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Oxidation states add up to 0 in a neutral compound or to the charge on an ion, with oxygen usually −2 and hydrogen +1. Free AQA notes and 15 questions answered.
Oxidation States is part of Oxidation, reduction and redox equations in A Level AQA Chemistry.
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Oxidation states add up to 0 in a neutral compound or to the charge on an ion, with oxygen usually −2 and hydrogen +1. Free AQA notes and 15 questions answered.
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Use this lesson to stay inside Oxidation, reduction and redox equations while connecting the detail back to the wider Physical chemistry topic.
Read the core explanations from Oxidation States before testing yourself from memory.
Imagine you are watching a reaction between iron and chlorine. Electrons are being transferred, but you cannot see them move. Oxidation states are the bookkeeping system chemists use to track exactly where those electrons go — and which atom gains or loses them.
At A Level, oxidation states appear everywhere: naming transition metal compounds, writing half-equations, balancing redox reactions, and interpreting electrode potentials. Getting this foundation right unlocks all of those topics.
The key insight is this: oxidation states are a convention, not a physical reality. We assign them using a fixed set of rules, and those rules always give a consistent answer. Once you know the rules, every formula becomes solvable.
There are seven rules, applied in priority order. When two rules appear to conflict, the higher-priority rule wins. Work through them in sequence and you will always reach the correct answer.
Rules 1 to 6 give you fixed values for specific atoms. Rule 7 is the algebraic tool that lets you find the oxidation state of the remaining unknown atom. In practice, you assign all the known atoms first, then use Rule 7 to solve for the unknown.
Start with the straightforward cases. In a binary compound — one containing only two elements — you assign the fixed-rule element first, then use the sum rule to find the other. The worked example below shows the method in full.
Worked example: Finding the Oxidation State of Sulphur in $SO_2$
Determine the oxidation state of sulphur in $SO_2$.
Answer: Sulphur has an oxidation state of $+4$ in $SO_2$.
The same four-step method works for every binary compound. Assign the fixed element, count atoms, write the sum rule equation, and solve. Now try a compound where hydrogen is involved.
Worked example: Finding the Oxidation State of Nitrogen in $NH_3$
Determine the oxidation state of nitrogen in $NH_3$.
Answer: Nitrogen has an oxidation state of $-3$ in $NH_3$.
Polyatomic ions carry an overall charge, and that charge changes the sum rule. Instead of setting the total equal to zero, you set it equal to the charge on the ion. This is where many students lose marks — they forget to account for the ion's charge.
The manganate(VII) ion, $MnO_4^-$, is a classic A-Level example. It appears in redox titrations and half-equations, so knowing how to find the oxidation state of manganese here is directly examinable.
Worked example: Finding the Oxidation State of Manganese in $MnO_4^-$
Determine the oxidation state of manganese in the manganate(VII) ion, $MnO_4^-$.
Answer: Manganese has an oxidation state of $+7$ in $MnO_4^-$.
Notice how the Roman numeral in the name 'manganate(VII)' directly tells you the oxidation state of manganese. AQA expects you to both read and write these names correctly, so the link between the calculation and the nomenclature is worth cementing now.
Two exceptions to the standard rules catch students out repeatedly in AQA papers. Both involve breaking the default assumption about oxygen or hydrogen — and both are tested directly.
| Compound | Formula | Exception | Oxidation state of O or H |
|---|---|---|---|
| Hydrogen peroxide | $H_2O_2$ | Peroxide: O-O bond present | O is $-1$ |
| Sodium peroxide | $Na_2O_2$ | Peroxide: O-O bond present | O is $-1$ |
| Oxygen difluoride | $OF_2$ | Fluorine takes priority (Rule 3) | O is $+2$ |
| Sodium hydride | $NaH$ | Metal hydride: H bonded to metal | H is $-1$ |
| Calcium hydride | $CaH_2$ | Metal hydride: H bonded to metal | H is $-1$ |
For hydrogen peroxide, $H_2O_2$: hydrogen is +1 (Rule 5, bonded to oxygen which is a non-metal), so using the sum rule: $2(+1) + 2(\text{O}) = 0$, giving oxygen as $-1$. The O-O bond is the structural clue that signals a peroxide.
For metal hydrides such as $NaH$: sodium is +1 (Rule 6). Using the sum rule: $(+1) + \text{H} = 0$, so hydrogen is $-1$. The rule of thumb is that hydrogen takes a negative oxidation state only when it is bonded directly to a metal.
Transition metals are the most common source of oxidation state questions at A Level because they can adopt multiple oxidation states. The method is identical — assign fixed elements first, then solve — but the numbers are sometimes larger.
Worked example: Finding the Oxidation State of Iron in $Fe_2O_3$
Determine the oxidation state of iron in iron(III) oxide, $Fe_2O_3$.
Answer: Iron has an oxidation state of $+3$ in $Fe_2O_3$.
Now consider a compound containing a polyatomic ion within a larger formula — for example, iron(II) sulphate, $FeSO_4$. Here you need to handle the sulphate ion carefully.
Worked example: Finding the Oxidation State of Sulphur in $FeSO_4$
Determine the oxidation state of sulphur in iron(II) sulphate, $FeSO_4$.
Answer: Sulphur has an oxidation state of $+6$ in $FeSO_4$.
Once you can calculate oxidation states, you can use them to identify what has been oxidised and what has been reduced in any reaction. The definitions are precise and must be stated exactly in AQA mark schemes.
Consider the reaction between zinc and copper(II) sulphate solution: $Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$. Zinc goes from 0 to +2 — its oxidation state increases, so zinc is oxidised. Copper goes from +2 to 0 — its oxidation state decreases, so copper is reduced.
The sulphate ion's oxidation states do not change throughout this reaction (sulphur remains +6, oxygen remains -2). Identifying which atoms change — and which do not — is a key analytical skill at A Level.
AQA oxidation state questions follow predictable patterns. Knowing the examiner's expectations means you can target marks efficiently rather than writing more than is needed.
When asked to 'state' an oxidation state, give the signed integer only — no working required. When asked to 'calculate' or 'determine', show the sum rule equation explicitly, even if the answer seems obvious. Method marks are available for the working, not just the final value.
For 'explain' questions about oxidation and reduction, the mark scheme expects you to state the change in oxidation state and link it to the definition. A full-mark response reads: 'Iron is oxidised because its oxidation state increases from +2 to +3, representing a loss of electrons.' Stating only 'iron loses electrons' without referencing oxidation states will typically score one mark out of two.
One final point on notation: AQA mark schemes accept oxidation states written as +3 or as the integer 3 in some contexts, but the safest approach is always to include the sign. An oxidation state of 3 without a sign is ambiguous; +3 is unambiguous. Make the sign explicit every time.
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Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
The oxidation state rules are: uncombined elements are 0; simple monatomic ions equal their charge; fluorine is always −1; oxygen is almost always −2; hydrogen is +1 with non-metals; Group 1 metals are +1 and Group 2 metals +2; and oxidation states sum to 0 in a compound or to the charge on an ion.
To work out an oxidation state, assign fixed values first, such as oxygen −2, hydrogen +1 and Group 1 metals +1, then use the sum rule: oxidation states add up to 0 in a neutral compound or to the charge on an ion. For SO₂, S + 2 × (−2) = 0, so sulphur is +4.
Oxygen has an oxidation state of −1 in hydrogen peroxide, H₂O₂, because it is a peroxide with an O-O bond. Hydrogen is +1, so 2(+1) + 2(O) = 0, giving oxygen −1. Oxygen is also an exception in OF₂, where it is +2 because fluorine takes priority.
Fluorine is always −1 in compounds. It is the most electronegative element and never takes a positive oxidation state, which is why it takes priority over oxygen, making oxygen +2 in OF₂.
The oxidation state of nitrogen in NH₃ is −3. Hydrogen is +1 when bonded to a non-metal, so three hydrogens total +3. Ammonia is neutral, so nitrogen + 3 = 0, giving −3. Negative oxidation states are common for non-metals bonded to hydrogen.
Manganese has an oxidation state of +7 in the manganate(VII) ion, MnO₄⁻. Oxygen is −2, so four oxygens give −8, and the sum must equal the ion's charge of −1: Mn + (−8) = −1, so Mn = +7. The Roman numeral VII in the name states this.
The oxidation state of iron in Fe₂O₃ is +3. Three oxygens at −2 total −6, and the compound is neutral, so 2 × (iron) = +6, giving iron +3. The name iron(III) oxide confirms this, since the Roman numeral states the oxidation state.
The oxidation state of sulphur in SO₂ is +4. Oxygen is −2 because SO₂ is not a peroxide and contains no fluorine, so the two oxygens total −4. The compound is neutral, so sulphur + (−4) = 0, giving sulphur +4.
The oxidation state of oxygen in OF₂ is +2. Fluorine takes priority over oxygen: it is the most electronegative element and is always −1 in compounds, so two fluorines total −2 and oxygen must be +2 for the neutral compound to sum to zero.
The oxidation state of sulphur in the sulphate ion, SO₄²⁻, is +6. Four oxygens at −2 total −8, and the sum must equal the ion's charge of −2, so sulphur is +6. In FeSO₄, iron(II) is +2, so +2 + S − 8 = 0 also gives sulphur +6.
Oxidation is an increase in oxidation state and reduction is a decrease in oxidation state. In Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s), zinc goes from 0 to +2 so it is oxidised, and copper goes from +2 to 0 so it is reduced, while sulphur stays at +6.
The oxidation state of hydrogen in a metal hydride, such as NaH or CaH₂, is −1. In NaH, sodium is +1 as a Group 1 metal, so the sum rule gives +1 + H = 0 and hydrogen is −1. Hydrogen takes a negative oxidation state only when bonded directly to a metal.
The Roman numeral in a compound name states the oxidation state of the element. For example, manganese is +7 in manganate(VII), MnO₄⁻, and iron is +3 in iron(III) oxide, Fe₂O₃. The name can be used as a cross-check on a calculated oxidation state.
The oxidation state of any uncombined element in its standard state is 0. This applies to Na(s), O₂(g), Cl₂(g), S₈(s) and all other pure elements. An oxidation state of zero is written as 0, not +0.
Oxidation states are written as a signed integer with the sign before the number, such as +3 or −2, and zero is written as 0. An oxidation state of 3 without a sign is ambiguous, while +3 is unambiguous, so the sign should be included every time.