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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