OCR A Level Chemistry A: Transition elements (H432)

This study guide teaches section 5.3.1, Transition elements, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. It covers every lettered outcome, 5.3.1(a) to (l), with its additional guidance. H432 is the full A level. Per the assessment overview, Module 5 content is assessed in component 01 and in component 03, Unified chemistry. No PAG is attached to 5.3.1; its test-tube reactions feed 5.3.2, which carries PAG4.

Course links: H432 chemistry hub, printable list of outcomes, free diagnostic quizzes.

Coverage of 5.3.1

Statement	Skill required
5.3.1(a)	Write sub-shell electron configurations for Sc–Zn atoms and ions
5.3.1(b)	Explain why Ti–Cu are transition elements
5.3.1(c)(i)–(iii)	Illustrate variable oxidation states, coloured ions and catalysis, using at least two elements
5.3.1(d)	Define and use “ligand”, including bidentate ligands
5.3.1(e)(i)–(ii)	Use “complex ion” and “coordination number”; give examples of each shape (M4.1, M4.2)
5.3.1(f)(i)–(ii)	Show cis–trans and optical isomerism in complexes with 3-D diagrams (M4.1–M4.3; HSW8)
5.3.1(g)	Describe cis-platin as an anti-cancer drug (HSW9)
5.3.1(h)(i)–(ii)	Write ligand substitution equations and colour changes for Cu²⁺ and Cr³⁺
5.3.1(i)	Explain the role of iron in haemoglobin, including O₂ and CO
5.3.1(j)(i)–(ii)	Give precipitation and complex formation with NaOH(aq) and NH₃(aq)
5.3.1(k)(i)–(iii)	Describe Fe, Cr and Cu redox interconversions with colours
5.3.1(l)	Interpret and predict unfamiliar reactions

5.3.1(a): Electron configurations

The 4s sub-shell fills before 3d, and 4s electrons are also lost first when an ion forms. Use full sub-shell notation, for example Fe: 1s²2s²2p⁶3s²3p⁶3d⁶4s².

Two atoms break the simple pattern:

• Cr is 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ (3d half full)
• Cu is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ (3d completely full)

Worked example 1

Give the configurations of V²⁺ and Cr³⁺.

1. V (Z = 23) is 1s²2s²2p⁶3s²3p⁶3d³4s². Remove both 4s electrons: V²⁺ is 1s²2s²2p⁶3s²3p⁶3d³.
2. Cr (Z = 24) is …3d⁵4s¹. Remove the 4s electron, then two 3d electrons: Cr³⁺ is 1s²2s²2p⁶3s²3p⁶3d³.

5.3.1(b): Which d-block elements count

OCR defines transition elements as d-block elements with an ion whose d-sub-shell is incomplete. In Period 4 that means Ti to Cu.

• Sc³⁺, scandium’s only ion, is [Ar] with an empty 3d sub-shell, so Sc is excluded.
• Zn²⁺, zinc’s only ion, is [Ar]3d¹⁰ with ten d electrons, so Zn is excluded too.
• Copper qualifies because Cu²⁺ is [Ar]3d⁹, even though Cu⁺ is 3d¹⁰.

5.3.1(c): Three typical properties

Illustrate each property with at least two elements. No detail of how colour arises is needed.

(i) More than one oxidation state. Manganese shows +2 (Mn²⁺), +4 (MnO₂) and +7 (MnO₄⁻). Vanadium shows +2, +3, +4 (VO²⁺) and +5 (VO₂⁺).

(ii) Coloured ions. In aqueous solution: Cu²⁺ is blue, Fe²⁺ pale green, Fe³⁺ yellow to orange-brown, Mn²⁺ very pale pink, MnO₄⁻ purple, Cr₂O₇²⁻ orange.

(iii) Catalysis. The specification’s practical examples:

• Cu²⁺ ions speed up the reaction of zinc with acid: Zn + 2H⁺ → Zn²⁺ + H₂.
• MnO₂ speeds up H₂O₂ breakdown: 2H₂O₂ → 2H₂O + O₂.

Industry uses iron in the Haber process and vanadium(V) oxide in the Contact process. As 3.2.2(d) explains, catalysts allow lower temperatures and so cut fossil-fuel use. No process detail is required. HSW9 weighs this against the toxicity of many transition metals.

5.3.1(d) and (e): Ligands, complexes and shapes

A ligand donates a lone pair of electrons to a central metal ion (or atom); it may be a molecule or an ion. The bond formed is coordinate (dative covalent). A complex ion is a central metal ion surrounded by ligands. The coordination number is the number of coordinate bonds to the central ion, not the number of ligands.

• Monodentate ligands use one lone pair: H₂O, Cl⁻, NH₃.
• Bidentate ligands use two: ethane-1,2-diamine, NH₂CH₂CH₂NH₂ (“en”), bonds through both nitrogen atoms.

So [Ni(en)₃]²⁺ has three ligands but a coordination number of 6. Other ligands may be introduced in exams.

Coordination number	Shape	Bond angle	Examples from the specification
6	octahedral	90°	[Cu(H₂O)₆]²⁺, [Fe(H₂O)₆]³⁺
4	tetrahedral	109.5°	[CuCl₄]²⁻, [CoCl₄]²⁻
4	square planar	90°	Pt(NH₃)₂Cl₂

Chloride ions are larger than water molecules, so only four fit around Cu²⁺ or Co²⁺. Square planar is typical of platinum complexes.

Worked example 2

For [Cr(en)₂Cl₂]⁺, find the oxidation number of Cr and the coordination number.

1. en is neutral and each Cl⁻ is −1. Oxidation number of Cr = +1 − (2 × −1) = +3.
2. Each en gives two bonds, each Cl⁻ one: 2 × 2 + 2 = 6, so the shape is octahedral.

5.3.1(f): Stereoisomerism in complexes

You must draw 3-D diagrams: wedges and dashed bonds for octahedral complexes, a flat square for square planar ones.

(i) Cis–trans isomerism. In square planar Pt(NH₃)₂Cl₂, the two Cl ligands can sit next to each other at 90° (cis) or opposite each other at 180° (trans). Octahedral complexes of type [MA₄B₂] show the same pattern, for example [Co(NH₃)₄Cl₂]⁺. Tetrahedral complexes cannot show cis–trans isomerism, because every position is adjacent to every other.

(ii) Optical isomerism. A complex with no plane of symmetry has a non-superimposable mirror image. [Ni(en)₃]²⁺ is the specification’s example: the three en ligands wrap around nickel like a propeller, either left-handed or right-handed. The cis form of [M(en)₂X₂] is also chiral; trans-[M(en)₂X₂] is not, because a mirror plane passes through it.

5.3.1(g): Cis-platin

Cis-platin is the cis isomer of Pt(NH₃)₂Cl₂, used as an anti-cancer drug. Inside a cell, the chloride ligands are replaced. The platinum then binds to nitrogen atoms on bases in DNA, usually two neighbouring guanines. This distorts the DNA so it cannot be copied properly, which prevents cell division. The trans isomer cannot form the same link.

HSW9: the drug kills cancer cells but also attacks healthy dividing cells, giving unpleasant side effects such as nausea, hair loss and kidney damage.

5.3.1(h): Ligand substitution

Use complexed formulae in these equations, as the guidance says.

(i) Copper(II) with ammonia. Excess aqueous ammonia replaces four of the six water ligands:

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

Pale blue solution → deep blue solution. A pale blue precipitate forms first (see (j)).

Copper(II) with chloride. Concentrated hydrochloric acid supplies Cl⁻:

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

Blue → yellow, often passing through green as both complexes are present. Coordination number falls from 6 to 4 and the shape changes from octahedral to tetrahedral.

(ii) Chromium(III) with ammonia.

[Cr(H₂O)₆]³⁺ + 6NH₃ → [Cr(NH₃)₆]³⁺ + 6H₂O

Violet (often seen as green in the lab) → purple, again after a precipitate first forms.

5.3.1(i): Iron in haemoglobin

Haemoglobin contains Fe²⁺ in a haem group. The Fe²⁺ is octahedral: four ring nitrogen atoms in a plane, one nitrogen from the protein, and a sixth site where O₂ binds as a ligand in the lungs and is released in tissues.

Carbon monoxide can replace O₂ at this site. It binds much more strongly and is not readily released, so that haemoglobin can no longer carry oxygen. This is why CO is toxic.

5.3.1(j): Precipitation with NaOH(aq) and NH₃(aq)

Hydroxide ions remove protons from water ligands, giving a neutral hydroxide that precipitates. Aqueous ammonia does the same at first, because NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

Ion	First drops of either reagent	Plenty of NaOH(aq)	Plenty of NH₃(aq)
Cu²⁺	pale blue ppt, Cu(OH)₂	no change	dissolves: deep blue solution, [Cu(NH₃)₄(H₂O)₂]²⁺
Fe²⁺	green ppt, Fe(OH)₂; turns orange-brown at the surface on standing in air	no change	no change
Fe³⁺	orange-brown ppt, Fe(OH)₃	no change	no change
Mn²⁺	light brown ppt, Mn(OH)₂; darkens on standing in air	no change	no change
Cr³⁺	grey-green ppt, Cr(OH)₃	dissolves: dark green solution, [Cr(OH)₆]³⁻	dissolves: purple solution, [Cr(NH₃)₆]³⁺

Either formula style is acceptable for precipitation, for example Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s). Complex formation:

• Cr(OH)₃ + 3OH⁻ → [Cr(OH)₆]³⁻
• Cr(OH)₃ + 6NH₃ → [Cr(NH₃)₆]³⁺ + 3OH⁻
• Cu(OH)₂ + 4NH₃ + 2H₂O → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2OH⁻

5.3.1(k): Redox reactions

You do not have to recall these equations, but you must build them from half-equations and oxidation numbers (5.2.3).

(i) Fe²⁺ ⇌ Fe³⁺. Acidified manganate(VII) oxidises Fe²⁺: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Purple is decolourised; pale green turns yellow. Iodide reduces Fe³⁺: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, and the solution turns brown as iodine forms.

(ii) Cr³⁺ ⇌ Cr₂O₇²⁻. Hot alkaline hydrogen peroxide oxidises Cr(III) to chromate(VI): 2Cr³⁺ + 3H₂O₂ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O (green → yellow). Acidifying converts chromate to dichromate: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O (yellow → orange). Zinc and acid reduce dichromate: Cr₂O₇²⁻ + 14H⁺ + 3Zn → 2Cr³⁺ + 7H₂O + 3Zn²⁺ (orange → green).

(iii) Copper. Iodide reduces Cu²⁺: 2Cu²⁺ + 4I⁻ → 2CuI + I₂, giving a white precipitate of copper(I) iodide in a brown solution. In water, Cu⁺ disproportionates: 2Cu⁺ → Cu²⁺ + Cu. Copper(I) oxide in dilute sulfuric acid gives a blue solution and a red-brown solid.

Worked example 3: building and using an equation

A bottle of FeSO₄·7H₂O (Mr = 277.9) has partly oxidised in air. A 3.40 g sample goes into dilute H₂SO₄ in a 100.0 cm³ volumetric flask, filled to the mark. Titrating 10.0 cm³ aliquots needs a mean 17.85 cm³ of 0.0125 mol dm⁻³ KMnO₄. Find the percentage of FeSO₄·7H₂O left.

1. Half-equations: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻. Multiply the iron one by 5, so the ratio MnO₄⁻ : Fe²⁺ is 1 : 5.
2. n(MnO₄⁻) = 0.0125 × 0.01785 = 2.231 × 10⁻⁴ mol.
3. n(Fe²⁺) in 10.0 cm³ = 5 × 2.231 × 10⁻⁴ = 1.116 × 10⁻³ mol; in 100.0 cm³ = 1.116 × 10⁻² mol.
4. Mass = 1.116 × 10⁻² × 277.9 = 3.10 g. Percentage = 3.10 ÷ 3.40 × 100 = 91.2%.

Iron(III) already present does not react, so only unoxidised salt is counted.

5.3.1(l): Unfamiliar reactions

For a metal or ligand you have not met, apply these rules:

1. Find the oxidation number of the metal from the overall charge.
2. Count coordinate bonds (bidentate = 2 each) to get the coordination number and shape.
3. Ligand substitution: the oxidation number does not change. Balance ligands, waters and charge.
4. Precipitation: the charge on the ion equals the number of OH⁻ needed for a neutral hydroxide.
5. Redox: if the oxidation number changes, write half-equations and balance electrons.

For example, cobalt(II) turning from pink to blue in concentrated HCl matches the copper pattern: [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O.

Common errors

• Removing 3d electrons before 4s.
• Calling Zn a transition element because it sits in the d-block.
• Giving the number of ligands as the coordination number for en complexes.
• Drawing [CuCl₄]²⁻ as octahedral, or forgetting the shape change.
• Saying excess NaOH redissolves Cu(OH)₂; chromium(III) hydroxide is the one that redissolves.

Next steps

• Condensed tables and a self-test: Transition elements revision notes
• Original questions with mark-by-mark answers: Transition elements practice questions
• Oxidation numbers and half-equations: Redox
• Catalysts and activation energy: Reaction rates

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 5.3.1 Transition elements, learning outcomes (a) to (l) with their additional guidance. Published by OCR.