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  4. The diamagnetic species among the following is

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Q. The diamagnetic species among the following is

KVPYKVPY 2019Chemical Bonding and Molecular Structure

Solution:

To be diamagnetic, no unpaired electron must be present.
Hint Write molecular orbital electronic configuration of each species.
For all options given,
$\sigma_{1 s}^{2} \sigma_{1 s}^{* 2} \sigma_{2 s}^{2} \sigma_{2 s}^{* 2} \sigma_{2 p_{z}}^{2} \pi_{2 p_{x}}^{2}=\pi_{2 p_{y}}^{2}$ is common.
After that, $O _{2}^{+}: \pi_{2 p_{x}}^{* 1}$
$O _{2}^{-}: \pi_{2 p_{x}}^{* 2}=\pi_{2 p_{y}}^{*^{1}}$
and $O _{2}: \pi_{2 p_{x}}^{* 1}=\pi_{2 p_{y}}^{* 1}$
all contain unpaired electron.
They are paramagnetic, but $O _{2}^{2-}: \pi_{2 p_{x}}^{* 2}=\pi_{2 p_{y}}^{*^{2}}$ has no unpaired electrons.
$\therefore $ It is diamagnetic.