Question

Compare the relative stability of the following species and indicate their magnetic properties :

$O_2,O_2^{+},{O_2}^{-}$ (superoxide) and ${O_2}^{2-}$ (peroxide).

Answer 1

(1) The electronic configuration of $O_2$ molecule is
$KK\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x{)}^2\left(\pi 2p_y{)}^2\right({\pi }^{\ast }2p_x{)}^1({\pi }^{\ast }2p_y{)}^1$
Its bond order is $\frac{8-4}{2}=2$.
It contains 2 unpaired electrons and is paramagentic.
(2) The electronic configuration of $O_2^{+}$ ion is
$KK\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x{)}^2\left(\pi 2p_y{)}^2\right({\pi }^{\ast }2p_x{)}^1$
Its bond order is $\frac{8-3}{2}=2.5$.
It contains 1 unpaired electron and is paramagnetic.
(3) The electronic configuration of $O_2^{-}$ ion is
$KK\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x{)}^2\left(\pi 2p_y{)}^2\right({\pi }^{\ast }2p_x{)}^2({\pi }^{\ast }2p_y{)}^1$
Its bond order is $\frac{8-5}{2}=1.5$.
It contains 1 unpaired electron and is paramagnetic.
(4) The electronic configuration of $O_2^{2-}$ ion is
$KK\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x{)}^2\left(\pi 2p_y{)}^2\right({\pi }^{\ast }2p_x{)}^2({\pi }^{\ast }2p_y{)}^2$
Its bond order is $\frac{8-6}{2}=1$.
It contains all paired electrons and is diamagnetic.
Higher is the bond order, stronger is the bond, and higher is the stability.

The decreasing order of the stability is $O_2^{+}>O_2>{O_2}^{-}>O_2^{2-}$.