A) \[{{B}_{2}}\]
B) \[{{H}_{2}}\]
C) \[L{{i}_{2}}\]
D) \[{{N}_{2}}\]
Correct Answer: D
Solution :
\[{{B}_{2}}(10)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},p2p_{x}^{1}\approx \pi 2p_{y}^{1}\] Since, two unpaired electrons are present,\[{{B}_{2}}\]is paramagnetic. All other species, due to the absence of unpaired electrons, are diamagnetic.
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