Ten Examples

**I. Rule Number One**

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

This includes the seven diatomic elements (such as O

_{2}) and the other molecular elements (P_{4}and S_{8}).

**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 Na_{2}O?

Since O = −2, the two Na must each be +1. Notice that we treat each sodium as an individual unit, we do not address it as Na_{2}^{2}¯.

**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 ClO_{2}¯?

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 ClO_{3}¯?

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 ClO_{4}¯?

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 SO_{4}^{2}¯

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 SO_{3}^{2}¯

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 K_{2}O exists. The O is −2 by defintion, therefore each K must be +1 in order to keep the K_{2}O 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 NaMnO_{4}?

Na = +1 because Na_{2}O 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.

**Bonus Example:** What is the oxidation number of N and of P in NH_{4}H_{2}PO_{4}?

**Solution:**

1) Let's split ammonium dihydrogen phosphate apart. It forms the following two polyatomic ions:

ammonium ---> NH_{4}^{+}

dihydrogen phosphate ---> H_{2}PO_{4}¯

2) Analysis of the ammonium ion:

We know that NH_{4}^{+}takes on a +1 charge because we know that NH_{4}Cl exists.We know that the chloride ion is −1, therefore ammonium is a +1.

Remember that we know chloride is a −1 because HCl exists and H is defined as a +1 (except in hydrides).

In NH

_{4}^{+}, the four hydrogens total to +4. (Ammonium is not a hydride.)Therefore, the nitrogen is a −3, in order to leave a +1 for the overall charge on the ion.

2) Analysis of the dihydrogen phosphate ion:

Two H are present ---> +2

Four O are present ---> −8In order for the entire ion to have a −1 charge, the P must be a +5.