Question

How do you find the oxidation numbers for compounds?

Answer 1

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_2O` molecule exists as a neutrally charged substance. You don't see it being written as `H_2O^+` or `H_2O^-` : 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 `NO_3^"-1"` ion is -1).

(3) All elements from Group 1A has an oxidation state of +1. (e.g. `Na^"+1"`, `Li^"+1"`)

(4) All Group 2A and 3B elements have an oxidation state of +2 and +3, respectively. (e.g. `Ca^"2+"`, `Mg^"2+"`,`Al^"3+"`)

(5) Oxygen always have a charge -2 except for peroxide ion (`O_2^"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 `HCl`) and always have a -1 charge if it is bonded with a metal (as in `AlH_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_2SO_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 : `color (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) + `color (red) y` + (-8) = 0 (based on Rule 1)

Thus,

`color (red) y` + (-6) = 0

`color (red) y` = +6

Therefore, the oxidation state of `S` is +6 or `S^"6+"`.