Question
The molecule having one unpaired electron is :
A.
$$NO$$
B.
$$CO$$
C.
$$C{N^ - }$$
D.
$${O_2}$$
Answer :
$$NO$$
Solution :
NOTE THIS STEP: Write the electronic configuration of each species according to molecular orbital theory.
$$NO\left( {15{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^0} \right.\,\,1\,{\text{unpaired}}\,{\text{electron}}.} \right.$$
$$CO\left( {14{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\sigma 2p_x^2} \right.\,\,\,\,\,{\text{no}}\,{\text{unpaired}}\,{\text{electron}}$$
$$C{N^ - }\left( {14{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\sigma 2p_x^2} \right.$$
$${O_2}\left( {16{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1;} \right.} \right.\,\,{\text{Two}}\,\,\,{\text{unpaired}}\,{\text{electrons}}{\text{.}}$$