Oxidation States Revision Notes | A Level AQA Chemistry

1. Why Oxidation States Matter

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.

2. The Rules for Assigning Oxidation States

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.

• Rule 1 — Uncombined elements: The oxidation state of any element in its standard state is 0. This applies to $Na(s)$, $O_2(g)$, $Cl_2(g)$, $S_8(s)$, and all other pure elements.
• Rule 2 — Simple monatomic ions: The oxidation state equals the ionic charge. $Na^+$ is +1; $Mg^{2+}$ is +2; $Cl^-$ is -1.
• Rule 3 — Fluorine: Fluorine is always -1 in compounds. It is the most electronegative element and never takes a positive oxidation state.
• Rule 4 — Oxygen: Oxygen is almost always -2 in compounds. The exceptions are peroxides (e.g. $H_2O_2$, where oxygen is -1) and compounds with fluorine (e.g. $OF_2$, where oxygen is +2 because fluorine takes priority).
• Rule 5 — Hydrogen: Hydrogen is +1 when bonded to non-metals (e.g. $HCl$, $H_2O$). Hydrogen is -1 in metal hydrides (e.g. $NaH$, $CaH_2$).
• Rule 6 — Group 1 and Group 2 metals: Group 1 metals are always +1 in compounds; Group 2 metals are always +2 in compounds.
• Rule 7 — The sum rule: The sum of all oxidation states in a neutral compound equals 0. The sum of all oxidation states in a polyatomic ion equals the overall charge on that ion.

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.

3. Applying the Rules: Simple Compounds

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 $SO_2$.

1. Oxygen is present. By Rule 4, oxygen is $-2$ in this compound (no peroxide, no fluorine). — Always assign fixed-rule elements first. Oxygen is -2 here because SO2 is not a peroxide and contains no fluorine.
2. $SO_2$: 1 sulphur atom, 2 oxygen atoms. — The subscript tells you how many atoms of each element are present.
3. $\text{oxidation state of S} + 2 \times (-2) = 0$ — The compound is neutral, so the sum of all oxidation states must equal zero (Rule 7).
4. $\text{oxidation state of S} - 4 = 0 \Rightarrow \text{oxidation state of S} = +4$ — Sulphur has an oxidation state of +4 in SO2. Notice the sign is written before the number.

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 $NH_3$.

1. Hydrogen is bonded to a non-metal (nitrogen). By Rule 5, hydrogen is $+1$. — Hydrogen is +1 when bonded to non-metals. Nitrogen is a non-metal, so this rule applies.
2. $NH_3$: 1 nitrogen atom, 3 hydrogen atoms. — The subscript 3 tells you there are three hydrogen atoms.
3. $\text{oxidation state of N} + 3 \times (+1) = 0$ — NH3 is a neutral compound, so the total must equal zero.
4. $\text{oxidation state of N} + 3 = 0 \Rightarrow \text{oxidation state of N} = -3$ — Nitrogen has an oxidation state of -3 in ammonia. Negative oxidation states are common for non-metals bonded to hydrogen.

Answer: Nitrogen has an oxidation state of $-3$ in $NH_3$.

4. Oxidation States in Polyatomic Ions

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, $MnO_4^-$.

1. The ion carries a charge of $-1$. The sum of all oxidation states must equal $-1$. — This is the critical step for ions. The sum rule target is the ion's charge, not zero.
2. Oxygen is $-2$ (Rule 4). There are 4 oxygen atoms: $4 \times (-2) = -8$. — MnO4- contains no peroxide linkage and no fluorine, so oxygen is -2.
3. $\text{oxidation state of Mn} + (-8) = -1$ — Set the total equal to the ion's charge of -1.
4. $\text{oxidation state of Mn} = -1 + 8 = +7$ — Manganese has an oxidation state of +7 in the manganate(VII) ion. This is why it is named manganate(VII) — the Roman numeral states the oxidation state.

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.

5. Exceptions and Traps: Peroxides and Metal Hydrides

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.

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.

6. Worked Examples: Transition Metals and Complex Formulae

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, $Fe_2O_3$.

1. Oxygen is $-2$ (Rule 4). There are 3 oxygen atoms: $3 \times (-2) = -6$. — No peroxide, no fluorine — oxygen is -2 here.
2. $2 \times (\text{oxidation state of Fe}) + (-6) = 0$ — Fe2O3 is a neutral compound, so the total equals zero. There are 2 iron atoms, so multiply the unknown by 2.
3. $2 \times (\text{oxidation state of Fe}) = +6 \Rightarrow \text{oxidation state of Fe} = +3$ — Divide both sides by 2. Iron is +3, consistent with the name iron(III) oxide.
4. The compound is named iron(III) oxide. The Roman numeral III confirms the oxidation state of iron is +3. — Use the IUPAC name as a cross-check whenever it is given.

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, $FeSO_4$.

1. The name iron(II) tells us iron is $+2$. Alternatively, iron is a transition metal — we will use the name here. — When the name gives the oxidation state directly, use it as a known value.
2. Oxygen is $-2$ (Rule 4). There are 4 oxygen atoms: $4 \times (-2) = -8$. — Standard assignment — no exceptions apply here.
3. $(+2) + \text{S} + (-8) = 0$ — FeSO4 is neutral. All three elements contribute to the sum.
4. $\text{S} = 0 - 2 + 8 = +6$ — Sulphur is +6 in the sulphate ion. This is consistent with the sulphate ion $SO_4^{2-}$ having sulphur at +6 (you can verify: +6 + 4(-2) = -2, which matches the ion's charge).

Answer: Sulphur has an oxidation state of $+6$ in $FeSO_4$.

7. Linking Oxidation States to Oxidation and Reduction

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.

8. Exam Technique and Mark-Winning Strategies

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.