A) \[{{O}_{2}}\]
B) \[C{{N}^{\odot }}\]
C) \[{{N}_{2}}\]
D) \[{{C}_{2}}\]
Correct Answer: A
Solution :
[a] From \[{{O}_{2}}\]the election is removed form anti-bonding orbital. \[K{{K}^{*}}\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2{{p}_{z}}^{2},(\pi 2{{p}_{y}}^{2}=\pi 2{{p}_{x}}^{2}){{\pi }^{*}}2{{p}_{y}}^{1}{{\pi }^{*}}2{{p}_{x}}^{1}\]
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