A) \[{{N}_{2}}\]
B) NO
C) CO
D) \[{{O}_{3}}\]
Correct Answer: B
Solution :
[b] The molecular orbital configuration of the molecules given is Total no. of electrons in \[NO=7\left( N \right)+8\left( O \right)\] \[=15\] Hence E.C. of NO \[=KK\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\] \[\pi {{(2{{p}_{x}})}^{2}}=\pi {{(2{{p}_{y}})}^{2}}\pi *{{(2{{p}_{x}})}^{1}}=\pi {{(2{{p}_{y}})}^{0}}\] Due to presence of one unpaired electron NO is paramagnetic. Except NO all are diamagnetic due to absence of unpaired electrons.
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