Using molecular orbital theory,compare the bond energy and magnetic character of $O_2^{+}$ and $O_2^{2-}$ species.

(N/A) According to molecular orbital theory,the electronic configurations are as follows:
$O_2^{+}: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x^2, \pi 2p_y^2) (\pi^* 2p_x^1)$
Bond order of $O_2^{+} = \frac{10-5}{2} = 2.5$
$O_2^{2-}: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x^2, \pi 2p_y^2) (\pi^* 2p_x^2, \pi^* 2p_y^2)$
Bond order of $O_2^{2-} = \frac{10-8}{2} = 1.0$
Since bond energy is directly proportional to bond order,$O_2^{+}$ has higher bond energy than $O_2^{2-}$.
$O_2^{+}$ is paramagnetic due to the presence of one unpaired electron,whereas $O_2^{2-}$ is diamagnetic as all electrons are paired.