First you need to know the most common oxidation states of several different elements. These are easy to remember if you know the groups that these elements belong to in the Periodic Table. Group I has Li, Na, K (the others are less important), and all these have the oxidation state of +1. Group II has Be, Mg, Ca and these have oxidation state +2. The transition metals, from Sc to Zn and below have different oxidation states that are useful to remember for the common elements. In group III, B, Al, Ga have ox. number +3. Gr. IV has C, Si, Ge, Sn, which have ox. no. +4. From Gr V on the oxidation numbers are negative (ox. no. = Gr. no. - 8). So Gr. V (N, P) has ox. no. -3. Gr. VI (O, S) has ox. no. -2. However, N is often considered +5, but sometime +3 and S can be either +6 or +4 (which one it is depends on the compond, and can be determined from the other elements in that compound). Gr. VII (F, Cl, Br) is always -1. And Gr. VIII has ox. no. 0. So, elements of Gr. VIII never form compounds.
To find the oxidation no. of a radical (like sulfite, nitrate, etc.) you simply add the oxidation states of the different elements in it.
Sulfite is SO3(2-). We know that each O is -2, so the 3 O's make up -6. So, SO3(2-) has a total charge of -2. Let the oxidation state of S be x. Then x + 3(-2) = -2. Or x-6=-2. Hence, x=4. The ox. no. of S is +4 in this case.
In a compound (or stable element), the total charge is always 0. So, the sum of positives must equal the sum of the negatives.
So, in sulfur dioxide (SO2) , the total ox. no. is 0. We know for sure that O is -2. And there are 2 O's, so the total negative charge is -4. So, for the compound to have zero total charge, S must be +4.
What would be the ox. no. of S in sulfur trioxide (SO3) ?
