OxfordAQA A-Level Chemistry: Transition metals (9620) – 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.

Practise section 3.2.5, Transition metals (3.2.5.1 to 3.2.5.6) from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for exams from May/June 2020. All of it is International A-level only. Where needed, take h = 6.63 × 10⁻³⁴ J s, L = 6.02 × 10²³ mol⁻¹ and c = 3.00 × 10⁸ m s⁻¹.

Revise first with the study guide or revision notes; the course hub, checklist and free diagnostics cover the rest.

Questions

1. State the four characteristic properties of the elements titanium to copper, and the feature of their electron structure that causes them. [2]

2. This question is about the complex ion [Cr(H₂O)₄Cl₂]⁺.

(a) State the co-ordination number of chromium. [1] (b) Deduce the oxidation state of chromium. [1] (c) Explain why the chloride ion is described as a ligand. [1]

3. Concentrated hydrochloric acid is added to a solution containing [Fe(H₂O)₆]³⁺ ions.

(a) Write an equation for the ligand substitution. [1] (b) State the shape of the iron complex before and after the reaction. [1] (c) Explain why the co-ordination number changes. [2]

4. EDTA⁴⁻ is added to hexaamminenickel(II) ions:

[Ni(NH₃)₆]²⁺ + EDTA⁴⁻ → [Ni(EDTA)]²⁻ + 6NH₃

Values for this change: ΔH = −1.8 kJ mol⁻¹, ΔS = +205 J K⁻¹ mol⁻¹.

(a) Explain why ΔS is large and positive. [1] (b) Explain why ΔH is close to zero. [1] (c) Calculate ΔG at 298 K and explain what your answer shows. [3]

5. This question is about haemoglobin.

(a) Describe how haemoglobin carries oxygen in the blood. [2] (b) Someone breathes air containing carbon monoxide. Explain why their blood then delivers less oxygen. [1]

6. This question is about the shapes of complexes.

(a) [Pt(NH₃)₂Cl₂] is square planar and has two isomers. Describe how they differ and identify cisplatin. [2] (b) [Co(H₂NCH₂CH₂NH₂)₂(C₂O₄)]⁺ has three ligands and two mirror-image forms. Explain why its co-ordination number is 6 and why the forms are different compounds. [2] (c) Give the shape of [CoCl₄]²⁻ and of [Ag(NH₃)₂]⁺, with the bond angle in each. [2]

7. Hexaaquacopper(II) ions are pale blue.

(a) Explain, in terms of d electrons and visible light, why the solution is coloured. [2] (b) A different complex absorbs light whose energy is 205 kJ mol⁻¹. Calculate the wavelength absorbed, in nm. [2] (c) Excess aqueous ammonia changes the colour of the copper(II) solution. Explain why. [1]

8. Manganese in 0.450 g of steel is oxidised to MnO₄⁻ and made up to 250 cm³. Colorimeter readings are below. Ar(Mn) = 54.9.

[MnO₄⁻] / mol dm⁻³	2.0 × 10⁻⁴	4.0 × 10⁻⁴	6.0 × 10⁻⁴	8.0 × 10⁻⁴	steel solution
Absorbance	0.46	0.92	1.38	1.84	0.69

(a) Suggest why the manganese is measured as MnO₄⁻ rather than as Mn²⁺(aq). [1] (b) Find the gradient of the calibration line and the concentration of MnO₄⁻ in the steel solution. [2] (c) Calculate the percentage by mass of manganese in the steel. [2]

9. A yellow vanadate(V) solution is shaken with zinc and dilute sulfuric acid.

(a) Give the formula of the blue species, and the colour seen when all the vanadium is V³⁺. [2] (b) Use half-equations to write an overall equation for the reduction of VO²⁺ to V³⁺ by zinc. [2] (c) A green colour appears between yellow and blue, although neither VO₂⁺ nor VO²⁺ is green. Explain this. [1]

10. Data: Ag⁺ + e⁻ ⇌ Ag, E⦵ = +0.80 V; [Ag(NH₃)₂]⁺ + e⁻ ⇌ Ag + 2NH₃, E⦵ = +0.37 V.

(a) Identify the stronger oxidising agent, and state what the data show about the effect of a ligand on redox potential. [2] (b) Tollens’ reagent is warmed with hexanal, and separately with hexan-2-one. Explain what is seen in each case. [2]

11. This question is about manganate(VII) titrations.

(a) Write the ionic equation for the reaction of acidified MnO₄⁻ with C₂O₄²⁻. [1] (b) 0.1675 g of sodium ethanedioate, Na₂C₂O₄ (Mr = 134.0), is dissolved in acid, warmed and titrated. It needs 25.00 cm³ of the KMnO₄ solution. Calculate the concentration of KMnO₄. [3] (c) A 0.1500 g sample of impure FeC₂O₄·2H₂O (Mr = 179.8) is dissolved in acid. Both Fe²⁺ and C₂O₄²⁻ are oxidised, and the titration needs 24.50 cm³ of the same KMnO₄.

(i) Show that 5 mol of FeC₂O₄ react with 3 mol of MnO₄⁻. [1] (ii) Calculate the percentage purity of the sample. [3]

12. This question is about transition metal catalysts.

(a) V₂O₅ reacts with sulfur dioxide in the Contact process yet is not used up. Explain, with equations and vanadium oxidation states. [3] (b) A plant spreads its catalyst on a ceramic support. An impurity in the feed gas sticks to the catalyst surface. Explain the purpose of the support, and the effect of the impurity on the plant’s costs. [2] (c) Fe³⁺ ions catalyse the reaction between S₂O₈²⁻ and I⁻ as well as Fe²⁺ ions do. Explain why the reaction is slow on its own, and write the two catalysed steps in order, starting from Fe³⁺. [3] (d) A colorimeter tracks [MnO₄⁻] as acidified MnO₄⁻ reacts with excess C₂O₄²⁻ at constant temperature. Describe and explain the shape of the concentration–time curve. [2]

Answers

1. Complex formation, coloured ions, variable oxidation state and catalytic activity (all four) [1]; an incomplete d sub-level in the atom or ion [1] [2] Examiner insight: “Being in the d block” is not enough; zinc is in the d block too.

2. (a) 6 (four H₂O plus two Cl⁻) [1] (b) Cr + 2(−1) = +1, so +3 [1] (c) It donates a lone pair of electrons to chromium, forming a co-ordinate bond [1] Examiner insight: In (b), water contributes 0 to the charge; only the chloride ligands count.

3. (a) [Fe(H₂O)₆]³⁺ + 4Cl⁻ → [FeCl₄]⁻ + 6H₂O [1] (b) Octahedral before, tetrahedral after [1] (c) Cl⁻ ligands are larger than H₂O molecules [1]; so only four fit around Fe³⁺ and the co-ordination number drops to 4 from 6 [1] Examiner insight: A product charge of 3− or 1+ is a common slip; add +3 to four lots of −1.

4. (a) Two particles become seven, so disorder increases [1] (b) Six Ni–N bonds are broken and six similar co-ordinate bonds to EDTA’s N and O donor atoms are formed [1] (c) ΔS = +0.205 kJ K⁻¹ mol⁻¹ [1]; ΔG = −1.8 − (298 × 0.205) = −62.9 kJ mol⁻¹ [1]; negative, so the substitution is feasible: the chelate effect, driven by entropy [1] Examiner insight: Using 205 instead of 0.205 gives about −61 000 kJ mol⁻¹; check units.

5. (a) Oxygen forms a co-ordinate bond [1]; to the iron(II) ion in haem, so it is carried round the body [1] (b) CO forms a co-ordinate bond to Fe(II) in haem in place of O₂, so less oxygen is carried [1] Examiner insight: Name the bond type and the iron oxidation state.

6. (a) In one isomer the two Cl⁻ ligands are next to each other at 90°; in the other they are opposite at 180° [1]; cisplatin is the cis isomer (Cl⁻ at 90°) [1] (b) Three bidentate ligands give 3 × 2 = 6 co-ordinate bonds [1]; the mirror images are non-superimposable: optical isomers [1] (c) [CoCl₄]²⁻: tetrahedral, 109.5° [1]; [Ag(NH₃)₂]⁺: linear, 180° [1] Examiner insight: In (b), “mirror images” alone is not enough; say they cannot be superimposed.

7. (a) Some visible wavelengths are absorbed as d electrons are promoted to a higher energy level (excited state) [1]; the remaining wavelengths are transmitted, and these give the colour [1] (b) ΔE per ion = 205 000 ÷ 6.02 × 10²³ = 3.405 × 10⁻¹⁹ J [1]; λ = hc/ΔE = 5.84 × 10⁻⁷ m = 584 nm [1] (c) The ligand changes, so ΔE changes and a different set of wavelengths is absorbed [1] Examiner insight: In (b), dividing kJ mol⁻¹ by L without first multiplying by 1000 gives an answer 1000 times too large.

8. (a) Mn²⁺(aq) is very pale and absorbs little light; MnO₄⁻ is intensely coloured [1] (b) Gradient = 0.46 ÷ 2.0 × 10⁻⁴ = 2300 dm³ mol⁻¹ [1]; [MnO₄⁻] = 0.69 ÷ 2300 = 3.00 × 10⁻⁴ mol dm⁻³ [1] (c) n(Mn) = 3.00 × 10⁻⁴ × 0.250 = 7.50 × 10⁻⁵ mol; mass = 7.50 × 10⁻⁵ × 54.9 = 4.12 × 10⁻³ g [1]; percentage = 4.12 × 10⁻³ ÷ 0.450 × 100 = 0.915% [1] Examiner insight: Show the gradient working.

9. (a) VO²⁺ [1]; green [1] (b) VO²⁺ + 2H⁺ + e⁻ → V³⁺ + H₂O, doubled to match Zn → Zn²⁺ + 2e⁻ [1]; 2VO²⁺ + 4H⁺ + Zn → 2V³⁺ + Zn²⁺ + 2H₂O [1] (c) A mixture of yellow VO₂⁺ and blue VO²⁺ while the first reduction is part-way [1] Examiner insight: Check charge balance in (b): +8 on each side.

10. (a) Ag⁺ (more positive E⦵) [1]; the ammonia ligand makes the silver(I) ion harder to reduce, lowering its redox potential [1] (b) Hexanal: a silver mirror forms, because the aldehyde is oxidised and Ag(I) is reduced to Ag [1]; hexan-2-one: no change, because ketones are not oxidised by this mild reagent [1] Examiner insight: Each observation needs its reason.

11. (a) 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O [1] (b) n(C₂O₄²⁻) = 0.1675 ÷ 134.0 = 1.25 × 10⁻³ mol [1]; n(MnO₄⁻) = 1.25 × 10⁻³ × 2/5 = 5.00 × 10⁻⁴ mol [1]; [KMnO₄] = 5.00 × 10⁻⁴ ÷ 0.02500 = 0.0200 mol dm⁻³ [1] (c)(i) Each FeC₂O₄ gives 3 e⁻ (1 from Fe²⁺, 2 from C₂O₄²⁻) and each MnO₄⁻ takes 5 e⁻, so 15 e⁻ gives 5 : 3 [1] (ii) n(MnO₄⁻) = 0.0200 × 0.02450 = 4.90 × 10⁻⁴ mol [1]; n(FeC₂O₄) = 4.90 × 10⁻⁴ × 5/3 = 8.17 × 10⁻⁴ mol [1]; mass = 8.17 × 10⁻⁴ × 179.8 = 0.147 g, so purity = 97.9% [1] Examiner insight: Using 2 : 5 in (c) ignores the iron(II); count electrons from both ions.

12. (a) V₂O₅ + SO₂ → V₂O₄ + SO₃ [1]; V₂O₄ + ½O₂ → V₂O₅ [1]; V is reduced from +5 to +4, then oxidised back to +5, so V₂O₅ is regenerated [1] (b) The support gives a large surface area from less catalyst, lowering cost [1]; the impurity poisons the catalyst by blocking active sites, so efficiency falls and replacement adds cost [1] (c) S₂O₈²⁻ and I⁻ are both negative and repel, giving a high activation energy [1]; first 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ [1]; then 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻ [1] (d) Shallow at first: no Mn²⁺ catalyst yet [1]; steeper as Mn²⁺ forms and autocatalyses, then levels off as MnO₄⁻ runs out [1] Examiner insight: In (a), the second equation must regenerate V₂O₅.

Where marks are usually lost

• Counting ligands, not co-ordinate bonds, in chelate complexes.
• Explaining the chelate effect without counting particles.
• Leaving λ in nm, or forgetting × 1000 for kJ.
• Missing a dilution or scale-up factor.
• Using 1 : 5 or 2 : 5 for iron(II) ethanedioate.
• Not showing that a catalyst is regenerated.

Next steps

• Revision notes: transition metals
• Study guide: transition metals
• Course hub for OxfordAQA A-Level Chemistry
• Topic checklist to print
• Book a free trial class
• All free 10-minute diagnostics

Official syllabus

Source: OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for exams from May/June 2020. Section 3.2.5 Transition metals (International A2).