OCR A Level Chemistry A: The halogens (H432)

This study guide teaches section 3.1.3, The halogens, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. It belongs to Module 3, Periodic table and energy, and its seven lettered learning outcomes, 3.1.3(a) to (g), are all taught below with worked examples. H432 is the full A level, and OCR assesses Module 3 in Periodic table, elements and physical chemistry (01) and Unified chemistry (03). Outcome 3.1.3(g) is linked to PAG4, qualitative analysis of ions.

Every H432 topic is linked from the OCR chemistry hub; tick off outcomes on the printable checklist. To find weak spots before you start, try the free diagnostics.

Outcomes in 3.1.3

Outcome	Skill	Notes from the specification
3.1.3(a)	State that halogens exist as diatomic molecules; explain the boiling point trend of Cl₂, Br₂ and I₂ using London forces	Links to 2.2.2(k)
3.1.3(b)	Use the s²p⁵ outer shell to explain why halogens gain one electron to form 1− ions	Explain redox by electron transfer and oxidation number change; write full and ionic equations (links to 2.1.5)
3.1.3(c)	Show the reactivity trend of Cl₂, Br₂ and I₂ by their reactions with other halide ions	Include colour changes in aqueous and organic solutions
3.1.3(d)	Explain the trend in (c) from the decreasing ease of forming 1− ions, using attraction, atomic radius and shielding	–
3.1.3(e)	Explain disproportionation using chlorine with water, chlorine with cold dilute NaOH, and analogous reactions	–
3.1.3(f)	Weigh the benefits of chlorinating water against the risks	HSW9, 10, 12: benefits against risks, people’s right to choose, other purification methods
3.1.3(g)	Use precipitation with Ag⁺(aq), then aqueous ammonia, to test for Cl⁻, Br⁻ and I⁻, with ionic equations	Ammonia complexes not required beyond observations; PAG4; HSW4; see 3.1.4(a)

The specification names chlorine, bromine and iodine in this section. Fluorine and astatine are not part of the listed reactions, so this guide does not teach them.

3.1.3(a): Diatomic molecules and boiling points

Each halogen atom has seven outer electrons. Two atoms share one pair to make a single covalent bond, so every halogen element exists as X₂ molecules: Cl₂, Br₂, I₂. The solids and liquids are simple molecular. When a halogen boils, the X–X covalent bond does not break. Only the weak forces between molecules are overcome.

Chlorine is a pale green gas under ordinary conditions. Bromine is a dense red-brown liquid giving off orange-brown vapour. Iodine is a shiny grey-black solid that gives a purple vapour when warmed gently.

Halogen	Electrons per molecule	Boiling point (approx.)	State at room temperature
Cl₂	34	−34 °C	gas
Br₂	70	59 °C	liquid
I₂	106	184 °C	solid

The only forces between these non-polar molecules are induced dipole–dipole interactions, called London forces. A temporary dipole in one molecule induces a dipole in its neighbour. More electrons in a molecule means larger temporary dipoles and stronger London forces, so more energy is needed to separate the molecules.

Worked example 1: a three-point boiling point answer

Question: Explain why iodine boils at a much higher temperature than chlorine.

1. Both are simple molecular substances made of diatomic molecules held together by London forces.
2. An I₂ molecule has 106 electrons; a Cl₂ molecule has 34.
3. More electrons give stronger London forces between I₂ molecules, so more energy is needed to overcome them.

Name the force, compare electron numbers, link to energy.

3.1.3(b): Gaining one electron

The outer shell of every halogen is s²p⁵: for example chlorine is [Ne]3s²3p⁵ and bromine is [Ar]3d¹⁰4s²4p⁵. One more electron fills the p sub-shell, so in many redox reactions a halogen atom gains one electron to form a 1− halide ion.

Cl₂ + 2e⁻ → 2Cl⁻

The halogen’s oxidation number falls from 0 to −1, so the halogen is reduced. Because it removes electrons from something else, a halogen acts as an oxidising agent. In this section, describe each redox reaction both ways: by electron transfer and by oxidation number change. Write the full equation and the ionic equation.

3.1.3(c): Displacement reactions and colours

A more reactive halogen oxidises the halide ion of a less reactive halogen. The reactivity order is Cl₂ > Br₂ > I₂.

Halogen added	to Cl⁻(aq)	to Br⁻(aq)	to I⁻(aq)
Cl₂(aq)	–	reacts: Br₂ forms	reacts: I₂ forms
Br₂(aq)	no reaction	–	reacts: I₂ forms
I₂(aq)	no reaction	no reaction	–

Colours let you see which halogen is present at the end. In water the colours can look alike at low concentration, so add cyclohexane (an organic solvent) and shake. Halogens dissolve better in it than in water, and it floats as a top layer that shows the colours clearly.

Halogen	Colour in water	Colour in cyclohexane
Cl₂	pale green	pale green
Br₂	orange	orange
I₂	brown	violet (purple)

Worked example 2: bromine water with two halide solutions

Bromine water is added to two test tubes, one holding potassium chloride solution and the other potassium iodide solution. Cyclohexane is then added and each tube is shaken.

• Potassium chloride: no reaction, because bromine is less reactive than chlorine. The organic layer is orange from the unreacted bromine.
• Potassium iodide: the aqueous mixture turns brown and the organic layer turns violet, because iodine forms.

Full equation: Br₂(aq) + 2KI(aq) → 2KBr(aq) + I₂(aq)

Ionic equation: Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq)

Oxidation numbers: bromine goes from 0 to −1 (reduced); iodine goes from −1 to 0 (oxidised). Each Br₂ gains two electrons; two I⁻ ions each lose one. The potassium ions are spectator ions, which is why they drop out of the ionic equation.

3.1.3(d): Why reactivity falls down the group

Halogen reactivity here means how readily the atom gains an electron to become a 1− ion. Going from chlorine to iodine:

• the atomic radius increases, because each element has an extra shell of electrons;
• shielding from inner shells increases;
• so the nuclear attraction felt by an incoming electron decreases, even though the nuclear charge is larger.

The incoming electron is attracted less strongly, so the 1− ion forms less easily. Iodine is the weakest oxidising agent of the three and chlorine the strongest. Always talk about the atom gaining an electron.

3.1.3(e): Disproportionation

Disproportionation is a redox reaction in which the same element is both oxidised and reduced.

Chlorine with water

Cl₂(aq) + H₂O(l) ⇌ HClO(aq) + HCl(aq)

Chlorine starts at 0. In HCl it is −1 (reduced) and in HClO, chloric(I) acid, it is +1 (oxidised). One chlorine atom of each Cl₂ goes each way. This reaction is the basis of chlorine treatment of drinking water.

Chlorine with cold, dilute sodium hydroxide

Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Ionic form: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O

Again one chlorine atom is reduced (0 to −1, in NaCl) and one is oxidised (0 to +1, in NaClO, sodium chlorate(I)). The solution of sodium chlorate(I) is household bleach.

Analogous reactions, outcome (e)(iii)

Bromine and iodine behave in the same way. Expect to write equations for a different halogen or a different alkali by analogy.

Worked example 3: bromine with potassium hydroxide

Question: Write an equation for bromine reacting with cold, dilute potassium hydroxide. Name the bromine-containing products and show that this is disproportionation.

Br₂ + 2KOH → KBr + KBrO + H₂O

Products: potassium bromide (Br −1) and potassium bromate(I) (Br +1). Bromine starts at 0, so one atom is reduced and one is oxidised: the same element is oxidised and reduced, which is disproportionation.

Worked example 4: making a bleach solution

A technician needs 500 cm³ of bleach containing 0.0800 mol dm⁻³ NaClO, made from chlorine and 2.00 mol dm⁻³ NaOH(aq). Use Ar: Cl = 35.5.

1. n(NaClO) = 0.500 × 0.0800 = 0.0400 mol
2. From the equation, 1 mol Cl₂ gives 1 mol NaClO, so n(Cl₂) = 0.0400 mol.
3. Mass of Cl₂ = 0.0400 × 71.0 = 2.84 g
4. n(NaOH) = 2 × 0.0400 = 0.0800 mol
5. Volume of NaOH(aq) = 0.0800 ÷ 2.00 = 0.0400 dm³ = 40.0 cm³

3.1.3(f): Chlorine in water treatment, benefits and risks

Benefit. Chloric(I) acid and chlorate(I) ions formed in the water kill bacteria. Chlorination has sharply reduced waterborne diseases such as cholera and typhoid. A small residual amount stays in the pipes and keeps protecting the water on its way to taps.

Risks.

• Chlorine gas is toxic. It irritates the respiratory system, so storage and transport need careful control.
• Chlorine may combine with dissolved organic compounds, giving chlorinated hydrocarbons; some of these are suspected carcinogens.

Weighing it up. Most public health bodies judge that the benefit of safe drinking water is far greater than the small risks. Some people argue that adding a chemical to everyone’s water removes their right to choose. Other ways of purifying water exist: ozone treatment and ultraviolet light both kill microorganisms, but neither leaves a residue that protects water in the pipes. Filtration removes solids but does not kill bacteria on its own.

A strong answer gives a benefit, a named risk and a judgement.

3.1.3(g): Testing for halide ions (PAG4)

1. Add dilute nitric acid to the solution. This removes ions such as carbonate, which would otherwise form a silver precipitate too.
2. Add aqueous silver nitrate. A precipitate of the silver halide forms.
3. Note the colour, then add aqueous ammonia to confirm, because the colours are close.

Ion tested	Colour of silver halide	What aqueous ammonia does
chloride	white (AgCl)	precipitate dissolves in dilute ammonia
bromide	cream (AgBr)	stays in dilute ammonia; dissolves in concentrated
iodide	yellow (AgI)	stays even in concentrated ammonia

The ionic equations all follow one pattern. For iodide, Ag⁺(aq) + I⁻(aq) → AgI(s). For bromide, swap in Br⁻ to give AgBr(s); for chloride, swap in Cl⁻ to give AgCl(s).

You need the observations with ammonia, not the formulae of the silver–ammonia complexes. In section 3.1.4(a) the halide test comes last in the sequence carbonate, sulfate, then halide, because Ag₂SO₄ and BaCO₃ are both insoluble.

Worked example 5: naming a halide and finding its concentration

50.0 cm³ of a sodium halide solution is acidified and treated with excess silver nitrate. A yellow precipitate forms, which does not dissolve in concentrated ammonia. After filtering, washing and drying it weighs 0.587 g. Use Ar: Ag = 107.9, I = 126.9.

1. Yellow and insoluble in concentrated ammonia: the halide is iodide, so the salt is NaI.
2. Mr(AgI) = 107.9 + 126.9 = 234.8
3. n(AgI) = 0.587 ÷ 234.8 = 0.00250 mol, and n(I⁻) = n(AgI).
4. Concentration of NaI = 0.00250 ÷ 0.0500 = 0.0500 mol dm⁻³

Common errors

• Saying the covalent bond breaks when a halogen boils.
• Giving “bigger molecules” without mentioning more electrons and stronger London forces.
• Writing iodine in cyclohexane as brown. It is violet.
• Claiming iodine displaces bromide. A less reactive halogen cannot oxidise the halide of a more reactive one.
• Explaining reactivity with “more shells” but leaving out the weaker attraction for the incoming electron.
• Calling the chlorine–water reaction “oxidation” only. Both changes must be shown for disproportionation.
• Writing NaClO₃ for bleach. Cold dilute alkali gives NaClO.
• Forgetting the dilute nitric acid step, or using hydrochloric acid, which adds chloride ions.

Next steps

• Condensed tables and a quick self-test: The halogens revision notes
• Original questions with marked answers: The halogens practice questions
• Oxidation numbers and redox language: Redox study guide
• London forces in more depth: Bonding and structure
• Mole calculations: Amount of substance
• Test yourself across topics: free diagnostics

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.1.3 The halogens, learning outcomes (a) to (g). Published by OCR.