The species having highest bond energy is :

${\mathrm{O}}_2^{2-}$ has highest bond energy.

Here, 

${\mathrm{O}}_2^{2-}={\mathrm{KK}}^{\mathrm{\prime }}\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{p}}_z^2,\pi 2{\mathrm{p}}_{\mathrm{z}}^2,\pi 2{\mathrm{p}}_{\mathrm{y}}^2,2{\mathrm{p}}_{\mathrm{x}}^2,\pi 2{\mathrm{p}}_{\mathrm{y}}^2$

$\text{ Bond order }=\frac{N_b-N_a}{2}=\frac{8-6}{2}=1$

${\mathrm{O}}_2^{+}=\mathrm{KK}\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{p}}_z^2,\pi 2{\mathrm{p}}_{\mathrm{K}}^2\approx \pi 2{\mathrm{p}}_{\mathrm{y}}^2,\pi 2{\mathrm{p}}_{\mathrm{r}}^1$

$\text{ Bond order }=\frac{8-3}{2}=2.5$

$\left({\mathrm{O}}_2^{-}={\mathrm{KK}}^{\mathrm{\prime }}\sigma 2{\mathrm{s}}^2\ast \sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{p}}_2^2,\pi 2{\mathrm{p}}_{\alpha }^2\approx \pi 2{\mathrm{p}}_{\mathrm{y}}^2\pi 2{\mathrm{p}}_{\mathrm{K}}^2{\approx }^{\ast }\pi 2{\mathrm{p}}_{\mathrm{y}}^1\right.$

$\text{ Bond order }=\frac{8-5}{2}=1.5$

${\mathrm{O}}_2={\mathrm{KK}}^{\mathrm{\prime }}\sigma 2{\mathrm{s}}^2;\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{p}}_z^2,\pi 2{\mathrm{p}}_{\pi }^2\approx \pi 2{\mathrm{p}}_{\mathrm{y}}^2\ast 2{\mathrm{p}}_{\mathrm{x}}^1{\approx }^{\ast }\pi 2{\mathrm{p}}_{\mathrm{y}}^1$

$\text{ Bond order }=\frac{8-4}{2}=2$

$\because \text{ Bond energy }\propto \frac{1}{\text{ Bond order }}$

$\therefore {\mathrm{O}}_2^{2-}$ has highest bond energy.

Hence correct option is D.