Among the following, the paramagnetic compoun | vedclass

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Question

$\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 \m

Vedclass · Accepted Answer

$\mathrm{KO}_2$

Vedclass · Answer

$\mathrm{O}_2^{2-}=\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_z^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi^* 2 \mathrm{p}_{\mathrm{x}}^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}^2$

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.

Among the following, the paramagnetic compound is

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