Using molecular orbital theory, compare the bond energy and magnetic character of ${\rm{O}}_2^ + $ and ${\rm{O}}_2^{2 - }$ species.

According to molecular orbital theory electronic configurations of $\mathrm{O}_{2}^{+}$and $\mathrm{O}_{2}^{-}$species are as follows : $\mathrm{O}_{2}^{+}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}^{2}, \pi 2 p_{y}^{2}\right)\left(\pi^{*} 2 p_{x}^{2}\right)$

Bond order of $\mathrm{O}_{2}^{+}=\frac{10-5}{2}=\frac{5}{2}=2.5$

$\mathrm{O}_{2}^{-}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}^{2}, \pi 2 p_{y}^{2}\right)$

$\left(\pi^{*} 2 p_{x}^{2},\left(\pi^{*} 2 p_{y}^{1}\right)\right.$

Bond order of $\mathrm{O}_{2}^{-}=\frac{10-7}{2}=\frac{3}{2}=1.5$

As Higher bond order of $\mathrm{O}_{2}^{+}$shows that it is more stable than $\mathrm{O}_{2}^{-}$. Both the species have unpaired electrons. So, both are paramagnetic in nature.