Answer:

Correct Answer: B

Solution:

Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration.

Molecular orbital configuration of the given species is as $ CO(6+8=14)=\sigma 1s^{2},\overset{*}\sigma,1s^{2}, $

$ \sigma 2s^{2},\overset{*}\sigma,2s^{2},\pi {2p _x}^{2} $

$\approx \overset{*}\pi,{2p _y}^{1} $ (All the electrons are paired so it is diamagnetic.)

$ O_2^{-}(8+8+1=17) $

$ =\sigma 1s^{2},\overset{*}\sigma,1s^{2},\sigma,2s^{2}$

$,\overset{*}\sigma,s^{2}, \sigma {2p_z}^{2} \pi {2p_x}^{2}, $

$ \pi,{2p _y}^{2},\overset{*}\pi,{2p _x}^{2}$

$\approx \overset{*}\pi,{2p _y}^{1} $

It contains one unpaired electron so it is paramagnetic.

$ C{N^{-}}(6+7+1=14)= $ same as CO $ N{O^{+}}(7+8-1)= $ same as CO Thus,

among the given species only $ O_2^{-} $ is paramagnetic.