OxfordAQA A-Level Chemistry: Group 7(17), the halogens (9620) – 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.

This page practises section 3.2.3, covering the Group 7(17) halogens, of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, namely point 3.2.3.1 on trends in properties, point 3.2.3.2 on the uses of chlorine and chlorate(I), and Required practical 3. All of it is International AS. It is set in the Unit 1 paper (Inorganic 1 and Physical 1); the Unit 5 paper (Practical and synoptic) can also draw on it.

Work through the study guide and the revision notes first. To plan your revision, use the course hub, the printable checklist or a free 10-minute diagnostic. Relative atomic masses to use: Ag = 107.9, Br = 79.9, Cl = 35.5, H = 1.0, Mg = 24.3, Na = 23.0, O = 16.0.

Questions

1. The H–Cl bond is more polar than the H–I bond. Explain why, in terms of electronegativity. [3]

2. A student writes: “Fluorine boils at a lower temperature than iodine because the F–F bond is weaker than the I–I bond.” Explain why the student is wrong and give the correct explanation. [3]

3. A colourless solution contains one sodium halide, Z. Bromine water and cyclohexane are added and the mixture is shaken. The upper organic layer turns violet.

(a) Identify the halide ion in Z. [1] (b) Write an ionic equation for the reaction. [1] (c) Explain why bromine reacts with the ion in Z but does not react with chloride ions. [2]

4. Concentrated sulfuric acid can be reduced to SO₂, S or H₂S by halide ions.

(a) State the oxidation state of sulfur in each of SO₂, S and H₂S. [1] (b) Name the sodium halide that reduces sulfuric acid as far as H₂S. [1] (c) Explain why sodium chloride gives only steamy fumes. [1]

5. Solid sodium bromide reacts with concentrated sulfuric acid to form sodium hydrogensulfate, bromine, sulfur dioxide and water.

(a) Write a balanced equation for this reaction, starting from NaBr. [2] (b) Use oxidation states to identify the element oxidised and the element reduced. [2]

6. A solution contains both chloride ions and iodide ions. It is acidified with dilute nitric acid and silver nitrate solution is added in excess. Excess dilute ammonia solution is then added.

(a) Describe what is seen after each addition. [2] (b) Explain the observation after adding the ammonia. [1] (c) Write an equation, with state symbols, for the change that takes place when ammonia is added. [1]

7. 20.0 cm³ of 0.0450 mol dm⁻³ magnesium chloride solution, MgCl₂, is acidified with dilute nitric acid.

(a) Calculate the volume of 0.120 mol dm⁻³ silver nitrate solution that reacts exactly with all the chloride ions. [2] (b) Calculate the mass of precipitate formed. [2]

8. A student claims that chlorine is simply reduced when it dissolves in water.

(a) Write an equation for the reversible reaction of chlorine with water that forms chloride ions and chlorate(I) ions. [1] (b) Use the oxidation states of chlorine in the products to show that the student is wrong, and name this type of reaction. [2] (c) In sunlight the reaction is 2Cl₂ + 2H₂O → 4H⁺ + 4Cl⁻ + O₂. Use oxidation states to explain why this is not the same type of reaction as in (b). [2]

9. A works absorbs 1.50 kg of chlorine in excess cold, dilute sodium hydroxide, making 40.0 dm³ of solution.

(a) Calculate the maximum concentration of sodium chlorate(I), NaClO, in this solution, in mol dm⁻³. [2] (b) State one use of the solution produced. [1]

10. A town council is deciding whether to chlorinate a new drinking-water supply.

(a) State why chlorine is added to drinking water. [1] (b) Give two disadvantages of adding chlorine. [2] (c) Explain why chlorination is still used despite these disadvantages. [1] (d) Fluoride is added to water in some places. Give one reason for adding it and one reason against. [2]

11. Four unlabelled bottles contain white solids: sodium carbonate, sodium sulfate, sodium iodide and ammonium chloride. Describe one test for each solid that would identify it, and give the positive result for each. [8]

12. A white solid, W, is a single sodium halide. With concentrated sulfuric acid, W gives steamy fumes and an orange-brown vapour, but no purple vapour, no yellow solid and no bad-egg smell.

(a) Identify W, explaining how the observations rule out the other halides. [2] (b) A solution of W is acidified and silver nitrate solution is added. Predict the colour of the precipitate and how it behaves in dilute and in concentrated ammonia. [2] (c) A 0.618 g sample of W is dissolved and treated with excess acidified silver nitrate. Calculate the mass of precipitate expected. [3] (d) A student records 1.16 g. Suggest one practical reason why this is higher than expected. [1] (e) State the role of the halide ion in W in its reaction with sulfuric acid. [1]

Answers

1. Chlorine is more electronegative than iodine [1]. The Cl atom is smaller, with fewer shells and less shielding, so the bonding pair is closer to its nucleus and attracted more strongly [1]. The difference in electronegativity between H and Cl is therefore larger, so the electron density is pulled further towards Cl, giving a larger δ− [1]. [3] Examiner insight: “Chlorine has a bigger nuclear charge” is wrong (iodine has more protons); credit goes to distance and shielding.

2. Boiling does not break the covalent X–X bond; only the forces between molecules are overcome [1]. I₂ has more electrons (106) than F₂ (18), so the van der Waals forces between I₂ molecules are stronger [1]. Separating I₂ molecules therefore takes more energy, so iodine boils at the higher temperature [1]. [3] Examiner insight: Say plainly that no covalent bond breaks; otherwise “why the student is wrong” is left unanswered.

3. (a) Iodide, I⁻ [1] (b) Br₂ + 2I⁻ → 2Br⁻ + I₂ [1] (c) Bromine is a stronger oxidising agent than iodine but a weaker one than chlorine [1]. So bromine can take electrons from iodide ions, but cannot take them from chloride ions [1]. [4] Examiner insight: In (c), compare bromine with both iodine and chlorine, not just one.

4. (a) SO₂ +4, S 0, H₂S −2 (all three for the mark) [1] (b) Sodium iodide [1] (c) Chloride ions are too weak a reducing agent to reduce sulfuric acid, so only the acid–base reaction forming HCl happens [1]. [3] Examiner insight: In (c), “chloride is unreactive” is too vague; the answer needs the idea that Cl⁻ is not a strong enough reducing agent.

5. (a) Correct formulae of all products from NaBr and H₂SO₄ [1]; 2NaBr + 3H₂SO₄ → 2NaHSO₄ + Br₂ + SO₂ + 2H₂O [1] (b) Bromine is oxidised, −1 → 0 [1]; sulfur is reduced, +6 → +4 [1]. [4] Examiner insight: One H₂SO₄ is reduced and two more form NaHSO₄; two H₂SO₄ cannot balance.

6. (a) After silver nitrate: a pale yellow / mixed precipitate (white AgCl and yellow AgI) [1]. After dilute ammonia: some precipitate dissolves and a yellow precipitate remains [1]. (b) AgCl dissolves in dilute ammonia but AgI does not [1]. (c) AgCl(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq) [1] [4] Examiner insight: For (c) the equation must be for the silver halide that dissolves; writing it for AgI contradicts the answer to (b).

7. (a) n(MgCl₂) = 0.0200 × 0.0450 = 9.00 × 10⁻⁴ mol, so n(Cl⁻) = 1.80 × 10⁻³ mol [1]; volume = 1.80 × 10⁻³ ÷ 0.120 = 0.0150 dm³ = 15.0 cm³ [1] (b) Mr(AgCl) = 107.9 + 35.5 = 143.4 [1]; mass = 1.80 × 10⁻³ × 143.4 = 0.258 g [1] [4] Examiner insight: Each MgCl₂ supplies two chloride ions; a 1 : 1 ratio halves both answers.

8. (a) Cl₂ + H₂O ⇌ HClO + HCl (or Cl₂ + H₂O ⇌ 2H⁺ + ClO⁻ + Cl⁻) [1] (b) Cl⁻ (HCl) −1 but ClO⁻ (HClO) +1, so chlorine is oxidised as well as reduced [1]; disproportionation [1] (c) Chlorine only goes from 0 to −1, so it is only reduced [1]; oxygen goes from −2 to 0 and is oxidised, so two different elements change oxidation state [1]. [5] Examiner insight: “Redox” on its own does not name the reaction in (b); the specific term disproportionation is what gains the mark.

9. (a) n(Cl₂) = 1500 ÷ 71.0 = 21.13 mol = n(NaClO) [1]; concentration = 21.13 ÷ 40.0 = 0.528 mol dm⁻³ [1] (b) Bleach (or a disinfectant for cleaning) [1] [3] Examiner insight: Convert 1.50 kg to grams before dividing by 71.0.

10. (a) To kill bacteria (and so prevent water-borne disease) [1] (b) Chlorine is toxic [1]; with organic matter in water it forms chlorinated organic compounds that may be harmful [1] (c) The health benefits of killing bacteria outweigh the toxic effects at the small amounts used [1] (d) For: helps prevent tooth decay [1]. Against: too much causes fluorosis (or people disagree about adding it to everyone’s water) [1]. [6] Examiner insight: In (a), “to clean the water” or “to purify it” is too vague; the answer needs bacteria or microorganisms.

11. Ammonium chloride: warm with sodium hydroxide solution and test the gas given off with damp red litmus paper [1]; litmus turns blue [1]. Sodium carbonate: add dilute acid and bubble the gas through limewater [1]; fizzing and limewater turns milky [1]. Sodium sulfate: dissolve and add acidified barium chloride solution [1]; white precipitate [1]. Sodium iodide: dissolve in water, acidify with dilute nitric acid and add silver nitrate solution [1]; yellow precipitate, which stays even in concentrated ammonia [1]. [8] Examiner insight: For the iodide, write “dilute nitric acid”, not just “acid”; an unnamed acid could be hydrochloric, which adds chloride ions, so the method is less likely to gain credit.

12. (a) Sodium bromide: orange-brown vapour is Br₂, so the halide was oxidised to bromine, which chloride or fluoride cannot do [1]. No purple vapour, sulfur or H₂S rules out iodide [1]. (b) Cream precipitate [1]; stays in dilute ammonia but dissolves when concentrated ammonia is added [1]. (c) Mr(NaBr) = 102.9; n = 0.618 ÷ 102.9 = 6.006 × 10⁻³ mol [1]; n(AgBr) = 6.006 × 10⁻³ mol [1]; Mr(AgBr) = 187.8, mass = 6.006 × 10⁻³ × 187.8 = 1.13 g [1] (d) The precipitate was not fully dried (water still present) [1]. (e) Reducing agent (electron donor) [1] [9] Examiner insight: In (a), naming bromide without using the missing iodide signs gains only part of the credit, because the question asks you to rule out the other halides.

Where marks are usually lost

• Explaining electronegativity with nuclear charge instead of distance and shielding.
• Leaving NaHSO₄ out of an equation that starts from the solid sodium halide.
• Calling every chlorine reaction disproportionation, including the sunlight one.
• Giving a test with no result, or a result with no reagent, in Required practical 3 questions.

Next steps

• Revision notes: the halogens
• Study guide: the halogens
• Oxidation, reduction and redox equations practice
• Course hub: OxfordAQA A-Level Chemistry
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Source: OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards; within it, section 3.2.3, the halogens of Group 7(17), at International AS: point 3.2.3.1, on trends in properties; point 3.2.3.2, on the uses of chlorine and chlorate(I); and Required practical 3.