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**

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

**III. Rule Number Three**

Each oxygen atom, 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 a few exceptions, which you will usually never see at an introductory level (ozone, O_{3} is an example), oxidation states can be assigned to all atoms in a formula. There are more extensive sets of rules (here is an example) 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.

By the way, the oxidation numbers for the atoms of oxygen in ozone will be dealt with using a concept called 'formal charge.' This concept is usually not discussed in a high school chemistry course.

**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 #1:** 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.

**Bonus Example #2:** (a) For Fe_{3}O_{4}, is the oxidation number for O_{4}, −2 or −8? (b) For H_{2}O_{2}, is the oxidation number for O_{2}, −2 or −4?

**Solution:**

This is a question I copied from Yahoo Answers because of the confusion shown by the student. Notice that he/she asks about the oxidation number of O_{4}. This is improper. Each oxygen atom should be treated individually, not as a group. In the case, the correct answer is that each oxygen atom has an oxidation number of −2.An important skill in this area is being able to recognize a given formula as representing an ionic compound or a molecular one. This leads to the second part of the problem.

One recognizes H

_{2}O_{2}as for formula for hydrogen peroxide and that, in this formula, the O_{2}is seen to be a legitimate polyatomic ion, named peroxide. It is written O_{2}^{2}¯ and the charge on theionis −2. The oxidation number on each oxygen atom is, therefore, −1.To me, the student (in this second part) is aware of the peroxide rule, but is not sure if the oxygen in H

_{2}O_{2}should be dealt with as a peroxide and as separate atoms, as would be done in an oxide.This type of confusion may happen to you. Be preapred!! Study, study, study!!

By the way, in Fe

_{3}O_{4}, the oxidation number of each iron atom is NOT 2.67. Each oxidation number must be an integer, which means that two of the Fe atoms are each +3 and one Fe atom is +2. This adds to +8, which offsets the −8 created by the four O atoms.