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.

Ionic bond The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as the electrovalent bond or ionic bond. e.g., the formation of NaCl from sodium and chlorine can be explained as

$$\underset{[\mathrm{Ne}] 3 \mathrm{~s}^1}{\mathrm{Na}} \longrightarrow \underset{[\mathrm{Ne}]}{\mathrm{Na}^{+}}+\mathrm{e}^{-}$$

$$\underset{[\mathrm{Ne}] 3 s^2 3 p^5}{\mathrm{Cl}}+\mathrm{e} \longrightarrow \underset{[\mathrm{Ne}] 3 s^2 3 p^6 \text { or }[\mathrm{Ar}]}{\mathrm{Cl}^{-}}$$

$$\mathrm{Na}^{+}+\mathrm{Cl}^{-} \longrightarrow \mathrm{NaCl} \text { or } \mathrm{Na}^{+} \mathrm{Cl}^{-}$$

Similarly, the formation of $\mathrm{CaF}_2$ may be shown as

$$\underset{[A r] 4 \mathrm{~s}^2}{\mathrm{Ca}} \longrightarrow \underset{[A r]}{\mathrm{Ca}^{2+}}+2 e^{-}$$

$$\underset{[\mathrm{He}] 2 s^2 2 p^5}{\mathrm{~F}}+\mathrm{e}^{-} \longrightarrow \underset{[\mathrm{He}] 2 s^2 2 p^6 \text { or }[\mathrm{Ne}]}{\mathrm{F}^{-}}$$

$$\mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \longrightarrow \mathrm{CaF}_2 \text { or } \mathrm{Ca}^{2+}\left(\mathrm{F}^{-}\right)_2$$

Covalent bond The bond formed between the two atoms by mutual sharing of electrons between them is called covalent bond. e.g., the formation of chlorine molecule can be explained as

Similarly, in the formation of HCl

Greater is the electronegativity difference between the two bonded atoms, greater is the ionic character.

Therefore, increasing order of ionic character of the given bonds is as follows

$$\mathrm{C}-\mathrm{H}<\mathrm{N}-\mathrm{H}<\mathrm{O}-\mathrm{H}<\mathrm{F}-\mathrm{H}$$