Question

$ KO_{2} $ exhibits paramagnetic behaviour. This is due to the paramagnetic nature of

• A. $ KO^{-} $
• B. $ K^{+} $
• C. $ O_{2} $
• D. $ O^{-}_{2} $

Answer 1

Answer: (D)

In $KO _{2}$, oxygen exists as $O _{2}^{-}$ ion. Its molecular orbital electronic configuration is as follows

$O _{2}^{-} =\sigma 1 s ^{2}, \overset{*}{\sigma} 1 s^{2}, \sigma 2 s^{2},{ } 2 s^{2}, \overset{*}{\sigma} 2 P_{z}^{2} $

$\pi 2 p_{x}^{2} \approx \pi^{2} p_{y}^{2}, \overset{*}{\pi} 2 p^{2}{ }_{x} \approx \overset{*}{\pi} 2 p^{-1}{ }_{y}$

Due to presence of $1$ unpaired electron in anti bonding molecular orbital, it is paramagnetic inbehaviour.