Question

Which of the following species is/are diamagnetic
(I) $O_2^{+}$ (II) $O_2$
(I) $O_2^{-}$ (II) $O_2^{2-}$

• A. $O_2^{-},O_2^{+}$
• B. $O_2$ and $O_2^{2-}$
• C. $O_2^{2-}$ only
• D. $O_2^{+},O_2^{-2}$

Answer 1

The correct option is C $O_2^{2-}$ only

$O_2$
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2pz}{)}^2\right({\pi }_{2px}{)}^2=\left({\pi }_{2py}{)}^2\right({\pi }_{2px}^{\ast }{)}^1=\left({\pi }_{2py}^{\ast }{)}^1\right({\sigma }_{2pz}^{\ast }{)}^0$
$O_2^{+}$
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2pz}{)}^2\right({\pi }_{2px}{)}^2=\left({\pi }_{2py}{)}^2\right({\pi }_{2px}^{\ast }{)}^1=\left({\pi }_{2py}^{\ast }{)}^0\right({\sigma }_{2pz}^{\ast }{)}^0$
$O_2^{-}$
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2pz}{)}^2\right({\pi }_{2px}{)}^2=\left({\pi }_{2py}{)}^2\right({\pi }_{2px}^{\ast }{)}^2=\left({\pi }_{2py}^{\ast }{)}^1\right({\sigma }_{2pz}^{\ast }{)}^0$
$O_2^{-2}$
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2pz}{)}^2\right({\pi }_{2px}{)}^2=\left({\pi }_{2py}{)}^2\right({\pi }_{2px}^{\ast }{)}^2=\left({\pi }_{2py}^{\ast }{)}^2\right({\sigma }_{2pz}^{\ast }{)}^0$

$\begin{array}{cc}& \text{Unpaired}{e}^{-}\\ O_2=& 2\\ O_2^{+}=& 1\\ O_2^{-}=& 1\\ O_2^2=& 0\end{array}$
The species having no unpaired electron is diamagnetic.
So, only $O_2^{2-}$ is diamagnetic.