Question

Which one of the following molecules is paramagnetic?

• A. $ F_{2} $
• B. $ B_{2} $
• C. $ Li_{2} $
• D. $ N_{2} $

Answer 1

The Correct Option is B

Molecules in which unpaired electrons are present, are paramagnetic while that have absence of unpaired electrons, are diamagnetic.
$ {{F}_{2}}(18)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}},\sigma 2{{s}^{2}}\overset{*}{\sigma}\,2{{s}^{2}},\sigma 2p_{z}^{2}, $
$ \pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\pi }\,2p_{x}^{2}\approx \overset{*}{\pi}\,2p_{y}^{2} $
(No unpaired electron, so diamagnetic)
$ {{B}_{2}}(10)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}}\sigma 2{{s}^{2}},\overset{*}{\sigma }\,2{{s}^{2}},\pi 2p_{x}^{1}\approx \pi 2p_{y}^{1} $
(Two unpaired electrons, paramagnetic)
$ L{{i}_{2}}(6)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}},\sigma 2{{s}^{2}} $
(No unpaired electrons, diamagnetic)
$ {{N}_{2}}(14)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\sigma }\,2{{s}^{2}},\pi 2p_{x}^{2} $
$ \approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2} $
(No unpaired electrons, diamagnetic)