Flashcards · 9701 · AS Topic 11

Group 17 (AS Level) flashcards

14 cards on AS Topic 11, subtopics 11.1 to 11.4. Pick a subtopic or study them all.

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Colours and states of chlorine, bromine and iodine at room temperature: Chlorine: pale yellow-green gas. Bromine: red-brown liquid. Iodine: grey-black solid. Colour darkens down the group.
Why does volatility decrease (boiling point increase) down Group 17?: The molecules have more electrons, so the instantaneous dipole–induced dipole forces between them are stronger.
Trend in X–X bond strength down Group 17, and why: Decreases (Cl–Cl 242, Br–Br 193, I–I 151 kJ mol⁻¹): the atoms are larger, so the shared pair is further from both nuclei. This is about the bond within the molecule, not the forces between molecules.

Trend in oxidising power of the halogens down the group, and why: Decreases: larger radius and more shielding, so an incoming electron is attracted less strongly.
Displacement: chlorine water added to KBr(aq) and to KI(aq): Cl₂ + 2KBr → 2KCl + Br₂: solution turns orange. Cl₂ + 2KI → 2KCl + I₂: solution turns brown. Iodine does not displace bromide.
Reactions of hydrogen with Cl₂, Br₂ and I₂: Cl₂: explosive in sunlight. Br₂: needs heat and a catalyst, less vigorous. I₂: slow, needs heat and a catalyst, and forms an equilibrium (does not go to completion).
Thermal stability of the hydrogen halides HCl, HBr and HI: Decreases down the group: HCl is stable to heat, HI decomposes significantly on strong heating, because the H–X bond gets weaker.

Trend in reducing power of the halide ions: Increases down the group: a larger ion loses an electron more easily.
Silver nitrate test for halide ions, with the ammonia follow-up: Acidify with dilute nitric acid, add AgNO₃(aq). Cl⁻: white, dissolves in dilute NH₃. Br⁻: cream, dissolves only in concentrated NH₃. I⁻: yellow, insoluble in both.
Products when concentrated H₂SO₄ is added to solid NaCl, NaBr and NaI: NaCl: HCl only (no redox). NaBr: HBr, then SO₂ and Br₂ (red-brown fumes). NaI: HI, then SO₂, S and H₂S (rotten-egg smell) and I₂.
Why do NaCl, NaBr and NaI give different products with concentrated sulfuric acid?: Increasing reducing power: Cl⁻ cannot reduce H₂SO₄; Br⁻ reduces it to SO₂; I⁻ reduces it further, to S and H₂S.

Chlorine with cold dilute NaOH: Cl₂ + 2NaOH → NaCl + NaClO + H₂O Disproportionation: Cl goes from 0 to −1 and to +1.
Chlorine with hot concentrated NaOH: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O Disproportionation: Cl goes from 0 to −1 and to +5.
Chlorine in water treatment: Cl₂ + H₂O ⇌ HCl + HClO. The chloric(I) acid (HClO) kills bacteria, which is why chlorine is used to treat water.