Question

Which one of the following $O_2^{-},O_2^{2-}$ may exhibit paramagnetism?

• A. $O_2^{-}$
• B. $O_2^{2-}$
• C. Both are paramagnetic
• D. Both are diamagnetic

Answer 1

The correct option is A $O_2^{-}$

The molecular electronic configuration of
$O_2^{-}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^2({\pi }_{2p_x}^{\ast }{)}^1$
$O_2^{2-}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^2({\pi }_{2p_x}^{\ast }{)}^2$
Since $O_2^{-}$ has unpaired electron in its antibonding molecular orbital, thus it is paramagnetic and oppositely $O_2^{2-}$ is diamagnetic