The big idea: The oxidation state (oxidation number) is a bookkeeping number that tracks how many electrons an atom has gained or lost compared with the free element.
It is the charge the atom would have if every bond were fully ionic — that is, if the more electronegative atom took the shared electrons.
- A more positive oxidation state = the atom has lost electron control. - A more negative oxidation state = the atom has gained electron control.
Why we bother: Oxidation states let us spot a redox (reduction–oxidation) reaction at a glance: if an atom's oxidation state changes during a reaction, electrons have moved, so it is redox.
Write oxidation states above the symbol of each element, e.g. the +1 above the H and the −2 above the O in H2O.
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Assign oxidation states by working through a short set of rules in order. The rules near the top win when two seem to clash.
| Rule (apply in order) | Oxidation state | Examples |
|---|---|---|
| A free, uncombined element | 0 | Na, O2, Cl2, S8 |
| A simple (monatomic) ion | = its charge | Na⁺ → +1, Mg²⁺ → +2, Cl⁻ → −1 |
| Oxygen in most compounds | −2 | (−1 in peroxides like H2O2) |
| Hydrogen in most compounds | +1 | (−1 in metal hydrides like NaH) |
| Group 1 / Group 2 metals | +1 / +2 | Na +1, Ca +2 |
| Fluorine in compounds | −1 | always −1 |
The two sums that finish the job: Once the 'known' atoms are fixed by the rules, the unknown atom is found from a simple sum — the oxidation states in any species must add up to its total charge.
| Species | The oxidation states must add up to… |
|---|---|
| A neutral compound | 0 |
| A polyatomic ion | the overall charge on the ion |
Worked example — sulfur in sulfate, SO₄²⁻
Deduce the oxidation state of sulfur in the sulfate ion, SO4²⁻.
Solution
- Fix the known atom: oxygen is −2 (the rule). There are 4 oxygens.
- Set up the sum: the four O plus the unknown S must add up to the ion's charge, −2:
- Solve for sulfur:
Final answer
Sulfur is in the +6 oxidation state.
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Oxidation vs reduction: Define them by how the oxidation state changes:
- Oxidation = an increase in oxidation state (electrons lost). - Reduction = a decrease in oxidation state (electrons gained).
Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).
Oxidising agent
- Takes electrons from something else.
- Is itself reduced (its oxidation state goes down).
- e.g. O2, Cl2, MnO4⁻.
Reducing agent
- Gives electrons to something else.
- Is itself oxidised (its oxidation state goes up).
- e.g. a reactive metal such as Zn, or H2.
The agent is the opposite of what happens to it: This is the classic trap: the species that gets oxidised is the reducing agent, and the species that gets reduced is the oxidising agent. The agent causes the other change.
Worked example — is this redox?
In the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, deduce the oxidation-state changes and name the oxidising and reducing agents.
Solution
- Zinc: Zn (free element) is 0, Zn²⁺ is +2 — an increase, so zinc is oxidised.
- Copper: Cu²⁺ is +2, Cu (free element) is 0 — a decrease, so copper is reduced.
- Name the agents: Cu²⁺ caused the oxidation, so it is the oxidising agent; Zn caused the reduction, so it is the reducing agent.
Final answer
Zn: 0 → +2 (oxidised, the reducing agent). Cu: +2 → 0 (reduced, the oxidising agent). Because oxidation states change, it is a redox reaction.
How this is tested: Oxidation states are tested in two ways:
- Paper 1A (MCQ): a quick 'deduce the oxidation state of X in this formula' or 'in which compound is the oxidation state +4?'. - Paper 2: 'show that this reaction is redox by reference to the oxidation-state changes of two elements', or deduce the state of an element across reactant and product.
Markers want each oxidation state stated with its sign (+/−) and, for redox, the explicit change (e.g. −1 → 0).
Don't drop the sign: An oxidation state is always written with a sign: write +6, not 6, and −2, not 2. A missing or wrong sign loses the mark — and never confuse the +2 of an oxidation state with the 2 in a formula subscript.
IB-style question — nitrogen across two species (a)
(a) Deduce the oxidation state of nitrogen in ammonia, NH3, and in the nitrate ion, NO3⁻. [2]
How to score the marks
- Ammonia, NH3: hydrogen is +1, and the molecule is neutral (sums to 0). So N + 3(+1) = 0, giving N = −3.
- Nitrate ion, NO3⁻: oxygen is −2, and the ion's charge is −1. So N + 3(−2) = −1, giving N = −1 + 6 = +5.
- State both clearly with their signs: N is −3 in NH3 and +5 in NO3⁻.
Final answer
Nitrogen is −3 in NH3 and +5 in NO3⁻.
IB-style question — explaining why it is redox (b)
(b) When iron(II) ions react with chlorine, 2Fe²⁺ + Cl2 → 2Fe³⁺ + 2Cl⁻. Explain, using oxidation states, why this is a redox reaction, and identify the oxidising agent. [3]
How to score the marks
- Iron: Fe²⁺ is +2 and Fe³⁺ is +3 — an increase, so iron is oxidised (loses an electron).
- Chlorine: Cl2 (free element) is 0 and Cl⁻ is −1 — a decrease, so chlorine is reduced.
- Conclusion: because oxidation states change (one up, one down), electrons are transferred — it is redox. Chlorine causes the oxidation of iron, so Cl2 is the oxidising agent.
Final answer
Fe: +2 → +3 (oxidised); Cl: 0 → −1 (reduced). The oxidation states change, so it is redox. Cl2 is the oxidising agent.