A) \[\sigma 2{{p}_{z}}\]
B) \[\pi 2{{p}_{x}}\approx 2{{p}_{y}}\]
C) \[\overset{*}{\mathop{\pi }}\,2{{p}_{x}}\approx \overset{*}{\mathop{\pi }}\,2{{p}_{y}}\]
D) \[\overset{*}{\mathop{\sigma }}\,2{{p}_{z}}\]
Correct Answer: A
Solution :
The molecular orbital configuration of nitrogen molecule is as \[{{N}_{2}}(14)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\pi 2p_{x}^{2}\] \[\approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] The electrons are filled in order of increasing energy hence, the maximum energy molecular orbital is \[\sigma 2{{p}_{z}}.\]
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