Question

The correct order of N – compounds in its decreasing order of oxidation state is:
A.${\text{HN}}{{\text{O}}_3},{\text{ N}}{{\text{H}}_4}{\text{Cl}},{\text{ NO}},{\text{ }}{{\text{N}}_2}$
B.${\text{HN}}{{\text{O}}_3},{\text{ NO}},{\text{ }}{{\text{N}}_2}{\text{, N}}{{\text{H}}_4}{\text{Cl}}$
C.${\text{N}}{{\text{H}}_4}{\text{Cl}},{\text{ }}{{\text{N}}_2},{\text{ NO , HN}}{{\text{O}}_3}$
D.${\text{HN}}{{\text{O}}_3},{\text{NO}},{\text{N}}{{\text{H}}_4}{\text{Cl}},{{\text{N}}_2}$

Answer 1

All the above molecules or compounds are neutral that there is neither a positive charge on them nor a negative charge. The oxidation state of nitrogen can be calculated by equating the sum of oxidation state in a compound, multiplied by the respective number of atoms to Zero.

Complete step by step answer:
Basically we have to find out the decreasing order of oxidation state of nitrogen in the four compounds that are ${\text{HN}}{{\text{O}}_3}$ that is nitric acid, ${\text{N}}{{\text{H}}_4}{\text{Cl}}$ that is ammonium chloride, ${\text{NO}}$ that is nitric oxide and ${{\text{N}}_2}$ nitrogen gas or dinitrogen. Let us calculate the oxidation state of nitrogen in each of the following compounds. We will write the oxidation state as O.S, just to avoid using long names repeatedly.
${\text{HN}}{{\text{O}}_3}$
${\text{O}}{\text{.S of H}} + {\text{O}}{\text{.S of N}} + 3 \times {\text{O}}{\text{.S of O}} = 0$
Rearranging the above equation we will get:
${\text{O}}{\text{.S of N}} = - 3 \times {\text{O}}{\text{.S of O}} - {\text{O}}{\text{.S of H}}$
The general oxidation state of H is 1 and the general oxidation state of O is -2 when in a bonded state. So using the above we will get oxidation state of nitrogen in${\text{HN}}{{\text{O}}_3}$ as:
${\text{O}}{\text{.S of N}} = - 3 \times - 2 - 1 = + 5$
Similarly we will calculate the oxidation state of ${\text{N}}{{\text{H}}_4}{\text{Cl}}$
${\text{O}}{\text{.S of N}} = - {\text{O}}{\text{.S of Cl}} - 4 \times {\text{O}}{\text{.S of H}}$
The oxidation state of chlorine is -1 in bonded state.
${\text{O}}{\text{.S of N}} = 1 - 4 \times 1 = - 3$
Oxidation state of nitrogen in ${\text{NO}}$ is:
${\text{O}}{\text{.S of N}} = - {\text{O}}{\text{.S of O}} = + 2$
The oxidation state of dinitrogen is zero, because it is a nonpolar covalent bond made out of the same atoms.

Hence the correct order is B.

Note:
Oxidation state or oxidation number is the hypothetical charge on an atom; it does not exist in real. It is calculated by assuming that all bonds in the compounds are ionic. Oxidation state is usually an integer, but it can also be a fractional value.