# How to find the oxidation number: 12 steps

In chemistry, the terms "oxidation" and "reduction" refer to reactions in which an atom (or a group of atoms) loses or gains electrons, respectively. Oxidation numbers are numbers assigned to atoms (or groups of atoms) that help chemists keep track of how many electrons are available for transfer and whether reactants are oxidized or reduced within a reaction. The process of assigning oxidation numbers to atoms can range from extraordinarily simple to somewhat complex, depending on the charge on the atoms and the chemical composition of the molecules of which they are a part. To complicate matters, some atoms can have more than one oxidation number. Fortunately, the assignment of oxidation numbers is governed by well-defined and easy-to-follow rules, although knowing basic chemistry and algebra will make it easier to use these rules.

## Steps

### Part 1 of 2: Assign the oxidation number based on the rules of chemistry

#### Step 1. Determine if the substance in question is elemental

Free and uncombined elemental atoms always have an oxidation number equal to 0. This is true for atoms whose elemental form is composed of a single atom and also for atoms whose elemental form is diatomic or polyatomic.

• For example, the Al(s) and the Cl2 they have an oxidation number of 0 because they are in their uncombined elemental form.
• Note that the elemental form of sulfide, S8 (or octosulfide), while irregular, it also has an oxidation number of 0.

#### Step 2. Determine if the substance in question is an ion

Ions have an oxidation number equal to that of their charge. This is true for ions that are not bound to any other element and also for those that are part of an ionic compound.

• For example, the ion Cl- it has an oxidation number of -1.
• The Cl ion continues to have an oxidation number of -1 when it is part of the NaCl compound. 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.

#### Step 3. For metal ions, note that multiple oxidation numbers are possible

Many metallic items have more than one charge. For example, the metal iron (Fe) can be an ion with a charge of +2 or +3. The charge of metal ions (and therefore their oxidation numbers) can be determined by considering the charge of the other atoms in the compound they form or when they are written in a text, observing the notation in Roman numerals (as in the sentence, “The iron (III) ion has a charge of +3”).

• For example, let's examine a compound that contains the metal ion aluminum. The compound AlCl3 has an average charge of 0. As we already know that Cl ions- have a -1 charge and there are 3 Cl ions- In the compound, the Al ion must have a charge of +3 for the average charge of all ions to add up to 0. Therefore, the oxidation number of Al is +3.

#### Step 4. Assign an oxidation number of -2 to oxygen (with exceptions)

In almost all cases, oxygen atoms have an oxidation number of -2. There are some exceptions to this rule:

• When oxygen is in its elemental state (O2), its oxidation number is 0, as is the case with 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 example, in the molecule H2OR2 (hydrogen peroxide), oxygen has an oxidation number (and charge) of -1.
• When oxygen is bound to fluorine, its oxidation number is +2. Read the fluoride rule later for more information.

#### Step 5. Assign an oxidation number of +1 to hydrogen (with exceptions)

Like oxygen, the oxidation number of hydrogen changes only in rare cases. Generally, hydrogen has an oxidation number of +1 (unless, as stated above, it is in its elemental form, H2). However, in the special case of hydride compounds, hydrogen has an oxidation number of -1.

• For example, in H2Or, we know that hydrogen has an oxidation number of +1 because oxygen has a charge of -1 and we need 2 charges +1 for the total charge of the compound to be 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 total charge of the compound to be 0, the charge (and therefore, the oxidation number) must be -1.

#### Step 6. Fluorine always has an oxidation number of -1

As mentioned above, the oxidation numbers of certain elements can vary depending on various factors (metal ions, oxygen atoms in peroxides, etc.). Fluorine, however, has an oxidation number of -1 that never changes. This is because fluorine is the most electronegative element or, in other words, it is the element that has the least possibility of delivering some of its own electrons and the one that has the most possibility of taking an electron from another atom. Therefore, its load does not change.

#### Step 7. Set the oxidation number of a compound equal to its charge

The sum of the oxidation numbers of all the atoms in a compound must have the same value as the charge on the compound. For example, if a compound has no charge, the sum of the oxidation numbers of its atoms must be 0; if the compound is a polyatomic ion with a -1 charge, the oxidation numbers must add up to -1, and so on.

### Part 2 of 2: Assigning Numbers to Atoms Without Using Oxidation Number Rules

#### Step 1. Find atoms to which the oxidation number rules do not apply

Some atoms do not have specific rules about the oxidation numbers they can acquire. If your atom does not appear in the rules described above and you are not sure what its charge is (for example, if it is part of a large compound and therefore individual charges are not shown), you can find its oxidation number using the process of elimination. First, you need to determine the oxidation of the other atoms in the compound, then simply solve for the unknown atom based on the total charge of the compound.

• For example, in the compound Na2SW4, the charge of sulfide (S) is not known (it is not in its elemental form, so it is not 0, but that is all we know). This is a good candidate for this algebraic method of determining an oxidation number.

#### Step 2. Find the known oxidation number of the other elements in the compound

Using the rules for assigning oxidation numbers, assign oxidation numbers to the other atoms in the compound. Stay tuned for rare cases for O, H, etc.

• In the Na2SW4We know that based on the rules, the Na ion has a charge (and therefore an oxidation number) of +1 and the oxygen atoms have an oxidation number of -2.

#### Step 3. Multiply the number of each atom by its oxidation number

Now that we know the oxidation numbers of all the atoms except the unknown, we must bear in mind that some of these atoms may appear more than once. Multiply the numerical coefficient of each atom (written in subscripts after the chemical symbol for the atom in the compound) by its oxidation number.

• In the Na2SW4We know that there are 2 Na atoms and 4 O atoms. We should multiply 2 times +1, the oxidation of Na, to obtain 2, and multiply 4 times -2, the oxidation of O, to obtain -8.

#### Step 4. Add the results

Adding the results of the products gives you the oxidation number of the compound without considering the oxidation number of the atom that you do not know.

• In our example the Na2SW4, we should add 2 to -8 to get -6.

#### Step 5. Calculate the unknown oxidation number based on the charge of the compound

Now you have everything you need to find the unknown oxidation number using simple algebra. Set up an equation that has the answer from the previous step added to the unknown oxidation number equal to the total charge of the compound. In other words: (Sum of known oxidation numbers) + (unknown oxidation number you want to get) = (component charge).

• In our example the Na2SW4 we must find the solution as follows:

• (Sum of known oxidation numbers) + (unknown oxidation number you want to obtain) = (component charge)
• -6 + S = 0
• S = 0 + 6
• S = 6. S has an oxidation number

Step 6. in the Na2SW4.