AQA A-Level Chemistry AS Paper 1, June 2023: Question 6

✅ Model Answer (Mark Breakdown)

• Mark 1: Down the group, atomic radius increases / atoms have more electron shells / there is more shielding (or the shared bonding pair is further from the nucleus).
• Mark 2: There is a weaker attraction between the nucleus and the bonding pair of electrons (or shared pair / electron density in the covalent bond).

💡 Key Knowledge

Electronegativity is the power of an atom to attract the shared pair of electrons in a covalent bond towards itself.

As you descend Group 7:

• Nuclear charge increases (+9 to +53), but...
• Extra electron shells increase atomic radius and shielding.
• The shielding and distance override the nuclear charge, reducing the electrostatic pull on the bonding pair.

❌ Common Errors & Examiner Traps

• Mentioning "outer electrons": The mark scheme explicitly states: "Ignore references to outer electrons". You must refer specifically to the bonding pair or shared pair of electrons. Referring to outer electrons describes ionisation energy, not electronegativity!
• Forgetting nuclear attraction: Stating that "shielding increases" only gets M1. You must state the effect on attraction (M2).

🧠 Exam Technique

Always structure trend explanations in two distinct steps:

1. The physical change: Larger atomic radius / more shielding / more shells.
2. The electrostatic consequence: Weaker electrostatic attraction between the positive nucleus and the bonding pair of electrons.

✅ Accepted Answers

One similarity (any 1 from):

• Both form hydrogen halides ( HCl / HBr ).
• Both undergo an acid-base (proton transfer) reaction.
• Misty white fumes are observed.
• Both produce sodium hydrogensulfate ( NaHSO₄ ) or sodium sulfate ( Na₂SO₄ ).
• Both reactions are exothermic / show effervescence.

One difference (any 1 from):

• Bromide undergoes a redox reaction / bromide ions are oxidised (chloride is not).
• Bromide ions reduce sulfur in sulfuric acid (chloride cannot).
• Red-brown (orange) fumes/liquid ( Br₂ ) are observed with bromide.
• Choking gas / sulfur dioxide ( SO₂ ) is produced with bromide.
• Different hydrogen halides are formed ( HCl vs HBr ).

💡 Key Knowledge: Why the Difference?

Concentrated H₂SO₄ acts as both an acid and an oxidising agent.

• Chloride (Cl⁻): A weak reducing agent. It only undergoes an acid-base reaction: NaCl + H₂SO₄ → NaHSO₄ + HCl
• Bromide (Br⁻): A stronger reducing agent due to its larger ionic radius. It first undergoes the acid-base step: NaBr + H₂SO₄ → NaHSO₄ + HBr Then reduces H₂SO₄ from (+6) to (+4) in SO₂ : 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

🧠 Exam Technique

When asked for a similarity and a difference:

• You can state an observable feature (e.g. misty white fumes seen in both; orange fumes only seen with NaBr).
• Or you can state a chemical / reaction type (both are acid-base; only NaBr undergoes redox).
• Keep answers brief and precise—do not contradict yourself by writing multiple conflicting statements.

❌ Common Errors

• Claiming NaCl undergoes redox: Cl⁻ is not a strong enough reducing agent to reduce concentrated H₂SO₄.
• Vague observations: Writing "gas produced" is too vague for the difference; name the specific product or colour (e.g. "red-brown fumes of bromine").

✅ Correct Equations (Mark Scheme)

Half-equation 1:

Half-equation 2:

Overall equation:

Also allowed: H₂SO₄ + 8HI → 4I₂ + H₂S + 4H₂O or sulfate forms. Ignore state symbols.

❌ Common Errors

• Incorrect H⁺ count: Forgetting that H₂SO₄ already contains 2 hydrogen atoms, mistakenly adding 10H⁺ to the left when using molecular H₂SO₄ .
• Unbalanced electrons: Adding half-equations directly without multiplying the iodide equation by 4 to eliminate electrons.
• Leaving electrons in the overall equation: Overall equations must never contain e⁻ .
• Confusing reduction products: Iodide is powerful enough to reduce S from +6 all the way to −2 ( H₂S , smell of bad eggs). Bromide can only reduce S to +4 ( SO₂ ).