# How is charge used to assign oxidation numbers to the elements in a polyatomic ion?

Sep 18, 2016

The sum of the oxidation numbers ALWAYS equals the charge on the ion.

#### Explanation:

Let's take 2 transition metal ions, dichromate, $C {r}_{2} {O}_{7}^{2 -}$, and permanganate, $M n {O}_{4}^{-}$.

Of course, oxygen normally takes a $- I I$ oxidation state, and it does so here. Given that the sum of the oxidation numbers equals the charge on the ion, then:

2xxCr^("ON")+7xxO^("ON")=-2;

2xxCr^("ON")+7xx-2=-2;

$2 \times C {r}^{\text{ON")=+12; Cr^("ON}} = V I +$.

For permanganate,

Mn^("ON")+4xxO^("ON")=-1;

Mn^("ON")+4xx(-2)=-1;

$M {n}^{\text{ON}} = + V I I$.

The oxidation state of other charged metal oxides are calculated the same way: $C r {O}_{4}^{2 -}$ $\equiv$ $C r \left(V I +\right) .$