Question

Which compound does nitrogen have oxidation number in fraction?

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

Answer 1

The correct option is D $N_{3}H$

Considering the oxidation state of $O$ as $-2$ and of $H$ as $+1$, oxidation number of $N$ ($x$) in each neutral molecule is calculated as:

1. $N{H}_{2}OH$: $x+2\times 1+(-2)+1=0x=-1$

2. $N_2{H}_4$: $2\times x+4\times 1=0x=-2$

3.$N{H}_3$:$x+3\times 1=0x=-3$

4.$N_{3}H$:$3\times x+1=0x=-1/3$

Therefore option $D$ is the correct answer.