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