A) \[{{N}_{3}}H\]
B) \[~N{{H}_{2}}OH\]
C) \[~{{N}_{2}}{{H}_{4}}\]
D) \[~N{{H}_{3}}\]
Correct Answer: A
Solution :
Let the oxidation state of nitrogen in the given compounds be \[x.\] [a]\[\overset{x}{\mathop{{{N}_{3}}}}\,\overset{+1}{\mathop{H}}\,\Rightarrow (x)3+(+1)=0\Rightarrow 3x=-1\] \[\Rightarrow \]\[x=-\frac{1}{3}\] [b] \[\overset{x}{\mathop{N}}\,\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2+1}{\mathop{OH}}\,\Rightarrow x+(+1)2+(-2)+(+1)=0\] \[x+2-2+1=0\Rightarrow x=-1\] [c] \[\overset{x}{\mathop{{{N}_{2}}}}\,\overset{+1}{\mathop{{{H}_{4}}}}\,\Rightarrow 2(x)+(+1)4=0\] \[\Rightarrow 2x=-4\Rightarrow x=-2\] [d] \[\overset{x}{\mathop{N}}\,\overset{+1}{\mathop{{{H}_{3}}}}\,\Rightarrow x+(+1)3=0\] \[\therefore \]\[x=-3\] \[\therefore \]Oxidation state of nitrogen is highest in \[{{N}_{3}}H.\]
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