AS Chemistry: Group 17 — The Halogens — 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 available free from your board.

Related: Group 17 revision notes

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Questions

1. State and explain the trend in volatility down Group 17. [3]

2. State and explain the trend in reactivity down Group 17. [3]

3. Chlorine is bubbled through potassium bromide solution.

(a) Write the ionic equation. [1] (b) State the observation. [1] (c) Explain, in terms of oxidising power, why the reaction occurs. [2]

4. Describe the test for halide ions using silver nitrate, giving the observation for chloride, bromide and iodide, and the effect of adding ammonia. [6]

5. Chlorine reacts with cold, dilute sodium hydroxide.

(a) Write the equation. [1] (b) Deduce the oxidation states of chlorine in the products and explain why this is disproportionation. [3]

6. Chlorine is added to drinking water.

(a) State the benefit. [1] (b) State one risk. [1] (c) Explain why the practice is generally considered justified. [2]

7. Solid sodium bromide is warmed with concentrated sulfuric acid.

(a) Name the initial gas produced. [1] (b) Describe what is then observed as a further reaction occurs. [2] (c) Explain, in terms of the halide ion’s reducing power, why sodium chloride does not give this further reaction under the same conditions. [2]

8. Chlorine reacts with hot, concentrated sodium hydroxide.

(a) Write the equation. [1] (b) State the oxidation states of chlorine in the two chlorine-containing products, and explain how this differs from the reaction with cold, dilute sodium hydroxide. [3]

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Answers

1. Volatility decreases down the group (boiling points increase) [1]. The molecules have more electrons down the group [1], so the instantaneous dipole–induced dipole forces between molecules are stronger and more energy is needed to separate them [1].

2. Reactivity decreases down the group [1]. A halogen atom must gain an electron; down the group the outer shell is further from the nucleus with more shielding [1], so the incoming electron is attracted less strongly [1].

3. (a) Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ [1]. (b) The solution turns orange / yellow-brown [1]. (c) Chlorine is a stronger oxidising agent than bromine [1], so it removes electrons from bromide ions, displacing bromine [1].

4. Add dilute nitric acid, then silver nitrate solution [1]. Chloride — white precipitate [1]; bromide — cream precipitate [1]; iodide — yellow precipitate [1]. The chloride precipitate dissolves in dilute ammonia [1]; the bromide dissolves only in concentrated ammonia, and the iodide is insoluble in both [1].

5. (a) Cl₂ + 2NaOH → NaCl + NaClO + H₂O [1].

(b) Chlorine goes from 0 to −1 in NaCl and from 0 to +1 in NaClO [1] [1]. The same element is both oxidised and reduced in the same reaction, which is disproportionation [1].

6. (a) It kills bacteria, preventing waterborne disease such as cholera and typhoid [1]. (b) Chlorine is toxic, and it can form chlorinated hydrocarbons which are suspected carcinogens [1]. (c) The risk from untreated water is far greater than the small risk from chlorination [1] — waterborne disease killed very large numbers before treatment was routine, so the benefit substantially outweighs the harm [1].

7. (a) Hydrogen bromide, HBr [1]. (b) Red-brown fumes are seen [1], as some of the HBr is further oxidised by the sulfuric acid to bromine (Br₂), with the sulfuric acid reduced to sulfur dioxide (SO₂) [1]. (c) The chloride ion is too weak a reducing agent to reduce sulfuric acid at all, so with sodium chloride only HCl gas is produced with no further reaction [1]; the bromide ion is a stronger reducing agent than chloride (larger ion, outer electron more easily lost), so it is able to reduce the sulfuric acid further [1].

8. (a) 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O [1]. (b) Chlorine is −1 in NaCl and +5 in NaClO₃ [1] [1]. With cold, dilute NaOH the products are NaCl and NaClO (chlorine at −1 and +1) [1]; the hot, concentrated reaction pushes the oxidised product to a higher oxidation state (+5 rather than +1), showing that the outcome of disproportionation depends on the reaction conditions.

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Where marks are usually lost

• Saying volatility and reactivity trend in the same direction.
• Explaining volatility by “bigger molecules” rather than number of electrons.
• Forgetting to acidify before adding silver nitrate.
• Not giving both oxidation state changes when justifying disproportionation.
• Forgetting that iodide, the strongest reducing halide ion, reduces concentrated sulfuric acid further than bromide does, giving a mixture of sulfur dioxide, sulfur, and hydrogen sulfide (H₂S, rotten-egg smell) rather than stopping cleanly at sulfur dioxide.
• Assuming cold and hot sodium hydroxide give the same products with chlorine — they do not; the oxidation state of the oxidised chlorine product depends on the conditions.
• Writing halogen molecules as single atoms (Cl rather than Cl₂) in equations — they exist as diatomic molecules.

The halide-ion reducing power ladder

The concentrated sulfuric acid test is a direct consequence of the trend in halide reducing power down the group, and is worth learning as a single ladder: chloride cannot reduce sulfuric acid at all (HCl gas only); bromide reduces it as far as sulfur dioxide (HBr, then red-brown Br₂ fumes and SO₂); iodide, the strongest reducing agent of the three, reduces it further still, giving a mixture of sulfur dioxide, sulfur, and hydrogen sulfide (HI, then black/purple iodine, a yellow deposit of sulfur, and the rotten-egg smell of H₂S). The same trend — increasing reducing power down the group — also explains the displacement reactions in question 3: a more reactive halogen higher up the group oxidises, and therefore displaces, a less reactive halide ion lower down. For the full reaction schemes, including the hot-versus-cold sodium hydroxide comparison, see the Group 17: The Halogens revision notes.