OCR A Level Chemistry A: Transition elements (H432) – Revision Notes

These revision notes condense 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. Every lettered outcome, (a) to (l), appears below with its additional guidance. H432 is the full A level, and the assessment overview places Module 5 in components 01 and 03. The Transition elements study guide has the full teaching and worked examples.

When you are ready, try the Transition elements practice questions. Other sections sit on the H432 chemistry hub. Tick off progress using the printable list of outcomes, and use the free diagnostic quizzes to spot weak areas.

Configurations and definition: (a) and (b)

• 4s fills before 3d; 4s empties first when ions form.
• Exceptions: Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹. Write the full form (1s²2s²2p⁶3s²3p⁶…) when asked for sub-shell notation.
• Transition element: a d-block element that forms an ion with an incomplete d-sub-shell. In Period 4 that is Ti to Cu.
• Sc³⁺ has 3d⁰ and Zn²⁺ has 3d¹⁰, so Sc and Zn are d-block but not transition elements.

Method: ion configuration

1. Write the atom (watch Cr and Cu).
2. Remove 4s electrons first.
3. Then remove 3d electrons for any remaining charge.

Quick reminder: Mn (Z = 25) is [Ar]3d⁵4s², so Mn²⁺ is [Ar]3d⁵ and Mn³⁺ is [Ar]3d⁴.

Properties: (c)

Use at least two elements for each property. No explanation of colour is required.

Element	Oxidation states with species and colour
Fe	+2 Fe²⁺ pale green; +3 Fe³⁺ yellow to orange-brown
Cr	+3 Cr³⁺ green (hexaaqua ion violet); +6 Cr₂O₇²⁻ orange, CrO₄²⁻ yellow
Mn	+2 Mn²⁺ very pale pink; +4 MnO₂ black-brown solid; +7 MnO₄⁻ purple
Cu	+1 Cu⁺ compounds such as CuI white; +2 Cu²⁺ blue

Catalysis (no process detail required):

• Cu²⁺ speeds up Zn + acid.
• MnO₂ speeds up 2H₂O₂ → 2H₂O + O₂.
• Industry: Fe (Haber process), V₂O₅ (Contact process). Catalysts lower the temperature needed and so cut energy use (3.2.2(d)). HSW9 risk: many transition metals are toxic.

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

Term	Meaning
Ligand	Molecule or ion that donates an electron pair to a metal ion or atom, forming a coordinate (dative covalent) bond
Monodentate	One coordinate bond: H₂O, Cl⁻, NH₃
Bidentate	Two coordinate bonds: NH₂CH₂CH₂NH₂ (en), through both N atoms
Complex ion	Central metal ion surrounded by coordinately bonded ligands
Coordination number	Number of coordinate bonds to the central metal ion

Coordination number	Shape	Angle	Specification examples
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

Method: oxidation number in a complex

1. Overall charge = metal oxidation number + sum of ligand charges.
2. Neutral ligands (H₂O, NH₃, en) add 0; Cl⁻ and OH⁻ add −1 each.

Quick reminder: in [Fe(CN)₆]⁴⁻, Fe = −4 − (6 × −1) = +2.

Isomerism and cis-platin: (f) and (g)

• Cis–trans: square planar [MA₂B₂] (e.g. Pt(NH₃)₂Cl₂) or octahedral [MA₄B₂]. Cis = same pair at 90°; trans = at 180°.
• Optical: no plane of symmetry, so non-superimposable mirror images. Key example: [Ni(en)₃]²⁺. The cis form of [M(en)₂X₂] is optical too; the trans form is not.
• Tetrahedral complexes do not show cis–trans isomerism.
• Draw in 3-D: wedge and dash bonds, with the en ligand drawn as a curved link spanning two adjacent positions.

Cis-platin: cis-Pt(NH₃)₂Cl₂. Cl⁻ ligands are replaced inside the cell; Pt binds to nitrogen atoms on DNA bases (often two adjacent guanines). The DNA is distorted, so it cannot be copied and the cell cannot divide. HSW9: benefit of chemotherapy against side effects such as nausea, hair loss and kidney damage.

Ligand substitution and haemoglobin: (h) and (i)

Change	Equation (complexed formulae)	Colour	Shape
Cu²⁺ + excess NH₃	[Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O	pale blue → deep blue	stays octahedral
Cu²⁺ + conc. HCl	[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O	blue → (green) → yellow	octahedral → tetrahedral
Cr³⁺ + excess NH₃	[Cr(H₂O)₆]³⁺ + 6NH₃ → [Cr(NH₃)₆]³⁺ + 6H₂O	violet (or green) → purple	stays octahedral

Haemoglobin: Fe²⁺, octahedral. Four N from the haem ring, one N from the globin protein, sixth site for O₂. CO substitutes for O₂, binds far more strongly and is not easily released, so less oxygen is carried.

Precipitation: (j)

Ion	First drops of either	More NaOH	More NH₃
Cu²⁺	pale blue ppt	no change	deep blue solution
Fe²⁺	green ppt (browns in air)	no change	no change
Fe³⁺	orange-brown ppt	no change	no change
Mn²⁺	light brown ppt (darkens in air)	no change	no change
Cr³⁺	grey-green ppt	dark green solution, [Cr(OH)₆]³⁻	purple solution, [Cr(NH₃)₆]³⁺

• Excess NaOH redissolves just one hydroxide, Cr(OH)₃. Excess NH₃ redissolves two: Cr(OH)₃ and Cu(OH)₂.
• Non-complexed or complexed formulae are both acceptable for precipitation, e.g. Mn²⁺(aq) + 2OH⁻(aq) → Mn(OH)₂(s).
• NH₃(aq) gives precipitates because NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

Redox: (k)

Change	Reagent	Colour change	Equation
Fe²⁺ → Fe³⁺	H⁺/MnO₄⁻	purple decolourised; pale green → yellow	MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Fe³⁺ → Fe²⁺	I⁻	yellow → brown (I₂)	2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
Cr³⁺ → CrO₄²⁻	H₂O₂/OH⁻, warm	green → yellow	2Cr³⁺ + 3H₂O₂ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O
CrO₄²⁻ → Cr₂O₇²⁻	acid	yellow → orange	2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O
Cr₂O₇²⁻ → Cr³⁺	Zn/H⁺	orange → green	Cr₂O₇²⁻ + 14H⁺ + 3Zn → 2Cr³⁺ + 7H₂O + 3Zn²⁺
Cu²⁺ → Cu⁺	I⁻	white ppt (CuI) in brown solution	2Cu²⁺ + 4I⁻ → 2CuI + I₂
Cu⁺ → Cu²⁺ + Cu	water (disproportionation)	blue solution, red-brown solid	2Cu⁺ → Cu²⁺ + Cu

You need not recall these equations, but you must be able to build them.

Method: combining half-equations

1. Write each half-equation; balance O with H₂O, H with H⁺, charge with e⁻.
2. Multiply so electrons lost = electrons gained.
3. Add, cancel species on both sides, check atoms and charge.

Worked reminder: copper by iodine release. Cu²⁺ reacts with excess I⁻, and the iodine is titrated with thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Since 2Cu²⁺ gives one I₂, and one I₂ uses 2S₂O₃²⁻, the overall ratio Cu²⁺ : S₂O₃²⁻ is 1 : 1. So a titre containing 3.00 × 10⁻³ mol thiosulfate means 3.00 × 10⁻³ mol Cu²⁺ was present.

Worked reminder: disproportionation. In 2Cu⁺ → Cu²⁺ + Cu, copper goes from +1 to +2 (oxidised) and from +1 to 0 (reduced). Copper is oxidised and reduced at once, which is what disproportionation means.

Unfamiliar reactions: (l)

• Oxidation number unchanged → substitution or precipitation.
• Oxidation number changed → redox: use half-equations.
• Neutral hydroxide needs as many OH⁻ as the ion’s charge.
• Bidentate ligand replaces two monodentate ligands.

Must-know distinctions

• Ligands vs coordination number: [Cu(en)₂(H₂O)₂]²⁺ has four ligands, coordination number 6.
• Substitution vs redox: CuCl₄²⁻ formation keeps Cu at +2; CuI formation reduces Cu to +1.
• Precipitate vs solution: “deep blue precipitate” is wrong; the Cu-ammonia complex is a solution.
• Cis–trans vs optical: trans-[M(en)₂X₂] has a mirror plane; cis does not.

Quick self-test

1. Write the full sub-shell configuration of Co²⁺ (Z = 27).
2. Why is scandium not a transition element?
3. Give the coordination number and shape of [CoCl₄]²⁻.
4. State the oxidation number of Cr in [Cr(OH)₆]³⁻.
5. Which hydroxides dissolve in excess aqueous ammonia?
6. State the colour change when excess concentrated HCl is added to aqueous copper(II) sulfate.
7. How many moles of Fe²⁺ are oxidised by 2.40 × 10⁻⁴ mol MnO₄⁻ in acid?
8. 3.00 × 10⁻³ mol Cu²⁺ reacts with excess I⁻. How many moles of I₂ form?
9. Name the type of isomerism shown by [Ni(en)₃]²⁺.
10. Why is CO toxic?
11. 0.0180 mol Fe³⁺ reacts completely with excess I⁻. How many moles of I₂ form?
12. Give the observation when Mn²⁺(aq) is treated with excess aqueous sodium hydroxide and left to stand.

Answers

1. 1s²2s²2p⁶3s²3p⁶3d⁷
2. Its only ion, Sc³⁺, has no d electrons, so no ion has an incomplete d-sub-shell.
3. 4; tetrahedral.
4. +3
5. Cu(OH)₂ and Cr(OH)₃.
6. Blue → yellow (green when both complexes are present).
7. 5 × 2.40 × 10⁻⁴ = 1.20 × 10⁻³ mol
8. 3.00 × 10⁻³ ÷ 2 = 1.50 × 10⁻³ mol
9. Optical isomerism.
10. CO replaces O₂ on the Fe²⁺ in haemoglobin, binds much more strongly and is not easily released, so less O₂ is carried.
11. 2Fe³⁺ gives one I₂, so 0.0180 ÷ 2 = 9.00 × 10⁻³ mol
12. A light brown precipitate that does not dissolve in excess, and darkens on standing in air.

Where marks are usually lost

• Taking 3d electrons off before 4s, giving Fe²⁺ as [Ar]3d⁴4s².
• Writing the Cr or Cu atom as 3d⁴4s² or 3d⁹4s².
• Quoting the number of en ligands as the coordination number.
• Forgetting the shape change to tetrahedral in [CuCl₄]²⁻.
• Saying Cu(OH)₂ dissolves in excess NaOH.
• Using uncomplexed formulae in ligand substitution equations, which the guidance says should be complexed.
• Describing Fe(OH)₂ as orange-brown straight away; it starts green.
• Leaving out the acidification step when turning chromate into dichromate.
• Missing the white CuI precipitate and describing only the brown colour.

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.