Part 1: Physical trends and displacement reactions

Explain why boiling points rise but oxidising power falls from chlorine to iodine, then predict displacement observations.

Diatomic molecules and London forces

Chlorine, bromine and iodine exist as Cl₂, Br₂ and I₂. At room conditions chlorine is a pale green gas, bromine a red-brown liquid and iodine a grey-black solid that gives purple vapour on heating.

Melting and boiling points rise down the group because the larger electron clouds are more polarisable, giving stronger London forces between molecules. More energy is needed to separate molecules. This trend does not mean the covalent X–X bond becomes harder to break on boiling.

Halogens gain an electron per atom

Halogen atoms have outer ns²np⁵ configurations and gain one electron to form halide ions. From Cl to I, larger radius and increased shielding reduce attraction for an incoming electron despite increasing nuclear charge. They gain electrons less readily and become weaker oxidising agents.

Chlorine oxidises bromide and iodide; bromine oxidises iodide but not chloride; iodine displaces neither chloride nor bromide. In Cl₂ + 2Br⁻, chlorine is reduced 0 → −1 and bromide is oxidised −1 → 0. The halogen is the oxidising agent, the halide the reducing agent.

Identify the free halogen, not the halide

Aqueous halide salts are normally colourless in these tests. The observed colour belongs to free Cl₂, Br₂ or I₂. Add a small amount of halogen water to a halide solution, mix and observe; an appropriate organic solvent can distinguish iodine’s purple colour from bromine’s orange. Use a fume cupboard where required and keep flammable solvents away from flames.

Predict every displacement before interpreting the colour

Rank the added free halogen as an oxidising agent: Cl₂ > Br₂ > I₂. A reaction is predicted when the added halogen is above the halogen belonging to the dissolved halide. A lower-ranked halogen cannot oxidise the higher-ranked halide under this comparison. Mixing the same halogen/halide pair causes no net displacement.

A “no reaction” tube may still be coloured because the added reagent already contains a coloured halogen. Record whether a new halogen forms; do not equate “orange” with “a reaction occurred”.

Build the equation from electron transfer

For bromine reacting with iodide, two iodide ions lose two electrons in total to form I₂. One Br₂ molecule gains those two electrons to form two bromide ions. Add the half-equations and cancel electrons. The final equation conserves both atoms and total charge −2.

If potassium iodide is used, K⁺ remains in solution. It belongs in a full formula equation but cancels from the ionic equation. Do not put the potassium ion into a redox half-equation when its oxidation state is unchanged.

Why boiling point and oxidising power move in opposite directions

Boiling separates intact X₂ molecules, so discuss London attractions between their electron clouds. Larger, more polarisable clouds down the group strengthen these attractions. Electron gain concerns attraction between a nucleus and an incoming electron, so discuss radius and shielding instead.

Thus iodine can have the highest boiling point of Cl₂, Br₂ and I₂ while being the weakest oxidising agent of those three. There is no contradiction: the two questions involve different particles and processes. “Iodine has stronger bonds so reacts less” fails to identify either process correctly.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources