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$\mathrm{KO}_2$ exhibits paramagnetic behaviour. This is due to the paramagnetic nature of
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Verified Answer
The correct answer is:
$\mathrm{O}_2^{-}$
In $\mathrm{KO}_2$, oxygen exists as $\mathrm{O}_2^{-}$ion. Its molecular orbital electronic configuration is as follows
$$
\begin{aligned}
\mathrm{O}_2^{-} & =\sigma 1 s^2, \tilde{\sigma} 1 s^2, \sigma 2 s^2, \dot{\sigma} 2 s^2, \sigma 2 P_z^2 \\
\pi 2 p_x^2 & \approx \pi^2 p_y^2, \pi 2 p^2{ }_x \approx \tilde{\pi} 2 p^{-1} y
\end{aligned}
$$
Due to presence of 1 unpaired electron in anti bonding molecular orbital, it is paramagnetic inbehaviour.
$$
\begin{aligned}
\mathrm{O}_2^{-} & =\sigma 1 s^2, \tilde{\sigma} 1 s^2, \sigma 2 s^2, \dot{\sigma} 2 s^2, \sigma 2 P_z^2 \\
\pi 2 p_x^2 & \approx \pi^2 p_y^2, \pi 2 p^2{ }_x \approx \tilde{\pi} 2 p^{-1} y
\end{aligned}
$$
Due to presence of 1 unpaired electron in anti bonding molecular orbital, it is paramagnetic inbehaviour.
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