Question

Among $K{O}_2,Al{O}_2^{-},Ba{O}_2$ and $N{O}_2^{+}$, unpaired electron is present in

• A. $N{O}_2^{+}$ and $Ba{O}_2$
• B. $K{O}_2$ and $Al{O}_2^{-}$
• C. Only $K{O}_2$
• D. Only $Ba{O}_2$

Answer 1

The correct option is C Only $K{O}_2$

Molecular orbital electronic configurations are-
$K{O}_2\left(O_2^{-}\right):\sigma 1s^2\stackrel{.}{\sigma }1s^2\stackrel{.}{\sigma }2s^2\stackrel{.}{\sigma }2p_x^2\left|\begin{array}{c}\pi 2p_y^2\\ \pi 2p_z^2\end{array}\right|\left|\begin{array}{c}\stackrel{.}{\pi 2p_y^2}\\ \stackrel{.}{\pi 2p_z^2}\end{array}\right|\stackrel{.}{\sigma }2p_x^0$
Has one unpaired electron in $\stackrel{.}{\pi }$ 2p orbital.
$Al{O}_2^{-}$ has both oxygen in $O^{2-}$ state, therefore, no unpaired electron is present.
$Ba{O}_2\left(O_2^{2-}\right)$
$\sigma 1s^2\stackrel{.}{\sigma }1s^2\stackrel{.}{\sigma }2s^2\stackrel{.}{\sigma }2x^2\sigma 2p_x^2\left|\begin{array}{c}\pi 2p_y^2\\ \pi 2p_z^2\end{array}\right|\left|\begin{array}{c}\stackrel{.}{\pi 2p_y^2}\\ \stackrel{.}{\pi 2p_z^2}\end{array}\right|\stackrel{.}{\sigma }2p_x^0$
Has no unpaired electron.
$N{O}_2^{+}$ has $\left[\begin{array}{c}O=\stackrel{+}{N}=0\end{array}\right]$ bonding, hence no unpaired electron.