# What is the oxidation number of nitrogen in the nitrate ion?

Nov 4, 2015

+5

#### Explanation:

Here are some of the rules for solving oxidation state of any element:

(1) The total charge of a stable compound is always equal to zero (meaning no charge).

For example, the ${H}_{2} O$ molecule exists as a neutrally charged substance.

(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the ion is the charge (i.e. oxidation state of $N {O}_{3}^{\text{-1}}$ ion is -1).

(3) All elements from Group 1A has an oxidation state of +1 (e.g. $N {a}^{\text{+1}}$, $L {i}^{\text{+1}}$). All Group 2A and 3A elements have an oxidation state of +2 and +3, respectively. (e.g. $C {a}^{\text{2+}}$, $M {g}^{\text{2+}}$, $A {l}^{\text{3+}}$)

(4) Oxygen always have a charge -2 except for peroxide ion (${O}_{2}^{\text{2-}}$) which has a charge of -1.

(5) Hydrogen always have a charge of +1 if it is bonded with a non-metal (as in the case of $H C l$) and always have a -1 charge if it is bonded with a metal (as in $A l {H}_{3}$).

So for your problem, according to rule 2, the overall oxidation state of $N {O}_{3}^{-}$ is -1.

$N {O}_{3}^{-}$ = -1

According to rule 4, the $O$ atom should have a -2 charge. But since the oxygen has a subscript in the formula, we need to multiply the charge by the subscript.

$\textcolor{red}{x}$ + [(-2) (3)] = -1 where $\textcolor{red}{x}$ is the oxidation state on $N$ atom

solving for $\textcolor{red}{x}$,

$\textcolor{red}{x}$ + (-6) = -1

$\textcolor{red}{x}$ = -1 + (+6)

$\textcolor{red}{x}$ = +5

Therefore, answer is ${N}^{\text{5+}}$