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.
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.
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
4s fills first — and empties first
Fe is [Ar]3d64s2; forming Fe2+ removes the two 4s electrons first, leaving [Ar]3d6 — a partially filled d sub-level, so Fe2+ qualifies.
- 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-style questions
Which shows the electron configuration of an atom of a transition metal? Tick (✓) one box.
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 number | Shape | Bond angle | Example |
|---|---|---|---|
| 6 | octahedral | 90° | [Fe(H2O)6]2+, [Co(NH3)6]2+ |
| 4 | tetrahedral | 109.5° | [CuCl4]2−, [CoCl4]2− |
| 4 | square planar | 90° | cisplatin [Pt(NH3)2Cl2] |
| 2 | linear | 180° | [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. Octahedral complexes with bidentate ligands (such as [Ni(en)3]2+) show optical isomerism — two non-superimposable mirror images.
Octahedral · 6 bonds · 90°
[Fe(H2O)6]2+ — six small ligands (H2O, NH3) fit around the metal.
Tetrahedral · 4 bonds · 109.5°
[CuCl4]2− — Cl− is larger, so only four fit.
Square planar · 4 bonds · 90°
Cisplatin, [Pt(NH3)2Cl2] — all four ligands in one plane.
Linear · 2 bonds · 180°
[Ag(NH3)2]+ — the complex in Tollens’ reagent.
Cis–trans · square planar
[Pt(NH3)2Cl2]: cis — the two Cl 90° apart; trans — 180° apart. Cisplatin is the cis isomer.
Cis–trans · octahedral
[Cr(NH3)4Cl2]+ — an octahedral MA4B2 complex (the Q6 pair).
Optical isomers · octahedral with bidentate ligands
[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.
Interactive — name the shape, number, angle & ligand
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.
| Complex | Metal ox. state | Overall charge | Co-ord. no. | Shape | Ligand |
|---|---|---|---|---|---|
| [Fe(H2O)6]2+ | +2 | 2+ | 6 | octahedral | H2O |
| [Cu(NH3)4(H2O)2]2+ | +2 | 2+ | 6 | octahedral | NH3, H2O |
| [CoCl4]2− | +2 | 2− | 4 | tetrahedral | Cl− |
| [Pt(NH3)2Cl2] | +2 | 0 | 4 | square planar | NH3, Cl− |
| [Ag(NH3)2]+ | +1 | 1+ | 2 | linear | NH3 |
🧪 Exam-style questions
Which will not act as a ligand in the formation of a complex ion? Tick (✓) one box.
Which statement is not correct? Tick (✓) one box.
Which shows the correct oxidation state and co-ordination number of cobalt in [Co(NH3)5Cl]Cl2? Tick (✓) one box.
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
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.
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+ + 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+ + 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+ + 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.
Interactive — add a reagent, watch the complex & colour change
1 · start with
2 · add
🧪 Exam-style questions
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
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.
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:
| Ion | Colour |
|---|---|
| [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
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
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.
Interactive — what each complex absorbs, and the ΔE behind it
🧪 Exam-style questions
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
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.
| Substitution | From | To | What 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
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 state | Species | Colour |
|---|---|---|
| +5 | VO2+ (vanadate(V)) | yellow |
| +4 | VO2+ | blue |
| +3 | V3+ | green |
| +2 | V2+ | 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
| Run | Initial / cm3 | Final / cm3 | Titre / cm3 |
|---|
🧪 Exam-style questions
Which compound decolourises acidified potassium manganate(VII) solution? Tick (✓) one box.
Which equation does not show the reduction of a transition metal? Tick (✓) one box.
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 = 0.162 ÷ 134.0 = 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 × 10−5 ÷ (23.85 × 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).
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
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+ catalyses S2O82− + 2I− → 2SO42− + I2 through an intermediate (Fe3+): each step is between oppositely charged ions.
🧪 Exam-style questions
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.
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
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.
- 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.
- 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.
Interactive — build the mark-scheme answer
“An excess of concentrated hydrochloric acid is added to a pink solution containing cobalt(II) ions. Give the formula, shape and cobalt oxidation state of the complex formed, and explain why the colour changes.” [5 marks] Select every statement that earns a mark — and nothing that doesn’t. Order doesn’t matter.
🧪 Capstone question
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.
- 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.