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.

According to molecular orbital theory, electronic configurations and bond order of $\mathrm{N}_2, \mathrm{~N}_2^{+}, \mathrm{O}_2$ and $\mathrm{O}_2^{+}$species are as follows

$$\begin{aligned} \mathrm{N}_2\left(14 e^{-}\right) & =\sigma 1 s^2, \stackrel{\star}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right), \sigma 2 p_z^2 \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-4)=3 \\ \mathrm{~N}_2^{+}\left(13 e^{-}\right) & =\sigma 1 s^2, \stackrel{\star}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right) \sigma 2 p_z^1 \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(9-4)=2.5 \\ \mathrm{O}_2\left(16 e^{-}\right) & =\sigma 1 s^2,{ }^{\star} \sigma 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2, \sigma 2 p_z^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right),\left({ }^{\star} \pi 2 p_x^1 \approx \stackrel{\star}{\pi} 2 p_y^1\right) \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-6)=2 \\ \mathrm{O}_2^{+}\left(15 e^{-}\right) & =\sigma 1 s^2, \stackrel{\star}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2, \sigma 2 p_z^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right),\left({ }^{\star} \pi 2 p_x^1 \approx \stackrel{\star}{\pi} 2 p_y\right) \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-5)=2.5 \end{aligned}$$

(a) $\underset{\text { B.O. }=3}{\mathrm{~N}_2} \longrightarrow \underset{\text { B.O. }=2.5}{\mathrm{~N}_2^{+}}+e^{-}$

Thus, bond order decreases.

(b) $\underset{\text { B.O = 2}}{\mathrm{O}_2} \longrightarrow \underset{\text { B.O = 2.5}}{\mathrm{O}_2^{+}}+\mathrm{e}^{-}$

Thus, bond order increases.

(a) A covalent bond is formed by the overlap of atomic orbitals. The direction of overlapping gives the direction of bond. In ionic bond, the electrostatic field of an ion is non-directional. Each positive ion is surrounded by a number of anions in any direction depending upon its size and vice-versa. That's why covalent bonds are directional bonds while ionic bonds are non-directional.

(b) In $\mathrm{H}_2 \mathrm{O}$, oxygen atom is $s p^3$ hybridised with two lone pairs. The four $s p^3$ hybridised orbitals acquire a tetrahedral geometry with two corners occupied by hydrogen atoms while other two by the lone pairs. The bond angle is reduced to $104.5^{\circ}$ from $109.5^{\circ}$ due to greater repulsive forces between Ip-Ip and the molecule thus acquires a V -shape or bent structure (angular structure).

In $\mathrm{CO}_2$ molecule, carbon atom is $s p$-hybridised. The two $s p$ hybrid orbitals are oriented in opposite direction forming an angle of $180^{\circ}$.

That's why $\mathrm{H}_2 \mathrm{O}$ molecule has bent structure whereas $\mathrm{CO}_2$ molecule is linear.

(c) In ethyne molecule, both the carbon atoms are $s p$ hybridised, having two unhybridised orbitals, i.e., $2 p_x$ and $2 p_y$. The two $s p$ hybrid orbitals of both the carbon atoms are oriented in opposite direction forming an angle of $180^{\circ}$.

That's why ethyne molecule is linear.