Using MO theory predict which of the following species has the shortest bond length?

detailed solution

Correct option is B

$$\begin{gathered} \mathrm{Molecularorbitalconfigurationof} {\mathrm{O}}_2^{+}: \\ {\mathrm{\sigma }}_{1{\mathrm{s}}^2}>{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{\ast }>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{\ast }>{\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}^{\ast }={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^0}^{\ast } \\ \mathrm{Bondorderof} {\mathrm{O}}_2^{+}=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}=\frac{10-5}{2}=2.5 \\ \mathrm{Molecularorbitalconfigurationof} {\mathrm{O}}_2^{-}: \\ {\mathrm{\sigma }}_{1{\mathrm{s}}^2}>{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{\ast }>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{\ast }>{\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}^{\ast }={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{z}}1}^{\ast } \\ \mathrm{Bondorderof} {\mathrm{O}}_2^{-}=\frac{10-7}{2}=1.5 \\ \mathrm{Molecularorbitalconfigurationof} {\mathrm{O}}_2^{2-} : \\ {\mathrm{\sigma }}_{1{\mathrm{s}}^2}>{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{\ast }>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{\ast }>{\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}^{\ast }={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{z}}2}^{\ast } \\ \mathrm{Bondorderof} {\mathrm{O}}_2^{2-}=\frac{10-8}{2}=1 \\ \mathrm{Molecularorbitalconfigurationof} {\mathrm{O}}_2^{2+} : \\ {\mathrm{\sigma }}_{1{\mathrm{s}}^2}>{\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{\ast }>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}>{\mathrm{\sigma }}_{2{\mathrm{s}}^2}^{\ast }>{\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}}}^0}^{\ast }={\mathrm{\pi }}_{2{{\mathrm{p}}_{\mathrm{z}}}^0}^{\ast } \\ \mathrm{Bondorderof} {\mathrm{O}}_2^{2+}=\frac{10-4}{2}=3 \\ \mathrm{Bondorder} \propto \frac{1}{ \mathrm{Bondlength} } \\ \mathrm{So}, {\mathrm{O}}_2^{2+} \mathrm{hastheshortestbondlength}. \end{gathered}$$