Question

The number of paramagnetic species among the following is
$O_{2},CO,N_{2},C_{2},CsO_{2},BaO_{2},CO_{2}$

Answer 1

The magnetic properties of a substance can be determined by examining its molecular orbital configuration, If it has unpaired electrons, then the substance is paramagnetic and if all electrons are paired, the substance is then diamagnetic.
Molecular orbital configuration of $O_{2}$ molecule : $\sigma 1s^{2}\sigma ^{\star}1s^{2}\sigma 2s^{2}\sigma ^{\star}2s^{2}\sigma 2p_{z}^{2}\pi 2p_{x}^{2}\equiv \pi 2p_{y}^{2}\pi ^{\star}2p_{x}^{1}\equiv \pi ^{\star}2_{y}^{1}$
$O_{2}$ molecule has 2 unpaired electrons, so it is paramagnetic
In $CsO_{2 }$ molecule, super oxide ion $O_{2}^{-}$ has one unpaired electron, therefore it is paramagnetic.
$O_{2}^{-}ion:\sigma 1s^{2}\sigma ^{\star}1s^{2}\sigma 2s^{2}\sigma ^{\star}2s^{2}\sigma 2p_{z}^{2}\pi 2p_{x}^{2}\equiv \pi 2p_{y}^{2}\pi ^{\star}2p_{x}^{2}\equiv \pi ^{\star}2_{y}^{1}$
$N_{2 }$ molecule is isoelctronic with $CO$ , there are no unpaired electrons in $N_{2 }$ molecule .
$N_{2 ,}CO,C_{2 ,}BaO_{2}$ and $CO_{2}$ are diamagnetic due to the absence of unpaired electrons.