Question

In which of the following compounds, nitrogen exhibits highest oxidation state?

• A. $N _{2} H _{4}$
• B. $NH _{3}$
• C. $N _{3} H$
• D. $NH _{2} OH$

Answer 1

Answer: (C)

Let the oxidation state of nitrogen in the given compounds be $x$.
(a) $\overset{x+1}{N _{2} H _{4}}$ :
$2 x+(+1) 4 =0$
$2 x =-4$
$x =-2$
(b) $\overset{x+1}{NH _{3}}$ :
$x+(+1) 3 =0$
$x =-3$
(c) $\overset{x+1}{N _{3} H}$ :
$3 x+(+1) =0$
$3 x =-1$
$x =-1 / 3$
(d) $\overset{x +1-2 +1}{NH _{2} OH}$ :
$x+(+1) 2+(-2)+(+1) =0$
$x+2-2+1 =0$
$x+1 =0$
$x =-1$
The oxidation state of nitrogen is highest in $N _{3} H$.
i.e. in minus, $-\frac{1}{3}$ is greater than $-3$.