AQA A-Level Chemistry: Group 7(17), the halogens (7405) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.2.3 Group 7(17), the halogens of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: 3.2.3.1 Trends in properties, 3.2.3.2 Uses of chlorine and chlorate(I), and Required practical 4. All of it is AS content, so nothing here is A-level only. The topic is examined on AS Paper 1 and A-level Paper 1, and A-level Paper 3 can test any content.

Learn the ideas first in the halogens study guide and the halogens revision notes. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. The half-equation for bromide reducing sulfuric acid to SO₂ is practised in the redox practice questions, so it is not repeated here. Questions 7 and 9 need a calculator. Use Ar: Ag = 107.9, Cl = 35.5, Na = 23.0, O = 16.0, H = 1.0.

Questions

1. Which ion is the strongest reducing agent?

• A F⁻
• B Cl⁻
• C Br⁻
• D I⁻

[1]

2. State and explain the trend in electronegativity from fluorine to iodine. [3]

3. Explain why bromine has a higher boiling point than chlorine. [3]

4. A student adds chlorine water to potassium bromide solution in one tube and to potassium iodide solution in another.

(a) State what the student sees in each tube. [2] (b) Write an ionic equation for the reaction in the potassium iodide tube. [1] (c) Explain why no reaction is seen when iodine solution is added to potassium chloride solution. [2]

5. Concentrated sulfuric acid is added to solid sodium iodide. One product is a yellow solid.

(a) Identify the yellow solid. [1] (b) Write a half-equation for the formation of the yellow solid from H₂SO₄. [2] (c) Write an equation for the reaction of hydrogen iodide with sulfuric acid to form iodine, the yellow solid and water. [1] (d) State the role of the iodide ions in this reaction. [1]

6. Three solutions, P, Q and R, each contain one sodium halide. Each is acidified with dilute nitric acid and then silver nitrate solution is added.

• P: yellow precipitate that does not dissolve in concentrated ammonia.
• Q: white precipitate that dissolves in dilute ammonia.
• R: no precipitate.

(a) Identify the halide ion in each solution. [3] (b) Write an ionic equation, with state symbols, for the reaction in Q. [1] (c) Explain why the solutions are acidified with dilute nitric acid. [1]

7. A 25.0 cm³ sample of sodium chloride solution is treated with excess acidified silver nitrate. The dried precipitate has a mass of 0.430 g. Calculate the concentration of the sodium chloride solution in mol dm⁻³. [3]

8. Chlorine reacts with water.

(a) Write an equation for the reaction that forms chloride ions and chlorate(I) ions. [1] (b) Give the oxidation state of chlorine in each chlorine-containing species in your equation, and name this type of reaction. [2] (c) In sunlight, chlorine water releases a gas. Write an equation for this reaction. [2]

9. Chlorine reacts with cold, dilute sodium hydroxide.

(a) Write an equation for the reaction. [1] (b) State one use of the solution formed. [1] (c) Calculate the mass of sodium hydroxide that reacts with 3.55 g of chlorine. [2]

10. Compare the reactions of concentrated sulfuric acid with solid sodium chloride and with solid sodium bromide. Include observations and equations, and explain the difference in terms of the halide ions. [6]

11. Chlorine is added to the drinking water supply. Evaluate this practice. [6]

12. Describe tests to show that a white solid is ammonium bromide. Give the result of each test. [4]

Answers

1. D [1]. Reducing ability of halide ions increases down the group. [1] Examiner insight: A is a common wrong answer from confusing this trend with the oxidising ability of the halogens, which is greatest for fluorine.

2. Electronegativity decreases from F to I [1]. Atomic radius increases and there is more shielding from extra inner shells [1]. So the bonding pair of electrons is further from the nucleus and is attracted less strongly [1]. [3] Examiner insight: “Nuclear charge decreases” is wrong and loses the explanation mark; nuclear charge increases but is outweighed.

3. Both are simple molecular, made of diatomic molecules [1]. Br₂ has more electrons (70) than Cl₂ (34), so the van der Waals forces between Br₂ molecules are stronger [1]. More energy is needed to overcome these forces [1]. [3] Examiner insight: Any mention of breaking Br–Br or Cl–Cl bonds contradicts the answer and can lose the last mark.

4. (a) KBr tube: solution turns yellow/orange [1]; KI tube: solution turns brown [1]. (b) Cl₂ + 2I⁻ → 2Cl⁻ + I₂ [1] (c) Iodine is a weaker oxidising agent than chlorine [1], so it cannot oxidise Cl⁻ ions (it cannot gain electrons from them) [1]. [5] Examiner insight: In (a) “a colour change” alone scores nothing; the colour must be named for each tube.

5. (a) Sulfur [1] (b) Correct species balanced with H⁺ and H₂O: H₂SO₄ + 6H⁺ → S + 4H₂O [1]; with electrons: H₂SO₄ + 6H⁺ + 6e⁻ → S + 4H₂O [1] (c) 6HI + H₂SO₄ → 3I₂ + S + 4H₂O [1] (d) Reducing agent (electron donor) [1]. [5] Examiner insight: A half-equation with the wrong number of electrons can still earn the first mark in (b) if atoms are balanced, but not the second.

6. (a) P: iodide [1]; Q: chloride [1]; R: fluoride [1]. (b) Ag⁺(aq) + Cl⁻(aq) → AgCl(s) [1] (c) To react with (remove) carbonate or hydroxide ions, which would otherwise form other silver precipitates [1]. [5] Examiner insight: State symbols are part of the mark in (b) when the question asks for them; AgCl without (s) does not score.

7. Mr(AgCl) = 107.9 + 35.5 = 143.4; n(AgCl) = 0.430 / 143.4 = 0.002999 mol [1]. n(NaCl) = n(AgCl) = 0.002999 mol (1 : 1) [1]. Concentration = 0.002999 / 0.0250 = 0.120 mol dm⁻³ [1]. [3] Examiner insight: Forgetting to convert 25.0 cm³ to dm³ gives an answer 1000 times too small; an error carried forward from a wrong Mr can still earn the later marks.

8. (a) Cl₂ + H₂O ⇌ HCl + HClO (or Cl₂ + H₂O ⇌ 2H⁺ + Cl⁻ + ClO⁻) [1] (b) Cl₂ 0; Cl⁻ (or HCl) −1; ClO⁻ (or HClO) +1 [1]; disproportionation [1]. (c) Correct species: Cl₂ and H₂O forming H⁺/HCl and O₂ [1]; balanced: 2Cl₂ + 2H₂O → 4HCl + O₂ [1]. [5] Examiner insight: In (b) all three oxidation states are needed for the mark; “redox” alone does not earn the name mark.

9. (a) Cl₂ + 2NaOH → NaCl + NaClO + H₂O [1] (b) Bleach (or disinfectant) [1] (c) n(Cl₂) = 3.55 / 71.0 = 0.0500 mol, so n(NaOH) = 0.100 mol [1]; mass = 0.100 × 40.0 = 4.00 g [1]. [4] Examiner insight: Using a 1 : 1 ratio instead of 1 : 2 gives 2.00 g and loses the first mark, but the second can be credited as an error carried forward.

10. Award one mark for each point.

• Both form steamy white fumes of hydrogen halide in an acid–base reaction [1].
• NaCl + H₂SO₄ → NaHSO₄ + HCl, and no further reaction occurs [1].
• With NaBr, HBr forms and then orange-brown fumes of bromine are seen [1].
• 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O; sulfur is reduced from +6 to +4 [1].
• Br⁻ is a stronger reducing agent than Cl⁻ [1].
• Br⁻ is larger, so its outer electrons are further from the nucleus and more shielded, and are lost more easily [1].

[6] Examiner insight: A six-mark comparison needs both halides covered; describing only the bromide reaction caps the answer.

11. Award one mark for each point, up to 6.

• Chlorine kills bacteria in water [1].
• This prevents water-borne disease, a large health benefit [1].
• Chlorine is toxic [1].
• It can react with organic matter to form chlorinated organic compounds that may be harmful [1].
• The amounts added are small and controlled [1].
• Judgement: the health benefits outweigh the toxic effects, so chlorination is justified [1].

[6] Examiner insight: “Evaluate” needs points on both sides and a final judgement; a list of benefits alone cannot reach full marks.

12. NH₄⁺: warm with sodium hydroxide solution and hold damp red litmus paper at the mouth of the tube [1]; litmus turns blue (ammonia given off) [1]. Br⁻: dissolve, add dilute nitric acid then silver nitrate solution [1]; cream precipitate, which does not dissolve in dilute ammonia but dissolves in concentrated ammonia [1]. [4] Examiner insight: Each ion is worth a method mark and a result mark, so a correct colour with no reagent named earns only half the credit for that ion.

Where marks are usually lost

• Confusing the oxidising trend of the halogens with the reducing trend of the halides.
• Saying covalent bonds break when halogens boil.
• Describing colour changes without naming the colour.
• Unbalanced charge in sulfuric acid half-equations.
• Missing the acid–base step with NaCl, or calling it redox.
• Acidifying with HCl before the silver nitrate test.
• Mixing up dilute and concentrated ammonia for AgBr.
• Not naming disproportionation for chlorine with water or NaOH.
• Using the wrong mole ratio for chlorine and sodium hydroxide.
• Ending an “evaluate” answer without a judgement.

Next steps

• Halogens revision notes
• Halogens study guide
• Oxidation, reduction and redox equations
• Periodicity practice questions
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.2.3 Group 7(17), the halogens (3.2.3.1 Trends in properties and 3.2.3.2 Uses of chlorine and chlorate(I)), including Required practical 4.