AQA A-Level Chemistry: Transition metals (7405)

This study guide covers section 3.2.5, Transition metals, of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. The whole of 3.2.5 is marked “A-level only”, so it is not part of AS. The specification lists inorganic chemistry (section 3.2) under A-level Paper 1, and Paper 3 can assess any content. It lists no required practical; required practical 11 belongs to section 3.2.6.

Use it with the revision notes and the practice questions. Redox titrations build on oxidation, reduction and redox equations and electrode potentials; the chelate effect uses ΔG = ΔH − TΔS from thermodynamics. For the rest of the course, see the AQA A-Level Chemistry hub and the printable checklist. To find your weak spots quickly, try the free diagnostics.

What this section covers

Spec section	What you must be able to do	Status
3.2.5.1 General properties	Link Ti–Cu properties to an incomplete d sub-level; define ligand, complex, co-ordination number	A-level only
3.2.5.2 Substitution reactions	Exchange of H₂O, NH₃ and Cl⁻; bidentate and multidentate ligands; haem and CO; explain the chelate effect using ΔH and ΔS	A-level only
3.2.5.3 Shapes of complex ions	Octahedral, tetrahedral, square planar and linear complexes; cis–trans and optical isomerism; cisplatin; Tollens’ reagent	A-level only
3.2.5.4 Formation of coloured ions	Explain colour; use ΔE = hν = hc/λ; why ΔE changes; colorimetry	A-level only
3.2.5.5 Variable oxidation states	Reduce vanadate(V) with zinc; effect of pH and ligand on redox potential; Tollens’ test; MnO₄⁻ titrations and calculations	A-level only
3.2.5.6 Catalysts	Heterogeneous and homogeneous catalysis; supports and poisoning; V₂O₅, Fe²⁺ and Mn²⁺ mechanisms with equations	A-level only

3.2.5.1 General properties

The elements titanium to copper have an incomplete d sub-level in their atoms or ions. That is the origin of their four characteristic properties:

• complex formation
• formation of coloured ions
• variable oxidation state
• catalytic activity.

Learn the three definitions word for word:

• Ligand: a molecule or ion that forms a co-ordinate bond with a transition metal by donating a pair of electrons.
• Complex: a central metal atom or ion surrounded by ligands.
• Co-ordination number: the number of co-ordinate bonds to the central metal atom or ion.

Worked example 1: is it a transition element?

Cu²⁺ is [Ar] 3d⁹: an incomplete d sub-level, so copper shows transition metal properties. Zinc’s only ion, Zn²⁺, is [Ar] 3d¹⁰, and the atom is [Ar] 3d¹⁰ 4s², so zinc does not. Ions lose 4s electrons before 3d.

3.2.5.2 Substitution reactions

H₂O, NH₃ and Cl⁻ are monodentate ligands: each forms one co-ordinate bond.

NH₃ and H₂O are similar in size and uncharged, so they swap without a change of co-ordination number:

[Co(H₂O)₆]²⁺ + 6NH₃ → [Co(NH₃)₆]²⁺ + 6H₂O
[Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

The copper reaction (with excess ammonia) shows incomplete substitution: four of the six water ligands are replaced, and the co-ordination number stays 6. The solution changes from blue to deep blue.

Cl⁻ is larger than NH₃ and H₂O, so fewer fit round the metal ion. Exchange of H₂O by Cl⁻ (for example with concentrated hydrochloric acid) can change the co-ordination number from 6 to 4:

[Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O     (blue → yellow)
[Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O     (pink → blue)
[Fe(H₂O)₆]³⁺ + 4Cl⁻ → [FeCl₄]⁻ + 6H₂O

Bidentate and multidentate ligands

• Bidentate ligands form two co-ordinate bonds: ethane-1,2-diamine, H₂NCH₂CH₂NH₂ (often written “en”), and ethanedioate, C₂O₄²⁻.
• Multidentate ligands form several: EDTA⁴⁻ forms six.

Haem is an iron(II) complex with a multidentate ligand. In haemoglobin, oxygen forms a co-ordinate bond to Fe(II), so oxygen can be carried in the blood. Carbon monoxide is toxic because it replaces the oxygen co-ordinately bonded to Fe(II), so less oxygen is transported.

The chelate effect

Bidentate and multidentate ligands replace monodentate ligands from complexes. This is the chelate effect. Explain it with the balance of enthalpy and entropy:

• ΔH is small: similar numbers of similar co-ordinate bonds are broken and made.
• ΔS is positive: the number of particles in solution increases.
• So ΔG = ΔH − TΔS is negative and the reaction is feasible.

Worked example 2: counting particles and finding ΔG

[Ni(H₂O)₆]²⁺ + 3H₂NCH₂CH₂NH₂ → [Ni(H₂NCH₂CH₂NH₂)₃]²⁺ + 6H₂O

1. Particles: 1 + 3 = 4 on the left; 1 + 6 = 7 on the right. Disorder increases, so ΔS is positive.
2. Illustrative data for this example (not measured values): ΔH = −8.0 kJ mol⁻¹, ΔS = +110 J K⁻¹ mol⁻¹.
3. Convert ΔS to kJ: +0.110 kJ K⁻¹ mol⁻¹.
4. At 298 K: ΔG = −8.0 − (298 × 0.110) = −40.8 kJ mol⁻¹. Negative, so feasible.

With EDTA⁴⁻, [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O goes from 2 particles to 7.

3.2.5.3 Shapes of complex ions

Shape	When it forms	Bond angle	Example
Octahedral	Small ligands such as H₂O and NH₃	90°	[Cu(H₂O)₆]²⁺
Tetrahedral	Larger ligands such as Cl⁻	109.5°	[CuCl₄]²⁻
Square planar	Some four-co-ordinate complexes	90°	cisplatin, [Pt(NH₃)₂Cl₂]
Linear	Ag⁺ with NH₃	180°	[Ag(NH₃)₂]⁺ (Tollens’ reagent)

Isomerism (draw the complexes in 3D with wedges):

• Octahedral, monodentate ligands: cis–trans isomerism, a special case of E–Z isomerism. In [Co(NH₃)₄Cl₂]⁺ the two Cl⁻ ligands are either at 90° (cis) or at 180° (trans).
• Octahedral, bidentate ligands: optical isomerism. [Ni(H₂NCH₂CH₂NH₂)₃]²⁺ exists as two non-superimposable mirror images.
• Square planar: cis–trans isomerism. Cisplatin is the cis isomer: both Cl⁻ ligands are on the same side, at 90°.

Worked example 3: co-ordination number and charge

[Cr(C₂O₄)₃]³⁻: three bidentate ligands make 3 × 2 = 6 co-ordinate bonds, so it is octahedral. Each C₂O₄²⁻ carries −2, so Cr + 3(−2) = −3, giving Cr = +3. Count bonds, not ligands.

3.2.5.4 Formation of coloured ions

Transition metal ions can be identified by their colour. Colour arises when some wavelengths of visible light are absorbed and the remaining wavelengths are transmitted or reflected. When light is absorbed, d electrons move from the ground state to an excited state. The energy gap is:

ΔE = hν = hc/λ

where h is the Planck constant, ν the frequency and λ the wavelength. Changes in oxidation state, co-ordination number and ligand alter ΔE, so the wavelength absorbed changes and so does the colour. That is why [Co(H₂O)₆]²⁺ (pink, octahedral) and [CoCl₄]²⁻ (blue, tetrahedral) differ.

Worked example 4: from wavelength to energy

A complex absorbs most strongly at 620 nm. Take h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and L = 6.02 × 10²³ mol⁻¹ (use the values a question gives).

1. λ = 620 × 10⁻⁹ m.
2. ΔE = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (620 × 10⁻⁹) = 3.21 × 10⁻¹⁹ J per ion.
3. Per mole: 3.21 × 10⁻¹⁹ × 6.02 × 10²³ = 1.93 × 10⁵ J = 193 kJ mol⁻¹.

Colorimetry

Absorption of visible light is used in spectroscopy; a simple colorimeter measures it.

1. Choose a filter that gives the colour of light the solution absorbs most.
2. Zero the colorimeter with a blank (the solvent only).
3. Measure the absorbance of several standard solutions of known concentration and plot absorbance against concentration (the calibration graph).
4. Measure the unknown under the same conditions and read its concentration from the graph.

Worked example 5: calibration graph

Copper(II) standards give a straight line through the origin with gradient 12 dm³ mol⁻¹ (absorbance 0.12 at 0.010 mol dm⁻³). An unknown reads 0.42.

1. c = 0.42 ÷ 12 = 0.0350 mol dm⁻³.
2. If the unknown was 250 cm³ of solution: n = 0.0350 × 0.250 = 8.75 × 10⁻³ mol; mass of Cu = 8.75 × 10⁻³ × 63.5 = 0.556 g.

3.2.5.5 Variable oxidation states

Transition elements show variable oxidation states. Zinc in acidic solution reduces vanadate(V) step by step:

Species	Oxidation state of V	Colour
VO₂⁺	+5	yellow
VO²⁺	+4	blue
V³⁺	+3	green
V²⁺	+2	violet

VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O
VO²⁺ + 2H⁺ + e⁻ → V³⁺ + H₂O
V³⁺ + e⁻ → V²⁺
Zn → Zn²⁺ + 2e⁻

Green appears briefly between yellow and blue (a mixture of both).

pH and ligand change the redox potential for a metal ion going from a higher to a lower oxidation state. Half-equations such as MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O contain H⁺, so their position depends on pH. In alkaline conditions, higher oxidation states form more easily: cobalt(II) in excess ammonia is oxidised by air to cobalt(III), but [Co(H₂O)₆]²⁺ is not.

Tollens’ reagent: [Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃. Aldehydes reduce it to a silver mirror; ketones do not, so it distinguishes them.

Redox titrations with manganate(VII)

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺           (1 : 5)
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂      (2 : 5)

MnO₄⁻ is purple and Mn²⁺ is almost colourless, so no indicator is needed: the end point is the first permanent pale pink. Use dilute sulfuric acid. Hydrochloric acid would be oxidised to chlorine, raising the titre.

Worked example 6: iron in a tablet

An iron(II) tablet is dissolved in dilute sulfuric acid and made up to 100 cm³. A 25.0 cm³ portion needs 11.60 cm³ of 0.00500 mol dm⁻³ KMnO₄. Ar of Fe = 55.8.

1. n(MnO₄⁻) = 0.00500 × 11.60/1000 = 5.80 × 10⁻⁵ mol.
2. n(Fe²⁺) in 25.0 cm³ = 5 × 5.80 × 10⁻⁵ = 2.90 × 10⁻⁴ mol.
3. In 100 cm³: × 4 = 1.16 × 10⁻³ mol.
4. Mass = 1.16 × 10⁻³ × 55.8 = 0.0647 g = 64.7 mg of iron per tablet.

The same steps give iron in steel, the Mr of a hydrated salt, or [H₂O₂] (2MnO₄⁻ : 5H₂O₂).

3.2.5.6 Catalysts

• A heterogeneous catalyst is in a different phase from the reactants; the reaction occurs at active sites on the surface. A support medium spreads the catalyst thinly to maximise surface area and minimise cost. Impurities can poison the catalyst by blocking active sites, reducing efficiency, which has a cost implication.
• A homogeneous catalyst is in the same phase as the reactants; the reaction proceeds through an intermediate species.
• Variable oxidation states matter because the catalyst can gain and lose electrons, giving a route with a lower activation energy, and is then regenerated.

V₂O₅ in the Contact process (heterogeneous):

SO₂ + V₂O₅ → SO₃ + V₂O₄         (V: +5 → +4)
V₂O₄ + ½O₂ → V₂O₅               (V: +4 → +5)

Fe in the Haber process is also heterogeneous.

Fe²⁺ with I⁻ and S₂O₈²⁻ (homogeneous). Uncatalysed, two negative ions repel, so the activation energy is high:

S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

Each step is between oppositely charged ions. Fe³⁺ works too, with the steps in the other order.

Mn²⁺ autocatalyses C₂O₄²⁻ and MnO₄⁻. Mn²⁺ is a product, so the reaction starts slowly, speeds up as Mn²⁺ forms, then slows as reactants run out:

4Mn²⁺ + MnO₄⁻ + 8H⁺ → 5Mn³⁺ + 4H₂O
2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂

Common errors

• Giving co-ordination number as the number of ligands for en, C₂O₄²⁻ or EDTA⁴⁻.
• Saying colour comes from light emitted. Light is absorbed; you see what is transmitted or reflected.
• Explaining the chelate effect with bond strength alone. You must say ΔS is positive because particles increase.
• Using 1 : 1 for MnO₄⁻ : Fe²⁺. It is 1 : 5.
• Calling [CuCl₄]²⁻ square planar. The specification gives Cl⁻ as the example of a larger ligand that forms tetrahedral complexes.

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards (AQA). Section 3.2.5 Transition metals (A-level only).