Question

The oxidation number of ‘$N$’ in ${N_3}H$ (hydrazoic acid) is:
A.$- \dfrac{1}{3}$
B.$+ 3$
C.$0$
D.$- 3$

Answer 1

Oxidation number is defined as the total number of electrons that an electron loses or gains in order to form a bond with another atom. The $3$ Nitrogen molecules are connected through covalent bonds hence the total charge on this ion will be $- 1$.

Complete step by step answer:

Let’s start with discussing the oxidation number for better understanding of the question. Oxidation number is defined as the total number of electrons that an electron loses or gains in order to form a bond with another atom. So if there are two atoms $A$ and $B$ and $A$ give $3$ electrons to $B$ form a bond then the oxidation number of $A$ will be $+ 3$.
Coming back to the question we are asked to calculate the oxidation number of $N$ in ${N_3}H$. It is to be noticed that hydrogen can give only one electron hence the charge of Hydrogen is $+ 1$ and the charge of ${N_3}$ molecules is $- 1$. A Nitrogen molecule is having $5$ electrons in its outermost shell and requires only $3$ to complete its octet. These $3$ Nitrogen molecules are connected through covalent bonds so the equation will be as follows
$\Rightarrow 3x + 1 = 0$
Simplifying the terms we get,
$\Rightarrow 3x = - 1$
Solving for $x$ we get,
$\Rightarrow x = \dfrac{{ - 1}}{3}$
Hence,
$\therefore$ The answer to this question is option A. $- \dfrac{1}{3}$.

Note:
As we know Oxidation number is one of the important properties of the atom which tells oxidation state of the atom. An atom can have varying oxidation states and each oxidation state tells how stable the atom is. It is mostly defined as the amount of electron that is donated or accepted by the atom, so in case of covalent bond the oxidation number is $0$.