Trends in properties — AQA A-Level Chemistry

Both explanations have to be built, not asserted. Electronegativity is the power of an atom to attract the pair of electrons in a covalent bond towards itself, which is not the same as its ability to gain an electron. So the explanation must name the increase in atomic radius and the increase in shielding, which together put the bonding pair further from, and less exposed to, the nucleus. The boiling point explanation must name the structure as simple molecular, name the intermolecular force as van der Waals or induced dipole-dipole, and then link the strength of that force to the number of electrons in each molecule. The commonest lost mark is saying that bonds are broken when a halogen boils; the second commonest is offering relative molecular mass as the cause with no mention of electrons.

A halogen acts as an oxidising agent by gaining electrons: X₂ + 2e⁻ → 2X⁻. Down the group the atom is larger and better shielded, so the incoming electron is attracted less strongly and the halogen is reduced less readily, which means oxidising power falls from F₂ to I₂. The consequence is that a halogen displaces the ions of any halogen below it. Chlorine added to aqueous sodium bromide gives an orange solution of bromine: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂. Chlorine added to aqueous sodium iodide gives a brown solution, or a black solid, of iodine: Cl₂ + 2I⁻ → 2Cl⁻ + I₂. Bromine displaces iodine but not chlorine, and iodine displaces neither. Shaking with cyclohexane confirms which halogen formed: bromine colours the organic layer orange and iodine colours it violet.

Reducing power runs the opposite way to oxidising power: I⁻ gives up an electron most readily because its outer electrons are furthest from the nucleus and most shielded, so reducing power increases down the group. Add concentrated sulfuric acid to the solid sodium halides and that difference shows. With NaCl there is no redox at all, only an acid and base reaction: NaCl + H₂SO₄ → NaHSO₄ + HCl, seen as misty white fumes of hydrogen chloride. With NaBr the HBr forms first and then reduces the acid: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O, giving brown bromine vapour and choking SO₂, with sulfur falling from +6 to +4. With NaI the reduction goes furthest, through SO₂ to yellow sulfur and finally to H₂S, which smells of bad eggs, and sulfur falls from +6 to −2.

Silver halides are insoluble and conveniently different in colour, which is what makes silver nitrate the standard test. Add dilute nitric acid first, then silver nitrate solution. Fluoride gives no precipitate, because silver fluoride is soluble. Chloride gives a white precipitate, bromide a cream one and iodide a yellow one, each by Ag⁺(aq) + X⁻(aq) → AgX(s). The precipitates darken on standing in light as silver is formed, which is the basis of photographic film. The acid matters: carbonate, hydroxide and sulfite ions all precipitate with silver ions and would be mistaken for a halide, and the nitric acid removes them. It must be nitric acid, because hydrochloric acid would introduce chloride ions and sulfuric acid would precipitate silver sulfate.

White and cream are hard to tell apart in a test tube, so ammonia is used to separate the three precipitates by solubility. Silver chloride dissolves in dilute ammonia. Silver bromide does not, but does dissolve in concentrated ammonia. Silver iodide is insoluble in both. Solubility in ammonia therefore falls down the group, matching the increasing insolubility of the silver halide itself. Dissolving is not a trick of the eye; the precipitate reacts to form a soluble complex ion, AgCl(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq). So a white precipitate that clears with dilute ammonia is chloride, a cream one that clears only with concentrated ammonia is bromide, and a yellow one that resists both is iodide.

Three separate 'why' questions hang on this one test. Silver nitrate is used because all nitrates are soluble, so nothing precipitates until a halide is present, and the silver halides that then form are insoluble and differ in colour. The solution is acidified with dilute nitric acid because carbonate, hydroxide and sulfite ions would give their own precipitates with silver ions and be misread as a halide; nitric acid removes them without adding an interfering ion, whereas hydrochloric acid would add chloride and sulfuric acid would precipitate silver sulfate. Ammonia is added afterwards to confirm which halide is present, because the three precipitates differ in how readily they dissolve to form the diamminesilver(I) complex.