Question
Which of the following is paramagnetic?
A.
$$CO$$
B.
$$O_2^ - $$
C.
$$C{N^ - }$$
D.
$$N{O^ + }$$
Answer :
$$O_2^ - $$
Solution :
Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration.
Molecular orbital configuration of the given species is as
$$CO\left( {6 + 8 = 14} \right)$$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$ $$\pi 2p_x^2 \approx \pi 2p_y^2,\sigma 2p_z^2$$
( All the electrons are paired so, it is diamagnetic ).
$$O_2^ - \left( {8 + 8 + 1 = 17} \right)$$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$ $$\sigma 2p_z^2,\pi 2p_x^2 \approx \pi 2p_y^2,\mathop \pi \limits^* 2p_x^2 \approx \mathop \pi \limits^* 2p_y^1$$
( It contains one unpaired electron so, it is paramagnetic. )
$$C{N^ - }\left( {6 + 7 + 1 = 14} \right) = {\text{same}}\,{\text{as}}\,CO$$
$$N{O^ + }\left( {7 + 8 - 1 = 14} \right) = {\text{same}}\,{\text{as}}\,CO$$
Thus, among the given species only $$O_2^ - $$ is paramagnetic.