Question

Maximum number of moles of electrons taken up by one mole of $N{O_3}^ -$ when it is reduced to
(A) $N{H_3}$
(B) $N{H_2}OH$
(C) $NO$
(D) $N{O_2}$

Answer 1

When a reduction reaction is carried out the oxidation state of elements involves changes and it takes a number of moles of electrons equal to change in the oxidation of the particular atom to reduce the atom. In this question we will see each reaction and see when $N{O_3}^ -$ reacts with all above-mentioned compounds how much change in oxidation state occurs.

Complete answer:
Let’s see the first option $N{O_3}^ -$ is reduced to $N{H_3}$ .
$N{O_3}^ - \to N{H_3}$
when nitrate ion is converted to ammonia the change in oxidation state is calculated as
oxidation state of ${\text{N}}$ in $N{O_3}^ -$ :
${\text{x}} + 3 \times \left( { - 2} \right) = - 1$ , where is assumed as oxidation state of ${\text{N}}$
${\text{x}} = + 5$ , oxidation state of ${\text{N}}$ in $N{O_3}^ -$ is $+ 5$ .
Now see the oxidation state of ${\text{N}}$ in $N{H_3}$ :
${\text{x}} + \left( {3 \times 1} \right) = 0$
${\text{x}} = - 3$
The oxidation state of ${\text{N}}$ in $N{H_3}$ is $- 3$ .
$5 - \left( { - 3} \right) = 8$ , total $8$ moles of electrons will be required to reduce $N{O_3}^ -$ to $N{H_3}$ .
So, option (A) is the correct answer.

Note:
Oxidation state of an atom in a molecule is calculated by adding the product of the number of other atoms in the molecule with their common oxidation state and equating it with the total charge on the molecule.
Similarly, for calculating the oxidation state of $N{H_2}OH$ .
$N{O_3}^ - \to N{H_2}OH$
oxidation state of ${\text{N}}$ in $N{H_2}OH$ :
${\text{x}} + \left( {3 \times 1} \right) - 2 = 0$
${\text{x}} = - 1$
The oxidation state of ${\text{N}}$ in $N{H_2}OH$ is $- 1$ .
$5 - \left( { - 1} \right) = 6$ , total six moles of electrons will be required to reduce $N{O_3}^ -$ to $N{H_2}OH$ .
Similarly, for calculating oxidation state of $NO$ :
${\text{x}} + \left( {1 \times - 2} \right) = 0$
${\text{x}} = 2$
The oxidation state of ${\text{N}}$ in $NO$ is $2$ .
$5 - 2 = 3$ , total three moles of electrons will be required to reduce $N{O_3}^ -$ to $NO$ .
Similarly, for calculating oxidation state of $N{O_2}$ :
${\text{x}} + \left( {2 \times - 2} \right) = 0$
${\text{x}} = 4$
The oxidation state of ${\text{N}}$ in $N{O_2}$ is $1$ .
$5 - 4 = 1$ , total one mole of electrons will be required to reduce $N{O_3}^ -$ to $NO$ .
So, $8$ moles of electrons will be required to reduce $N{O_3}^ -$ to $N{H_3}$ , which are the most among the mentioned compounds so option (A) is the correct answer.