AQA A-Level Chemistry Paper 1, 2022: Question 6

✅ Correct Answer

Chlorine is toxic / poisonous (or too much chlorine causes death / harmful side-effects).

❌ Common Errors & Misconceptions

• Vague answers like "it makes water taste bad" or "it's dangerous" without stating that it is toxic or poisonous.
• Confusing the elemental risk of Cl₂ with chlorination by-products (e.g., chlorinated hydrocarbons/cancer risk) when a simple statement of toxicity is required.

✅ Correct Answer

Equation:

Cl₂ + H₂O → HCl + HClO

(Also allowed: Cl₂ + H₂O → 2H⁺ + Cl⁻ + ClO⁻ )

Explanation:

Chlorine gains electron(s) (to form Cl⁻ ) AND loses electron(s) (to form ClO⁻ ).

🧠 Exam Technique & Traps

• Crucial: The prompt explicitly states "with reference to electrons". Stating only oxidation state changes (from 0 to -1 and +1) or simply writing "it is disproportionation" scores 0 for the explanation mark!
• Always state: "Chlorine gains electrons to form Cl⁻ (reduction) and loses electrons to form ClO⁻ / HClO (oxidation)".

✅ Correct Answer

Observation: Brown solution formed (or black solid forms).

Ionic Equation:

Cl₂ + 2I⁻ → 2Cl⁻ + I₂

❌ Common Errors

• Observation trap: Writing purple . Iodine is only purple as a vapour or in an organic non-polar solvent (like cyclohexane). In aqueous solution, it forms a brown solution (or a black precipitate if concentrated). The mark scheme strictly states: "do not accept purple".
• Writing a full molecular equation ( Cl₂ + 2NaI → 2NaCl + I₂ ) when the prompt specifically requested an ionic equation. Spectator ions ( Na⁺ ) must be cancelled out.

✅ Correct Answer

Two gaseous sulfur-containing products: Sulfur dioxide ( SO₂ ) and Hydrogen sulfide ( H₂S ).

Equation 1 (for SO₂):

H₂SO₄ + 2H⁺ + 2I⁻ → SO₂ + 2H₂O + I₂

(Or full: 2H₂SO₄ + 2NaI → Na₂SO₄ + SO₂ + 2H₂O + I₂ )

Equation 2 (for H₂S):

H₂SO₄ + 8H⁺ + 8I⁻ → H₂S + 4H₂O + 4I₂

(Or full: 5H₂SO₄ + 8NaI → 4Na₂SO₄ + H₂S + 4H₂O + 4I₂ )

Role of sulfuric acid: Oxidising agent (or electron acceptor).

📐 Half-Equation Balancing Strategy

Construct these complex redox equations effortlessly using half-equations:

1. Oxidation: 2I⁻ → I₂ + 2e⁻
2. Reduction to SO₂ (+6 to +4, needs 2e⁻):
   H₂SO₄ + 2H⁺ + 2e⁻ → SO₂ + 2H₂O
   Combine 1:1 → H₂SO₄ + 2H⁺ + 2I⁻ → SO₂ + 2H₂O + I₂
3. Reduction to H₂S (+6 to -2, needs 8e⁻):
   H₂SO₄ + 8H⁺ + 8e⁻ → H₂S + 4H₂O
   Multiply iodide half-equation by 4 and add:
   H₂SO₄ + 8H⁺ + 8I⁻ → H₂S + 4H₂O + 4I₂

✅ Correct Answers

Identity of sodium halide: NaF (or sodium fluoride)

Identity of the gas: CO₂ (or carbon dioxide)

Ionic equation:

CO₃²⁻ + 2H⁺ → CO₂ + H₂O

🧠 Step-by-Step Deduction Logic

1. Deducing the gas: The gas turns limewater (calcium hydroxide) cloudy (forms white precipitate CaCO₃ ). Therefore, the gas is carbon dioxide ( CO₂ ).
2. Deducing the impurity: Carbonate ions ( CO₃²⁻ ) react with the acid in acidified silver nitrate ( HNO₃ ) to effervesce CO₂ .
3. Deducing the sodium halide: The resultant solution is colourless with no precipitate. All silver halides are precipitates ( AgCl white, AgBr cream, AgI yellow) except AgF, which is water-soluble! Hence, the halide must be sodium fluoride (NaF).

✅ Correct Answer

Shape: Bent / V-shaped / non-linear

Explanation: Lone pair–lone pair repulsion is greater than bond pair–bond pair repulsion (lp–lp > bp–bp), OR lone pairs repel to be as far apart as possible.

Bond angle: 104° to 106° (any value in range 95° – 106° accepted).

📐 Electron Pair Count Strategy

1. Central atom: Chlorine (Group 7) = 7 valence electrons.
2. Positive charge (+1): Subtract 1 electron = 6 electrons remaining.
3. Surrounding atoms: 2 Fluorine atoms form 2 single covalent bonds (2 electrons shared).
4. Total bonding electrons: 2 bonding pairs (bp).
5. Remaining non-bonding electrons: 6 - 2 = 4 electrons = 2 lone pairs (lp).
6. Total pairs: 4 electron pairs → tetrahedral arrangement.
7. Molecular shape: 2 bp + 2 lp gives a bent / V-shaped geometry (isoelectronic with H₂O ).
8. Angle deduction: Standard tetrahedral is 109.5°. Each lone pair reduces bond angle by ~2.5°: 109.5° - 2(2.5°) = 104.5° .

✅ Correct Answer

TiCl₄ + 2Mg → 2MgCl₂ + Ti

💡 Key Knowledge

• Titanium cannot be extracted with carbon because it forms brittle titanium carbide ( TiC ).
• Instead, TiO₂ is converted to TiCl₄ and reduced by Mg in an argon atmosphere at around 1000 °C.
• Ensure charges balance: Ti is +4, so it requires 2 moles of Mg (which forms Mg²⁺ ).