MCQ
Which of the following is paramagnetic ?
- A$CN^-$
- B$NO^+$
- C$CO$
- ✓$O_{2}^-$
✓
Answer
Correct option: D.
$O_{2}^-$
d
$\mathrm{CN}^{-}, \mathrm{CO}$ and $\mathrm{NO}^{+}$ are isoelectronic with 14 electrons each and there is no unpaired electrons in the MO configuration of these species.
$\mathrm{CN}^{-}, \mathrm{CO}$ and $\mathrm{NO}^{+}$ are isoelectronic with 14 electrons each and there is no unpaired electrons in the MO configuration of these species.
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)
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