Variable oxidation states (A-level only) — AQA A-Level Chemistry
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Variable oxidation states (A-level only) explained
The 4s and 3d sub-levels lie close in energy, so once 4s electrons are removed, 3d electrons can be lost without a large jump in ionisation energy.
Read the full explanation
This gives a ladder of accessible oxidation states. Iron shows +2 and +3, copper +1 and +2, chromium +2, +3 and +6, and manganese runs from +2 to +7. The highest states survive only when bonded to oxygen or fluorine, which is why manganese(VII) appears as MnO₄⁻ rather than a simple cation. Deducing an oxidation state from a formula is a standard task: oxygen counts as −2 and the sum equals the overall charge. Crucially, positive oxidation states must always include an explicit positive sign (e.g., +7, not 7).
Vanadium species in oxidation states IV, III and II are formed by the reduction of vanadate(V) ions by zinc in acidic solution.
Zinc in acid is a strong enough reducing agent to walk vanadium down the whole ladder, and each step has its own colour. Yellow VO₂⁺, vanadium(V), becomes blue VO²⁺, vanadium(IV), then green V³⁺ and finally violet V²⁺. The half-equations need the acid: VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O, then VO²⁺ + 2H⁺ + e⁻ → V³⁺ + H₂O, then the simpler V³⁺ + e⁻ → V²⁺. Zinc supplies the electrons through Zn → Zn²⁺ + 2e⁻, so one mole of zinc reduces two moles of vanadium at each step. A green colour part way through can be genuine V³⁺, or a mixture of blue vanadium(IV) with yellow vanadium(V), so watch the sequence rather than one snapshot.
The redox potential for a transition metal ion changing from a higher to a lower oxidation state is influenced by pH and by the ligand.
A standard electrode potential applies only under standard conditions, and two things routinely shift it. The first is pH: many reduction half-equations consume H⁺, as in MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Raising the acid concentration shifts that equilibrium to the right, making the potential more positive and manganate(VII) a stronger oxidising agent; in alkaline conditions it is far weaker and is reduced only as far as MnO₂. The second is the ligand: replacing water changes how readily the metal gives up an electron. [Co(H₂O)₆]²⁺ is barely oxidised by air, but add ammonia and the resulting cobalt(II) ammine complex is oxidised to cobalt(III) readily, because ammonia stabilises the higher oxidation state.
Your focus
- Deduce the oxidation state of a transition metal from the formula of any of its compounds or ions.
- Explain variable oxidation states in terms of the similar energies of the 4s and 3d sub-levels.
- Identify which element is oxidised and which is reduced in a given transition metal redox equation.
Show all 9 objectives
- Write the half-equation for each reduction step from vanadium(V) to vanadium(II), balancing with H⁺ and H₂O.
- State the colour of each vanadium species and predict the sequence seen as zinc reacts with acidified vanadate(V).
- Combine the vanadium and zinc half-equations into a balanced overall equation.
- Explain how changing pH alters the electrode potential of a half-equation that includes H⁺.
- Predict the effect of adding ammonia on the ease of oxidation of a cobalt(II) or iron(II) complex.
- Justify why a reaction may occur under given conditions even though standard potentials suggest otherwise.
Variable oxidation states (A-level only) exam tips
Marking Points
- Correct oxidation state of the metal in a given species, ensuring explicit signs are used (e.g., +7 in MnO₄⁻).
- Explaining variable oxidation states by the similar energies of the 4s and 3d sub-levels.
- Identifying the change in oxidation state during a named redox reaction.
- Using oxidation state changes to balance a half-equation with electrons.
- one mark for the correct sequence of colours, yellow then blue then green then violet
- one mark for each correct half-equation, balanced with H⁺ and H₂O as well as electrons
- one mark for the oxidation half-equation Zn → Zn²⁺ + 2e⁻
- one mark for combining the half-equations so that the electrons cancel in the overall equation
- one mark for quoting a half-equation that contains H⁺ to show the pH dependence
- one mark for applying Le Chatelier's principle to the effect of changing the H⁺ concentration on that equilibrium
- one mark for stating the direction in which the electrode potential moves and what that does to oxidising power
- one mark for a ligand example, such as ammonia making cobalt(II) far easier to oxidise than water does
Examiner Tips
- 💡Set the arithmetic out: the sum of the oxidation states equals the overall charge on the species.
- 💡Always include the sign for oxidation states, explicitly writing '+' for positive values.
- 💡Balance oxygen with water first, then hydrogen with H⁺, then charge with electrons.
- 💡Scale the half-equations so the electrons cancel exactly before adding them together.
- 💡Learn the four colours as a sequence, because questions often ask you to predict the next change.
- 💡Look for H⁺ in the half-equation, because if it is there then pH will change the potential.
- 💡Use the phrase the equilibrium shifts to the right when arguing the effect of added acid.
- 💡Say explicitly whether the species becomes a stronger or a weaker oxidising agent.
Common Mistakes
- Reading the overall charge on MnO₄⁻ as the oxidation state of the manganese.
- Omitting the positive sign on a positive oxidation state (writing 7 instead of +7).
- Assuming every transition metal reaches +7, when only manganese does.
- Forgetting that oxygen is −2 in an oxoanion when working out the metal's oxidation state.
- writing V⁵⁺ and V⁴⁺ as simple ions instead of the oxoions VO₂⁺ and VO²⁺
- omitting H⁺ and H₂O from the half-equations, so leaving oxygen unbalanced
- giving the wrong charge on VO²⁺, often writing VO⁺ by forgetting that oxygen contributes −2
- assuming a green solution is always V³⁺ rather than possibly a mixture of the yellow and blue species
- assuming standard electrode potentials apply whatever the pH or the ligand
- claiming acid always makes a species a better reducing agent, when for an oxoanion it does the opposite
- saying a reaction cannot happen because standard potentials forbid it, while ignoring the non-standard conditions given
- confusing the effect of a ligand on colour with its effect on redox potential