Question

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

Answer 1

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

Explanation:

Let's take 2 transition metal ions, dichromate, `Cr_2O_7^(2-)`, and permanganate, `MnO_4^-`.

Of course, oxygen normally takes a `-II` 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;`

`2xxCr^("ON")=+12; Cr^("ON")=VI+`.

For permanganate,

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

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

`Mn^("ON")=+VII`.

The oxidation state of other charged metal oxides are calculated the same way: `CrO_4^(2-)` `-=` `Cr(VI+).`