Question

One mole of $HN{O}_3$ absorbs $5$ moles of electrons in a redox reaction. The reduced product of this reaction is :

• A. $1$ mole $N_2$
• B. $1$ mole $N_{2}O$
• C. $1$ mole $N{O}_2$
• D. $1$ mole $N{H}_3$

Answer 1

The correct option is A $1$ mole $N_2$

The oxidation state of $N$ in $HN{O}_3$ is $+5$.
If it takes $5$ moles of electrons, then oxidation state of N will be zero.
So, product formed will be $N_2$ (0 oxidation state of N) and $0.5$ mole of it will be formed.
Compound Oxidation state
$N_{2}O$ +2
$N{O}_2$ +4
$N{H}_3$ -3