Calculating oxidation numbers
In a nutshell
The oxidation number represents the oxidation state of an element in a specific compound. Roman numerals can be used to specify the oxidation number. The oxidation number can be calculated both from the chemical formula and from the systematic name for a compound.
Oxidation numbers
The oxidation number of a compound tells the hypothetical charge of an atom in a particular compound or ion.
Type of particle | Oxidation number | Example |
Neutral | The oxidation number of an atom which is neutral and only contains one element is zero. | Ar, H2, S8 |
Single ion | The oxidation number of single ions is the same as the total charge on the ion. | Mg2+, Na+ |
Molecular ion | For a molecular ion the total sum of all oxidation numbers will equal the overall charge of the ion. | (SO42−), (NO3−) |
Neutral compound | For a neutral compound the overall charge is zero. If the compound is made of more than one element, each element may have its own oxidation number. | CaCl2, MgO |
Oxidation numbers of hydrogen and oxygen
The oxidation number of oxygen is nearly always −2. Except in O2 where the oxidation number is zero and in peroxides, such as H2O2, where the oxidation number of oxygen is −1.
The oxidation number of hydrogen is nearly always +1. Except in metal hydrides, such as BH3, where the oxidation number is −1 and in H2 which has an oxidation number of zero.
Roman numerals
Certain elements, especially transition metals like iron (Fe) can have several stable oxidation numbers. In this case roman numerals are used to show the oxidation number.
Roman Numeral | Oxidation NUMber |
| |
| |
| |
The roman numeral is written after the element which is corresponds to.
Example
Compound name | Oxidation number | Example |
| | Iron (III) oxide |
| | Iron (II) oxide |
Compounds ending in -ate and their oxidation numbers
Compounds that end in -ate such as: nitrate, sulfate, chlorate and carbonate, contain oxygen and at least one other element. For example, nitrates contain oxygen and nitrogen, and sulfates contain sulfur and oxygen. The other elements in this case, the nitrogen and sulfur, can have multiple oxidation numbers.
Example
Compound name | Oxidation number | Example |
Sulfate (IV) | | (SO32−) |
Sulfate (VI) | | (SO42−) |
Calculating oxidation numbers
The oxidation number of elements in compounds can be calculated from the both the formulae and the systematic name of the compound.
Procedure table
1. | Look to see if the molecule has a charge. The sum of the oxidation numbers will be equal to the total charge on the molecule. |
2. | Find if there are any elements which have known oxidation numbers (for example, oxygen is typically −2, or hydrogen is +1). |
3. | Sum together the known oxidation numbers until there is one element left with an unknown oxidation number. |
4. | Calculate the oxidation number of the remaining element. |
Example 1
What is the oxidation number of iron in Fe2O3?
First check the charge of the compound:
No charge. The sum of the oxidation numbers of iron and oxygen equals zero.
Next, write down the oxidation number of oxygen:
Oxidation number for oxygen = −2
There are three oxygen atoms present in the molecule. Find the total oxidation numbers of the three oxygen atoms:
−2 × 3 = −6
The compound is neutral so the oxidation numbers must balance out. There are two iron atoms and they must have a positive oxidation number to balance out the negative oxygen atoms. Find the oxidation number of each iron atom:
26 = +3
The oxidation number of iron in Fe2O3 is +3.
Example 2
What is the oxidation number of chlorine in ClO2−?
First check the charge of the compound:
A charge of −1. The sum of the oxidation numbers of chlorine and oxygen must equal −1.
Next, write down the oxidation number of oxygen:
Oxidation number for oxygen = −2
There are two oxygen atoms present in the molecule. Find the total oxidation number of the two oxygen atoms:
−2 × 2 =−4
The overall compound has a charge of −1. Given that the oxidation numbers of the two oxygen atoms and chlorine atom must sum to −1, find the oxidation number of chlorine:
(−4 + x) = −1
x = +3
The oxidation number of chlorine in ClO2− is +3.