In the following the correct bond order sequence is:

$$\begin{gathered} {\mathrm{O}}_2=\left({\mathrm{\sigma }}_{1\mathrm{s}}^2\right)\left({\mathrm{\sigma }}_{1\mathrm{s}}^{*2}\right)\left({\mathrm{\sigma }}_{2\mathrm{s}}^2\right)\left({\mathrm{\sigma }}_{2\mathrm{s}}^{*2}\right)\left({\mathrm{\sigma }}_{2\mathrm{pz}}^2\right)({\mathrm{\pi }}_{2\mathrm{px}}^2={\mathrm{\pi }}_{2\mathrm{py}}^2)({\mathrm{\pi }}_{2\mathrm{px}}^{*1}={\mathrm{\pi }}_{2\mathrm{py}}^{*1}) \\ \mathrm{bond} \mathrm{order}=\frac{(\mathrm{bonding}-\mathrm{antibonding})}{2}=\frac{10-6}{2}=2 \end{gathered}$$

Now as the electron on the atom increases bond order decreases.

The bond order of the following is: ${\mathrm{O}}_2^{2-}=\frac{10-8}{2}=1; {\mathrm{O}}_2^{-}=\frac{10-7}{2}=1.5; {\mathrm{O}}_2=2; {\mathrm{O}}_2^{+}=\frac{10-5}{2}=2.5$

Hence, the correct option is: (D)${\mathrm{O}}_2^{+}>{\mathrm{O}}_2>{\mathrm{O}}_2^{-}>{\mathrm{O}}_2^{2-}$