Question
Which of the following species exhibits the diamagnetic behaviour?
A.
$$NO$$
B.
$$O_2^{2 - }$$
C.
$$O_2^ + $$
D.
$${O_{2'}}$$
Answer :
$$O_2^{2 - }$$
Solution :
Diamagnetic species have no unpaired electrons
$$\,O_2^{2 - }\, \Rightarrow \sigma 1{s^{ + 2}},{\sigma ^*}1{s^2},\sigma 2{s^{2,}}{\sigma ^*}2{s^2},\sigma 2p_x^2,$$ $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^2} \right.} \right.$$
Whereas paramagnetic species has one or more unpaired electrons as in
$$\,{O_2} \to \,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$ $$\,\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1 - 2} \right.\,\,{\text{unpaired}}\,{\text{electrons}}} \right.$$
$$O_2^ + \to \sigma 1{s^2}\,,\,{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$ $$\,\left\{ {\pi 2p_y^2 = \pi 2p_z^2\,\,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^0 - 1} \right.\,\,{\text{unpaired}}\,{\text{electrons}}} \right.$$
$$NO \to \sigma 1{s^2},{\sigma ^ * }1{s^2},\sigma 2{s^2},{\sigma ^ * }2{s^2},\sigma 2p_x^2,$$ $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2,} \right.\left\{ {{\pi ^ * }2p_y^1 = {\pi ^ * }2p_z^0 - 1} \right.$$ $${\text{unpaired electron}}$$