A) \[N{{H}_{2}}OH\]
B) \[{{N}_{2}}{{O}_{3}}\]
C) \[N{{O}_{2}}\]
D) \[{{N}_{2}}{{H}_{6}}C{{l}_{2}}\]
Correct Answer: C
Solution :
The total valence electron present in\[N{{O}_{2}}\]are 17 which is an odd number. Therefore, it will be paramagnetic in character. All other have even number (14 in\[N{{H}_{2}}OH,\]28 in\[{{N}_{2}}{{O}_{3}}\]and 30 in\[{{N}_{2}}{{H}_{6}}C{{l}_{2}}\]) of electrons and are diamagnetic.
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