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Among the following, the paramagnetic compound is
$\mathrm{Na}_2 \mathrm{O}_2$
$\mathrm{O}_3$
$\mathrm{N}_2 \mathrm{O}$
$\mathrm{KO}_2$
Solution
$\mathrm{O}_2^{2-}=\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_z^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi^* 2 \mathrm{p}_{\mathrm{x}}^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}^2$
Number of unpaired electrons $=0$.
$\mathrm{N}=\mathrm{N} \longrightarrow \mathrm{O} \quad$ Number of unpaired electrons $=0$
$Image$ Number of unpaired electrons $=0$
$\mathrm{O}_2^{-}=\sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_z^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi^* 2 \mathrm{p}_{\mathrm{x}}^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}^1$
Number of unpaired electrons $=1$
Thus $\mathrm{O}_2^{-}$is paramagnetic.
Hence $(D)$ is correct.
