### Rules for assigning oxidation numbersTen Examples

I. Rule Number One

All free, uncombined elements have an oxidation number of zero.

This includes the seven diatomic elements (such as O2) and the other molecular elements (P4 and S8).

II. Rule Number Two

Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one)

III. Rule Number Three

Oxygen, in all its compounds except peroxides and superoxides, has an oxidation number of -2 (negative two).

IV. Rule Number Four

For single ions (in other words, not polyatomic), the charge on the ion is taken to be the oxidation number.

V. Rule Number Five

When no charge is indicated in a given formula, the total charge is taken to be zero.

With only a very few exceptions (which you will never see at an introductory level), oxidation states can be assigned to all atoms in a formula. There are more extensive sets of rules and, for the most part, they derive from the five rules above.

As an example, this rule is sometimes seen:

The oxidation state of fluorine is -1 in all of its compounds.

Here is an example. You might want to consider searching for redox rules to see how others have approached this topic.

Example #1: What is the oxidation number of Cl in HCl?

Since H = +1, the Cl must be -1 (minus one).

Example #2: What is the oxidation number of Na in Na2O?

Since O = -2, the two Na must each be +1.

Example #3: What is the oxidation number of Cl in ClO¯?

The O is -2, but since a -1 must be left over, then the Cl is +1.

Example #4: What is the oxidation number of Cl in ClO2¯?

Two O is -4 (from -2 x 2), but since a -1 must be left over, then the Cl is +3.

Example #5: What is the oxidation number of Cl in ClO3¯?

Three O is -6 (from -2 x 3), but since a -1 must be left over, then the Cl is +5.

Example #6: What is the oxidation number of Cl in ClO4¯?

Four O is -8 (from -2 x 4), but since a -1 must be left over, then the Cl is +7.

Example #7: What is the oxidation number of S in SO42¯

O = -2. There are four oxygens for -8 total. Since -2 must be left over, the S must = +6.

Example #8: What is the oxidation number of S in SO32¯

O = -2. There are three oxygens for -6 total. Since -2 must be left over, the S must = +4.

Example #9: What are the oxidation numbers in KCl?

K = +1 because K2O exists. The O is -2 by defintion, therefore each K must be +1 in order to keep the K2O formula at zero charge.

Cl = -1 because HCl exists. H is a +1 by definition, therefore Cl must be a -1. You can also say that Cl is a -1 because K must be a +1 and we need to have a zero charge for the formula.

Example #10: What is the oxidation number for each element in NaMnO4?

Na = +1 because Na2O exists. We know the O = -2, so each Na is a +1. (We could also use this: we know NaCl exists and that Cl is a -1. We know this because HCl exists. Therefore, the Na is a +1.)

O = -2 by definition

Mn = +7. There are 4 oxygens for a total of -8, K is +1, so Mn must be the rest.