A) \[{{N}_{3}}H\]
B) \[{{N}_{2}}{{O}_{4}}\]
C) \[N{{H}_{2}}OH\]
D) \[N{{H}_{3}}\]
Correct Answer: B
Solution :
\[{{N}_{3}}H:\] Oxidation number of\[N=-\frac{1}{3}\] \[{{N}_{2}}{{O}_{4}}:\] Oxidation number of \[N=+4\] \[N{{H}_{2}}OH:\] Oxidation number of\[N=-1\] \[N{{H}_{3}}:\] Oxidation number of \[N=-3\] Hence, in\[{{N}_{2}}{{O}_{4}}\]the oxidation number of nitrogen is highest.
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