A) an odd electron molecule
B) paramagnetic in nature
C) very reactive
D) All of the above
E) None of the above
Correct Answer: D
Solution :
\[NO\,(7+8=15)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},2p_{z}^{2},\]\[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\,\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\] From the MO configuration, it is clear that NO contains odd electron, thus paramagnetic and very reactive.
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