OCR A-Level Chemistry Unified chemistry (03), June 2025: Question 4

✅ Correct Answer

sulfate(IV) ion (or sulphate(IV))

❌ Common Errors

• Writing an oxide name: sulfur trioxide or sulfur(IV) oxide (both refer to neutral covalent gases, SO₃, not ions).
• Omitting the oxidation state when using systematic sulfate naming (e.g. simply writing 'sulfate', which denotes SO₄²⁻).

✅ Correct Box Entries

• Reaction with H⁺(aq)/H₂S(g):
  • Yellow solid = S (or S₈ )
  • Colourless neutral liquid = H₂O
• Acidic solution containing three ions:
  • Cl⁻ , H⁺ (or H₃O⁺ ), SO₄²⁻ (or HSO₄⁻ )
• Top-left precipitate (with Ag⁺):
  • AgCl (white precipitate; allow Ag₂SO₄)
• Top-right precipitate (with Ba²⁺):
  • BaSO₄ (white precipitate)

💡 Chemical Principles Behind the Flowchart

• Comproportionation: SO₃²⁻ (S has +4) oxidises H₂S (S has -2) to elemental sulfur, S (0), and water:
  SO₃²⁻ + 2H₂S + 2H⁺ → 3S + 3H₂O
• Oxidation by chlorine water: Aqueous Cl₂ oxidises sulfite to sulfate while being reduced to chloride:
  SO₃²⁻ + Cl₂ + H₂O → SO₄²⁻ + 2Cl⁻ + 2H⁺
  This creates an acidic solution containing H⁺, Cl⁻, and SO₄²⁻.
• Precipitation tests:
  • Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
  • Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

📐 Step-by-Step Balancing Method

1. Oxidation half-equation (Sulfite to Sulfate):
   SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻
2. Reduction half-equation (Dichromate(VI) to Chromium(III)):
   Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
3. Multiply to equate electrons:
   Multiply oxidation half-equation by 3 (giving 6e⁻):
   3SO₃²⁻ + 3H₂O → 3SO₄²⁻ + 6H⁺ + 6e⁻
4. Combine and cancel spectators (3 H₂O and 6 H⁺):
   3SO₃²⁻ + Cr₂O₇²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

✅ Final Equation & Mark Breakdown

3SO₃²⁻ + Cr₂O₇²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

📐 Counting Atoms from the Structural Diagram

• Central metal: 1 × Ni
• Two axial water ligands: 2 × H₂O = H₄O₂
• Two amino acid chelate rings: each ring has -NH₂-CH(CH₃)-COO⁻
  Formula per ligand = C₃H₆NO₂
  Two ligands = C₆H₁₂N₂O₄
• Total molecular formula:
  Ni: 1 | C: 6 | H: 12 + 4 = 16 | N: 2 | O: 4 + 2 = 6
  → C₆H₁₆N₂NiO₆
• Empirical formula check: Since Ni has an index of 1, the ratio (6:16:2:1:6) cannot be simplified further.

✅ Correct Answer

C₆H₁₆N₂NiO₆

(Elements can be in any order, but must contain all 5 elements in the simplest whole-number ratio).

Structure of Amino Acid B (Alanine)

By removing the Ni²⁺ ion and protonating the carboxylate group, the free ligand is alanine (2-aminopropanoic acid):

CH₃-CH(NH₂)-COOH

Accepted as skeletal, displayed, or structural formula.

3D Structure of Complex Ion C: [Ni(H₂O)₆]²⁺

Must show 3D octahedral geometry around the central Ni²⁺ ion:

• 3D representation: 2 bonds in plane, 2 wedged bonds (pointing forward/out), 2 dashed bonds (pointing back/into paper).
• Ligands: 6 water molecules coordinated through the oxygen atom ( Ni-OH₂ ).
• Overall charge: Must show 2+ outside brackets or adjacent to the complex.

🧠 Identifying the Stereoisomer

In the original Complex A:

• The axial ligands are both H₂O (trans to each other).
• In the equatorial plane, both oxygen atoms are coordinated along dashed bonds (pointing back), and both nitrogen atoms are coordinated along wedged bonds (pointing forward). The two O atoms are cis to each other in the equatorial plane.

To draw a stereoisomer while keeping the axial water ligands in place:

Swap the positions of the donor atoms on one of the bidentate ligands so that one O is dashed and the other O is wedged (trans arrangement of O atoms in the equatorial plane).

✅ Required 3D Diagram

On the provided template (Ni with axial OH₂ above and below):

• Left side: Connect dashed bond to NH₂ and wedged bond to O (inverted relative to original).
• Right side: Keep original connectivity (dashed bond to O , wedged bond to NH₂ ) — or invert both to achieve the trans-donor isomer.
• Examiner tip: "Watch for adjacent Os". Having one O wedged and one O dashed creates the stereoisomer.