(a) The applications of dipole moment are

(i) The dipole moment helps to predict whether a molecule is polar or non-polar. As $\mu=q \times d$, greater is the magnitude of dipole moment, higher will be the polarity of the bond. For non-polar molecules, the dipole moment is zero.

(ii) The percentage of ionic character can be calculated as

$$\text { Percentage of ionic character }=\frac{\mu_{\text {observed }}}{\mu_{\text {ionic }}} \times 100$$

(iii) Symmetrical molecules have zero dipole moment although they have two or more polar bonds (in determination of symmetry).

(iv) It helps to distinguish between cis and trans isomers. Usually cis-isomer has higher dipole moment than trans isomer.

(v) It helps to distinguish between ortho, meta and para isomers. Dipole moment of para isomer is zero. Dipole moment of ortho isomer is greater than that of meta isomer.

(b)

Formation of $\mathrm{N}_2$ molecule Electronic configuration of N - atom ${ }_7 \mathrm{~N}=1 s^2, 2 s^2, 2 p_x^1, 2 p_y^1, 2 p_z^1$

$N_2$ molecule $=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x{ }^2 \approx \pi 2 p_y{ }^2, \sigma 2 p_z^2$

Bond order $=\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-4)=3$.

Bond order value of 3 means that $\mathrm{N}_2$ contains a triple bond.

Formation of $\mathrm{F}_2$ molecule, ${ }_9 \mathrm{~F}=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^2, 2 p_z^1$

$$\mathrm{F}_2 \text { molecule }=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^{\star} 2 p^2 x \approx \pi^{\star} 2 p^2 y$$

$$\text { Bond order }=\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-8)=1$$

Bond order value 1 means that $F_2$ contains single bond.

Formation of $\mathrm{Ne}_2$ molecule ${ }_{10} \mathrm{Ne}=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^2, 2 p_z^2$

$$\mathrm{Ne}_2 \text { molecule }=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z{ }^2, \pi 2 p_x^2 \approx \pi 2 p_y{ }^2, \pi^{\star} 2 p^2 x \approx \pi^{\star} 2 p_y^2, \sigma^{\star} 2 p_z^2$$

Bond order $=\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-10)=0$

Bond order value zero means that there is no formation of bond between two Ne -atoms. Hence, $\mathrm{Ne}_2$ molecule does not exist.

Valence bond theory (VBT) was introduced by Heitler and London (1927) and developed further by Pauling and other. VBT is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, 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 $e_B$. 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

i.e., $$N_A-e_A \text { and } N_B-e_B$$

(ii) nucleus of one atom and electron of other atom

i.e., $$N_A-e_B, N_B-e_A$$

Similarly, repulsive forces arise between

(i) electrons of two atoms like $e_A-e_B$

(ii) nuclei of two atoms like $N_A-N_B$

Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.

Experimentally, we have been found that the magnitude of new attractive force is more than the new repulsive forces. As a result two atoms approach each other and potential energy decreases.

Hence, a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage, two H -atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm .

Since, the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of isolated hydrogen atoms.

The energy so released is called as bond enthalpy, which is corresponding to minimum in the curve depicted in the given figure. Conversely 435.8 kJ of energy is required to dissociate one mole of $\mathrm{H}_2$ molecule.

$$\mathrm{H}_2(g)+435.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)$$

The potential energy curve for the formation of $\mathrm{H}_2$ molecule as a function of internuclear distance of the H -atoms. The minimum in the curve corresponds to the most stable state or $\mathrm{H}_2$.

Formation of PCl$$_5$$

In $\mathrm{PCl}_5$, phosphorus is $s p^3 d$ hybridised to produce a set of five $s p^3 d$ hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal. These five $s p^3 d$ hybrid orbitals overlap with singly occupied $p$-orbitals of Cl -atoms to form five $\mathrm{P}-\mathrm{Cl}$ sigma bonds.

Three P-Cl bonds lie in one plane and make an angle of $120^{\circ}$ with each other. These bonds are called equatorial bonds. The remaining two $\mathrm{P}-\mathrm{Cl}$ bonds one lying above and other lying below the plane make an angle of $90^{\circ}$ with the equatorial plane. These bonds are called axial bonds. Axial bonds are slightly longer than equatorial bonds because axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.

Formation of $\mathrm{SF}_6$

In $\mathrm{SF}_6$, sulphur is $s p^3 d^2$ hybridised to produce a set of $s i x ~ s p^3 d^2$ hybrid orbitals which are directed towards the six corners of a regular octahedron. These six $s p^3 d^2$ hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds. Thus, $\mathrm{SF}_6$ molecule has a regular octahedral geometry and all $\mathrm{S}-\mathrm{F}$ bonds have same bond length.