
Calculate first the bond order which is as follows
O2  2, O2
+
 2.5, O2
-
 1.5, O2
2-
1 & then arrange according to increasing bond order.
Prob 11. Arrange the following:
(i) N2, O2, F2, Cl2 in increasing order of bond dissociation energy.
(ii) Increasing strength of hydrogen bonding (X – H – X):
O, S, F, Cl, N
Sol. (i) F2 < Cl2 < O2 < N2 (ii) Cl < S < N < O < F
Prob 12. Explain the following

o - hydroxy benzaldehyde is liquid at room temperature while p - hydroxy benzaldehyde is high melting
solid.
Sol. There is intramolecular H bonding in o - hydroxy benzaldehyde while intermolecular hydrogen bonding
in p-hydroxy benzaldehyde
CH=O---H
o-hydroxy benzaldehyde
CH=O---H—O
O—H----OCH
CHO
OH
O
Prob 14. Explain why ClF2
-
is linear but ClF2
+
is a bent molecular ion?
Sol. Chlorine atom lies in sp
3
d hybrid state. Three lone pairs are oriented along the corners of triangular
plane
F Cl F
2
ClF
 
 
Chlorine atom lies in sp
3
hybrid state. Two lone pairs are oriented along two corners of tetrahedral
Cl
F
F
[ClF2
+
]
Prob 16. AlF3 is ionic while AlCl3 is covalent.
Sol. Since F
–
is smaller in size, its polarisability is less and therefore it is having more ionic character.
Whereas Cl being large in size is having more polarisability and hence more covalent character.
Prob 18. Write down the resonance structure of nitrous oxide  
2
N O
Sol.
N N O N N O
Prob 20. Explain why BeH2 molecule has zero dipole moment although the BeH bonds are polar.
Sol. BeH2 is a linear molecule (HBeH) with bond angle equal to 180
o
. Although the BeH bonds are polar
on account of electronegativity difference between Be and H atoms, the bond polarities cancel with
each other. The molecule has resultant dipole moment of zero.
1. What are the factors influencing ionic bond formation?
2. Out of MgO and NaCl, whech has higher lattice energy and why?
3. BaSO4 being an electrovalent compound and still it does not pass into solution state in water.
4. Why an ionic bond is formed between two elements having large difference in their
electronegativity?
COVALENT BOND :

31
6. Which compound has 3 carbon atoms, 4 double bonds and the total no. of atoms are five?
7. Which compound has 3 carbon atoms. 4 and 4 bonds but not 4 double bonds and the total
no. of atoms are 5. Writes its IUPAC
8. Element A has 3 electrons in the valency shell and its principal quantum no. for the last electron
is 3 and element B has 4 electrons in the valency shell and its principal quantum to. For the last
electrons is 2. Identify the compound and it‘s nature of bonding.
9. The compound X has 8 atoms, 4  bonds, no  bonds, no. ionic bonds, no coordinate bonds, no
H-bond. Explain its structure.
10. The compound X has 8 atoms, 0  bonds, no ionic or  bonds. Explain its structure.
11. SnCl4 has melting point –15ºC where as SnCl2 has melting point 535ºC. Why?
12. Inorganic benzene is more reactive than organic benzene. Why?
FAJAN’S RULE :
13. SnCl2 is white but SnI2 is red. Why?
14. Explain the least melting point and highest solubility in H2O.
(i) LiCl NaCl KCl
(ii) NaCl NaBr Nal
(iii) LiCl BeCl2 BCl3
(iv) beSO4 MgSO4 CaSO4 SrSO4 BaSO4
(v) CaF2 CaCl2 CaBr2
15. Which one has highest and lowest melting point and why?
NaCl KCl RbCl CsCl
16. LiOH and carbonates decomposes on heating in I-group. Other hydroxides and carbonates of
this group will not. Why?
LEWIS STRUCTURE AND FORMAL CHARGE :
17. Draw the Lewis structures of the following molecules and ions.
PH3, H2S, BeF2, SiCl4, N2O4, H2SO4, O2
2
, IO6
5-
.
V.B.T. & HYBRIDISATION :
20. Explain hydrbidisation in
(1) XeF2 (2) XeF6 (3) PCl3 (4) IF3
(5) IF5 (6) IF7 (7) CCl4 (8) SiCl4
(9) SlH4 (10) H2O
22. PH5 is not possible but PCI5 is possible. Why?
VSEPR THEORY :
24. Which one has highest and least bond angle in the following -
(1) NH3 PH3 AsH3 SbH3
(2) CH4 PH3 AsH3 SbH3
(3) H2O H2S H2Se H2Te
(4) CH4 PH3 AsH3 H2Te
(5) CH4 SiH4 CCl4 SiCl4
(6) NCl3 PCl3 NBr3 PBr3
(7) PF3 PH3
(8) As3F3 AsH3
MOT :
29. Super oxide are coloured and paramagnetic why?
30. Write the important conditions required for the linear combination of atomic orbitals to form
molecular orbitals.
33. Exalain why NO+
is more stable towards dissociation into its atoms than NO, where as CO+
is
less stable than CO.
BACK BONDING AND HYDROLYSIS :

40. Arrange the following boron trihalides in the increasing order of their ease of hydrolysis. Also
give the reason for the same.
BF3, BCl3, BBr3
DIPOLE MOMENT :
53. Why NH3 is having more dipole moment than NF3.
54. Why is having more dipole moment than /
55. Why p-dichloro benzene in having zero dipole moment while hydroquinone is having some
dipole moment?
56. Why CH3 Cl is having high dipole moment than CH3F?
57. Why ortho fluoro Phenol have greater dipole moment than ortho chloro phenol?
58. Why trans–1, 2 sichloro ethene in having zero dipole moment than cis form?
59. Write the order of dipole moments of 1, 2–1.3– and 1, 4–dichlorobenzene.
60. Arrange in increasing order of dipole moment ; H2O, H2S, BF3.
61. BcF2 has zero dipole moment whereas H2O has a dipole moment?
62. CCl4 having zero dipole momnet but CHCl3 having some dipole moment?
63. While down the resonance structure (S) for :
(i) N3
–
(ii) O3 (iii) CO2
(iv) N2O4 (v) SCO2
2–
1. Arrange the following in increasing order of property given-
(i) O, F, S, Cl, N strength of H-bonding (X-H_X)
(ii) N2, O2, F2, Cl2 bond dissociation energy
(iii) MCl, MCl2, MCl3 ionic nature
(iv) HI, HBr, HCl, HF dipole moment
(v) AsH3, PH3, NH3 bond angle
5. Explain the structure hybridisation and oxidation state of S i sulphuric acid, marshall‘s acid,
caro‘s acid, oleum.
6. Boric acid is monobasic acid. Why?
7. Boron has exceptionally high melting point. Why?
8. BCl3 is more acidic than BF3. Why?
9. CCl4 is not dissolved in H2O but SiCl4 dissolves. Why?
10. Trimethylamine (CH3)3 N, is pyramidal but trisylyamine (SiH3)3 N is planer. Why?
11. SnCl4 has melting point –15ºC Where as SnCl2 has melting point 535ºC.Why?
12. PbCl4 is possible but PbBr4 and Pbl4 are not. Why?
13. Pb+4
, Bi+5
and Tl+3
act as oxidising agent. Why?
14. NCl3 & PCl3 on hydrolysis will give different products. Why?
15. ClO2 does not forms dimer but NO2 forms. Why?
16. How many  and  bonds are presents in hexacyanoethane and tetra cyanoethylene?
17. Explain the structure of ClF3 on the basis of bent rule.
18. All bonds length of PCl5 are not equal but PF5 has same bond lengths. Why?
19. The experimentally determined N – F bond length in NF3 is greater than sum of single bond
covalent radii of N and F.
24. The dipole moment of HBr is 7.95 debye and the intermolecular separation is 1.94 × 10-10
m
Find the % ionic character in HBr molecule.
25. HBr has dipole moment 2.6 × 10-30
cm. If the ioninc character of the bond is 11.5%, calculate
the interatomic spacing.
26. Dipole moment of LiF was experimentally determined and was found to be 6.32 D. Calculate
percentage ionic character in LiF molecule Li – F bond length is 0.156 pm.
27. A diatomic molecule has a dipole moment of 1.2 D. If bond length is 1.0 A, what percentage of
an electronic charge exists on each atom.
SECTION (B) FILL IN THE BLANKS :

33
1. Two atoms of similar electronegativity are expected to form __________ compound.
2. When two atoms approach each other, potential energy ______ and a ___________ is formed
between them.
3. Conversion of a neutral atom into a cation is _________ process.
4. The strongest hydrogen bond is formed between _________ and hydrogen.
5. Low ionization potential of electropositive element and high electron affinity of electronegative
element favours the formation of __________ bond.
6. NaCl is soluble in water due to its low ___________ energy.
7. The ________ value of lattice energy of a crystal favours the formation of an ionic compound.
8. Solid NaCl ___________ conductor of electricity.
9. For dissolution of an ionic solid, ________ energy should be low and hydration energy should
be _________________.
10. ___________ cation and ___________ anion favour covalency.
11. Anhydrous AlCl3 is a _____________ compound while hydrated AlCl3 is ___________.
12. Covalent compounds are generally _________ conductors of electricity.
13. There are ___________  bonds in a nitroge molecule.
14. A double bond is shorter than ___________ bond.
15. Axial overlapping of half-filled atomic orbitals results in __________ bond.
16. ____________ and ___________ bonds are present in N2O5 molecule.
17. The angle between two covalent bonds is maximum in _________ (CH4, H2O, CO2).
18. ___________ hybrid orbital of nitroger atom are involved in the formation of ammonium ion.
19. The hybridization state of oxygen in water molecule is ___________.
20. In the ion [Cu (H2O)4]++
, copper is in dsp2
state of hybridization. The shape of the ion is
________.
TRUE OR FALSE :
5. A diatomic molecule has a dipole moment of 1.2 D. If bond length is 1.0 a, what percentage of
an electronic charge exists on each atom.
6. The size of negative ion decreases with increasing magnitude of negative charge.
7. Ca2+
is smaller in size than K+
because the effective nuclear charge is greater.
8. The higher the latteice energy of an ionic solid, the greater will be its stability.
9. Molten sodium chloride conducts electricity due to the presence of free ions.
10. Linear overlap of two atomic p-rbitals leads to a -bond.
11. The H-N bond angle in NH3 is greater than the H-AS-H bond angle in AsH3.
12. sp2
hybrid orbitals have equal s and p-character.
13. The tetrahedral geometry in SiF4 is due to sp3
hybridization of Si atom.
14. SnCl2 is a non-linear molecule.
15. There are seven electron bond pairs in lF7 molecules.
16. Dipole moment of CHF3 is greater than CHCl3.
17. Dipole moment of NF3 is lesser than NH3.
18. Among HF, HCl, HBr and HI, HF has highest dipole moment.
19. All molecules with polar bonds have dipole moment.
20. The presence of polar bonds in a polyatomic molecule suggests that the molecule has non zero
dipole moment.
21. AgCl is more covalent than NaC.
REASONING AND ASSERTION :
Direction: These quaestions consist of two statements each printed as Assertion and Reason.While
answering these questions you are required to choose any one of the following four responses
to encircle (A, B, C, D) as per instructions given below:
(A) If both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(B) If both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(C) If Assertion is true but Reason is false.
(D) If Assertion is false but Reason is true.
3. Assertion : Na2SO4 is soluble in water while BaSO4 is insoluble.
Reason : Lattice energy of BaSO4 exceeds its hydration energy.
(a) A (b) b (c) C (d) D

4. Assertion : py and pz cannot combine to give molecular orbital.
Reason : Both py and pz are dumb-bell shaped.
(a) A (b) b (c) C (d) D
7. Assertion : Although PF5, PCl5 and PBr5 are known, the pentahalides of nitrogen have not been
observed.
Reason : Phosphorus has lower electronegativity than nitrogen.
(a) A (b) b (c) C (d) D
9. Assertion : NO3
–
is planar while NH3 is pyramidal.
Reason : N in NO3
–
is sp2
and in NH3 it is sp3
hybridized.
(a) A (b) b (c) C (d) D
11. Assertion : s-orbital cannot accommodate more than two electrons.
Reason : s-orbitals are extremely poor shielders.
(a) A (b) b (c) C (d) D
MATCH THE FOLLOWING :
1. Column - I Column - II
(A) Size of secondary layer of hydrated ions (P)Maximum in solid and minimum in gaseous
state.
(B) Magnitude of hydrogen bonding (Q) Strength of ion - dipole attraction
(C) Mobility of ions in water (R) Inversely proportional to the size of metal
ion
(D) Degree of polarity of a bond (S) Dipole moment
(T) Directly proprotional to the size of metal ion
(A) SO3 (gas) (P)Polar with p - d bonds and identical S – O
bond, lengths.
(B) OSF4 (Q) One lone pair and p – d bond.
(C) SO3F–
(R) non - polar with p –p and p – d bonds.
Identical S–O bod lengths.
(D) ClOF3 (S) Polar with p – d bond.
(T) Hybridisation of central atom in ClO2F3.
Molecule/ion Hybridisation of central atom
(A) IO2F2
–
(P) sp3
d
(B) F2SeO (Q) sp3
(C) ClOF3 (R) sp2
(D) XeF5
+
(S) sp2
ONLY ONE ANSWER CORRECT:
1. In which molecule (s) is/are the vander waals force likely to be most important in determining
m.p. and b.p.
(a) ICI (b) Br2
(c) H2S (d) CO
2. Which is the most ionic
(a) LiF (b) Li2O (c) Li3N (d) All same
3. The correct order of the increasing ionic character is-
(a) BeCl2 < MgCl2 < CaCl2 < BaCl2 (b) BeCl2 < MgCl2 < BaCl2 < CaCl2
(c) BeCl2 < BaCL2 < MgCl2 < CaCl2 (d) BaCl2 < MgCl2 < CaCl2 < BeCl2
4. An ionic bond A+
B–
is most likely to be formed when :
(a) the ionization energy of A is high and the electron affinity of B is low
(b) the ionization energy of A is high and the electron affinity of B is low
(c) the ionization energy of A and the electron affinity of B is high
(d) the ionization energy of A and the electron affinity of B is low
5. Which of the following compounds of elements in group IV is expected to the most ionic?

35
(a) PbCl2 (b) PbCl4 (c) CCl4 (d) SiCl4
6. The molecule BF3 borth are covalent compounds. But BF, is non,polar and NF3 is polar. The
reason is that-
(a) boron is a metal and nitrogen is a gas in uncombined state
(b) B – F bond have no dipole moment whereas N – F bond have dipole moment
(c) atomic size of boron is smaller than that of nitrogen
(d) BF3 is symmetrical molecule where as NF3 is unsymmetrical
7. Least melting point is shown by the compound-
(a) PbCl2 (b) SnCl4 (c) NaCl (d) AlCl3
8. Which of the following is in rder of increasing covalent character?
(a) CCl4 < BeCl2 < BCl3 < LiCl (b) LiCl < CCl4 < BeCl2 < BCl3
(c) LiCl < BeCl2 < BCl3 < CCl4 (d) LiCl < BeCl2 < CCl4 < BCl3
9. Which of the following compounds contain/s both ionic and covalent bonds?
(a) NH4Cl (b) KCN (c) CuSO4.5H2O (d) NaOH
10. The tyes of bonds present in CuSO4.5H2O are
(a) electrovalent and covalent (b) electrovalent and coordinate
(c) covalent and coordinate (d) electrovalent, covalent and coordinate
11. Three centre-two electron bonds exist in :
(a) B2H6 (b) Al2(CH3)6 (c) BeH2(s) (d) BeCl2(s)
Octet Rule :
12. Example of super octet molecule is :
(a) SF6 (b) PCl5 (c) IF7 (d) All of these
13. To which of the following species octet rule is not applicable :
(a) BrF5 (b) SF6 (c) IF7 (d) CO
14. NH3 and BF3 combine readily because of the formation of :
(a) a covalent bond (b) hydrogen bond (c) acoordinate bond (d) an ionic bond
15. Which of the following species cotain covalent and ccordincate bond.
(a) AlCl3 (b) CO (c) [Fe(CN)6]4–
(d) N–
3
FAJANS RULE
16. Which of the following combination of ion will have highest polarisation
(a) Fe2+
, Br–
(b) Ni4+
, Br–
(c) Ni2+
, Br–
(d) Fe, Br–
17. Which of the following cannot be explained on the basis of Fajan‘s Rules.
(a) Ag2S is much less soluble than Ag2O
(b) Fe(OH)3 is much less soluble than Fe(OH)2
(c) BaCO3 is much less soluble than MgCO3
(d) Melting point of AlCl3 is much less than that of NaCl
18. The correct order of decreasing polarizability of ion is :
(a) Cr–
, Br–
, I–
, F–
(b) F–
, I–
, Br–
, Cl–
(c) I–
, Br–
, Cl–
, F–
(d) F–
, Cl–
, Br–
, I–
19. Which ion has a higher polarising power
(a) Mg2+
(b) Al3+
(c) Ca2+
(d) Na+
20. Which combination will show maxinum polarising power & maxinum polarisability
(a) Mn2+
.F–
(b) Mn7+
, I–
(c) Mn2+
, I–
(d) Mn7+
, F–
LEWIS STRUCTURE :
22. Pick out among the following species isoelectronic with CO2.
(a) N–
3 (b) (CNO)–
(c) (NCN)2–
(d) NO–
2
V.B.T. & HYBRIDISATION :
23. The strength of bonds by s – s, p – p, s – p overlap is in the order :
(a) s – s < s – o < p – p (b) s – s < p – p < s – p
(c) s – p < s – s < p – p (d) p – p < s – s < s – p.
24. Number and ype of bonds between two carbon stoms in CaC2 are :

(a) one sigm3 () and one pi () bond (b) one  and two  bonds
(c) one  and one and a half  bond (d) one  bond
25. In the context of carbon, which of the follwoing is arranged in the correct order of
electronegativity:
(a) sp > sp2
> sp3
(b) sp3
> sp2
> sp (c) sp2
> sp > sp3
(d) sp3
> sp > sp2
26. Which of the following oxyacids of sulphur contain ‗S – S bonds?
(a) H2S2O8 (b) H2S2O6 (c) H2S2O4 (d) H2S2O5
27. Which of the following represent the given mode of hybridisation sp2
– sp2
– sp – sp from left to
right
(a) H2C = C = c = CH2 (b) HC  C – C  CH
(c) H2C = CH – C  N (d) H2C = CH – C  CH
28. For BF3 molecule which of the following is true.
(a) B-atom is sp2
hybridised
(b) There is a P –P back bonding in this molecule
(c) Observed B–F bond length is found to be less than the expected bond length.
(d) All of these
29. Which of the following is isoelectronic as well as isostructura‘ with N2O
(a) N3H (b) H2O (c) NO2 (d) CO2
30. The hybridization state of B in B2H6 is –
(a) sp (b) sp2
(c) sp3
(d) sp3
d
31. What is not true for SiH4 molecule –
(a) Tetrahedral hybridisation (b) 109º angle
(c) 4 bond (d) 4-lone pair of electrons
32. Which of the following has a geometry different from the other three species (having the same
geometry)?
(a) (b) (c) XeF4 (d)
33. In C – C bond C2H6 undergoes heterolytic fission, the hybridisation of two resulting carbon
atoms is/are
(a) sp2
both (b) sp3
both (c) sp2
, sp3
(d) sp, sp2
34. Which of the following statements are not correct?
(a) Hybridization is the mixing of atomic orbitals of large energy difference.
(b) sp2
– hybrid orbitals are formed form two p – atomic orbitals and one s-atomic orbitals
(c) dsp2
– hybrid orbitals are all at 90º to one another
(d) d2
sp3
– hybrid orbitals are directed towards the corners of a regular octahedron
35. p – d back bonding occurs between oxygen and
(a) phosphorus in P4O10 (b) chlorine in HClO4
(c) nitrogen in N2O5 (d) carbon in CO2
36. In which of the following groups all the the members have linear shape
(a) NO2
+
, N3
–
, H – C – H (b) N3
–
, I3
–
, H – C – H
(c) XeF2, C2H2, SO2 (d) CO2, BeCl2, SnCl2
37. Consider the following molecules :
H2O H2S H2Se H2Te
I II III IV
Arrange these molecules in increasing order of bond angles.
(a) I < II < III < IV (b) IV < III < II < I (c) I < II < IV < III (d) II < IV < III < I
38. Which has the smallest bond angle (X – O – X) in the given molecules?
(a) OSF2 (b) OSCl2 (c) OSBr2 (d) OSI2
39. Consider the following iodides :
PI3 AsI3 SbI3
102º 100.2º 99º
The bond angle is maximum in Pl3, which is
(a) due to small size of phosphorus (b) due to more bp–bp repulsion in PI3
(c) due to less electronegativity of P (d) none of these

37
BONDS ANGLES & BOND LENGTH :
40. The correct order of increasing X – O – X bond angle is (X = H, F or Cl):
(a) H2O > Cl2 > F2O (b) Cl2O > H2O > F2O
(c) F2O > Cl2O > H2O (d) F2O > H2O > Cl2O
41. Which of the following compounds have bond angle as nearly 90º?
(a) NH3 (b) H2S (c) H2O (d) SF6
42. Number of non bonding electrons in N2 is :
(a) 4 (b) 10 (c) 12 (d) 14
43. Which of the following species is paramagnetic ?
(a) NO–
(b) (c) CN–
(d) CO
44. Tho bond order depends on the number of electrons in the bonding and non bonding orbitals.
Which of the following statements is/are correct about bond order?
(a) Bond order cannot have a negative value
(b) It always has an integral value
(c) It is a nonzero quantity
(d) It can assume any value-positive or negative, integral or fractional, including zero
45. In the formation of from N2, the electron is removed from :
(a)  orbital (b)  orbital (c) * orbital (d) * orbital
46. Which of the following has fractional bond order :
(a) (b) (c) (d)
47. How many unpaired electrons are present in :
(a) 1 (b) 2 (c) 3 (d) 4
48. Which of the following have identical bond order?
(a) (b) (c) NO (d)
49. Which of the following have identical bond order?
(a) (b) NO+
(c) CN–
(d) CN+
50. Given the species : N2, CO, CN–
and NO+
, Which of the following statements are true for these
(a) All species are paramagnetic (b) The species are isoelectronic
(c) All the species have dipole moment (d) All the species are linear
51. Which of the following are paramangetic?
(a) B2 (b) O2 (c) N2 (d) He2
52. Which of the following species have a bond order of 3?
(a) CO (b) CN–
(c) NO+
(d)
53. Which of the following pairs have identical values of bond order?
(a) and (b) F2 and Ne2 (c) O2 and B2 (d) C2 and N2
54. Find out the bond order of :
(a) H2 (b) (c) He2 (d) Li2 (f) B2
55. The correct order of boiling point is :
(a) H2O < H2S < H2Se < H2Te (b) H2O > H2Se > H2Te > H2S
(c) H2O > H2S > H2Se > H2Te (d) H2O > H2Te > H2Se > H2S
56. Which of the following models best describes the bonding within a layer of the graphite
structure?
(a) metallic bonding (b) ionic bonding
(c) non-metallic covalent bonding (d) van der Waals forces
57. The critical temperature of water is higher than that of O2 because the H2O molecule has :
(a) fewer electrons than O2 (b) two covalet bonds
(c) V - shape (d) dipole moment
58. Arrange the following in order of decreasing boiling point :
(I) n-Butane (II) n-Butanol (III) n-Butyl chloride (IV) Isobutane
(a) IV > III > II > I (b) IV > II > III > I (c) I > II > III > IV (d) II > III > I > IV
59. For H2O2, H2S, H2O and HF, the correct order of increasing extent of hydrogen bonding is :
(a) H2O > HF > H2O2 > H2S (b) H2O > HF > H2S > H2O2
(c) HF > H2O > H2O2 > H2S (d) H2O2 > H2O > HF > H2S
60. Which one of the following does not have intermolecular H-bonding?

(a) H2O (b) o-nitro phenol (c) HF (d) CH3COOH
61.
(a) has intermolecular H - bonding (b) has intramolecular H-bonding
(c) has low boilng point (d) is steam-volatile
62. Which of the following is/are observed in metallic bonds?
(a) Mobile valence electrons (b) Overlapping valence orbitals
(c) Highly directed bond (d) Delocalized electrons
63. Intermolecular hydrogen bonding increases the enthalpy of vapourization of a liquid due to the
(a) decrease in the attraction betwwen molecules
(b) increase in the attraction between molecules
(c) decrease in the molar mass of unassociated liquid molecules
(d) increase in the effective molar mass of hydrogen - bonded molecules
64. Of the following molecules, the one, which has permanent diple moment, is-
(a) SiF4 (b) BF3 (c) PF3 (d) PF5
65. The dipole moments of the given molecules are such that-
(a) BF3 > NF3 > NH3 (b) NF3 > BF3 > NH3 (c) NH3 > NF3 > BF3 (d) NH3 > BF3 > NF3
66. Which of the following has the least dipole moment
(a) NF3 (b) CO2 (c) SO2 (d) NH3
67. Which of the following compounds possesses zero dipole moment?
(a) Water (b) Benzene
(c) Carbon tetrachloride (d) Boron trifluoride
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