Question

Which of the following has been arranged in the order of increasing oxidation number of nitrogen?
A.$\text{N}{{\text{H}}_{3}}$ < ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ < $\text{NO}$ < ${{\text{N}}_{2}}$
B. $\text{N}{{\text{O}}_{\text{2}}}^{\text{+}}$ < $\text{N}{{\text{O}}_{3}}^{\text{-}}$ < $\text{N}{{\text{O}}_{\text{2}}}^{-}$ < ${{\text{N}}_{3}}^{-}$
C.$\text{N}{{\text{H}}_{4}}^{\text{+}}$ < ${{\text{N}}_{\text{2}}}{{\text{H}}_{4}}$ < $\text{N}{{\text{H}}_{\text{2}}}\text{OH}$ < ${{\text{N}}_{\text{2}}}\text{O}$
D.$\text{N}{{\text{O}}_{\text{2}}}$ < $\text{Na}{{\text{N}}_{3}}$ < $\text{N}{{\text{H}}_{4}}^{\text{+}}$ < ${{\text{N}}_{\text{2}}}\text{O}$

Answer 1

Hint:The oxidation state of an element can be defined as the total number of electrons that an element has to lose to form a chemical bond with another atom.

Complete step by step answer:
The oxidation state of nitrogen in ammonia ${\text{N}}{{\text{H}}_3}$ is $\left( { - 3} \right)$, in nitrogen peroxide ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is $\left( { + 2} \right)$, in nitric oxide ${\text{NO}}$is $\left( { - 2} \right)$, in nitrogen molecule it is 0, ${{\text{N}}_2}$ in ${\text{N}}{{\text{O}}_{\text{2}}}^{\text{ + }}$ is $\left( { + 5} \right)$, in nitrate ${\text{N}}{{\text{O}}_3}^{\text{ - }}$ in $\left( { + 5} \right)$, in nitrite is ${\text{N}}{{\text{O}}_{\text{2}}}^ - $ $\left( { + 3} \right)$, in azide ion,${{\text{N}}_3}^ - $ it is $\left( { - 1} \right)$ (the total charge of the three nitrogen atoms is $\left( { - 1} \right)$ )
In ${\text{N}}{{\text{H}}_4}^{\text{ + }}$ it is $\left( { - 3} \right)$, in hydrazine ${{\text{N}}_{\text{2}}}{{\text{H}}_4}$ it is $\left( { - 2} \right)$, in hydroxylamine ${\text{N}}{{\text{H}}_{\text{2}}}{\text{OH}}$ it is $\left( { - 1} \right)$, in nitrous oxide ${{\text{N}}_{\text{2}}}{\text{O}}$ it is $\left( { + 1} \right)$, in nitrogen dioxide ${\text{N}}{{\text{O}}_{\text{2}}}$ it is $ + 4$, in sodium azide also it is $\left( { - 1} \right)$ (in total for the three nitrogen atoms).
From all the above values of oxidation states, the correct arrangement of oxidation number of nitrogen in the order of increasing order is correct in:
$\text{N}{{\text{H}}_{4}}^{\text{+}}$ < ${{\text{N}}_{\text{2}}}{{\text{H}}_{4}}$ < $\text{N}{{\text{H}}_{\text{2}}}\text{OH}$ < ${{\text{N}}_{\text{2}}}\text{O}$

So, the correct answer is option C.

Note:
The oxidation number of a free element is always zero.The dissimilarity between the oxidation state and valency is that valency does not have a sign while oxidation state does.
The oxidation number of a less electronegative element is always considered positive in respect to a less electronegative element.