AQA A-Level Chemistry Paper 1, June 2025: Question 5

Correct Answer

• M1: Chlorine is smaller / has fewer electrons / has fewer electron shells / has a lower relative molecular mass (M_r).
• M2: Consequently, the van der Waals forces between chlorine molecules are weaker, requiring less energy to overcome.

Key Knowledge

• Halogens exist as diatomic molecules (Cl₂, Br₂).
• The only intermolecular forces acting between these non-polar molecules are van der Waals forces (induced dipole-dipole forces).
• The strength of van der Waals forces increases as the size of the electron cloud increases (more electrons).

Exam Technique

Always specify that the intermolecular forces are between molecules. Examiners look for this exact terminology to ensure you aren't implying that covalent bonds are being broken during boiling.

Common Errors

❌ "Chlorine has weaker covalent bonds than bromine." — Zero marks! Boiling molecular substances does not break covalent bonds; it only overcomes intermolecular forces.

❌ Failing to mention "molecules" when describing where the van der Waals forces act.

Correct Answer

• Oxidation state of Br: +3
• Oxidation state of F: -1

Step-by-Step Determination

1. Rule 1: Fluorine is the most electronegative element and always has an oxidation state of -1 in its compounds.
2. Rule 2: The sum of oxidation states in a neutral molecule (BrF₃) must equal 0 .
3. Calculation:
   Br + 3(F) = 0
   Br + 3(-1) = 0
   Br - 3 = 0
   Br = +3

Exam Technique

Always include the sign ( + or - ) for oxidation states. Writing just "3" instead of "+3" can result in lost marks depending on the examiner's strictness.

Correct Answer & Explanation

Shape Drawing: T-shaped geometry.

Explanation (M2): Electron pairs (or lone pairs and bonding pairs) repel each other to be as far apart as possible to minimise repulsion.

Key Knowledge

• Bromine has 7 outer shell electrons. It forms 3 single covalent bonds with F, using 3 electrons.
• This leaves 4 non-bonding electrons, which form 2 lone pairs.
• Total electron pairs = 3 bonding pairs + 2 lone pairs = 5 pairs (based on a trigonal bipyramidal parent arrangement).
• To minimise repulsion, the 2 lone pairs occupy equatorial positions, resulting in a T-shaped molecular geometry.

Exam Technique

When explaining shapes, always state that electron pairs repel to get as far apart as possible. Mentioning "minimising repulsion" is highly recommended by examiners to secure the second mark.

Common Errors

❌ Drawing a trigonal planar shape because you forgot to calculate the lone pairs on the central bromine atom.

❌ Forgetting to draw the lone pairs when the question explicitly asks to "include any lone pairs of electrons that influence the shape".

Correct Answers

Acid-Base Reaction:

• Equation (M1): H₂SO₄ + NaBr → NaHSO₄ + HBr
  (OR H₂SO₄ + 2NaBr → Na₂SO₄ + 2HBr )
• Observation (M2): White, misty, or steamy fumes.

Redox Reaction:

• Equation (M3): H₂SO₄ + 2H⁺ + 2Br⁻ → SO₂ + Br₂ + 2H₂O
  (OR H₂SO₄ + 2HBr → SO₂ + Br₂ + 2H₂O )
• Observation (M4): Orange or brown fumes/gas (allow choking gas for SO₂).

Key Knowledge

• Concentrated sulfuric acid acts as both an acid (proton donor) and an oxidising agent.
• Bromide ions ( Br⁻ ) are strong enough reducing agents to reduce sulfuric acid ( H₂SO₄ , sulfur state +6) to sulfur dioxide ( SO₂ , sulfur state +4).
• The bromide ions themselves are oxidised to elemental bromine ( Br₂ ), which produces the characteristic orange-brown fumes.

Exam Technique

Make sure your observations match the specific reaction. Misty white fumes are due to HBr gas dissolving in moisture in the air (acid-base). Orange-brown fumes are due to Br₂ vapour (redox).

Common Errors

❌ Writing H₂S or S as the reduction product for bromide. Bromide is not a strong enough reducing agent to reduce sulfur to 0 or -2; only iodide ( I⁻ ) can do that!

❌ Forgetting to balance the redox equation. Ensure charges and atoms balance on both sides.

Correct Answer

2Cu²⁺ + 4I⁻ → 2CuI + I₂

Key Knowledge

• Iodide ions ( I⁻ ) are oxidized to iodine ( I₂ ).
• Copper(II) ions ( Cu²⁺ ) are reduced to copper(I) ions ( Cu⁺ ), which precipitate out of solution as white solid copper(I) iodide ( CuI ).
• The final mixture appears brown due to the presence of dissolved I₂ masking the white precipitate.

Exam Technique

This is a classic inorganic equation that frequently appears in AQA exams. Memorise the stoichiometry: 2 : 4 → 2 : 1. Double-check that both the atoms and the net charges balance (net charge of +2 on both sides).