JEE Main 2022 — Chemical Bonding and Molecular Structure Question with Solution

Bond order $=\frac{\mathrm{number} \mathrm{of} \mathrm{bonding} {\mathrm{e}}^{-}-\mathrm{number} \mathrm{of} \mathrm{antibonding} {\mathrm{e}}^{-}}{2}$

Electronic configuration of ${\mathrm{O}}_2, {\mathrm{O}}_2^{+}, {\mathrm{O}}_2^{-} \mathrm{and} {\mathrm{O}}_2^{-2}$ are:

$$\begin{gathered} {\mathrm{O}}_2={\mathrm{\sigma }}_{1{\mathrm{s}}^2}{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{{\mathrm{p}}_{\mathrm{x}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^2}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^1}^{*}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^1}^{*} (\mathrm{Bond} \mathrm{order}=2) \\ {\mathrm{O}}_2^{+}={\mathrm{\sigma }}_{1{\mathrm{s}}^2}{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{{\mathrm{p}}_{\mathrm{x}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^2}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^1}^{*}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^0}^{*} (\mathrm{Bond} \mathrm{order}=2.5) \\ {\mathrm{O}}_2^{-}={\mathrm{\sigma }}_{1{\mathrm{s}}^2}{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{{\mathrm{p}}_{\mathrm{x}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^2}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^2}^{*}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^1}^{*} (\mathrm{Bond} \mathrm{order}=1.5) \\ {\mathrm{O}}_2^{-2}={\mathrm{\sigma }}_{1{\mathrm{s}}^2}{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{*}{\mathrm{\sigma }}_{2{{\mathrm{p}}_{\mathrm{x}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^2}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^2}{\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{y}}}^2}^{*}={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^2}^{*} (\mathrm{Bond} \mathrm{order}=1) \end{gathered}$$