WJEC A-Level Chemistry Unit 3, June 2025: Question 10

✅ Correct Answer

Modification: Use concentrated nitric acid, HNO₃ (instead of hydrochloric acid).

Reasoning: Hydrochloric acid causes precipitation of insoluble silver chloride (AgCl) and lead(II) chloride (PbCl₂), preventing all metal ions from entering solution. In contrast, all metal nitrates are soluble.

🧠 Exam Technique

• Identify which elements in electrum (Au, Ag) and leaded tin bronze (Sn, Pb) cause problems with concentrated HCl.
• Recall core solubility rules: chloride ions form precipitates with Ag⁺ and Pb²⁺ (AgCl and PbCl₂).
• Nitric acid is an oxidising acid whose nitrates are unconditionally water-soluble, yielding a homogeneous solution suitable for titration.

✅ Correct Answer

Amphoteric Metal: Zinc (Zn) or Tin (Sn).

Definition: An amphoteric substance / metal reacts with both acids and bases (or alkalis).

❌ Common Errors

• Writing that it acts as "an acid and a base" without mentioning that it reacts with both acids and bases/alkalis.
• Selecting nickel or copper, which are purely basic oxide formers and not amphoteric.

✅ Correct Equations & Deduction

Reaction 1:
2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq)

Reaction 2:
I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)

Stoichiometric Deduction:
2 mol Cu²⁺ produces 1 mol I₂.
1 mol I₂ reacts with 2 mol S₂O₃²⁻.
Therefore: 2 mol Cu²⁺ ≡ 1 mol I₂ ≡ 2 mol S₂O₃²⁻.
Dividing by 2 gives: 1 mol Cu²⁺ ≡ 1 mol S₂O₃²⁻.

💡 Key Knowledge

• In reaction 1, Cu²⁺ is reduced to Cu⁺ in copper(I) iodide, CuI (an off-white precipitate). Iodide is oxidised to iodine, I₂.
• In reaction 2, iodine oxidises thiosulfate (S₂O₃²⁻) to tetrathionate (S₄O₆²⁻).
• Starch indicator is typically added near the end-point (when the solution is pale straw yellow) turning blue-black, then discharging to colourless.

❌ Common Errors & Calculation Traps

• Forgetting the ×10 scaling factor: Forgetting that only 25.0 cm³ of the 250 cm³ volumetric flask was titrated.
• Rounding too early: Rounding moles to 2 sig figs in early stages leads to rounding errors in final percentage.
• Incorrect alloy matching: 61.0% falls strictly inside cupronickel (60–90%), but outside brass (min 65%).

✅ Model Answer

1. d-Orbital Splitting: The approach of water ligands causes the partially filled 3d orbitals to split into two higher energy levels and three lower energy levels.

2. Light Absorption & Electron Promotion: Electrons absorb specific wavelengths/frequencies of visible light corresponding to the energy gap (ΔE = hf) and are promoted from lower to higher d-orbitals (d–d transition).

3. Transmitted Colour: The remaining non-absorbed frequencies are transmitted (or red/orange light is absorbed and the complementary colour, pale blue, is observed).

❌ Major Misconceptions

• Stating that "electrons emit light when falling back down" – this explains flame tests/emission spectra, not complex ion colour!
• Failing to mention that d-orbitals are split by ligands.
• Saying "blue light is absorbed" instead of blue light being transmitted/reflected.

✅ Correct Answer

Copper(I) has a fully filled 3d subshell (configuration [Ar] 3d¹⁰). Because all d-orbitals are completely full, no electrons can be promoted to a higher energy d-orbital (no d–d transitions are possible).

💡 Key Knowledge

For a complex ion to be coloured via d–d transitions, the metal ion must have a partially filled d-subshell (d¹ to d⁹). Species with d⁰ (e.g. Sc³⁺, Ti⁴⁺) or d¹⁰ (e.g. Cu⁺, Zn²⁺) are colourless/white.

✅ Selected Wavenumber & Justification

Any of the following three regions is acceptable (must include valid comparison):

• Option A (Best choice): Select a wavenumber between 14000 and 16000 cm⁻¹ (e.g., 15000 cm⁻¹).
  Reason: [CoCl₄]²⁻ has a strong absorption peak here, whereas [Co(H₂O)₆]²⁺ has zero (or negligible) absorbance. Monitoring this peak allows direct tracking of the product forming.
• Option B: Select around 20000 cm⁻¹ (18000–22000 cm⁻¹).
  Reason: [Co(H₂O)₆]²⁺ absorbs strongly, while [CoCl₄]²⁻ does not absorb.
• Option C: Select around 8000 cm⁻¹ (6000–10000 cm⁻¹).
  Reason: [Co(H₂O)₆]²⁺ absorbs here, while [CoCl₄]²⁻ does not absorb.

🧠 Exam Technique

To receive both marks, you must make a comparison between both complexes. Simply stating "[CoCl₄]²⁻ absorbs at 15000 cm⁻¹" is not enough – you must explicitly state that the other complex does not absorb significantly at that wavenumber.

✅ Model Answer

The reaction is: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)

Relevant half-equations:

• Reduction: Cu⁺ + e⁻ → Cu (E° = +0.52 V)
• Oxidation: Cu⁺ → Cu²⁺ + e⁻ (E° = -0.16 V)

Cell EMF Calculation:
E°cell = E°(reduction) − E°(oxidation) = +0.52 V − (+0.16 V) = +0.36 V

Because E°cell is positive (E°cell > 0), the disproportionation reaction is energetically feasible / spontaneous.

💡 Alternative Explanation

The standard electrode potential for the Cu⁺/Cu system (+0.52 V) is more positive than that for the Cu²⁺/Cu⁺ system (+0.16 V). Therefore, Cu⁺ is a strong enough oxidising agent to oxidise another Cu⁺ ion into Cu²⁺ while itself being reduced to Cu.

✅ (I) Final Thallium Species (2 Marks)

Species: Tl⁺ / thallium(I) ions.

Explanation:

• For Tl³⁺ → Tl⁺ using Cu⁺ → Cu²⁺:
  E°cell = +1.25 V − (+0.16 V) = +1.09 V (Positive, reaction feasible).
• For Tl⁺ → Tl using Cu⁺ → Cu²⁺:
  E°cell = −0.34 V − (+0.16 V) = −0.50 V (Negative, reaction not feasible).

Thus, Cu⁺ is powerful enough to reduce Tl³⁺ to Tl⁺, but cannot reduce Tl⁺ to Tl metal.

✅ (II) Predicted Stable Oxidation State (1 Mark)

Stable state: Thallium(I) / +1

Reason: The inert pair effect becomes increasingly pronounced down Group 3 (Group 13). The 6s² electron pair remains unshared, favouring the +1 oxidation state (group number − 2) over the +3 oxidation state.