4. Chemical bonding and molecular structure — Chemistry STD 11 Science — Question

So these are diamagnetic. $\mathrm{O}_{2}^{*}$ is paramagnetic due to the presence of one unpaired electron.

$\left(\mathrm{CN}^{-}, \mathrm{CO}, \mathrm{NO}^{+}\right) 14=\sigma \mathrm{s}^{2}, \sigma^{*} \mathrm{L}^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2} \sigma 2 p_{x}^{2}$

$\pi 2 p_{y}^{2} \approx \pi 2 p_{z}^{2} \quad$ (No unpaired electron, diamagnetic)

$\mathrm{O}_{2}^{-}=\sigma \mathrm{s}^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}$

$\pi 2 p_{x}^{2} \approx \pi 2 p_{z}^{2}, \pi^{*} 2 p_{y}^{2}=\pi^{*} 2 p_{z}^{1}$

(one unpaired electron, paramagnetic)