Question

Which of the following is paramagnetic ?

• A. $N{O}^{+}$
• B. $CO$
• C. $O_2^{-}$
• D. $C{N}^{-}$

Answer 1

Answer: (C)

Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration.
Molecular orbital configuration of the given species is as
$CO(6+8=14)=\sigma 1s^2{,}_{\sigma }^{\ast }1s^2,$
$\sigma 2s^2{,}_{\sigma }^{\ast }2s^2,\pi 2P_X^2\approx \pi 2P_Y^2,\sigma 2P_Z^2$
(All the electrons are paired so, it is diamagnetic).
$O_2^{-}(8+8+1=17)$
$=\sigma 1s^2{.}_{\sigma }^{\ast }1s^2,\sigma 2s^2{,}_{\sigma }^{\ast }2s^2,\sigma 2P_Z^2,\pi 2P_X^2\approx \pi 2P_Y^2,$
${}_{\pi }^{\ast }2P_X^2{\approx }_{\pi }^{\ast }2P_Y^1$
(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.