OCR A Level Chemistry A: The halogens (H432) – 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.

Here you practise section 3.1.3, The halogens, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, covering outcomes 3.1.3(a) to (g) from Module 3, Periodic table and energy. H432 is the full A level. The last three questions are longer and mix halogen chemistry with mole calculations.

Revise first with the halogens study guide and the halogens revision notes. Other topics are on the OCR chemistry hub; mark off outcomes on the printable checklist. To test other topics too, use the free diagnostics.

Where a calculation needs data, it is given in the question.

Questions

1. This question is about bromine.

(a) Give the full electron configuration of a bromine atom. [1] (b) Write a half-equation for bromine molecules forming bromide ions. State, with a reason, whether bromine is oxidised or reduced. [2]

2. Chlorine, bromine and iodine all exist as diatomic molecules.

(a) State the physical state of bromine and of iodine at room temperature. [1] (b) Explain the difference in physical state between chlorine and iodine under room conditions. [3]

3. A student adds chlorine water to sodium bromide solution in one tube and to sodium iodide solution in another. She then shakes each tube with cyclohexane.

(a) State the colour of the organic layer in each tube. [2] (b) Write the full equation for the reaction in the sodium iodide tube. [1] (c) Write the ionic equation for the reaction in the sodium bromide tube. [1]

4. Explain why chlorine is a stronger oxidising agent than iodine. [3]

5. An aqueous halogen, Y, is either chlorine, bromine or iodine. Y is added separately to solutions of sodium chloride, sodium bromide and sodium iodide. A reaction takes place only with sodium iodide.

(a) Identify Y. [1] (b) Write the ionic equation for the reaction. [1] (c) Use oxidation numbers to identify the species oxidised and the species reduced. [2]

6. Chlorine is bubbled into water at a treatment works.

(a) Write an equation for the reaction of chlorine with water. [1] (b) Use oxidation numbers to show that this is a disproportionation reaction. [2] (c) Name the chlorine-containing acid that kills bacteria. [1]

7. Bromine reacts in a similar way to chlorine.

(a) Give the equation for bromine reacting with water. [1] (b) Give the equation for bromine reacting with cold, dilute sodium hydroxide. [1] (c) Name the two bromine-containing products in (b) and give the oxidation number of bromine in each. [2]

8. When chlorine is passed into hot, concentrated potassium hydroxide, this reaction occurs:

3Cl₂ + 6KOH → 5KCl + KClO₃ + 3H₂O

(a) Give the oxidation number of chlorine in KClO₃. [1] (b) Explain why this reaction is a disproportionation. [2] (c) State the ratio of chlorine atoms reduced to chlorine atoms oxidised, and show that the electrons gained equal the electrons lost. [2]

9. Three colourless solutions, in bottles labelled J, K and L, contain potassium chloride, potassium bromide and potassium iodide in some order.

(a) Describe how you would identify each solution using silver nitrate and aqueous ammonia. Give the result for each halide. [5] (b) Give an ionic equation, including state symbols, for the precipitate that forms with potassium chloride. [1]

10. A factory makes bleach by passing 7.10 kg of chlorine into cold sodium hydroxide solution of concentration 4.00 mol dm⁻³. Use Ar: Na = 23.0, Cl = 35.5, O = 16.0.

(a) Give a balanced equation for the reaction. [1] (b) Calculate the amount, in mol, of chlorine used. [1] (c) Calculate the minimum volume, in dm³, of sodium hydroxide solution needed. [2] (d) The product is made up to 250 dm³ with water. Calculate the concentration of sodium chlorate(I), in mol dm⁻³. [2] (e) State one use of the solution. [1]

11. A small island community is choosing how to treat its drinking water. One proposal is to use ultraviolet light alone. Another is to use ultraviolet light and then add a small amount of chlorine before the water enters the pipes.

Evaluate the second proposal. Refer to the chemistry of chlorine in water, the benefits, the risks and the alternatives. [6]

12. A 20.0 cm³ sample of a solution contains sodium chloride and sodium bromide. It is acidified with dilute nitric acid and excess silver nitrate solution is added. The dried precipitate weighs 0.2373 g. This precipitate is then shaken with excess dilute aqueous ammonia, and the solid left is filtered off and dried. It weighs 0.0939 g. Use Ar: Ag = 107.9, Cl = 35.5, Br = 79.9.

(a) Identify the solid left after the ammonia treatment and explain why it remains. [2] (b) Calculate the concentrations of sodium chloride and sodium bromide in the sample, in mol dm⁻³. [4] (c) Calculate the minimum volume of 0.0500 mol dm⁻³ silver nitrate solution needed to precipitate all the halide ions, in cm³. [1]

Answers

1. (a) 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁵ [1] (b) Br₂ + 2e⁻ → 2Br⁻ [1]; reduced, because bromine gains electrons (oxidation number 0 → −1) [1] [3] Examiner insight: Electrons must appear on the left of the half-equation; “reduced” without a reason in terms of electrons or oxidation number is unlikely to gain the second mark.

2. (a) Bromine is a liquid and iodine is a solid (both needed) [1] (b) Both are simple molecular, held together by London forces between molecules [1]. I₂ has 106 electrons and Cl₂ has 34 [1]. So I₂ has stronger London forces, which need more energy to overcome; at room temperature there is not enough energy to separate I₂ molecules [1] [4] Examiner insight: Any mention of breaking covalent bonds during melting or boiling usually costs the explanation mark, even if the rest is correct.

3. (a) Sodium bromide tube: orange [1]; sodium iodide tube: violet (purple) [1] (b) Cl₂ + 2NaI → 2NaCl + I₂ [1] (c) Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ [1] [4] Examiner insight: The question asks for the organic layer, so “brown” for iodine is the aqueous colour and would not be accepted here.

4. Chlorine atoms are smaller, with fewer electron shells, than iodine atoms [1]. There is less shielding in chlorine [1]. So the nucleus attracts an incoming electron more strongly and chlorine forms Cl⁻ more easily, meaning it gains electrons from other species more readily [1] [3] Examiner insight: Each of radius, shielding and attraction is a separate point; “chlorine is higher up the group” alone describes, it does not explain.

5. (a) Bromine [1] (b) Br₂ + 2I⁻ → 2Br⁻ + I₂ [1] (c) Iodide ions are oxidised: −1 → 0 [1]. Bromine is reduced: 0 → −1 [1] [4] Examiner insight: Identify species, not just elements: “I⁻ is oxidised” is clearer than “iodine is oxidised”, which could mean I₂.

6. (a) Cl₂ + H₂O ⇌ HClO + HCl (→ also allowed) [1] (b) Chlorine in Cl₂ is 0; in HClO it is +1, so it is oxidised; in HCl it is −1, so it is reduced [1]. The same element is both oxidised and reduced, so this is disproportionation [1] (c) Chloric(I) acid, HClO [1] [4] Examiner insight: Both changes, with both values, are needed for (b); a single oxidation number change only shows ordinary redox.

7. (a) Br₂ + H₂O ⇌ HBrO + HBr [1] (b) Br₂ + 2NaOH → NaBr + NaBrO + H₂O [1] (c) Sodium bromide, −1 [1]; sodium bromate(I), +1 [1] [4] Examiner insight: For analogous equations, copy the chlorine pattern exactly; writing NaBrO₃ for a cold, dilute reaction is a common slip that loses the mark.

8. (a) +5 [1] (b) Chlorine starts at 0. It goes to −1 in KCl (reduced) [1] and to +5 in KClO₃ (oxidised); the same element is both oxidised and reduced [1] (c) Five Cl atoms are reduced for every one oxidised, ratio 5 : 1 [1]. Electrons gained: 5 × 1 = 5; electrons lost: 1 × 5 = 5, so they balance [1] [5] Examiner insight: An unfamiliar equation is still tested by oxidation numbers; show the arithmetic for the electron count rather than just stating “they balance”.

9. (a) Take a sample of each and add dilute nitric acid [1], then aqueous silver nitrate [1]. Chloride gives a white precipitate, bromide cream, iodide yellow [1]. Add dilute ammonia: only the white precipitate (chloride) dissolves [1]. Add concentrated ammonia to the other two: the cream one (bromide) dissolves; the yellow one (iodide) does not [1] (b) Ag⁺(aq) + Cl⁻(aq) → AgCl(s) [1] [6] Examiner insight: Colours alone are close, so the ammonia results are what make the identification secure; in (b), include all four state symbols because the question asks for them.

10. (a) Cl₂ + 2NaOH → NaCl + NaClO + H₂O [1] (b) n(Cl₂) = 7100 ÷ 71.0 = 100 mol [1] (c) n(NaOH) = 2 × 100 = 200 mol [1]; volume = 200 ÷ 4.00 = 50.0 dm³ [1] (d) n(NaClO) = 100 mol [1]; concentration = 100 ÷ 250 = 0.400 mol dm⁻³ [1] (e) Bleach (or disinfectant) [1] [7] Examiner insight: Change 7.10 kg into grams first; skipping that step gives 0.100 mol, and every later answer is then a thousand times too small.

11. Indicative points, one mark each, up to 6:

• Chlorine reacts with water to form HClO and HCl: Cl₂ + H₂O ⇌ HClO + HCl [1]
• HClO (and ClO⁻) kills bacteria, reducing diseases such as cholera [1]
• Ultraviolet kills microorganisms at the works but leaves no residue, while chlorine keeps protecting water in the pipes [1]
• Risk: chlorine gas is toxic, so storing and handling it needs care [1]
• Risk: chlorine can react with organic matter to form chlorinated hydrocarbons, some suspected of causing cancer [1]
• Judgement: the small amount added makes the benefit outweigh the risk, so the second proposal is better; some residents may object on the grounds of right to choose [1] [6] Examiner insight: “Evaluate” needs a judgement; a list of facts with no conclusion comparing the two proposals does not answer the command word.

12. (a) Silver bromide, AgBr [1]. AgCl dissolves in dilute ammonia but AgBr does not [1] (b) n(AgBr) = 0.0939 ÷ 187.8 = 5.00 × 10⁻⁴ mol [1]; mass of AgCl = 0.2373 − 0.0939 = 0.1434 g, n(AgCl) = 0.1434 ÷ 143.4 = 1.00 × 10⁻³ mol [1]. c(NaBr) = 5.00 × 10⁻⁴ ÷ 0.0200 = 0.0250 mol dm⁻³ [1]; c(NaCl) = 1.00 × 10⁻³ ÷ 0.0200 = 0.0500 mol dm⁻³ [1] (c) n(Ag⁺) = 1.50 × 10⁻³ mol; volume = 1.50 × 10⁻³ ÷ 0.0500 = 0.0300 dm³ = 30.0 cm³ [1] [7] Examiner insight: Find the AgCl mass by subtraction first; dividing the total mass by one Mr mixes the two halides and gives wrong values for both.

Where marks are usually lost

• Linking boiling point to “mass” or “size” instead of the number of electrons and London forces.
• Giving aqueous colours when the question asks about the organic layer.
• Leaving spectator ions (Na⁺, K⁺) in an ionic equation.
• Explaining the reactivity trend without the weaker attraction for an incoming electron.
• Showing only one oxidation number change in a disproportionation answer.
• Writing NaClO₃ or KBrO₃ for reactions with cold, dilute alkali.
• Missing the dilute nitric acid step, or using hydrochloric acid to acidify.
• Saying silver bromide dissolves in dilute ammonia.
• Forgetting to convert kg to g, or cm³ to dm³, in bleach and precipitate calculations.
• Ending an “evaluate” answer without a judgement.

Next steps

• The halogens revision notes
• The halogens study guide
• Redox practice questions
• OCR A Level Chemistry hub
• Printable OCR chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.1.3 The halogens, learning outcomes (a) to (g). Published by OCR.