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

Bond order is calculated using the following expression:

$\mathrm{Bond} \mathrm{order}=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}$

N^b is the number of bonding electrons and N_a is the number of anti-bonding electrons.

For ${\mathrm{O}}_2^{2+},$ the number of bonding electrons is 10 and the number of anti-bonding electrons is 4. Bond order of ${\mathrm{O}}_2^{2+}$ is,

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& \frac{10-4}{2}\\ & =& 3\end{array}$

For ${\mathrm{O}}_2^{+},$ the number of bonding electrons is 10 and the number of anti-bonding electrons is 5. Bond order of ${\mathrm{O}}_2^{+}$ is:

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& \frac{10-5}{2}\\ & =& 2.5\end{array}$

For ${\mathrm{O}}_2^{-},$ the number of bonding electrons is 10 and the number of anti-bonding electrons is 7. Bond order of ${\mathrm{O}}_2^{-}$ is:

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& \frac{10-7}{2}\\ & =& 1.5\end{array}$

For ${\mathrm{O}}_2^{2-},$ the number of bonding electrons is 10 and the number of anti-bonding electrons is 8. Bond order of ${\mathrm{O}}_2^{-}$ is:

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& \frac{10-8}{2}\\ & =& 1\end{array}$

The more is the bond order, the more is the bond strength and the small is the bond length.

$\mathrm{Bond} \mathrm{order}\infty \mathrm{Bond} \mathrm{strength}\infty \frac{1}{\mathrm{Bond} \mathrm{length}}$

Among the given species, ${\mathrm{O}}_2^{2+}$ has the highest bond order, thereby has the smallest (or shortest) bond length.