A) \[{{N}_{2}}{{O}_{4}}\]is paramagnetic in nature
B) \[N{{O}_{2}}\]dimerises to \[{{N}_{2}}{{O}_{4}}\]
C) \[N{{O}_{2}}\]is diamagnetic
D) Itself ionises as \[N{{O}^{+}},\,NO_{3}^{-}\]
E) None of the above
Correct Answer: B
Solution :
\[N{{O}_{2}}\] is paramagnetic due to the presence of one unpaired electron. But \[N{{O}_{2}}\] get dimerises to \[{{N}_{2}}{{O}_{4}},\] due to which paramagnetism is lost because sharing of unpaired electron between two N-atoms occurs. Thus. \[{{N}_{2}}{{O}_{4}}\] is diamagnetic. \[\underset{Paramagnetic}{\mathop{2N{{O}_{2}}}}\,\underset{Diamagnetic}{\mathop{{{N}_{2}}{{O}_{4}}}}\,\]
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