Edexcel UK AS 8CH0 / A-Level 9CH0 · Topic 4 · Year 12 / AS

Part 3: Halogen trends, displacement and chlorine

Reviewed 9 October 2026.

Connect physical trends to intermolecular forces and reactivity trends to electron gain, then use oxidation numbers to explain chlorine’s useful reactions.

Melting and boiling temperatures generally increase down Group 7 because the molecules become larger and their electron clouds more polarisable. Stronger London attractions between molecules require more energy to overcome. The covalent X–X bonds do not need to be broken to melt or boil the halogen.

Electronegativity decreases down the group: increased distance and shielding weaken attraction to a bonding pair, despite more protons. This is a different phenomenon from the increasing intermolecular attraction that raises boiling temperature. One trend concerns nuclear attraction to bonding electrons, the other attraction between molecules.

Halogens at ordinary room conditions
ElementAppearance/stateTrend explanation
Fluorine, F₂Very pale yellow gasSmallest and least polarisable molecule in the group
Chlorine, Cl₂Pale green gasStronger London attractions than F₂
Bromine, Br₂Red-brown liquid; orange-brown vapourMore electrons and stronger London attractions
Iodine, I₂Grey-black solid; purple vapour on warmingLarge, readily polarised electron cloud

Halogens are oxidising agents because they gain electrons

A halogen molecule can gain two electrons to form two halide ions: X₂ + 2e⁻ → 2X⁻. The ability to oxidise other species decreases from chlorine to bromine to iodine. Down the group, the incoming electron is attracted less strongly because of increased atomic radius and shielding.

This means a halogen can oxidise the halide ion of a less reactive halogen. Chlorine reacts with bromide and iodide; bromine reacts with iodide; iodine does not displace chloride or bromide. Do not swap the trends: halogen oxidising power decreases down the group, while halide reducing ability increases.

With metals, the halogen is reduced and the metal is oxidised. For Group 1, 2Na + Cl₂ → 2NaCl; Na goes 0 → +1. For Group 2, Mg + Br₂ → MgBr₂; Mg goes 0 → +2. In both, each halogen atom goes 0 → −1.

Observe the product in the correct phase

Use small volumes of aqueous halogen and aqueous halide solution under the prescribed risk assessment. Mix, record the aqueous colour, then add a small amount of a suitable organic solvent such as cyclohexane, mix gently and allow the layers to separate. A coloured halogen often partitions into the organic layer, making its identity easier to distinguish.

Aqueous iodine is brown, while iodine in cyclohexane is violet/purple. Bromine is orange in an organic solvent and yellow-orange to brown in water, depending on concentration. Chlorine is pale green or nearly colourless when very dilute. Name the layer being described; purple is not the expected description for ordinary aqueous iodine.

Cyclohexane is less dense than water and forms the upper layer; a different solvent may have a different density, so do not memorise ‘organic always on top’. Keep flammable organic solvents away from flames and use ventilation appropriate to the halogen. Compare with controls containing the original halogen and solvent, since a colour can come from unreacted reagent as well as product.

Worked evidence chain: adding chlorine water to iodide solution produces brown aqueous iodine; after extraction the organic layer is violet. The reaction Cl₂ + 2I⁻ → 2Cl⁻ + I₂ shows iodine ions lose electrons to chlorine. Thus chlorine is a stronger oxidising agent than iodine. The 2023 AS Q3(b)(ii) report identifies missing solution conditions and observations/inferences as weaknesses.

Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)
Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq)
Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq)
Aqueous displacement predictions
Added halogenCl⁻Br⁻I⁻
Cl₂No displacementBr₂ formedI₂ formed
Br₂No displacementNo displacementI₂ formed
I₂No displacementNo displacementNo displacement

Chlorine in water undergoes disproportionation

Chlorine reacts reversibly with water to form hydrochloric acid and chloric(I) acid, HClO (also called hypochlorous acid). Chlorine begins at 0 and becomes −1 in chloride/HCl and +1 in HClO. The same starting element is reduced and oxidised, so this is disproportionation.

The solution is acidic and has an oxidising/bleaching action. Damp indicator may first show acidity and then lose its colour; bleaching is a chemical reaction, not proof that the solution has become neutral. Chlorine-based water treatment reduces the risk from microorganisms through active chlorine species such as HClO.

A balanced evaluation weighs pathogen control against chlorine’s toxicity, unwanted by-products, taste and the need to control dose. Chlorination is useful when carefully managed; it is not a reason to claim that unrestricted chlorine exposure is harmless or that treatment removes every possible contaminant.

Cl₂(aq) + H₂O(l) ⇌ HCl(aq) + HClO(aq)

Conditions determine the oxidation product

Cold, dilute aqueous sodium hydroxide gives sodium chloride and sodium chlorate(I), NaClO, an active component of bleach. Chlorine changes 0 → −1 and 0 → +1, so the chloride:chlorate(I) ratio is 1:1. The halogen colour fades as it is consumed.

Hot alkali gives chloride and chlorate(V), ClO₃⁻. One chlorine atom changes 0 → +5 and loses five electrons; five others change 0 → −1 and each gain one. The resulting chloride:chlorate(V) ratio is 5:1. Do not retain the cold-product formula ClO⁻ when the conditions or supplied product require ClO₃⁻.

Apply the same accounting to an analogous bromine or iodine reaction when the product is specified. The balanced generalized hot-alkali form is 3X₂ + 6OH⁻ → 5X⁻ + XO₃⁻ + 3H₂O. Product stability varies between halogens, so the conditions and information in the question matter; an analogy is not permission to assume every halogen behaves identically.

Bleach must not be mixed with acids or ammonia: different reactions can release harmful gases. In practical questions, tie risk controls to the actual reagent and reaction rather than giving a generic ‘be careful’ statement.

Cl₂ + 2NaOH → NaCl + NaClO + H₂O (cold, dilute aqueous alkali)
3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O (hot aqueous alkali)

Make predictions and state their limits

Following the group pattern, fluorine is expected to be a very strong oxidising agent and form fluoride salts with metals. Its extreme reactivity means it reacts with water, so the familiar aqueous chlorine/bromine/iodine displacement procedure cannot simply be repeated with F₂. Fluorine in its compounds is −1 and should not be assigned a positive halogen oxidation number by analogy with chlorate.

An extrapolation predicts astatine to be less electronegative, less effective as an oxidising agent and more likely to be a solid at room temperature than iodine. Astatide would be expected to act as a stronger reducing agent than iodide in the simple group model. These are stated as trend-based predictions: astatine is scarce and radioactive, and detailed behaviour is not established by the simple trend alone.

For an unfamiliar compound, separate the reliable inference (likely ion charge or electron-transfer direction) from a precise physical value that has not been supplied. Do not invent numerical boiling temperatures, measured colours or a safe classroom procedure for a rare radioactive sample.

Quick checks

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

Q1. Why is iodine solid while chlorine is gaseous at room temperature?Show answer

I₂ has a larger, more polarisable electron cloud and stronger London attractions between molecules. More energy is needed to separate them. This does not require stronger I–I covalent bonds than Cl–Cl bonds.

Q2. Write the ionic equation for bromine added to aqueous iodide and predict the colour after extraction into cyclohexane.Show answer

Br₂ + 2I⁻ → 2Br⁻ + I₂. The cyclohexane layer becomes violet/purple because of iodine. The aqueous iodine colour is brown; identify the phase explicitly.

Q3. Explain why chlorine’s reaction with cold dilute alkali is disproportionation.Show answer

Chlorine starts at 0 in Cl₂ and forms −1 in Cl⁻ and +1 in ClO⁻. Some chlorine is reduced and some oxidised from the same starting species. The ionic equation is Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.

Q4. Why does the hot-alkali chlorate(V) equation form five chloride ions for each chlorate(V) ion?Show answer

One Cl atom going 0 → +5 loses five electrons. Each atom going 0 → −1 gains one, so five reduced atoms balance one oxidised atom. This gives 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O.

Q5. Give one useful prediction about astatine and one reason not to repeat the aqueous chlorine experiment with fluorine.Show answer

Astatine is predicted to be a weaker oxidising agent than iodine from the down-group trend. Fluorine reacts with water itself, so an aqueous displacement experiment would introduce a different reaction and severe hazards rather than providing a simple like-for-like comparison.

Sources

Sources and examiner guidance (reviewed 9 October 2026)

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