How To Find Oxidation Numbers 12 Steps With Pictures

For example, Al(s) and Cl2 both have oxidation numbers of 0 because they are in their uncombined elemental forms. Note that sulfur’s elemental form, S8, or octasulfur, though irregular, also has an oxidation number of 0.

For instance, the ion Cl- has an oxidation number of -1. The Cl ion still has an oxidation number of -1 when it’s part of the compound NaCl. Because the Na+ ion, by definition, has a charge of +1, we know that the Cl- ion has a charge of -1, so its oxidation number is still -1.

For example, let’s examine a compound containing the metallic aluminum ion. The compound AlCl3 has an overall charge of 0. Because we know that Cl- ions have a charge of -1 and there are 3 Cl- ions in the compound, the Al ion must have a charge of +3 so that the overall charge of all the ions adds to 0. Thus, Al’s oxidation number is +3 in this compound.

When oxygen is in its elemental state (O2), its oxidation number is 0, as is the case for all elemental atoms. When oxygen is part of a peroxide, its oxidation number is -1. Peroxides are a class of compounds that contain an oxygen-oxygen single bond (or the peroxide anion O2-2). For instance, in the molecule H2O2 (hydrogen peroxide), oxygen has an oxidation number (and a charge) of -1. When oxygen is part of a superoxide, its oxidation number is -1⁄2. Superoxides contain the superoxide anion O2-. When oxygen is bound to fluorine, its oxidation number is +2. See fluorine rule below for more info. However, there is an exception: in (O2F2), the oxidation number of oxygen is +1.

For instance, in H2O, we know that hydrogen has an oxidation number of +1 because oxygen has a charge of -2 and we need two +1 charges to make the compound’s charges add up to zero. However, in sodium hydride, NaH, hydrogen has an oxidation number of -1 because the Na+ ion has a charge of +1 and, for the compound’s total charge to equal zero, hydrogen’s charge (and thus oxidation number) must equal -1.

This is a good way to check your work - if the oxidation in your compounds don’t add up to the charge of your compound, you know that you have assigned one or more incorrectly.

For example, in the compound Na2SO4, the charge of sulfur (S) is unknown - it’s not in its elemental form, so it’s not 0, but that’s all we know. This is a good candidate for this method of algebraic oxidation number determination.

In Na2SO4, we know, based on our set of rules, that the Na ion has a charge (and thus oxidation number) of +1 and that the oxygen atoms have oxidation numbers of -2.

In Na2SO4, we know there are 2 Na atoms and 4 O atoms. We would multiply 2 × +1, the oxidation number of Na, to get an answer of 2, and we would multiply 4 × -2, the oxidation number of O, to get an answer of -8.

In our Na2SO4 example, we would add 2 to -8 to get -6.

In our Na2SO4 example, we would solve as follows: (Sum of known oxidation numbers) + (unknown oxidation number you are solving for) = (charge of the compound) -6 + S = 0 S = 0 + 6 S = 6. S has an oxidation number of 6 in Na2SO4.