# How do you find the oxidation numbers for compounds?

Oct 30, 2015

By remembering some important rules.

#### Explanation:

(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. You don't see it being written as ${H}_{2} {O}^{+}$ or ${H}_{2} {O}^{-}$ : these molecules simply does not exist.

(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the in 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. $N {a}^{\text{+1}}$, $L {i}^{\text{+1}}$)

(4) All Group 2A and 3B elements have an oxidation state of +2 and +3, respectively. (e.g. $C {a}^{\text{2+}}$, $M {g}^{\text{2+}}$,$A {l}^{\text{3+}}$)

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

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

If you can remember and apply these rules, assigning an oxidation state for a given element would be easy for you.

Let's try an example. Find the oxidation state of $S$ in the formula ${H}_{2} S {O}_{4}$.

Following the rules above and converting the atoms according to their oxidation states,

H : (+1)(2 atoms) = 2

This based on Rule 6 wherein the $H$ atom is bonded to nonmetallic $S$ atom; and based on the subscript, there are a total of two $H$ atoms in the formula.

S : $\textcolor{red}{y}$ or unknown
O : (-2) (4 atoms) = -8

Again, based on the Rule 5. The oxidation state is multiplied by the number of atoms based on the subscript.

Getting the sum, we have

(+2) + $\textcolor{red}{y}$ + (-8) = 0 (based on Rule 1)

Thus,

$\textcolor{red}{y}$ + (-6) = 0

$\textcolor{red}{y}$ = +6

Therefore, the oxidation state of $S$ is +6 or ${S}^{\text{6+}}$.