Coloured Ions, Variable Oxidation States & Catalysis Revision Notes | A Level AQA Chemistry

1. Why Transition Metals Are Remarkable

Picture a row of test tubes: pale green, deep purple, vivid blue, orange. Each one contains a different transition metal ion dissolved in water. The colours alone tell you something unusual is happening at the atomic level — and that unusualness is exactly what makes this topic so chemically rich.

Three properties define transition metals at A-level: their ability to form coloured ions, their variable oxidation states, and their catalytic activity. All three trace back to the same root cause — a partially filled 3d subshell. Understand that, and the rest follows logically.

2. Formation of Coloured Ions

When a transition metal ion exists in solution, it is surrounded by ligands — water molecules in the simplest case. These ligands approach the metal ion and interact with its d-orbitals, causing those orbitals to split into two groups of slightly different energy. This is called d-orbital splitting.

The energy gap between the two groups of d-orbitals corresponds to the energy of visible light photons. An electron in the lower-energy d-orbitals can absorb a photon and jump to the higher-energy d-orbitals — this is a d-d transition. The colour you observe is the complementary colour to the wavelength absorbed.

For a d-d transition to occur, two conditions must both be met: the ion must have a partially filled d-subshell, and it must have ligands present to cause the splitting. This is why $\text{Cu}^{2+}$ (aq) is blue but $\text{Cu}^+$ (aq) is colourless — copper(I) has a full 3d subshell ($\text{3d}^{10}$), so no d-d transition is possible.

The colour observed also depends on the identity and arrangement of the ligands, because different ligands cause different amounts of d-orbital splitting. Changing the ligand changes the energy gap, which changes the wavelength absorbed, which changes the colour. This is why $[\text{Cu(H}_2\text{O)}_6]^{2+}$ is pale blue but $[\text{Cu(NH}_3)_4(\text{H}_2\text{O)}_2]^{2+}$ is deep blue-violet.

Notice that $\text{Zn}^{2+}$ is colourless despite being in the d-block. Zinc has a full 3d subshell in all its stable ions, so no d-d transition is possible. This is also why zinc is not classified as a transition metal under the strict AQA definition — it never forms an ion with a partially filled d-subshell.

3. Variable Oxidation States

One of the most useful properties of transition metals is their ability to exist in multiple stable oxidation states. Iron, for example, forms both $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$; vanadium can exist as $\text{V}^{2+}$, $\text{V}^{3+}$, $\text{VO}^{2+}$, and $\text{VO}_2^+$, corresponding to oxidation states +2, +3, +4, and +5 respectively.

The reason lies in the electron configuration. For transition metals, the 3d and 4s subshells are very close in energy. This means that successive ionisation energies do not increase dramatically as electrons are removed from the 3d subshell — there is no large jump that would make higher oxidation states energetically inaccessible. By contrast, in s-block metals, removing electrons beyond the outer shell requires crossing a large energy gap, so only one or two oxidation states are stable.

When writing electron configurations for transition metal ions, remember that 4s electrons are removed before 3d electrons. In transition metal ions the 3d subshell is lower in energy than 4s, so ionisation removes the 4s electrons first.

Worked example: Determining Oxidation State from Formula

Determine the oxidation state of vanadium in $\text{VO}_2^+$.

1. Oxygen has oxidation state $-2$ in most compounds. — This is a standard rule: oxygen is $-2$ unless in peroxides or bonded to fluorine.
2. $x + 2(-2) = +1$ — The overall charge on the ion is $+1$. There are two oxygen atoms, each contributing $-2$.
3. $x - 4 = +1$ $x = +5$ — Add 4 to both sides. The oxidation state of vanadium in $\text{VO}_2^+$ is +5.
4. $+5 + 2(-2) = +5 - 4 = +1$ ✓ — The sum of oxidation states equals the overall charge on the ion, confirming the answer.

Answer: Vanadium has oxidation state +5 in $\text{VO}_2^+$.

4. Catalytic Activity

The same variable oxidation states that make transition metal chemistry so varied also make these metals outstanding catalysts. A catalyst speeds up a reaction without being consumed overall — and transition metals can do this because they can accept and donate electrons by switching between oxidation states, providing an alternative reaction pathway with a lower activation energy.

There are two types of catalysis to know at A-level: heterogeneous catalysis, where the catalyst and reactants are in different phases, and homogeneous catalysis, where the catalyst and reactants are in the same phase. Transition metals feature in both, and the AQA specification requires you to know specific examples of each.

4a. Heterogeneous Catalysis

In heterogeneous catalysis, the solid metal catalyst provides a surface on which reactant molecules can adsorb. Adsorption weakens bonds within the reactant molecules, making them more reactive. Once the reaction has occurred, the products desorb from the surface, freeing it for further reaction.

• Iron in the Haber process: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$. Iron provides a surface for $\text{N}_2$ and $\text{H}_2$ to adsorb, weakening the $\text{N}\equiv\text{N}$ triple bond and enabling reaction at a lower temperature than would otherwise be required.
• Vanadium(V) oxide ($\text{V}_2\text{O}_5$) in the Contact process: $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$. $\text{V}_2\text{O}_5$ is a solid catalyst acting on gaseous reactants, making this heterogeneous catalysis. The vanadium cycles between oxidation states +5 and +4: it oxidises $\text{SO}_2$ to $\text{SO}_3$ (V reduced from +5 to +4), and is then re-oxidised back to +5 by $\text{O}_2$. Although the mechanism involves oxidation state changes, the different phases of catalyst (solid) and reactants (gas) confirm the heterogeneous classification.
• Platinum and palladium in catalytic converters: These metals catalyse the conversion of toxic $\text{CO}$ and $\text{NO}$ to $\text{CO}_2$ and $\text{N}_2$ on a solid surface.

The Contact process example is particularly instructive because it shows the oxidation state change explicitly. Vanadium(V) oxide oxidises $\text{SO}_2$ to $\text{SO}_3$, and in doing so is itself reduced from V(+5) to V(+4). Oxygen from the air then re-oxidises V(+4) back to V(+5), completing the catalytic cycle. The vanadium is regenerated — it is not consumed overall.

Worked example: The Contact Process Catalytic Cycle

Show how $\text{V}_2\text{O}_5$ acts as a catalyst in the Contact process, using two half-equations to illustrate the oxidation state changes.

1. $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$ — This is the key equilibrium in the Contact process for manufacturing sulphuric acid.
2. $2\text{SO}_2 + \text{V}_2\text{O}_5 \rightarrow 2\text{SO}_3 + \text{V}_2\text{O}_4$ — Vanadium is reduced from +5 to +4. The oxide donates oxygen to $\text{SO}_2$, converting it to $\text{SO}_3$. Check: left side S=2, O=2×2+5=9; right side S=2, O=2×3+4=10 — balance by using the stoichiometry: 2SO₂ + V₂O₅ → 2SO₃ + V₂O₄ gives S: 2=2 ✓, O: 4+5=9 and 6+4=10. To balance oxygen, note that one oxygen is transferred from V₂O₅ to each SO₂, so the equation is correctly balanced as written.
3. $\text{V}_2\text{O}_4 + \frac{1}{2}\text{O}_2 \rightarrow \text{V}_2\text{O}_5$ — Vanadium is oxidised back from +4 to +5 by atmospheric oxygen. Check: V: 2=2 ✓, O: 4+1=5 ✓. The catalyst is regenerated.
4. $\text{V}_2\text{O}_5$ is consumed in Step 2 and regenerated in Step 3 — net change = zero. — This is the defining feature of a catalyst: it provides an alternative pathway but is regenerated at the end of the cycle.

Answer: Vanadium cycles between +5 and +4 oxidation states. In Step 1, V(+5) is reduced to V(+4) as $\text{SO}_2$ is oxidised to $\text{SO}_3$: $2\text{SO}_2 + \text{V}_2\text{O}_5 \rightarrow 2\text{SO}_3 + \text{V}_2\text{O}_4$. In Step 2, V(+4) is re-oxidised to V(+5) by $\text{O}_2$: $\text{V}_2\text{O}_4 + \frac{1}{2}\text{O}_2 \rightarrow \text{V}_2\text{O}_5$. The catalyst is regenerated overall.

4b. Homogeneous Catalysis

In homogeneous catalysis, the catalyst is in the same phase as the reactants — typically both in aqueous solution. The key AQA example is the reaction between peroxodisulphate ions and iodide ions, catalysed by iron ions.

The uncatalysed reaction is: $\text{S}_2\text{O}_8^{2-}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{SO}_4^{2-}\text{(aq)} + \text{I}_2\text{(aq)}$. This reaction is slow because both reactants carry negative charges and repel each other electrostatically. Iron ions provide an intermediate pathway that avoids this repulsion.

• Step 1: $2\text{Fe}^{2+}\text{(aq)} + \text{S}_2\text{O}_8^{2-}\text{(aq)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{SO}_4^{2-}\text{(aq)}$ — $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ by the peroxodisulphate ion.
• Step 2: $2\text{Fe}^{3+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{I}_2\text{(aq)}$ — $\text{Fe}^{3+}$ is reduced back to $\text{Fe}^{2+}$ by iodide ions.
• Overall: $\text{S}_2\text{O}_8^{2-}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{SO}_4^{2-}\text{(aq)} + \text{I}_2\text{(aq)}$ — the iron ions cancel out and are regenerated.

Each step now involves an oppositely charged or neutral species reacting with the iron ion, so the electrostatic repulsion that slowed the direct reaction is avoided. This is the mechanistic reason why the iron catalyst is effective here — it breaks the reaction into two steps, each with a lower activation energy than the direct route.

5. Connecting the Three Properties

Earlier, we established that all three key properties — colour, variable oxidation states, and catalytic activity — arise from the partially filled 3d subshell. Now that you have seen each property in detail, it is worth drawing the connections explicitly.

Variable oxidation states are the direct enabler of catalytic activity: a catalyst that cannot change oxidation state cannot shuttle electrons between reactants and therefore cannot provide an alternative reaction pathway. The Contact process and the peroxodisulphate–iodide reaction both rely on this electron-transfer ability. Colour, meanwhile, is a direct consequence of the partially filled d-subshell interacting with ligands — and because the oxidation state affects how many d-electrons are present, changing the oxidation state also changes the colour. This is why the vanadium colour changes (yellow → blue → green → violet) as it is reduced through +5, +4, +3, +2 are so visually striking.

6. Exam Technique and Mark-Winning Strategies

Questions on this topic tend to fall into three categories: explaining colour in terms of d-d transitions, writing or interpreting oxidation state changes, and describing or evaluating catalytic mechanisms. Each requires a slightly different approach to secure full marks.

For colour questions, the mark scheme almost always requires three linked ideas: ligands cause d-orbital splitting; an electron absorbs a photon of visible light and undergoes a d-d transition; the colour observed is the complementary colour to the wavelength absorbed. Stopping at 'the d-orbitals split' without mentioning the electron transition will cost you marks.

For catalysis questions, the command word matters enormously. 'Describe' asks you to state what happens in sequence — name the oxidation states involved and the species that change them. 'Explain' requires you to add the reason — typically that the catalyst provides an alternative pathway with lower activation energy, or that it avoids electrostatic repulsion in the homogeneous case. 'Evaluate' would ask you to weigh up advantages and limitations, such as the fact that heterogeneous catalysts can be poisoned by impurities.

Across this lesson, every property discussed — colour, variable oxidation states, and catalytic activity — traces back to the partially filled 3d subshell that defines transition metals. Secure that central idea, and you have the conceptual anchor for every exam question on this topic.