A) \[{{N}_{2}}\]
B) NO
C) CO
D) \[{{O}_{3}}\]
Correct Answer: B
Solution :
The molecular orbital configuration of the molecules given is Total no. of electrons in NO = 7(N) + 8(O) = 15 Hence E.C. of \[NO=KK{{[\sigma (2s)]}^{2}}{{[\sigma *(2s)]}^{2}}{{[\sigma 2{{p}_{z}}]}^{2}}\] \[{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[\pi *(2{{p}_{x}})]}^{1}}\] 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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