# Question #ed336

Nov 3, 2015

$N$ = +5; $O$ = -2; and the whole ion $N {O}_{3}^{-}$ is -1.

#### Explanation:

Here are a few rules to remember when solving for oxidation state.

(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 $A l {H}_{3}$).

Based on rule 2, the oxidation state of the $N {O}_{3}^{-}$ ion is -1.

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

based on rule 4, the oxidation state of $O$ atom is -2.

$x$ + [(3) (-2)] = -1 where $x$ is the oxidation state of $N$.

Notice that since $O$ atom has subscript, I needed to multiply the oxidation state by 3. Also, there are no rules for the oxidation state of $N$ so we need to solve it long hand.

$x$ + (-6) = -1
$x$ = -1 + (+6)
$x$ = +5

Therefore your oxidation states are $N$ = +5; $O$ = -2 and the whole ion $N {O}_{3}^{-}$ is -1.