Chemical Bond
The force of attraction which holds various constituents (atoms or ions) together in different chemical
species called chemical Bond.
Chemical Bond
Strong Bond
Weak Bond
Ionic Bond
Hydrogen Bond
Covalent Bond
Coordinate Bond
Metallic Bond
Lewis Symbols : Simple notations to represent the valence electrons of an atom are called Lewis
symbols. In these symbols, the valence electrons are represented by placing dots (•) or crosses (×)
•
around the symbols. e.g : Li,
•
•
B •,
Be,
•
•
•
••
:
:N• • , :Ne
••
Q. Write Lewis dot symbols for atoms of the following elements: Mg,Na, B, O,N, Br.
Element
No. of valence electrons
Lewis Dot structure
Mg
2
Mg
Na
1
Na
B
3
•
•B•
O
6
••
••
[NCERT Exercise]
••
N
5
••
•
Sol.
••
Br
7
••
Br
•
••
•
O
••
••
N
Octet Rule : Atoms of various elements can combine either by transference of valence electron from
one atom to another (gaining or losing) or by mutual sharing of valence electrons in order to have to
attain the configuration of eight electrons in outermost shell. This is called octet rule.
Covalent Bond → Bond formed by mutual sharing of electrons.
Important conditions for Lewis dot structure
• Each bond is formed as a result of sharing of an electron pair
between the atoms.
• Each combining atom contributes at least one electron to the
shared pair.
• The combining atoms attain the outer-shell noble gas configurations
as a result of the sharing of electrons.
APNI KAKSHA
1
Formation of Cl2 :
Cl → [Ne]3s2 3p5 →
it has 1 electron less for stable configuration.
Cl — Cl
If the number of shared electrons between two atoms are 2, 4 or 6 then a covalent bond is called
single, double or triple bond respectively.
..
..
..
O
..
O
..
..
..
:O + O
H—H
(single bond)
.. ..
O2 molecule :
or
..
H• •H
..
H2 molecule :
O
O
:N
+ N
..
N2 molecule :
…
double bond
…
•
Lewis Dot structures:
(Practice Question in the End, Q.3.5)
Step 1: Count total number of electrons required for writing the structures by adding the valence es
of combining atoms.
Step 2: In case of anions, add one electron to the total number of valence electrons for each negative
charge and for cation, subtract one electron from total number of valence electrons for each +ve
charge.
Step 3: Recall the chemical symbols of the combining atoms
and skeletal structure of the compound (known or guessed
intelligently), and try to distribute the total no. of electrons as
bonding shared pairs between the atoms in proportion to total
bonds.
Step 4: Generally, the least electronegative atoms occupy the
central position in the molecule/ion and is called central atom.
Step 5: After distributing the shared pair of
electrons for single bonds, the remaining electron
pairs are either utilised for multiple bonding or
remain as such in form of lone pairs, provide that
each combining atom gets an octet of electrons.
APNI KAKSHA
2
Molecule/ion
Lewis Representation
H2
H : H+
H—H
O2
O
O
O
O
O3
O
NF3
+
O+
–
O
O
F N F
F
HNO3
O
C
O
–
F N F
F
2–
O
CO32-
O
2–
O
O
O
+
C
O
+
O N O H
O
O N O H
O
–
–
Q. Draw the Lewis structures for the following molecules and ions:
H2 S, SiCl4 , BeF2 , CO3 2−,HCOOH.
Atom/Ions
[NCERT Exercise]
Lewis Dot structure
S
H2 S
H
H
Cl
SiCl4
BeF2
CO32−
Si
F
Be
O
C
F
O
2–
O
H
C
O
H
HCOOH
O
Formal Charge
Formal charge = no. of valence electrons in free atom – total no. of non-bonding (lone pair) electrons
–
total no. of bonding electrons
2
APNI KAKSHA
3
(a)
(b)
(c)
Sol. Carbonate ion CO 3 2– ion’s Lewis structure:
1
On Carbon = 4 − 0 −  (8) = 0
2
1
On Oxygen no. (1) = 6 − 4 − (4) = 0
2
1
On Oxygen no. (2) = 6 − 6 − (2) = −1
2
1
On Oxygen no. (3) = 6 − 6 − (2) = −1
2
Limitation of Octet Rule:
••
••
Q.
••
••
Let us consider the ozone molecule (O 3 ). The Lewis structure of O 3 may be drawn as :
1
••
The atoms have been numbered as 1, 2 and 3. The formal charge on:
O
1
The central O atom marked 1 = 6 − 2 − (6) = +1
2
•• 3
2
O
O
••
••
1
The end O atom marked 2 = 6 − 4 − (4) = 0
+
2
••
O
1
The end O atom marked 3 = 6 − 6 − (2) = –1
2
•• –
O
Hence, we represent O 3 along with the formal charges as follows:
O
••
••
Calculate the formal charge on each atoms of carbonate ion?
[NCERT Exercise]
2–
(1)
O
(2)
O
C
(3)
O
••
(i) Electron deficient molecules: There are some molecules in which the central atom is surround by
less than eight electrons, i.e their octet is incomplete but they exist. E.g. BeCl2 , BeH 2 etc.
Cl
••
Li : Cl
H : Be : H
Cl •• B •• Cl
(ii) Odd electron molecules: In molecules with an odd number of electrons like nitric oxide, NO and
nitrogen dioxide, NO 2 , the octet rule is not satisfied for all the atoms.
•+
••
••
••
•• –
N
O
O
N
O
••
•
••
••
(iii) The expanded octet: Elements of the third and higher periods of periodic table because of
availability of d-orbitals can expand their covalency and can accommodate more than eight valence
••
••
electrons around central atom.
F
F
e.g. PF5 → 10 electrons around P-atom.
••
••
••
F
F
F
SF6 → 12 electrons around S-atom.
••
••
••
••
F
P
S
••
••
••
••
F
F
F
••
••
••
F
F
Other drawbacks of the octet theory
••
••
(iv) It does not tell anything about the shapes of molecules and their relative stabilities.
(v) It is based on inertness of noble gases but some noble gases like xenon and krypton form several
compounds with oxygen and fluorine like XeF 2 , XeF6 , XeOF4 , XeO 2 F2 , KrF2 etc.
(vi) It fails to explain paramagnetic behaviour of oxygen. (which should be diamagnetic according to
this rule but it is infect paramagnetic).
Diamagnetic → no unpaired electron.
••
••
••
••
••
••
••
••
••
••
••
••
••
••
••
Paramagnetic → atleast one paired electron
APNI KAKSHA
4
IONIC OR ELECTROVALENT BOND
The bond formed as a result of electrostatic force of attraction between two oppositely charged ions
(positive and negative ions).
Complete transfer of electrons takes place here.
••
•• −
e.g. Na • + • Cl•• ⎯⎯
→ Na + • Cl••
••
••
Ionisation Energy: The amount of energy required to remove the most loosely bond electron from
an isolated gaseous atom to form a +ve ion.
Electron Gain Enthalpy: The amount of energy released when an e is added to an isolated gaseous
atom to form a negative ion.
The formation of ionic compounds would primarily depend upon:
•
•
The ease of formation of the positive and negative ions from the respective neutral atoms;
The arrangement of the positive and negative ions in the solid, that is, the lattice of the crystalline
compound.
The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom
and that of the negative ion involves the addition of electron(s) to the neutral atom.
M(g)
⎯⎯
→ M+ (g) + e–
Ionization enthalpy
→ X – (g)
X(g) + e– ⎯⎯
Electron gain enthalpy
→ MX(s) Lattice energy
M+(g) + X – (g) ⎯⎯
Now, for Ionic Bond:
•
•
Lesser the Ionisation energy, more tendency to lose e  and hence greater tendency to form ionic
bonds.
Higher the value of negative electron gain enthalpy greater will be the case of formation of ionic
compound.
Lattice Enthalpy: The amount of energy required to completely separate one mole of a solid Ionic
compound in to gaseous ions.
→ M+(g) + X – (g)
MX(s) ⎯⎯
•
Charge ↑  L.E. ↑
+
−
2+
−1
: Na C l  Ca Cl2
2+ 2−
 Mg O 2
APNI KAKSHA
5
•
Size ↑  L.E. ↓
: NaCl > NaBr
BOND PARAMETERS
Bond Length: The equilibrium distance between the nuclei of two bonded atoms in a molecule is
called bond length.
Bond length (d) = rA + rB
Covalent radius = rA = rB =
d
2
••
••
Bond Angle: Angle between the orbitals containing bonding
electron pairs around central atom in a molecule/complex ion
is called the bond angle. Expressed in degrees.
O
104.5°
H
H
Bond Enthalpy: The amount of energy required to break one mole of bonds of a particular type
between two atoms in gaseous state. Unit of bond enthalpy is kJ mol–1 .
For breaking H 2 bond (H — H bond):
→ H(g) + H(g) ; ΔH° = 435.8 kJ mol–1 .
H 2 (g) ⎯⎯
Bond order: The number of bonds formed between the two atoms in a molecule is called the bond
order.
Molecule
Bond order
H─H
1
N≡N
3
O═O
2
C≡O
3
 Bond order  Bond enthalpy 
1
Bond length
Resonance
Only one lewis dot structure cannot explain the properties of the molecules as more than on lewis
dot structure may be possible for a molecule, these different lewis dot structure are known as
APNI KAKSHA
6
Resonating structures or canonical forms. The actual structure is in between of all these contributing
structures and is known as resonance hybrid. This phenomenon is called Resonance.
Single Lewis dot structure based on the presence of two single bonds and one double bond between
carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal
bonds.
According to experimental findings, all carbon to oxygen bonds in CO32− are equivalent. Therefore,
the carbonate ion is best described by resonance hybrid of canonical forms.
•
•
The canonical forms are only imaginary structures i.e. do not exist in real.
The molecule does not exist for a certain fraction of time in one canonical form and for other
fractions of time in other canonical forms.
The difference between energy of most stable resonating structure and resonance hybrid is known
as Resonance Energy.
(Practice Question in the End, Q.6.)
VSEPR Theory
•
•
•
•
•
•
Valence Shell Electron Pair Repulsion (VSEPR) Theory explains the shape of molecule. Main
postulates of this theory:
The number of valence shell electron pairs (bonded or non-bonded) present around the central atom
decides the shape of the molecules. (The shared electron pairs are called bond pairs and unshared or
non-bonding electrons are called lone pairs).
Electron pairs of valence shell repel one another because their electron clouds are negatively
charged.
These electron pairs arrange themselves in such a way so that there is minimum repulsion and
maximum distance in between them.
Valence shell is considered as sphere in which electron pairs are localised on spherical surface at
maximum distance from one another.
A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a
multiple bond are treated as a single super pair.
A lone pair occupies more space than a bonding pair, since it lies closer to central atom.
[L.P → lone pair, BP → Bone pair]
⸫
Order of Repulsion: (LP – LP) > (B.P. – L.P.) > (B.P. – B.P.)
APNI KAKSHA
7
Shape of Molecule on basis of VSEPR Theory: It is depending on (L.P. + B.P.) present around the
central atom of a molecule as:
APNI KAKSHA
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APNI KAKSHA
9
Q. Although geometries of 𝐍𝐇𝟑 and 𝐇𝟐 𝐎 molecules are distorted tetrahedral, bond angle in
water is less than that of ammonia. Discuss.
[NCERT Exercise]
Sol.
No. of lone pair =
1
2
More no. of lone pair of central atom ⟹ Higher will be repulsion ⟹ Less bond angle
Polarity of Bond
The existence of a hundred percent ionic or covalent bond represents an ideal situation. In reality no
bond or a compound is either completely covalent or ionic.
In HF, the shared electron pair between the two atoms gets displaced more towards fluorine since the
electronegativity of fluorine is far greater than that of hydrogen. The resultant covalent bond is a polar
covalent bond.
H
F

+
–
H
F
Dipole moment
(Practice Question in the End, Q.2.)
Dipole moment is the measure of degree of polarity and is defined as the product of the magnitude
of the charge and the distance between the centres of positive and negative charge.
Symbol → µ (mu).
Dipole moment (µ) = charge (Q) × distance of separation (r)
APNI KAKSHA
10
Unit → Debye [1D = 3.3 × 10–30 C m]
It is a vector quantity and is represented by the crossed arrow ( +⎯⎯
→ ) painting towards more
electronegative atom.
•
•
•
In case of diatomic polar molecules, the dipole moment of the molecule is equal to the dipole
moment of polar bond. eg. The dipole moment of HCl molecule is same as that of H — Cl bond.
In case of polyatomic molecules, dipole moment not only depend upon the dipole moment of
individual bonds but also upon the arrangement of bonds present in the molecule. In such molecules,
net dipole moment is the vector sum of the dipole moments of various bonds.
In case of symmetrical molecules like BF 3 , CH 4 , CCl4 etc the molecular dipole moment is zero.
F
F  (
Be
+
)=0
F
F

B
(
)=0
+
F
NH3 Vs NF3
In case of NH 3 the orbital dipole due to lone pair
is in the same direction as the resultant dipole
moment of the N – H bonds, whereas in NF3 the
orbital dipole is in the direction opposite to
the resultant dipole moment of the three N–F
bonds. The orbital dipole because of lone
pair decreases the effect of the resultant N – F
bond moments, which results in the low dipole
moment of NF3 as represented below :
Type of
Molecule
Molecule (AB)
Example
Geometry
HF
HCl
HBr
HI
H2
Dipole Moment,
µ(D)
1.78
1.07
0.79
0.38
0
Molecule (AB 2 )
H2O
H2S
CO 2
1.85
0.95
0
bent
bent
linear
Molecule (AB 3 )
NH 3
NF3
BF3
1.47
0.23
0
trigonal-pyramidal
trigonal-pyramidal
trigonal-planar
Molecule (AB 4 )
CH 4
CHCl3
CCl4
0
1.04
0
tetrahedral
tetrahedral
tetrahedral
APNI KAKSHA
linear
linear
linear
linear
linear
11
Q. Although both 𝐂𝐎𝟐 and 𝐇𝟐 𝐎 is triatomic molecules, the shape of 𝐇𝟐 𝐎 molecule is bent while
that of 𝐂𝐎𝟐 is linear. Explain this on the basis of dipole moment.
[NCERT Exercise]
Sol. Experimentally it is found that the dipole moment of carbon dioxide is zero. This is possible
because CO2 molecule is linear so that the dipole moments of C – O bonds are equal and
opposite to nullify each other.
Resultant μ  0 D
Resultant μ = 0 D
H2 O has some dipole moment, it suggests that the structure of water H2 O molecule is bent.
Fajans’ Rule
All covalent bonds have some partial ionic character, the ionic bonds also have some partial covalent
character.
Polarization: When a cation approaches to an anion closely, it attracts the electron of anion and
same time it repels the positively charge nucleus of the anion, due to which electron charge density
of the anion no longer remain symmetrical, but is elongated towards the cation, Which is known as
distortion or polarization of anion by cation.
•
•
•
•
Small size of cation, greater covalent character of ionic bond.
Covalent character : Li+ > Na+ > K+ > Rb+ > Cs+.
Large size of Anion, greater covalent character of ionic bond.
Covalent character : F– < Cl– < Br– < I – .
Greater the charge on cation, greater the covalent character of ionic bond.
Covalent character : Na+ < Mg2+ < Al3+.
In case of two cations having same size and charge, the one with pseudo noble gas configuration
(with 18 es in outermost shell) has greater polarising power as compared to the other with noble
gas configuration (with 8 electrons in outermost shell).
Covalent character : Cu+ > Na+ .
Covalent character
Small size of cation
Large size of anion
APNI KAKSHA
High charge on
cation and anion.
Cations having
18-electron shell.
12
Valence Bond Theory (VBT)
It is based on the knowledge of atomic orbitals electronic configuration of elements, the criteria of
overlapping of atomic orbitals and the hybridisation of atomic orbitals and the principles of variation
and superposition.
Consider two hydrogen atoms A and B approaching each other having nuclei N A and N B and
electrons present in them are represented by e A and eB. When the two atoms are at large distance
from each other, there is no interaction between them. As these two atoms approach each other, new
attractive and repulsive forces begin to operate.
Attractive forces arise between:
(i) nucleus of one atom and its own electron that is N A – eA and N B– eB.
(ii) nucleus of one atom and electron of other atom i.e., N A– eB, N B – eA.
Similarly, repulsive forces arise between
(i) electrons of two atoms like eA – eB,
(ii) nuclei of two atoms N A – NB.
Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to
push them apart.
Bond enthalpy: Since the energy gets released when the bond is formed between two hydrogen
atoms, the energy so released is called as bond enthalpy, which is corresponding to minimum in the
curve. The hydrogen molecule is more stable than that of isolated hydrogen atoms.
H2 (g) + 435.8 kJ mol–1 → H(g) + H(g)
APNI KAKSHA
13
Overlapping of Atomic orbitals
When orbitals of two atoms come close to form bond, their overlap may be positive, negative or zero
depending upon the sign (phase) and direction of orientation of amplitude of orbital wave function in
space.
Positive / in phase overlap
Negative / out of phase overlap
Zero overlap (out of phase due to different orientation direction of approach)
Orbital Overlap Concept : According to this concept, a covalent bond is formed by the overlapping
of two half-filled atomic orbitals, of valence shell having electrons with opposite spins. Greater, the
overlapping of atomic orbitals. Stronger is the bond formed between two atoms.
Types of overlapping and nature of covalent bonds :
(I) Sigma (σ) bond : This type of bond is formed by end to end (head -on) overlap of bonding orbitals
along the internuclear axis. It is also known as head on overlap or axial overlap. This can be formed
by any one of the following types of combinations of atomic orbitals:
(a) s–s overlapping : Overlap of two half-filled s-orbitals along the internuclear axis.
APNI KAKSHA
14
(b) s–p overlapping : This overlap takes place between half-filled s-orbitals of one atom and half-filled
p-orbitals of another atom.
(c) p–p overlapping : Occurs between half-filled p-orbitals of the two approaching atoms.
(II) Pi (π) bond : Atomic orbitals overlap in such a way that their axes remain parallel to each other and
perpendicular to internuclear axis. The orbitals formed due to side wise overlapping of two saucer
type charged clouds above and below the plane of participating atoms.
π-bond is a weaker bond as compared to σ-bond due to smaller extent of overlapping. In the
formation of multiple bonds (double or triple bonds) between two atoms of a molecule, π-bond(s) is
formed in addition to a sigma bond. e.g: a double bond contains one σ and one π-bond, whereas a
triple bond contains one σ and two π-bonds.
(Practice Question in the End, Q.1, 7, 9)
Q. Distinguish between sigma and pi bond
Sol.
[NCERT Exercise]
Sigma () Bond
Pi () Bond
(a) It is formed by the end to end overlap of
orbitals. And along to principle axis.
It is formed by the lateral overlap of orbitals.
The axis will be perpendicular to the principle
axis.
(b) The orbitals involved in the overlapping
are s – s, s – p, or p – p.
These bonds are formed by the overlap of p – p
orbitals only.
(c) It is a strong bond (according to VBT)
It is a weak bond (according to VBT).
(d) The electron cloud is symmetrical about
the line joining the two nuclei.
The electron cloud is not symmetrical.
(e) It consists of one electron cloud, which is
symmetrical about the internuclear axis.
There are two electron clouds lying above and
below the plane of the atomic nuclei.
(f) Free rotation about  bonds is possible.
Rotation is restricted in case of pi-bonds.
APNI KAKSHA
15
Hybridisation
The atomic orbitals mix together to generate a new set of equivalent orbitals, called the hybrid orbitals
or hybridised orbitals. These orbitals are used in bond formation. The phenomenon is called
hybridisation.
Salient features of hybridisation:
• The number of hybrid orbitals generated is equal to the number of the atomic orbitals that
hybridised.
• The hybridised orbitals are always equivalent in energy and shape.
• The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
• These hybrid orbitals arrange themselves in space in such a direction so that there is minimum
repulsion between electron pairs and thus a stable arrangement is obtained. Therefore, the type of
hybridisation indicates the geometry of the molecules.
Important conditions for hybridisation:
• Only valence shell orbitals of an atom take part in hybridisation.
• The orbitals undergoing hybridisation should have almost equal energy.
• Promotion of electron is not essential condition prior to hybridisation.
• It is not necessary that only half-filled orbitals involve in the hybridisation. In some cases, even
filled orbital of valence shell can also involve in hybridisation.
Types of hybridisation:
Steric number
Hybridisation
Shape
Example
2
sp
linear
CO2 , BeF2 , C2 H2
3
sp2
trigonal planar
BCl3 , C2 H4
4
sp3
tetrahedral
CH 4 , C2 H6
5
dsp2
square planar
PtCl4 2–
6
sp3 d
trigonal
bipyramidal
PCl5
7
sp3 d 2
octahedral
SF6
Steric number = No. of lone pairs + No. of side atoms
For CO 2 , steric no. = 0 + 2 = 2, So Hybridisation at carbon atom is sp.
For BF3 , steric no. = 0 + 3 = 3, So Hybridisation at carbon atom is sp 2 .
For NH 3 , steric no. = 1 + 3 = 4, So Hybridisation at carbon atom is sp 3 .
[I] sp Hybridisation: It occurs when one S and one p-orbitals mix together and result in the formation
of two equivalent sp-hybrid orbitals. Each sp hybrid orbitals. Each sp hybrid orbitals is associated
with 50% s-character and 50% p-character.
APNI KAKSHA
16
BeCl2 : The around state electronic configuration of Be is 1s2 2s2 . In the excited state one electron
from 2s orbital is promoted to vacant 2p orbital to make it bivalent. One 2s and 2p orbitals gets
hybridised to form two sp hybridised orbitals. Oriented at an angle of 180°. Each of the sp hybridised
orbitals overlap with the 2p-orbital of chlorine axially and forms two Be — Cl sigma bonds.
sp hybridisation is C 2 H2 :
[II] sp2 Hybridisation: when one s and two p-orbitals mix together, they result in formation of three
equivalent sp2 hybridised orbitals.
BCl3 : In it the ground state electronic configuration of central boron atom is 1s 2 2s2 2p1 . In the
excited state, one of the 2s electrons is promoted to vacant 2p-orbital so that boron has three unpaired
electrons.
These three orbitals (one 2s and two 2p) undergo hybridisation to give three sp 2 hybrid orbitals
oriented in a trigonal planar arrangement. Three sp 2 hybrid orbitals overlap with 2p-orbitals of
chlorine to form three B — Cl bonds.
Therefore, geometry of BCl3 is trigonal planar with Cl — B — Cl bond angle of 120°.
sp2 Hybridisation in C 2 H4 :
APNI KAKSHA
17
[II] sp3 Hybridisation: In this hybridisation, one s and three p-orbitals of the valence shell intermix to
form four sp3 hybrid orbitals of equivalent energies and shape. Each sp 3 hybrid orbital contains 25%
s-character and 75% p-character. The angle between sp3 hybrid orbitals is 109°5 eg.
Formation of (CH4 ) : Formation of sp3 hybrids by the combination of s, px, py and pz atomic
orbitals of carbon and the formation of CH 4 molecule.
Structure of NH3 :
Ammonia NH 3
Structure of H2 O:
Hybridisation of Elements involving d-orbitals:
(a) Formation of PCl5 (sp3 d)
P (ground state)
3s
3p
3d
P (excited state)
PCl5
Cl
Cl Cl Cl
Cl
Thus, the five orbitals (i.e. one s, and three p and one d -orbitals) undergo hybridisation and give a
set of five sp3 d hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal.
Three P — Cl bonds lie in one place at an angle of 120° with each other and are termed as equatorial
bond. Out of the remaining two P—Cl bonds, one lie above and the other lie below the equatorial
plane, making an angle of 90° with the plane. These bonds are called axial bonds. It has been found
APNI KAKSHA
18
that axial bond pair suffer greater repulsion from equatorial bond pairs and therefore, the axial bonds
have been found to be slightly longer and hence, slightly weaker than the equatorial bonds which
makes PCl5 molecule more reactive and hence, it readily split into PCl3 and Cl2 .
PCl5 ⎯⎯
→ PCl3 + Cl2
(b) Formation of SF6 (sp3 d2 )
S (ground state)
3s
3p
3d
F
F F F
S (excited state)
SF6
Shape of molecules/ions
F F
Hybridisation type
Atomic orbitals
Examples
Square planar
dsp2
d + s + p (2)
[Ni(CN)4 ]2– , [PtCl4 ]2–
Trigonal bipyramidal
sp3 d
s + p (3) + d
PF5 , PCl5
Square pyramidal
sp3 d 2
s + p (3) + d (2)
BrF5
Octahedral
sp3 d 2
s + p (3) + d (2)
SF6 , [CrF6 ]3–
d 2 sp3
d (2) + s + p (3)
[Co(NH 3 )6 ]3+
Q. Is there any change in the hybridisation of B and N atoms as a result of the following
reaction?
𝐁𝐅𝟑 + 𝐍𝐇𝟑 → 𝐅𝟑 𝐁 ← 𝐍𝐇𝟑
[NCERT Exercise]
Sol.
N has lone pair, so NH3 donate its lone pair to BF3 and, an adduct F3 B ← NH3 is formed as the
hybridization of ‘B’ changes to sp3 . However, the hybridization of ‘N’ remains unaffected. In aduct
NH3 , BF3 both have the same hybridisation (sp3 ).
MOLECULAR ORBITAL THEORY
The salient features of this theory are :
(Practice Question in the End, Q.8. 10, 11.)
(i) The electrons in a molecule are present in the various molecular orbitals as the electrons of atoms
are present in the various atomic orbitals.
(ii) The atomic orbitals of comparable energies and proper symmetry combine to form molecular
orbitals.
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(iii) While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is
influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an
atomic orbital is monocentric while a molecular orbital is polycentric.
(iv) The number of molecular orbital formed is equal to the number of combining atomic orbitals.
When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding
molecular orbital while the other is called antibonding molecular orbital.
(v) The bonding molecular orbital has lower energy and hence greater stability than the
corresponding antibonding molecular orbital.
(vi) Just as the electron probability distribution around a nucleus in an atom is given by an atomic
orbital, the electron probability distribution around a group of nuclei in a molecule is given by
a molecular orbital.
(vii) The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle
obeying the Pauli’s exclusion principle and the Hund’s rule.
Linear Combination of Atomic Orbitals (LCAO):
Consider the hydrogen molecule consisting of two atoms A and B. Each hydrogen atom in the ground
state has one electron in 1s orbital. The atomic orbitals of these atoms may be represented by the wave
functions A and B. The formation of molecular orbitals may be described by the linear combination
of atomic orbitals that can take place by addition and by subtraction of wave functions of individual
atomic orbitals as shown below:
MO = A ± B
Therefore, the two molecular orbitals σ and σ* are formed as : σ = A + B
σ* = A – B
Conditions for the Combination of Atomic Orbitals:
1. The combining atomic orbitals must have the same or nearly the same energy.
2. The combining atomic orbitals must have the same symmetry about the molecular axis.
3. The combining atomic orbitals must overlap to the maximum extent.
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Q. Write the significance of a plus and a minus sign shown in representing the orbitals.
[NCERT Exercise]
Sol. Molecular orbitals are represented by wave functions. A plus sign in an orbital indicates a positive
wave function while a minus sign in an orbital represents a negative wave function.
Stability of Molecules: If N b is the number of electrons occupying bonding orbitals and N a the
number occupying the antibonding orbitals, then
(i) the molecule is stable if N b is greater than N a, and
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(ii) the molecule is unstable if N b is less than N a.
In (i) more bonding orbitals are occupied and so the bonding influence is stronger and a stable
molecule result. In (ii) the antibonding influence is stronger and therefore the molecule is unstable.
Bond order
Bond order is defined as one half the difference between the number of electrons present in the
bonding and the antibonding orbitals i.e.,
Bond order = ½ (N b – Na).
The rules discussed above regarding the stability of the molecule can be restated in terms of bond
order as follows: A positive bond order (i.e., N b > Na) means a stable molecule while a negative (i.e.,
Nb < Na) or zero (i.e., N b = Na) bond order means an unstable molecule.
Nature of the bond
Integral bond order values of 1, 2 or 3 correspond to single, double or triple bonds respectively as
studied in the classical concept.
Bond-length
The bond order between two atoms in a molecule may be taken as an approximate measure of the
bond length. The bond length decreases as bond order increases.
Magnetic nature
If all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic (repelled
by magnetic field).
However, if one or more molecular orbitals are singly occupied it is paramagnetic (attracted by
magnetic field), e.g., O 2 molecule.
BONDING IN SOME HOMONUCLEAR DIATOMIC MOLECULES
1. Hydrogen molecule (H 2 ): It is formed by the combination of two hydrogen atoms.
H 2 : (σ1s)2
Bond order =
N b − Na 2 − 0
=
=1
2
2
2. Helium molecule (He2 ) : The electronic configuration of helium atom is 1s2 .
H 2 : (σ1s)2 (σ*1s)2
Bond order of He2 is ½ (2 – 2) = 0
He2 molecule is therefore unstable and does not exist.
3. Lithium molecule (Li2 ) : Li2 : (σ1s)2 (σ*1s)2 (σ2s)2
(
4. Carbon molecule (C 2 ) : C2 : (σ1s)2 (σ*1s)2 (σ*2s)2 2p2x = 2p2y
)
The bond order of C 2 is ½ (8 – 4) = 2 and C2 should be diamagnetic.
It is important to note that double bond in C 2 consists of both pi bonds because of the presence of four
electrons in two pi molecular orbitals.
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5. Oxygen molecule (O 2 ) : O 2 : (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2
( 2p
2
x
)(
 2p2y  *2p1x   *2p1y
)
1
1
[N b − Na ] = [10 − 6] = 2
2
2
Bond order =
O 2 molecule should be paramagnetic, a prediction that corresponds to experimental observation.
Q. Use molecular orbital theory to explain why the 𝐁𝐞𝟐 molecule does not exist.
[NCERT Exercise]
2
2
Sol. The electronic configuration of Beryllium is 1s 2s .
The molecular orbital electronic configuration for Be2 molecule : 𝜎1𝑠 2 < 𝜎 ∗ 1𝑠 2 < 𝜎2𝑠 2 < 𝜎 ∗ 2𝑠 2
1
Bond order =
2
(𝑁𝑏 − 𝑁𝑎 )
Nb = Number of electrons in bonding orbitals
Na =Number of electrons in anti-bonding orbitals
1
Bond order of Be2 = 2 (4 − 4) = 0
Zero bond order means that Be2 molecule does not exist.
Q. Compare the relative stability of the following species and indicate their magnetic
properties; 𝐎𝟐 , 𝐎𝟐+ , 𝐎𝟐− (superoxide),𝐎𝟐𝟐− (peroxide)
Sol. 𝐎𝟐 ∶
2
[𝜎(1𝑠)]2 [𝜎 ∗ (1𝑠)]2 [𝜎(2𝑠)]2 [𝜎 ∗ (2𝑠)]2 [𝜎(2𝑝𝑧 )]2 [𝜋(2𝑝𝑥 )]2 [𝜋(2𝑝𝑦 )] [𝜋 ∗ (2𝑝𝑥 )]1 [𝜋 ∗ (2𝑝𝑦 )]
1
Bond order = 2 (𝑁𝑏 − 𝑁𝑎 ) =
1
2
1
(8 − 4) = 2
Magnetic property ⟹ Paramagnetic due to presence of 2 unpaired electrons
2
𝐎𝟐+ ∶ [𝜎(1𝑠)]2 [𝜎 ∗ (1𝑠)]2 [𝜎(2𝑠)]2 [𝜎 ∗ (2𝑠)]2 [𝜎(2𝑝𝑧 )]2 [𝜋(2𝑝𝑥 )]2 [𝜋(2𝑝𝑦 )] [𝜋 ∗ (2𝑝𝑥 )]1
Bond order of O2+ =
1
2
(𝑁𝑏 − 𝑁𝑎 ) = 1 (8 − 3) = 2.5
2
Magnetic property ⟹ Paramagnetic due to presence of 1 unpaired electron
𝐎𝟐− ∶
2
[𝜎(1𝑠)]2 [𝜎 ∗ (1𝑠)]2 [𝜎(2𝑠)]2 [𝜎 ∗ (2𝑠)]2 [𝜎(2𝑝𝑧 )]2 [𝜋(2𝑝𝑥 )]2 [𝜋(2𝑝𝑦 )] [𝜋 ∗ (2𝑝𝑥 )]2 [𝜋 ∗ (2𝑝𝑦 )]
Bond order of O2− =
1
2
1
(𝑁𝑏 − 𝑁𝑎 ) = 1 (8 − 5) = 1.5
2
Magnetic property ⟹ Paramagnetic due to presence of 1 unpaired electron
𝐎𝟐𝟐−
2
∶ [𝜎 (1𝑠)]2 [𝜎 ∗ (1𝑠)]2 [𝜎(2𝑠)]2 [𝜎 ∗ (2𝑠)]2 [𝜎(2𝑝𝑧 )]2 [𝜋(2𝑝𝑥 )]2 [𝜋(2𝑝𝑦 )] [𝜋 ∗ (2𝑝𝑥 )]2 [𝜋 ∗ (2𝑝𝑦 )]
2
1
1
Bond order of O22− = 2 (𝑁𝑏 − 𝑁𝑎 ) = 2 (8 − 6) = 1
Magnetic property ⟹ Diamagnetic due to presence of paired electrons
The higher the bond order, the greater will be the stability.
The order of stability : O2+ > O2 > O2− > O22−
Hydrogen Bonding
(Practice Question in the End, Q.4.)
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule
with the electronegative atom (F, O or N) of another molecule.
– – – H + — F– – – – H + — F– – – – H+ — F–
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Cause of Formation of Hydrogen Bond
When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between
the two atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes highly
electropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards
X, the hydrogen acquires fractional positive charge (+) while ‘X’ attain fractional negative charge
(– ). This results in the formation of a polar molecule having electrostatic force of attraction which
can be represented as :
– – – H + — X– – – – H + — X – – – – H + — X–
The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the
solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong influence on the
structure and properties of the compounds.
Types of H-Bonds
There are two types of H-bonds
(i) Intermolecular hydrogen bond
(ii) Intramolecular hydrogen bond
(1) Intermolecular hydrogen bond : It is formed between two different molecules of the same or
different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.
(2) Intramolecular hydrogen bond : It is formed when hydrogen atom is in between the two highly
electronegative (F, O, N) atoms present within the same molecule. For example, in o-nitrophenol the
hydrogen is in between the two oxygen atoms.
O
N
O
O
+
H
–
•
•O••
H
+
H
+
H
+
H
–
•
•O••
H
–
•
•O••
H
+
+
“When competition becomes hard, Just compete with yourself”
Notes End
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Important NCERT Questions
Q1: Isostructural species are those which have the same shape and hybridisation.
Among the given species identify the isostructural pairs.
[NCERT Exemplar]
–
+
(a) [NF3 and BF3 ]
(b) [BF4 and NH 4 ]
(d) [NH 3 and NO 3 – ]
(c) [BCl3 and BrCl3 ]
Sol: (b) NF3 is pyramidal whereas BF3 is planar triangular.
BF4− and NH +4 ions both are tetrahedral.
BC13 is triangular planar whereas BrCl3 is pyramidal.
NH 3 is pyramidal whereas NO 3 is triangular planar.
Q2. Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of
the constituent atoms and shape of a molecule. Which of the following has the highest dipole
moment?
[NCERT Exemplar]
(a) CO 2
(b) HI
(c) H2 O
(d) SO 2
Sol: (c) H 2 O will have highest dipole moment due to maximum difference in electronegativity of H
and O atoms.
Q3. Write Lewis symbols for the following atoms and ions: S and S 2−; Al and Al3+; H and H−.
Atom / Ions
No. of valence electrons
S
6
••
••
S2–
6+2=8
••
••
[NCERT Exercise]
Atom/Ions
No. of valence electrons
Lewis Dot structure
Al
3
•
•
Sol. (i)
Lewis Dot structure
Al
Al+3
0
[Al]3+
Atom/Ions
No. of valence electrons
Lewis Dot structure
H
1
H•
H–
2
[H]
S
••
S
2–
••
•
(ii)
(iii)
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Q4. Hydrogen bonds are formed in many compounds e.g., H 2 O, HF, NH 3 . The boiling point of such
compounds depends to a large extent on the strength of hydrogen bond and the number of
hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :
(a) HF > H 2 O > NH 3
(b) H2 O > HF > NH 3 [NCERT Exemplar]
(c) HH 3 > HF > H 2 O
(d) NH 3 > H2 O > HF
Sol: (b) Strength of H-bonding depends on the electronegativity of the atom which follows the
order: F > O > N .
Strength of H-bond is in the order:
H……. F > H…….. O > H…….. N
But each H 2 O molecule is linked to 4 other H 2 O molecules through H-bonds whereas each HF
molecule is linked only to two other HF molecules.
Hence, correct decreasing order of the boiling points is H 2 O > HF > NH 3 .
Q5. Write the Lewis structure of the nitrite ion, NO−2 .
[NCERT Exemplar]
Sol. Step 1. Count the total number of valence electrons of the nitrogen atom, the oxygen atoms and
the additional one negative charge (equal to one electron).
N(2s2 2p3 ), O (2s2 2p4 )
5 + (2 × 6) +1 = 18 electrons
Step 2. The skeletal structure of NO−2 is written as : O N O
Step 3. Draw a single bond (one shared electron pair) between the nitrogen and each of the
oxygen atoms completing the octets on oxygen atoms. This, however, does not complete the
octet on nitrogen if the remaining two electrons constitute lone pair on it.
N
••
O
••
••
••
••
••
••
••
O
••
–
Hence we have to resort to multiple bonding between nitrogen and one of the oxygen atoms (in
this case a double bond). This leads to the following Lewis dot structures.
–
or
••
O
••
••
N
••
O
••
–
or
••
O
••
••
••
O
••
••
N
••
••
••
••
••
••
O
••
••
N
–
••
O
••
Q6. In NO3− ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are
(a) 2, 2
(b)
3, 1
(c)
1, 3
(d)
4, 0
[NCERT Exemplar]
Sol: (d) In N-atom, number of valence electrons = 5
Due to the presence of one negative charge, number of valence
electrons = 5 + 1 = 6. One O-atom forms two bonds (= bond) and
two O-atom are shared with two electrons of N-atom.
••
 ••
O = N − O ••

NO3− =  •• | ••
• ••
 •O
••






−
Thus, 3 O-atoms are shared with 8 electrons of N-atom.
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Number of bond pairs (or shared pairs) = 4
Number of lone pairs = 0
Q7. Which of the following species has tetrahedral geometry?
(a) BH−4
(b)
NH −2
(c)
CO32−
[NCERT Exemplar]
(d)
H3O+
Sol: (a) BH−4 = tetrahedral, NH −2 = V-shape, CO32− = triangular planar, H 3 O+ = triangular planar.
Q8. Which molecule/ion out of the following does not contain unpaired electrons?
(a) N+2
(b)
O2
(c)
O22−
(d)
B2
[NCERT Exemplar]
Sol: (c) N+2 = KK2s2 *2s22p2x = 2p2y2p1z
O2 = KK2s2 *2s22pz22p2x = 2p2y  *2p1x =  *2p1y
O22− = KK2s2 *2s22pz2 2p2x = 2p2y  *2p2x =  *2p2y
B2 = KK2s2 *2s22p1x = 2p1y
Thus, O22− has no unpaired electrons.
Q9. In which of the following molecule/ion all the bonds are not equal?
(a) XeF4
(b)
BF4−
(c)
C2 H 4
(d)
[NCERT Exemplar]
SiF4
Sol: (c) C2 H 4 has one double bond and four single bonds. Bond length of double bond (C = C) is
smaller than single bond (C – H).
H−C = C−H
|
|
H H
Q10. Which of the following order of energies of molecular orbitals of N 2 is correct?
(a) (π2py ) < (σ2pz) < (π*2px )  (π*2py )
[NCERT Exemplar]
(b) (π2py ) > (σ2pz) > (π*2px )  (π*2py )
(c) (π2py ) < (σ2pz) > (π*2px )  (π*2py )
(d) (π2py ) > (σ2pz) < (π*2px )  (π*2py )
Sol: (a) (π2py ) < (σ2pz) < (π*2px )  (π*2py )
Molecules like B 2 , C2 & N2 have 1 to 3 electrons in p orbital energy of σ 2p molecular orbital
are greater than that of π2px and π2py molecular orbitals.
Q11. Which of the following statement is not correct from the view point of molecular orbital
theory?
[NCERT Exemplar]
(a) Be2 is not a stable molecule.
(b) He2 is not stable but He +2 is expected to exist.
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(c) Bond strength of N 2 is maximum amongst the homonuclear diatomic molecules belonging
to the second period.
(d) The order of energies of molecular orbitals in N 2 molecule is
σ2s < σ*2s < σ2pz < (π2px = π2py ) < (π*2px = π*2py ) < σ*2pz
Sol: (d)
(a) Be2 (4 + 4 = 8) = σ1s2 , σ*1s2 , σ1s2 , σ*2s2
1
Bond order (BO) =
[Number of bonding electron (N b ) – Number of anti-bonding electrons N a]
2
4−4
=
=0
2
Here, bond order of He2 is zero. Thus, it does not exist.
(b) He2 (2 + 2 = 4) σ1s2 , σ*1s2
2−2
BO =
=0
2
Here, bond order of He2 is zero. Thus, it does not exist.
He+2 (2 + 2 −1 = 3) = 1s2 ,  *1s1
2 −1
= 0.5
2
Since, the bond order is not zero, this molecule is expected to exist.
BO =
2
2
2
(c) N2 (7 + 7 = 14) = σ1s2 , σ*1s2 , σ2s2 , σ*2s2 , 2p x  2p y , 2pz
0 − 4
=3
2
Thus, dinitrogen (N 2 ) molecule contains triple bond and no other molecule of second period
have more than double bond. Hence, bond strength of N 2 is maximum amongst the
homonuclear diatomic molecules belonging to the second period.
(d) It is incorrect. The correct order of energies of molecular orbitals in N 2 molecule is
σ2s < σ*2s < (σ2px  π2py ) < π2py ) < σ2pz < π*2px  π*2py < σ*2pz
Sol: (d) Bond energy of two (–O – H) bonds in H 2 O will be different.
BO =
“When competition becomes hard, Just compete with yourself”
Ab Phod Do!
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