A) \[{{N}_{2}}\]
B) NO
C) CO
D) \[{{O}_{3}}\]
Correct Answer: B
Solution :
Electronic configuration of NO =\[KK\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\]\[{{\pi }^{*}}{{(2{{p}_{x}})}^{1}}\] \[\because \]NO has one unpaired electron in anti bonding orbital. \[\therefore \] It is paramagnetic.
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