Question

Out of $K{O}_2,N{O}_2,NO$ and $K_2{O}_2$ unpaired electron are present in:

• A. $K{O}_2,N{O}_2$ only
• B. $K{O}_2,N{O}_2,NO,K_2{O}_2$
• C. $K{O}_2,N{O}_2,NO$
• D. $NO,K_2{O}_2$ only

Answer 1

The correct option is C $K{O}_2,N{O}_2,NO$

The elctronic configuration of $O_2=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2pz}{)}^2\right({\pi }_{2px}{)}^2\left({\pi }_{2py}{)}^2\right({\pi }_{2px}^{\ast }{)}^1({\pi }_{2py}^{\ast }{)}^1$
$K{O}_2$ is a superoxide i.e. $O_2^{-}$, hence it has an unpaired electron but $K_2{O}_2$ is a peroxide i.e. $O_2^{2-}$, so it has no unpaired electron.
$NO$ has a total of 15 electrons, since it is an odd number, it also has unpaired electrons. Similarly for the same reason $N{O}_2$ also has unpaired electrons.