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3.2.5 · Paper 1

Transition Metals

The d-block elements and why they are special: complexes and their shapes, ligand substitution and the chelate effect, where their colour comes from, variable oxidation states and redox titrations, and how they act as catalysts.

Joe White AQA A-Level Chemistry (7405) AQA ref: 3.2.5
AQA 7404/7405 Paper 1 A-level only

Transition Metals

Complexes
  1. General properties
  2. Shapes of complexes
  3. Ligand substitution
Colour & redox
  1. The origin of colour
  2. Colour changes in substitution
  3. Variable oxidation states
  4. Catalysis
Synoptic
  1. Two habits

Whole topic · A-level only (Year 13)

Transition-metal chemistry has a reputation for being one of the hardest parts of A-level chemistry, not because any single idea is difficult, but because it leans on several earlier ones at once. Three are worth refreshing before you start.

💡 Where this sits

This topic builds on three AS ideas. Electron configuration and the d sub-level (the 4s/3d filling order and the chromium and copper anomalies), because every property below comes from an incomplete d sub-level. The co-ordinate (dative) bond, a shared pair donated by one atom, because that is exactly how a ligand attaches to a metal. And the shapes of molecules from electron-pair repulsion, which carry straight over to the shapes of complex ions. If any of these feels shaky, revisit it first.

General properties

The transition metals are the d-block elements titanium to copper. What sets them apart is a specific electron feature.

You have already met these ions, if only briefly. At GCSE you knew that iron forms Fe2+ and Fe3+, and that copper(II) sulfate solution is blue. A-level explains what those descriptions were hiding: in water a metal ion is never “bare”. Each one is surrounded by small molecules or ions, so the blue solution you remember is really the ion [Cu(H2O)6]2+ (a copper ion wrapped in six water molecules) and its colour comes from that arrangement. The surrounding species are called ligands, the whole assembly is a complex, and in the same way “Fe2+(aq)” is really [Fe(H2O)6]2+. Almost everything in this topic follows from that one picture, so the precise definitions come next.

📖 Key definitions

A transition metal is a d-block element that forms at least one ion with a partially filled (incomplete) d sub-level.

A ligand is a molecule or ion that forms a co-ordinate bond with a transition metal by donating a pair of electrons.

A complex is a central metal atom or ion surrounded by ligands.

The co-ordination number is the number of co-ordinate bonds to the central metal atom or ion.

Because the definition needs a partially filled d sub-level in an ion, two d-block elements are not transition metals: scandium (its only ion Sc3+ is [Ar], empty d) and zinc (its only ion Zn2+ is [Ar]3d10, full d).

Electron configurations fill 4s before 3d, with two anomalies for extra stability: chromium is [Ar]3d54s1 and copper is [Ar]3d104s1. On forming ions, the 4s electrons are removed first (e.g. Fe is [Ar]3d64s2; Fe2+ is [Ar]3d6).

The d block: who counts as a transition metal

ScTiVCrMnFeCoNiCuZnthe transition metals: Ti → CuSc³⁺: 3d⁰Zn²⁺: 3d¹⁰ions with a partially filled 3d sub-level

4s fills first, and empties first

Fe3d4sFe²⁺3d4s4s emptied first− 2e⁻

Fe is [Ar]3d64s2; forming Fe2+ removes the two 4s electrons first, leaving [Ar]3d6: a partially filled d sub-level, so Fe2+ qualifies.

A transition metal must form an ion with a partially filled d sub-level: Sc3+ (empty 3d) and Zn2+ (full 3d) rule scandium and zinc out.
✅ The four characteristic properties
  • Complex formation: they form complexes with ligands.
  • Coloured ions: their compounds and solutions are coloured.
  • Variable oxidation states: they form ions in more than one oxidation state.
  • Catalytic activity: the metals and their compounds catalyse many reactions.

All four arise from the incomplete, close-in-energy d sub-level.

🧪 Exam questions
Q1 [1 mark]

Which shows the electron configuration of an atom of a transition metal? Tick (✓) one box.

[Ar]4s23d8 is nickel, a transition metal. [Ar]4s23d0 is calcium; [Ar]4s23d10 is zinc (full d, not a transition metal); [Ar]4s23d104p1 is gallium (a p-block element).
Q2 [1 mark]

State the meaning of the term transition metal complex.

Show answer

A central metal ion (atom) surrounded by ligands, each bonded to it by a co-ordinate (dative) bond. 1 mark

Define the complex, not the transition metal. A common lost mark.

Source: AQA A-Level Chemistry past papers.

Shapes of complex ions

The shape follows from the co-ordination number, which usually follows from the size of the ligand. Small ligands pack six around the metal; larger ligands fit only four. The shapes themselves come from the same electron-pair repulsion you used for simple molecules at AS: the ligands arrange themselves as far apart as possible around the central ion.

Co-ordination numberShapeBond angleExample
6octahedral90°[Fe(H2O)6]2+, [Co(NH3)6]2+
4tetrahedral109.5°[CuCl4]2−, [CoCl4]2−
4square planar90°cisplatin [Pt(NH3)2Cl2]
2linear180°[Ag(NH3)2]+ (Tollens’)

Isomerism. Octahedral complexes of the type MA4B2 (and square planar MA2B2) show cis–trans isomerism, a special case of E–Z isomerism. Cisplatin is the cis isomer (the two chlorides adjacent); the trans isomer is not an effective drug: the two chlorides have to be next to each other to cross-link two guanine bases in DNA. Octahedral complexes with bidentate ligands (such as [Ni(en)3]2+) show optical isomerism: two non-superimposable mirror images.

Octahedral · 6 bonds · 90°

90°FeOH₂OH₂H₂OOH₂H₂OOH₂

[Fe(H2O)6]2+: six small ligands (H2O, NH3) fit around the metal.

Tetrahedral · 4 bonds · 109.5°

109.5°CuClClClCl

[CuCl4]2−: Cl− is larger, so only four fit.

Square planar · 4 bonds · 90°

90°PtClClH₃NNH₃

Cisplatin, [Pt(NH3)2Cl2]: all four ligands in one plane.

Linear · 2 bonds · 180°

180°AgH₃NNH₃

[Ag(NH3)2]+: the complex in Tollens’ reagent.

Cis–trans · square planar

PtClClH₃NNH₃PtClClH₃NNH₃cis (cisplatin, the drug)trans

[Pt(NH3)2Cl2]: cis, the two Cl 90° apart; trans, 180° apart. Cisplatin is the cis isomer.

Cis–trans · octahedral

CrClClNH₃H₃NNH₃H₃NCrClClH₃NNH₃H₃NNH₃cis: Cl 90° aparttrans: Cl 180° apart

[Cr(NH3)4Cl2]+: an octahedral MA4B2 complex (the Q6 pair).

Optical isomers · octahedral with bidentate ligands

NiNNNNNNNiNNNNNNMIRROR

[Ni(en)3]2+ (en = H2NCH2CH2NH2): each N–C–C–N chain, drawn skeletally, is one bidentate ligand bonding through both N atoms. The two forms are non-superimposable mirror images.

Cis–trans and optical are both forms of stereoisomerism: cis–trans is about which positions the ligands take, optical is about mirror images. Do not confuse them.

Interactive: name the shape, number, angle & ligand

✅ Describe any complex: the five facts

For any named complex, work out the five facts in turn (oxidation state, overall charge, co-ordination number, shape, ligand) so you always give the charge, ligand count and shape the mark scheme wants.

ComplexMetal ox. stateOverall chargeCo-ord. no.ShapeLigand
[Fe(H2O)6]2++22+6octahedralH2O
[Cu(NH3)4(H2O)2]2++22+6octahedralNH3, H2O
[CoCl4]2−+22−4tetrahedralCl−
[Pt(NH3)2Cl2]+204square planarNH3, Cl−
[Ag(NH3)2]++11+2linearNH3
🧪 Exam questions
Q3 [1 mark]

Which will not act as a ligand in the formation of a complex ion? Tick (✓) one box.

A ligand must have a lone pair to donate. CH4 has no lone pair, so it cannot be a ligand. CO, H2O and NH3 each have a lone pair.
Q4 [1 mark]

Which statement is not correct? Tick (✓) one box.

[CuCl4]2− is tetrahedral (four large Cl− ligands), not square planar, so this statement is not correct. The other three are right.
Q5 [1 mark]

Which shows the correct oxidation state and co-ordination number of cobalt in [Co(NH3)5Cl]Cl2? Tick (✓) one box.

The complex ion is [Co(NH3)5Cl]2+ with two Cl− counter-ions. Inside it: five NH3 (neutral) + one Cl− give six co-ordinate bonds (co-ordination number 6), and for an overall 2+ the cobalt must be +3.
Q6 [3 marks]

The complex ion [Cr(NH3)4Cl2]+ shows isomerism. Draw the two isomers and state the type of isomerism shown.

Show answer

Octahedral cations drawn with the two Cl on adjacent positions (cis) and on opposite positions (trans). 2 marks

Type: cis–trans (E–Z) isomerism. 1 mark

Q7 [1 mark]

Which pair of reagents reacts to form a tetrahedral complex? Tick (✓) one box.

Cl− is the large ligand, so only four fit round the metal: excess concentrated HCl converts Fe3+ into the tetrahedral [FeCl4]−. Ammonia and water are small, so both ammonia reactions keep six ligands and stay octahedral, and ethanedioate is bidentate: three of them still give six co-ordinate bonds, so that complex is octahedral too.

Source: AQA A-Level Chemistry past papers.

Ligand substitution

One ligand can be replaced by another. Whether the co-ordination number changes depends on the size of the incoming ligand.

NH3 and H2O are similar in size and both uncharged, so they exchange with no change of co-ordination number (both stay 6). With copper the substitution is incomplete (only four of the six waters are replaced) and the colour deepens (pale blue → deep blue):

[Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O

Cl− is larger, so fewer fit: the co-ordination number changes from 6 to 4 and the colour changes (blue → yellow):

[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O

The same 6→4 change happens with cobalt (pink [Co(H2O)6]2+ → blue [CoCl4]2−) and iron(III) ([Fe(H2O)6]3+ → [FeCl4]−).

Ligands can donate more than one lone pair: bidentate ligands (ethane-1,2-diamine H2NCH2CH2NH2, and the ethanedioate ion C2O42−) donate two, and multidentate ligands (EDTA4−) donate several. Haem is an iron(II) complex with a multidentate ligand; oxygen bonds co-ordinately to the Fe(II) in haemoglobin to be carried in the blood, and carbon monoxide is toxic because it replaces that oxygen.

✅ The chelate effect

Bidentate and multidentate ligands replace monodentate ligands from a complex. The reason is entropy: replacing several small ligands with one big multidentate ligand releases more particles than it takes in, so ΔS is positive. The enthalpy change is close to zero (similar bonds are made and broken), so ΔG = ΔH − TΔS is negative: the substitution is favourable. This is the same entropy and free-energy reasoning you meet in physical chemistry (3.1.8).

Swap H2O for NH3: co-ordination number stays 6

[Cu(H₂O)₆]²⁺pale blue solution+ 4NH₃co-ord. number stays 6[Cu(NH₃)₄(H₂O)₂]²⁺deep blue solution

[Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O: similar-sized uncharged ligands; the substitution is incomplete.

Swap H2O for Cl−: 6 → 4

[Cu(H₂O)₆]²⁺pale blue solution+ 4Cl⁻co-ord. number 6 → 4[CuCl₄]²⁻yellow solution

[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O. Cl− is larger, so the co-ordination number falls to 4 (tetrahedral).

The chelate effect: count the particles

[Cu(H₂O)₆]²⁺enenen+ 3en4 particles(1 complex + 3 ligands)[Cu(en)₃]²⁺+ 6H₂O released7 particlesmore particles → ΔS positive

[Cu(H2O)6]2+ + 3en → [Cu(en)3]2+ + 6H2O: 4 particles become 7, so ΔS is positive; ΔH ≈ 0 because similar bonds break and form, so ΔG = ΔH − TΔS is negative.

The chelate effect is an entropy story: more particles are released than taken in.

Interactive: add a reagent, watch the complex & colour change

1 · start with

2 · add

🧪 Exam questions
Q8 [3 marks]

When ethanedioate ions (C2O42−) are added to [Fe(H2O)6]3+, all six water ligands are replaced. Explain, in terms of enthalpy and entropy, why this substitution is favourable.

Show answer

The enthalpy change is small / close to zero (similar Fe–O bonds are broken and made). 1 mark

There is an increase in entropy, because more particles are released than are taken in (e.g. 6 H2O released for 3 C2O42− added: 4 particles → 7). 1 mark

ΔG = ΔH − TΔS is negative (because TΔS is positive and ΔH ≈ 0), so the reaction is feasible. 1 mark

Q9 [1 mark]

Give an equation for the ligand-substitution reaction when an excess of concentrated hydrochloric acid is added to [Cu(H2O)6]2+.

Show answer

[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O. 1 mark

The co-ordination number changes from 6 to 4 (Cl− is larger); the colour changes from blue to yellow.

Source: AQA A-Level Chemistry past papers.

The origin of colour

In a complex, the d orbitals split into two energy levels a small gap apart. When white light passes through, d electrons absorb the wavelengths whose energy matches that gap and are promoted from the ground state to an excited state: a d–d transition. The energy of the gap is:

ΔE = hν = hc/λ

The colour you see is the light that is not absorbed: the remaining wavelengths that are transmitted or reflected. A complex looks yellow because it reflects/transmits yellow light, not because it emits it.

✅ The colour logic chain

Anything that changes the size of the d-orbital gap ΔE changes the wavelength absorbed, and so changes the colour. Three things do this:

  • the ligand (different ligands split the d orbitals by different amounts);
  • the oxidation state of the metal;
  • the co-ordination number / shape.

ligand / oxidation state / co-ordination number → ΔE → wavelength absorbed → colour.

The colours of some common ions:

IonColour
[Cu(H2O)6]2+pale blue
[Cu(NH3)4(H2O)2]2+deep blue
[CuCl4]2−yellow
[Fe(H2O)6]2+pale green
[Fe(H2O)6]3+yellow-brown (pale violet)
[Co(H2O)6]2+pink
[CoCl4]2−blue

The absorption of visible light is the basis of spectroscopy. Colorimetry applies it to measure concentration: a ligand is added to give an intense colour, a filter of the complementary colour is chosen, and the absorbance is measured. A calibration graph of absorbance against concentration (from standard solutions) then converts the unknown’s absorbance into its concentration.

A d electron jumps the gap

energyground stateexcited stateΔE = hν =hcλthe size of the gap setswhich λ is absorbedlight absorbed

In a complex the d orbitals split into two levels. A d electron absorbs a photon whose energy matches the gap and is promoted from the ground state to the excited state, a d–d transition.

The colour you see is what is left

absorbed400500600700wavelength / nmthe rest is transmitted / reflectedseen: pale blue[Cu(H₂O)₆]²⁺

Only the matching wavelengths are absorbed. The transmitted / reflected remainder reaches your eye. For [Cu(H2O)6]2+, red-orange light is absorbed and the solution looks pale blue.

Say the light is absorbed and the rest reflected, never that colour is “emitted”.

Interactive: what each complex absorbs, and the ΔE behind it

🧪 Exam questions
Q10 [3 marks]

Explain why complexes formed from transition metal ions are coloured.

Show answer

The d orbitals split into two levels; a d electron absorbs (visible) light and is promoted from the ground state to an excited state. 1 mark

The energy absorbed equals the gap, ΔE = hc/λ, so only certain wavelengths are absorbed. 1 mark

The colour seen is the remaining wavelengths, which are transmitted / reflected (not absorbed). 1 mark

Q11 [3 marks]

Calculate the energy, in J, gained by each electron that absorbs light of wavelength 490 nm. (c = 3.00 × 108 m s−1; h = 6.63 × 10−34 J s.)

Show answer

ΔE = hc/λ = (6.63 × 10−34 × 3.00 × 108) / (490 × 10−9) 1 mark

= 1.989 × 10−25 / 4.90 × 10−7 1 mark

= 4.06 × 10−19 J 1 mark

Convert 490 nm to 490 × 10−9 m first.

Source: AQA A-Level Chemistry past papers.

Colour changes in ligand substitution

The colour chain now explains the colour changes you met earlier under ligand substitution. Whenever a reaction swaps one ligand for another (or changes the co-ordination number), it alters the d-orbital gap ΔE, so a different wavelength of light is absorbed and the colour shifts.

SubstitutionFromToWhat changes (→ new ΔE)
[Cu(H2O)6]2+ + 4NH3 pale blue deep blue ligand (H2O → NH3); co-ordination number stays 6
[Cu(H2O)6]2+ + 4Cl− pale blue yellow ligand (H2O → Cl−) and co-ordination number 6 → 4
[Co(H2O)6]2+ + 4Cl− pink blue ligand (H2O → Cl−) and co-ordination number 6 → 4

In each case the reason is the same: a new ligand (and sometimes a new co-ordination number) gives a new ΔE, so a different wavelength is absorbed and a different colour is transmitted. A colour change on adding a reagent therefore signals that the ligand or co-ordination number has changed. A colour change from a change of oxidation state works for the same underlying reason (a new ΔE) and is covered next.

Interactive: colour-change quiz, both directions

⚠️ Precision points

Explain a substitution colour change through the chain: the ligand (and/or co-ordination number) changes → ΔE changes → a different wavelength is absorbed → a different colour is seen. Naming the colour change alone does not earn the explanation marks.

Don’t be thrown if you see green credited for the copper + concentrated HCl observation. The complex [CuCl4]2− itself is yellow, but the substitution is an equilibrium, so the solution in the test tube usually holds both blue [Cu(H2O)6]2+ and yellow [CuCl4]2−, and that mixture looks green. Name the complex’s colour as yellow; for what you see, green (or yellow-green) is also accepted.

Variable oxidation states & redox

Transition metals form ions in several oxidation states, each usually a different colour. Vanadium is the classic example: acidified vanadate(V) is reduced step by step by zinc, passing through four colours.

Oxidation stateSpeciesColour
+5VO2+ (vanadate(V))yellow
+4VO2+blue
+3V3+green
+2V2+violet
+5VO₂⁺yellow solutionZn / H⁺+4VO²⁺blue solutionZn / H⁺+3V³⁺green solutionZn / H⁺+2V²⁺violet solutionreduced one step at a time. Stop early and you catch the intermediate colours (often mixtures)
Learn the colour and oxidation state together: +5 yellow, +4 blue, +3 green, +2 violet.

Interactive: reduce vanadium(V) step by step with zinc

Whether one oxidation state can be reduced to another depends on the electrode potentials, and those are influenced by pH and by the ligand. (Tollens’ reagent, [Ag(NH3)2]+, uses this: it is reduced to silver by an aldehyde but not a ketone.)

Redox titrations

Manganate(VII), MnO4−, is a powerful oxidising agent and is self-indicating: it is decolourised as it reacts, so the end point is the first permanent trace of purple/pink. Two standard titrations:

5Fe2+ + MnO4− + 8H+ → 5Fe3+ + Mn2+ + 4H2O

2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O

Run the redox titration below. Because MnO4− is self-indicating, no indicator is added: every drop is decolourised as it reacts, and the end point is the drop that leaves the first permanent pale pink.

Interactive: run the redox titration and build your own titre table

0 50 Clamp Clamp stand Burette (MnO₄⁻) read to 0.05 cm³ Tap Conical flask reducing agent White tile

RunInitial / cm3Final / cm3Titre / cm3

🧪 Exam questions
Q12 [1 mark]

Which compound decolourises acidified potassium manganate(VII) solution? Tick (✓) one box.

Fe2+ (in FeSO4) is oxidised to Fe3+, reducing MnO4− to colourless Mn2+, so the purple colour is removed. Fe2(SO4)3 already contains Fe3+; the others are not reducing agents.
Q13 [1 mark]

Which equation does not show the reduction of a transition metal? Tick (✓) one box.

In CoO and in [CoCl4]2− cobalt is +2 throughout: no electrons are transferred, so nothing is reduced. In the others the metal’s oxidation state falls: Ti +4→0, Fe +3→+2, Mn +4→+2.
Q14 [4 marks]

A standard solution is made by dissolving 162 mg of Na2C2O4 (Mr = 134.0) and making up to 250 cm3. A 25.0 cm3 portion is acidified and titrated with KMnO4, requiring a mean titre of 23.85 cm3. Using 2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 8H2O + 10CO2, calculate the concentration of the KMnO4.

Show answer

n(Na2C2O4) total = 0.162134.0 = 1.21 × 10−3 mol; in 25.0 cm3: × 25/250 = 1.21 × 10−4 mol. 1 mark

Mole ratio MnO4− : C2O42− = 2 : 5, so n(MnO4−) = (2/5) × 1.21 × 10−4 = 4.84 × 10−5 mol. 2 marks

Concentration = 4.84 × 10−523.85 × 10−3 = 2.03 × 10−3 mol dm−3. 1 mark

At every step of the working, say which species the moles value belongs to and which volume it refers to (e.g. “n(C2O42−) in 25.0 cm3”), and warm the flask before titrating (the reaction is slow at first).

Q15 [1 mark]

State the colour change seen at the end point of a titration of a colourless reducing agent with potassium manganate(VII).

Show answer

Colourless to pink (the first permanent pale purple/pink). 1 mark

Not just “purple”, and not “clear” for colourless.

Source: AQA A-Level Chemistry past papers.

Catalysis

Transition metals and their compounds catalyse many reactions, and their variable oxidation states are the key: the metal can gain and lose electrons, cycling between states as it provides an easier route.

Heterogeneous catalysts

A heterogeneous catalyst is in a different phase from the reactants, and the reaction happens at active sites on its surface (reactants adsorb, react, then the products desorb). A support medium is used to spread the catalyst thinly, maximising the surface area and minimising the cost of the metal. Active sites can be poisoned by impurities that block them, reducing efficiency (a cost implication).

  • V2O5 catalyses the Contact process (SO2 → SO3), working through two steps as vanadium cycles between +5 and +4:

V2O5 + SO2 → V2O4 + SO3

2V2O4 + O2 → 2V2O5

  • Fe catalyses the Haber process (N2 + 3H2 ⇌ 2NH3).

Homogeneous catalysts

A homogeneous catalyst is in the same phase as the reactants, and the reaction proceeds through an intermediate. The classic example is Fe2+ catalysing the reaction between the two negative ions S2O82− and I−, which otherwise repel each other and react slowly:

S2O82− + 2Fe2+ → 2SO42− + 2Fe3+

2Fe3+ + 2I− → 2Fe2+ + I2

Each step is between oppositely charged ions, so the activation energy is lower, and the Fe2+ is regenerated. Mn2+ behaves as an autocatalyst in the MnO4−/C2O42− titration: it is a product of the reaction, so the reaction starts slowly and then speeds up as Mn2+ builds up:

4Mn2+ + MnO4− + 8H+ → 5Mn3+ + 4H2O

2Mn3+ + C2O42− → 2Mn2+ + 2CO2

Heterogeneous: different phase

solid catalyst: active sites on the surface1 adsorb2 react3 desorbpoisoned

Reactants adsorb at active sites on the surface, react, and the products desorb. Impurities that bind to the sites poison the catalyst.

Homogeneous: same phase

Fe²⁺Fe³⁺S₂O₈²⁻ in2SO₄²⁻ outstep 12I⁻ inI₂ outstep 2the catalyst is regenerated every cycle

Fe2+ catalyses S2O82− + 2I− → 2SO42− + I2 through an intermediate (Fe3+): each step is between oppositely charged ions.

Homogeneous catalysis works because the metal can switch oxidation state and be regenerated.
🧪 Exam questions
Q16 [2 marks]

In the Contact process, sulfur(IV) oxide is converted into sulfur(VI) oxide using vanadium(V) oxide as a catalyst. Give two equations to show how the vanadium(V) oxide acts as a catalyst.

Show answer

V2O5 + SO2 → V2O4 + SO3. 1 mark

2V2O4 + O2 → 2V2O5. 1 mark

The catalyst (V2O5) is regenerated in the second step.

Q17 [3 marks]

Explain how iron acts as a heterogeneous catalyst in the Haber process, and give one factor that affects its efficiency.

Show answer

It is in a different phase (solid) from the gaseous reactants. 1 mark

The reactants adsorb onto active sites on the iron surface, react, and the product (NH3) desorbs. 1 mark

Efficiency depends on the surface area (a support medium is used to maximise it) and is reduced by poisoning, when impurities block the active sites. 1 mark

Q18 [6 marks]

A student determines the concentration of ethanedioate ions in an acidified solution by titration with potassium manganate(VII) solution.

2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O

The mixture is warmed before the addition of potassium manganate(VII) solution because the reaction is slow at first. When more potassium manganate(VII) solution is added, the mixture goes colourless quickly due to the presence of an autocatalyst. Explain the meaning of the term autocatalyst. Explain, using equations where appropriate, why the reaction is slow at first and then goes quickly.

Show answer

Autocatalysis: a product of the reaction catalyses the reaction. 1 mark

Slow at first: the reacting ions are both negative, so they repel each other. 1 mark

This gives a high activation energy. 1 mark

Once the catalyst is present there is attraction between oppositely charged ions: the negative reactant ions and the positive Mn2+ / Mn3+. 1 mark

4Mn2+ + MnO4− + 8H+ → 5Mn3+ + 4H2O 1 mark

2Mn3+ + C2O42− → 2Mn2+ + 2CO2 1 mark

“A reactant catalyses the reaction” is not accepted for the first mark. “The catalyst reduces the activation energy” is allowed as an alternative for the high-activation-energy mark, but not for the attraction mark. State symbols are ignored in the equations.

Q19 [3 marks]

The reaction between peroxodisulfate ions and iodide ions in aqueous solution can be catalysed by Co2+ ions.

S2O82− + 2I− → 2SO42− + I2

The table below gives relevant standard electrode potentials. Use the electrode potential data to suggest how Co2+ catalyses the reaction.

Electrode half-equationE⦵ / V
S2O82−(aq) + 2e− → 2SO42−(aq)+2.01
Co3+(aq) + e− → Co2+(aq)+1.82
I2(aq) + 2e− → 2I−(aq)+0.54
Show answer

The cobalt interchanges between Co2+ and Co3+ and back to Co2+. 1 mark

E⦵ for S2O82−/SO42− is greater than E⦵ for Co3+/Co2+ (+2.01 V > +1.82 V), so the S2O82− ions oxidise Co2+: 2Co2+ + S2O82− → 2Co3+ + 2SO42−, Ecell = +0.19 V. 1 mark

E⦵ for Co3+/Co2+ is greater than E⦵ for I2/I− (+1.82 V > +0.54 V), so the Co3+ ions oxidise I−: 2Co3+ + 2I− → 2Co2+ + I2, Ecell = +1.28 V. 1 mark

Either the E⦵ comparison or the equation with its Ecell value scores. If neither of the last two marks is earned, one mark is available for stating both that Co2+ reduces S2O82− and that Co3+ oxidises I−, but not with a negative Ecell value.

Source: AQA A-Level Chemistry past papers.

Two habits for every question

Every transition-metal question rewards the same two habits: state the five facts for any complex, and run the colour chain for any colour change.

✅ The two methods
  • For any complex, state, in order: metal oxidation state → overall charge → co-ordination number → shape → ligand → any isomerism.
  • For any colour change, identify what changed (ligand, oxidation state or co-ordination number) → it alters ΔE → a different wavelength is absorbed → a different colour is seen.
⚠️ Precision points
  • Examiner reports flag: define a complex (not a transition metal); put the charge on every complex ion and the correct ligand count; do not confuse cis–trans with optical isomerism.
  • For colour, the light is absorbed and the rest reflected/transmitted, it is not emitted.
  • In redox-titration calculations, write next to every moles step which species it is and which volume it refers to (e.g. “n(Fe2+) in 25.0 cm3”), and use the correct mole ratio.

🎯 Capstone quiz: six past-paper questions

Six real AQA multiple-choice questions on this topic, in the style that opens Paper 3. Pick one answer each. The reasoning appears once you commit.

Q1 [1 mark]

Which will not act as a ligand in the formation of a complex ion?

  1. CH4
  2. CO
  3. H2O
  4. NH3
A ligand must own a lone pair to donate into a vacant metal orbital. That is what makes the co-ordinate bond. In methane all four of carbon’s electron pairs are tied up in C–H bonds, so there is nothing to donate. CO is the tempting choice because its carbon looks fully bonded too, but it does carry a lone pair, which is exactly why carbon monoxide binds to the iron in haemoglobin.
Q2 [1 mark]

Cobalt(II) chloride solution changes colour when an excess of concentrated hydrochloric acid is added. What type of reaction takes place?

  1. hydrolysis
  2. ligand substitution
  3. precipitation
  4. redox
The chloride ions replace the water ligands: pink [Co(H2O)6]2+ becomes blue [CoCl4]2−: a change of ligand and of co-ordination number. Redox is the tempting answer because the colour change looks so dramatic, but cobalt is +2 on both sides: no electrons are transferred, so nothing is oxidised or reduced.
Q3 [1 mark]

Which shows the electron configuration of an atom of a transition metal?

  1. [Ar] 4s23d0
  2. [Ar] 4s23d8
  3. [Ar] 4s23d10
  4. [Ar] 4s23d104p1
A transition metal is a d-block element that forms at least one stable ion with a partially filled d sub-shell. [Ar] 4s23d8 is nickel, and Ni2+ is 3d8: partially filled, so it qualifies. The trap is [Ar] 4s23d10: that is zinc, which sits in the d block but only ever forms Zn2+ (3d10, completely full), so it is not a transition metal.
Q4 [1 mark]

Which statement is not correct?

  1. CuCl42− is square planar.
  2. NH4+ is tetrahedral.
  3. [Co(H2NCH2CH2NH2)3]2+ is octahedral.
  4. [Fe(H2O)6]2+ is octahedral.
Chloride is a bulky ligand, so only four fit round the metal, and four-co-ordinate chloro complexes such as CuCl42− are tetrahedral, not square planar, which is why that is the incorrect statement. The ethane-1,2-diamine complex tempts people because the ligand looks huge, but it is bidentate: three of them supply six donor atoms, giving six co-ordination and an octahedral shape.
Q5 [1 mark]

Which process does not involve a heterogeneous catalyst?

  1. catalytic cracking of alkanes
  2. Contact process
  3. decomposition of ozone
  4. Haber process
Heterogeneous means the catalyst is in a different phase from the reactants. Ozone in the upper atmosphere is broken down by gaseous radicals such as Cl• and NO: catalyst and reactants both gases, so that catalysis is homogeneous. Cracking is the tempting distractor, but the zeolite there is a solid acting on vaporised alkanes, which is heterogeneous, exactly like the iron in the Haber process and the V2O5 in the Contact process.
Q6 [1 mark]

Which equation does not show the reduction of a transition metal?

  1. TiCl4 + 2Mg → Ti + 2MgCl2
  2. 2FeCl3 + 2KI → 2FeCl2 + 2KCl + I2
  3. MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
  4. CoO + 4HCl → [CoCl4]2− + H2O + 2H+
Work out the metal’s oxidation state on each side. Cobalt is +2 in CoO and still +2 in [CoCl4]2− (four Cl− ligands, overall 2−), so no electrons move. This is ligand chemistry dressed up as a redox equation. The MnO2 line is the one that catches people out: it looks like chloride is simply being swapped in, but manganese goes from +4 to +2 while chloride is oxidised to Cl2.

Source: AQA A-Level Chemistry past papers.

🧪 Capstone question
Q20 [3 marks]

Excess concentrated hydrochloric acid is added to a pale blue solution of [Cu(H2O)6]2+, turning it yellow. Name the type of reaction, give the formula and shape of the new complex, and explain in terms of ΔE why the colour changes.

Show answer

Ligand substitution: [Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O; the product is tetrahedral (co-ordination number 4). 2 marks

Changing the ligand (and co-ordination number) changes ΔE, so a different wavelength of light is absorbed and a different colour (yellow) is transmitted/reflected. 1 mark

Source: AQA A-Level Chemistry past papers.

🔗 Next, put this to work identifying ions: the reactions of metal ions in aqueous solution (3.2.6): their acidity and the coloured precipitates they form with bases. These are the observations behind Required practical 11.

📋 3.2.5 Transition metals: Quick-reference summary
  • Transition metal: d-block element (Ti–Cu) that forms at least one ion with an incomplete d sub-level (Sc and Zn do not). Properties: complexes, colour, variable oxidation states, catalysis.
  • Shapes: octahedral (6, H2O/NH3), tetrahedral (4, Cl−), square planar (4, cisplatin), linear (2, [Ag(NH3)2]+). Cis–trans (octahedral & square planar); optical (octahedral with bidentate ligands).
  • Substitution: NH3/H2O swap with no change of co-ordination number; larger Cl− changes it (6→4). Bidentate/multidentate ligands replace monodentate ones: the chelate effect, driven by a positive entropy change.
  • Colour: d electrons absorb visible light and are promoted (a d–d transition); ΔE = hν = hc/λ. The colour seen is the light not absorbed. Changing ligand, oxidation state or co-ordination number changes ΔE and so the colour.
  • Redox: vanadium goes VO2+ (yellow) → VO2+ (blue) → V3+ (green) → V2+ (violet) on reduction by zinc. Redox titrations use MnO4− (self-indicating, colourless→pink).
  • Catalysis: heterogeneous (different phase, active sites: V2O5 Contact, Fe Haber) and homogeneous (same phase, via an intermediate: Fe2+ for S2O82−/I−; Mn2+ autocatalysis). Variable oxidation states make it possible.

Understand this topic

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All 10 lessons in this topic
  • Transition metals and complex ions
  • Shapes and isomerism of complex ions
  • Ligand substitution
  • Multidentate ligands and the chelate effect
  • Colour in transition-metal ions
  • Colorimetry
  • Variable oxidation states
  • Redox titration calculations
  • Heterogeneous catalysis
  • Homogeneous catalysis and autocatalysis

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Reactions of Ions in Aqueous Solution

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3.1.11 · feasibility from electrode potentials

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Related topics: Oxidation, reduction & redox

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