A) \[{{N}_{2}}{{H}_{4}}\]
B) \[N{{H}_{3}}\]
C) \[{{N}_{3}}H\]
D) \[N{{H}_{2}}OH\]
Correct Answer: C
Solution :
Let the oxidation state of nitrogen in the given compounds be x. |
[a]\[{{\overset{x}{\mathop{N}}\,}_{2}}{{\overset{x1}{\mathop{H}}\,}_{4}}\] |
\[2(x)+(+1)4=0\] |
\[2x=-4\] |
\[\therefore \] \[x=-2\] |
[b]\[\overset{x}{\mathop{N}}\,{{\overset{+1}{\mathop{H}}\,}_{3}}\] |
\[x+(+1)3=0\] |
\[\therefore \] \[x=-3\] |
[c]\[{{\overset{x}{\mathop{N}}\,}_{3}}\overset{+1}{\mathop{H}}\,\] |
\[(x)3+(+1)=0\] |
\[3x=-1\] |
\[\therefore \] \[x=-\frac{1}{3}\] |
[d]\[\overset{x}{\mathop{N}}\,\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2+1}{\mathop{OH}}\,\] |
\[x+(+1)2+(-2)+(+1)=0\] |
\[x+2-2+1=0\] |
\[x+1=0\] |
\[x=-1\] |
Thus, oxidation state of nitrogen is highest in\[{{N}_{3}}H.\] |
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