OCR A Level Chemistry A: Transition elements (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set practises 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. Each lettered outcome, 5.3.1(a) to (l), is tested at least once. H432 is the full A level. Questions 8, 9 and 11 are longer, combined problems.

Before starting, work through the study guide and revision notes. For other sections, use the H432 chemistry hub, the outcome checklist to print and the diagnostic quizzes.

Data: Ar values O = 16.0, Cu = 63.5, Zn = 65.4.

Questions

1. This question is about electron configurations.

(a) Give the full sub-shell electron configuration of an Fe³⁺ ion (Z = 26). [1] (b) Give the full sub-shell electron configuration of a copper atom (Z = 29). [1] (c) Explain why copper is a transition element but zinc is not. [2]

2. 2,2’-Bipyridine, “bipy”, is a neutral ligand that bonds through lone pairs on two nitrogen atoms. Iron(II) forms [Fe(bipy)₃]²⁺.

(a) Explain the term ligand. [1] (b) State the oxidation number of iron in this complex. [1] (c) State the coordination number and the shape of the complex. [2] (d) Name the type of stereoisomerism shown by [Fe(bipy)₃]²⁺ and explain why it occurs. [2]

3. Using manganese and one other transition element, illustrate:

(a) more than one oxidation state, with the formula and colour of one species in each state [3] (b) the catalytic behaviour of a transition element compound, with an equation for the catalysed reaction. [2]

4. Cis-platin is used to treat some cancers.

(a) Draw or describe the two stereoisomers of Pt(NH₃)₂Cl₂ and explain why they exist. [2] (b) Describe how cis-platin prevents cancer cells dividing. [3] (c) State one risk of using cis-platin as a drug. [1]

5. Aqueous copper(II) sulfate is split between two tubes.

(a) Excess concentrated hydrochloric acid is added to the first tube. Write an equation using complexed formulae, state the final colour, and describe the change in shape and coordination number. [4] (b) Excess concentrated aqueous ammonia is added to the second tube. Write an equation for the formation of the final complex from the hexaaqua ion and state its colour. [2]

6. Explain how haemoglobin carries oxygen, and why carbon monoxide is toxic. [3]

7. Solutions P, Q and R each contain one of Cu²⁺, Fe²⁺, Fe³⁺, Mn²⁺ or Cr³⁺.

Solution	NaOH(aq), a few drops then excess	NH₃(aq), a few drops then excess
P	green ppt, unchanged in excess; surface browns on standing	green ppt, unchanged in excess
Q	grey-green ppt; dark green solution in excess	grey-green ppt; purple solution in excess
R	light brown ppt, unchanged in excess; darkens on standing	light brown ppt, unchanged in excess

(a) Identify the cation in each solution. [3] (b) Give the ionic equation for P’s precipitate forming, including state symbols. [1] (c) Write equations for the two reactions of Q’s precipitate in excess reagent. [2]

8. A student has 20.0 cm³ of 0.150 mol dm⁻³ Cr³⁺(aq).

(a) Aqueous sodium hydroxide is added dropwise until in excess. Describe what is seen. [2] (b) Hydrogen peroxide is added and the mixture is warmed. The half-equations are:

[Cr(OH)₆]³⁻ + 2OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻ H₂O₂ + 2e⁻ → 2OH⁻

State the colour change and the change in oxidation number of chromium, then construct the overall equation. [3] (c) Calculate the minimum volume, in cm³, of 0.250 mol dm⁻³ H₂O₂ needed to oxidise all the chromium. [2] (d) The yellow solution is then acidified. State the colour change and give the equation. [2] (e) The resulting dichromate(VI) is reduced back to Cr³⁺ with zinc and acid: Cr₂O₇²⁻ + 14H⁺ + 3Zn → 2Cr³⁺ + 7H₂O + 3Zn²⁺. Calculate the minimum mass of zinc needed. [2]

9. 0.305 g of brass (copper and zinc) is dissolved in nitric acid. Excess acid is removed and excess KI(aq) added. The iodine formed reacts with 32.50 cm³ of 0.0950 mol dm⁻³ Na₂S₂O₃(aq).

2Cu²⁺ + 4I⁻ → 2CuI + I₂ I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

(a) Describe what is seen when KI(aq) is added. [2] (b) Calculate the percentage by mass of copper in the brass. Give your answer to 3 significant figures. [4] (c) Suggest why excess nitric acid must be removed before adding iodide. [1]

10. Copper(I) oxide, Cu₂O, reacts with dilute sulfuric acid: Cu₂O + H₂SO₄ → CuSO₄ + Cu + H₂O.

(a) Explain, using oxidation numbers, how this reaction shows disproportionation. [2] (b) State what is seen. [1] (c) Calculate the mass of copper formed from 1.716 g of Cu₂O. [2]

11. Green aqueous [Ni(H₂O)₆]²⁺ gives a violet solution of [Ni(en)₃]²⁺ with excess ethane-1,2-diamine (en), and a green precipitate with aqueous sodium hydroxide. With less en, [Ni(en)₂(H₂O)₂]²⁺ can form.

(a) Write equations, using complexed formulae, for the formation of [Ni(en)₃]²⁺ and of the green precipitate. [2] (b) Classify each reaction in (a) and explain why neither is a redox reaction. [2] (c) [Ni(en)₂(H₂O)₂]²⁺ exists as three stereoisomers. Describe them and the types of isomerism involved. [3]

Answers

1. (a) 1s²2s²2p⁶3s²3p⁶3d⁵ [1] (b) 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ [1] (c) Cu forms Cu²⁺, 3d⁹, which has an incomplete d-sub-shell [1]. Zn forms only Zn²⁺, 3d¹⁰, a full d-sub-shell, so no ion has an incomplete d-sub-shell [1] Examiner insight: The definition concerns an ion, so answers about the atom’s configuration miss the point.

2. (a) A species with a lone pair that it gives to a central metal ion, forming a coordinate bond [1] (b) +2 [1] (c) Coordination number 6 (three bidentate ligands) [1]; octahedral [1] (d) Optical isomerism [1]. The complex has no plane of symmetry, so it has non-superimposable mirror images [1] Examiner insight: Each bidentate ligand forms two coordinate bonds, so 3 is wrong.

3. (a) Mn: +2, Mn²⁺, very pale pink; +7, MnO₄⁻, purple [1]. Second element, e.g. Fe: +2, Fe²⁺, pale green; +3, Fe³⁺, yellow to orange-brown [1]. Each element shown in two or more oxidation states with correct formulae [1] (b) MnO₂ catalyses the decomposition of hydrogen peroxide [1]: 2H₂O₂ → 2H₂O + O₂ [1] Examiner insight: The question needs manganese and a second element, so an answer using manganese alone limits the credit available in (a).

4. (a) Square planar: cis has the two Cl ligands at 90° (adjacent), trans has them at 180° (opposite) [1]. The flat arrangement is fixed, and there are two pairs of different ligands [1] (b) Chloride ligands are replaced inside the cell [1]. Pt bonds to nitrogen atoms on DNA bases [1]. The distorted DNA cannot be replicated, so the cell cannot divide [1] (c) Any one: damages healthy dividing cells; nausea; hair loss; kidney damage [1] Examiner insight: In (a) draw clear square planar structures; a tetrahedral sketch does not show cis–trans isomers and is unlikely to gain credit.

5. (a) [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O [1]. Final colour yellow (green in between) [1]. Shape changes from octahedral to tetrahedral [1]. Coordination number falls from 6 to 4 because Cl⁻ ions are larger than H₂O molecules [1] (b) [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O [1]; deep blue solution [1] Examiner insight: The guidance asks for complexed formulae in substitution equations; Cu²⁺ + 4Cl⁻ → CuCl₄²⁻ is weaker.

6. Fe²⁺ is bonded to four haem-ring N atoms and one protein N, leaving a sixth site [1]. O₂ binds there as a ligand in the lungs and is released in tissues [1]. CO replaces O₂, binds much more strongly and is not readily released, so less O₂ is carried [1] Examiner insight: Say “replaces O₂” or “ligand substitution”; “blocks the blood” is too vague for credit.

7. (a) P = Fe²⁺ [1]; Q = Cr³⁺ [1]; R = Mn²⁺ [1] (b) Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s) [1] (c) Cr(OH)₃ + 3OH⁻ → [Cr(OH)₆]³⁻ [1]; Cr(OH)₃ + 6NH₃ → [Cr(NH₃)₆]³⁺ + 3OH⁻ [1] Examiner insight: P’s brown surface comes from air oxidation; the green precipitate seen first identifies Fe²⁺.

8. (a) Grey-green precipitate [1], which dissolves in excess to give a dark green solution of [Cr(OH)₆]³⁻ [1] (b) Green → yellow; Cr changes from +3 to +6 [1]. Multiply the chromium half-equation by 2 and the peroxide one by 3 so 6 electrons cancel [1]. 2[Cr(OH)₆]³⁻ + 3H₂O₂ → 2CrO₄²⁻ + 2OH⁻ + 8H₂O [1] (c) n(Cr) = 0.150 × 0.0200 = 3.00 × 10⁻³ mol; n(H₂O₂) = 3.00 × 10⁻³ × 3/2 = 4.50 × 10⁻³ mol [1]. Volume = 4.50 × 10⁻³ ÷ 0.250 = 0.0180 dm³ = 18.0 cm³ [1] (d) Yellow → orange [1]. 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O [1] (e) n(Cr₂O₇²⁻) = 3.00 × 10⁻³ ÷ 2 = 1.50 × 10⁻³ mol; n(Zn) = 3 × 1.50 × 10⁻³ = 4.50 × 10⁻³ mol [1]. Mass = 4.50 × 10⁻³ × 65.4 = 0.294 g [1] Examiner insight: In (b), cancel OH⁻ and H₂O that appear on both sides before giving the final equation.

9. (a) A white (off-white) precipitate of CuI [1] in a brown solution, from iodine [1] (b) n(S₂O₃²⁻) = 0.0950 × 0.03250 = 3.0875 × 10⁻³ mol [1]. Ratio Cu²⁺ : I₂ : S₂O₃²⁻ = 2 : 1 : 2, so n(Cu) = 3.0875 × 10⁻³ mol [1]. Mass = 3.0875 × 10⁻³ × 63.5 = 0.196 g [1]. Percentage = 0.196 ÷ 0.305 × 100 = 64.3% [1] (c) Nitric acid would also oxidise iodide to iodine, making the titre and percentage too high [1] Examiner insight: Show the 2 : 1 : 2 chain; a bare 1 : 1 ratio with no reasoning is easy to misapply.

10. (a) Copper in Cu₂O is +1; in CuSO₄ it is +2 (oxidised) and in Cu it is 0 (reduced) [1]. Copper is simultaneously oxidised and reduced [1] (b) A blue solution and a red-brown solid (copper) [1] (c) n(Cu₂O) = 1.716 ÷ 143.0 = 0.0120 mol = n(Cu) [1]. Mass = 0.0120 × 63.5 = 0.762 g [1] Examiner insight: Only half the copper in Cu₂O becomes metal, so do not double the moles.

11. (a) [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ + 6H₂O [1]. [Ni(H₂O)₆]²⁺ + 2OH⁻ → Ni(OH)₂(H₂O)₄ + 2H₂O [1] (b) The first is ligand substitution; the second is precipitation [1]. Nickel stays at +2, so no oxidation number changes [1] (c) A trans isomer, H₂O ligands opposite [1]. Two cis isomers, H₂O ligands adjacent [1]. The cis pair are non-superimposable mirror images (optical isomers); trans has a mirror plane, so the complex shows cis–trans and optical isomerism [1] Examiner insight: Check the oxidation number first; it shows whether redox is involved.

Where marks are usually lost

• Removing 3d electrons before 4s in ion configurations.
• Counting ligands instead of coordinate bonds.
• Writing “deep blue precipitate” for the copper–ammonia complex.
• Claiming Cu(OH)₂ or Fe(OH)₃ dissolves in excess NaOH.
• Leaving OH⁻ or H₂O uncancelled after combining half-equations.
• Using 2 : 1 for Cu²⁺ : S₂O₃²⁻ instead of tracing the ratio through iodine.

Next steps

• Condensed tables: Transition elements revision notes
• Full explanations: Transition elements study guide
• Half-equations and oxidation numbers: Redox
• Every H432 section: chemistry hub
• Track progress: printable checklist
• Find gaps quickly: all free 10-minute diagnostics
• Book a free trial class

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.