In the figure (I), area of ++ overlap is equal to + - overlap, so net overlap is zero, while in figure (II), there is no overlap due to different symmetry.

$\mathrm{PCl}_5-$ The ground state and the excited state outer electronic configurations of phosphorus $(Z=15)$ are represented below

In $\mathrm{PCl}_5$, P is $s p^3 d$ hybridised, therefore, its shape is trigonal bipyramidal. $\mathrm{IF}_5$-The ground state and the excited state outer electronic configurations of iodine $(Z=53)$ are represented below.

In $\mathrm{IF}_5, \mathrm{I}$ is $s p^3 d^2$ hybridised, therefore, shape of $\mathrm{IF}_5$ is square pyramidal.

Dimethyl ether has greater bond angle than that of water, however in both the molecules central atom oxygen is $s p^3$ hybridised with two lone pairs. In dimethyl ether, bond angle is greater $\left(111.7^{\circ}\right)$ due to the greater repulsive interaction between the two bulky alkyl (methyl) groups than that between two H -atoms.

Actually C of $\mathrm{CH}_3$ group is attached to three H -atoms through $\sigma$-bonds. These three $\mathrm{C}-\mathrm{H}$ bond pair of electrons increases the electronic charge density on carbon atom.

The Lewis structure of the following compounds and formal charge on each atom are as

(i) $\mathrm{HNO}_3$

Formal charge on an atom in a Lewis structure $=[$ total number of valence electrons in free atom $]$ $-$ [total number of non-bonding (lone pairs) electrons] $-\frac{1}{2}$ [total number of bonding or shared electrons]

Formal charge on $\mathrm{H}=1-0-\frac{1}{2} \times 2=0$

Formal charge on $\mathrm{N}=5-0-\frac{1}{2} \times 8=1$

Formal charge on $\mathrm{O}(1)=6-4-\frac{1}{2} \times 4=0$

Formal charge on $\mathrm{O(2)=6-4-\frac{1}{2} \times 4=0}$

Formal charge on $\mathrm{O(3)=6-6-\frac{1}{2} \times 2=-1}$

(ii) NO$$_2$$

Formal charge on $\mathrm{O(1)=6-4-\frac{1}{2} \times 4=0}$

Formal charge on $\mathrm{N}=5-1-\frac{1}{2} \times 6=+1$

Formal charge on $\mathrm{O(2)=6-6-\frac{1}{2} \times 2=-1}$

(iii) $\mathrm{H}_2 \mathrm{SO}_4$

Formal charge on $\mathrm{H}(1)$ or $\mathrm{H}(2)=1-0-\frac{1}{2} \times 2=0$

Formal charge on $O(1)$ or $O(3)=6-4-\frac{1}{2} \times 4=0$

Formal charge on $\mathrm{O}(2)$ or $\mathrm{O}(4)=6-6-\frac{1}{2} \times 2=-1$

Formal charge on $S=6-0-\frac{1}{2} \times 8=+2$

Electronic configuration of N -atom $(Z=7)$ is $1 s^2 2 s^2 2 p_x^1 2 p_y^1 2 p_z^1$. Total number of electrons present in $\mathrm{N}_2$ molecule is 14,7 from each N -atom. From the view of various rules for filling of molecular orbitals, the electronic configuration of $\mathrm{N}_2$ molecule will be

$$\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2$$

Comparative study of the relative stability and the magnetic behaviour of the following species

(i) $\mathrm{N}_2$ molecule $\sigma 1 s^2, \sigma^* s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2$

Here, $N_b=10, N_a=4$.

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

Hence, presence of no unpaired electron indicates it to be diamagnetic.

(ii) $\mathbf{N}_2^{+}$ions $\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^1$

Here, $N_b=9, N_a=4$ so that $B O=\frac{1}{2}(9-4)=\frac{5}{2}=2.5$

Further, as $\mathrm{N}_2^{+}$ion has one unpaired electron in the $\sigma\left(2 p_2\right)$ orbital, therefore, it is paramagnetic in nature.

(iii) $\mathbf{N}_2^{-}$ions $\sigma 1 s^2, \sigma^* s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2, \pi^* 2 p_x^1$

Here, $N_b=10, N_a=5$ so that $\mathrm{BO}=\frac{1}{2}(10-5)=\frac{5}{2}=2.5$

Again, as it has one unpaired electron in the $\pi^*\left(2 p_x\right)$ orbital, therefore, it is paramagnetic.

(iv) $\mathbf{N}_2^{2+}$ ions $\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2$

Here, $N_b=8, N_a=4$. Hence, $\mathrm{BO}=\frac{1}{2}(8-4)=2$

Presence of no unpaired electron indicates it to be diamagnetic in nature.

As bond dissociation energies are directly proportional to the bond orders, therefore, the dissociation energies of these molecular species in the order.

$$\mathrm{N}_2>\mathrm{N}_2^{-}=\mathrm{N}_2^{+}>\mathrm{N}_2^{2+}$$

As greater the bond dissociation energy, greater is the stability, the stability of these species is also in the above order.