The oxidation number of an atom shows how many electrons it has gained or lost in a compound. You assign it by applying a short list of rules in a fixed order, then solving for the one unknown.
This lesson is part of SPM Chemistry redox equilibrium. The next lesson, identifying oxidation and reduction, uses the numbers you find here.
What rules do you apply, and in what order?
Apply them in this order and stop when only one unknown is left.
- An uncombined element is 0.
- A simple ion has an oxidation number equal to its charge, for example Mg²⁺ is +2.
- Oxygen is −2 and hydrogen is +1 in most compounds.
- Group 1 metals are +1 and Group 2 metals are +2 in compounds.
- The numbers add up to zero for a neutral compound, or to the charge for an ion.
Worked examples
Sulfur in H₂SO₄. Hydrogen: 2 × (+1) = +2. Oxygen: 4 × (−2) = −8.
Let S = x. Then 2 + x − 8 = 0, so x = +6.
Manganese in KMnO₄. Potassium is +1 and four oxygens give −8.
Let Mn = x. Then +1 + x − 8 = 0, so x = +7.
Nitrogen in NH₄⁺. Four hydrogens give +4.
Let N = x. The ion has a charge of +1, so x + 4 = +1 and x = −3.
Chromium in Cr₂O₇²⁻. Seven oxygens give −14. Two chromium atoms together are 2x.
The ion has a charge of −2, so 2x − 14 = −2, which gives 2x = 12 and x = +6.
The mistake that costs marks
In the chromium example, the common slip is to write x − 14 = −2 and get x = +12. That treats one chromium atom as carrying all the positive charge.
| Step | Wrong | Right |
|---|---|---|
| Equation | x − 14 = −2 | 2x − 14 = −2 |
| Solve | x = +12 | 2x = 12 |
| Answer | +12 | +6 per Cr atom |
The cure is to count atoms before writing the equation. The oxidation number belongs to one atom, so multiply by the number of atoms of that element in the formula.
A quick check that catches errors
Put your answer back into the sum. For Cr₂O₇²⁻: 2(+6) + 7(−2) = 12 − 14 = −2. This matches the charge on the ion, so the answer is consistent.
Check yourself
Find the oxidation number of (a) N in NO₃⁻ and (b) S in Na₂S₂O₃.
Answer
(a) Three oxygens give −6. Let N = x. The ion charge is −1, so x − 6 = −1 and x = +5.
(b) Two sodiums give +2. Three oxygens give −6. Two sulfurs are 2x. Then 2 + 2x − 6 = 0, so 2x = 4 and x = +2 per S atom.
Check (b): +2 + 2(+2) + (−6) = 0.
What to study next
Move on to identifying oxidation and reduction, where a change in oxidation number decides what is oxidised and what is reduced. Test the full chapter with the redox practice set.
If you want a teacher to go through your arithmetic on longer formulas, see online one-to-one Chemistry tuition.