Question

Paramagnetic behaviour of ${KO}_2$ is due to the fact that unpaired electron is present in:

• A. $\pi$2p molecular orbital
• B. ${\pi }^{\ast }$2p molecular orbital
• C. ${\sigma }^{\ast }$2p molecular orbital
• D. $\sigma$2p molecular orbital

Answer 1

Answer: (B)

${\pi }^{\ast }$2p molecular orbital

$K{O}_2$ is Potassium superoxide with $O$ atom existing as $O_2^{-}$

By molecular orbital explanation,

Electronic configuration of $O_2^{-}$ is $\sigma \left(1s^2\right){\sigma }^{\ast }\left(1s^2\right)\sigma \left(2s^2\right){\sigma }^{\ast }\left(2s^2\right)\sigma \left(2p_x^2\right)\pi \left(2p_y^2\right)=\pi \left(2p_z^2\right)$

${\pi }^{\ast }\left(2p_y^2\right)={\pi }^{\ast }\left(2p_z^1\right)$

So, it has got an unpaired electron in ${\pi }^{\ast }2p_z$ orbital which is responsible for its paramagnetic behavior.