Question

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

• A. $N_2{H}_4$
• B. $N{H}_3$
• C. $N_{3}H$
• D. $N{H}_{2}OH$

Answer 1

Answer: (C)

Let the oxidation state of nitrogen in the given compounds be $x$.
(a) $\stackrel{x+1}{N_2{H}_4}$ :
$2x+(+1)4=0$
$2x=-4$
$x=-2$
(b) $\stackrel{x+1}{N{H}_3}$ :
$x+(+1)3=0$
$x=-3$
(c) $\stackrel{x+1}{N_{3}H}$ :
$3x+(+1)=0$
$3x=-1$
$x=-1/3$
(d) $\stackrel{x+1-2+1}{N{H}_{2}OH}$ :
$x+(+1)2+(-2)+(+1)=0$
$x+2-2+1=0$
$x+1=0$
$x=-1$
The oxidation state of nitrogen is highest in $N_{3}H$.
i.e. in minus, $-\frac{1}{3}$ is greater than $-3$.