What is the effect of the following processes on the bond order in $\mathrm{N}_{2}$ and $\mathrm{O}_{2}$ ?

$(A)$ ${{\rm{N}}_2} \to {\rm{N}}_2^ +  + {{\rm{e}}^ - }$

$(B)$ ${{\rm{O}}_2} \to {\rm{O}}_2^ +  + {{\rm{e}}^ - }$

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

$\mathrm{N}_{2}\left(14 e^{-}\right)=\sigma 1 s^{2}, \sigma 1 s^{2}, \sigma 2 s^{2}, \sigma 2 s^{2},\left(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}\right), \sigma 2 p_{z}^{2}$

Bond order $=\frac{1}{2}\left[\mathrm{~N}_{b}-\mathrm{N}_{a}\right]=\frac{1}{2}(10-4)=3$

$\mathrm{N}_{2}^{+}\left(13 e^{-}\right)=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma 2 s^{2},\left(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}\right), \sigma 2 p_{z}^{1}$

Bond order $=\frac{1}{2}\left[\mathrm{~N}_{b}-\mathrm{N}_{a}\right]=\frac{1}{2}(9-4)=2.5$

$\mathrm{O}_{2}\left(16 e^{-}\right)=\sigma 1 s^{2}, \sigma 1 s^{2}, \sigma 2 s^{2},  2 s^{2}, \sigma 2 p_{z^{\prime}}^{2}\left(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}\right),\left(\pi 2 p_{x}^{1} \approx \pi 2 p_{y}^{1}\right)$

Bond order $=\frac{1}{2}\left[\mathrm{~N}_{b}-\mathrm{N}_{a}\right]=\frac{1}{2}(10-6)=2$

$\mathrm{O}_{2}^{+}\left(15 e^{-}\right)=\sigma 1 s^{2}, \sigma 1 s^{2}, \sigma 2 s^{2}, \sigma 2 s^{2}, \sigma 2 p_{z}^{2},\left(\pi 2 p_{x}^{2},=\pi 2 p_{y}^{2}\right),\left(\pi 2 p_{x}^{1}=\pi 2 p_{y}\right)$

Bond order $=\frac{1}{2}\left[\mathrm{~N}_{b}-\mathrm{N}_{a}\right]=\frac{1}{2}(10-5)=2.5$

$(a)$ $\mathrm{N}_{2} \rightarrow \mathrm{N}_{2}^{+}+e^{-}$

$\text { B.O. }=3 \quad \text { B.O. }=2.5$

Thus, bond order decreases.

$(b)$ $\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{+}+e^{-}$

$\text { B.O. }=3 \quad \text { B.O. }=2.5$

Thus, bond order increases.