The halogens — OCR A-Level Chemistry

This row is a study-guidance heading, not an assessed answer target. Use it to organise your reading of the halogens (Group 17: fluorine, chlorine, bromine, iodine). As you read, build a table of the characteristic physical properties across the group: appearance and physical state at room temperature (fluorine pale yellow gas, chlorine greenish-yellow gas, bromine red-brown liquid, iodine grey-black solid), and the trends of increasing melting and boiling point, increasing density and decreasing volatility down the group. Connect these trends to increasing strength of intermolecular forces as the molecules get larger. Note also the colour of the vapour and the trend in colour intensity. Treat the table as your own revision tool; the specification does not prescribe a fixed table format, so record the properties accurately and be ready to describe and explain the trends.

Halogens exist as diatomic molecules (F₂, Cl₂, Br₂, I₂) because two halogen atoms share one pair of electrons in a single covalent bond, giving each atom a full outer shell. The trend to teach is the rise in boiling point from Cl₂ to Br₂ to I₂. All three are simple molecular substances held together in the liquid or solid by induced dipole–dipole interactions, also called London forces. These arise from temporary, fluctuating dipoles in the electron clouds. As the molecules get larger down the group, they have more electrons and a larger, more polarisable electron cloud, so the induced dipoles are stronger and more energy is needed to overcome them. The covalent bonds are not broken during boiling.

This row is a study guide for the redox chemistry of halogens and their compounds. Read it alongside the specification's halogen section and your textbook. Focus on how halogen reactivity decreases down Group 17 and how this shows in displacement reactions: a more reactive halogen displaces a less reactive halide from solution, for example chlorine displacing bromide. Track oxidation states to identify what is oxidised and what is reduced, and note that a halogen is itself reduced as it acts as an oxidising agent. Link this to reactions of halide ions with silver nitrate and to the reducing power of halide ions. Use the guidance to select worked examples and practice questions that test these ideas.

A halogen atom has seven electrons in its outer shell, written as s²p⁵, so its outer shell is one electron short of a full octet. In many redox reactions each halogen atom gains one electron and is reduced, forming a halide ion with a single negative charge, such as F⁻, Cl⁻, Br⁻ or I⁻. For example, chlorine in Cl₂ gains one electron per atom when it reacts with sodium, giving two Cl⁻ ions and sodium ions. Because the halogen gains electrons, the halogen itself is the oxidising agent. Recognise that the s²p⁵ configuration explains the 1⁻ charge: one electron added fills the p subshell to s²p⁶.

Reactivity decreases down Group 17 from chlorine to bromine to iodine. A more reactive halogen displaces a less reactive halide ion from solution: chlorine displaces bromide and iodide ions, bromine displaces iodide ions, but iodine displaces neither chloride nor bromide ions. For example, chlorine water added to potassium bromide solution turns the solution orange-brown as bromine forms, while adding iodine to potassium chloride gives no reaction. The trend arises because the atoms get larger down the group, so the incoming electron is gained less readily and the halogen is a weaker oxidising agent.

Down Group 17 the halogens become less reactive because it becomes harder to gain an electron and form a 1⁻ ion. A halogen atom gains an electron into its outer shell, and the ease of this gain depends on the attraction between the incoming electron and the nucleus. Going down the group the atomic radius increases, so the outer shell is further from the nucleus and the attraction is weaker. At the same time each successive element has more inner shells, so electron shielding increases and further reduces the effective nuclear pull on the incoming electron. Together, the larger radius and greater shielding outweigh the rising nuclear charge, so the tendency to form a 1⁻ ion decreases and reactivity falls. For example, fluorine gains an electron most readily and is the most reactive halogen, while iodine gains an electron least readily.

Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced. In such a reaction one portion of the element increases its oxidation number while another portion of the same element decreases its oxidation number. The halogens illustrate this well. When chlorine reacts with cold dilute aqueous sodium hydroxide, chlorine is both oxidised and reduced: some chlorine atoms move to an oxidation number of 1⁻ in chloride ions while others move to an oxidation number of 1⁺ in chlorate(I) ions. Similarly, chlorine with water forms hydrochloric acid and chloric(I) acid, again with chlorine both oxidised and reduced. To explain disproportionation, identify the element, compare its oxidation numbers in the reactants and products, and show that one value rises while another falls for that same element.

Chlorine added to water undergoes disproportionation: chlorine is oxidised to chloric(I) acid and reduced to chloride ions. The equation is Cl₂ + H₂O ⇌ HClO + HCl, or in ionic form Cl₂ + H₂O ⇌ HClO + H⁺ + Cl⁻. In HClO the chlorine is in oxidation state 1⁺; in Cl⁻ it is 1⁻; in Cl₂ it is 0, so one chlorine atom rises and one falls. The mixture is reversible and the chloric(I) acid is the active disinfectant: it oxidises and destroys bacteria and other microorganisms, which is why chlorine is used in treating drinking water and in swimming pools. Learners should recognise the oxidation states, identify the disproportionation, and explain that HClO rather than Cl⁻ provides the antimicrobial action. A common exam question asks why the reaction is a disproportionation or which species kills bacteria.

When chlorine reacts with cold, dilute aqueous sodium hydroxide, disproportionation again occurs. The equation is Cl₂ + 2NaOH → NaCl + NaClO + H₂O, or in ionic form Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Chlorine is oxidised to chlorate(I) (hypochlorite) ions, ClO⁻, where chlorine has oxidation state 1⁺, and reduced to chloride ions, Cl⁻, with oxidation state 1⁻; in Cl₂ it is 0. The sodium chlorate(I) produced is the active ingredient of household bleach, which oxidises stains and microorganisms. The conditions matter: cold, dilute sodium hydroxide gives the chlorate(I) product, whereas hot, concentrated alkali gives chlorate(V). Learners should write the equation, identify the disproportionation and explain the link to bleach.

This clause extends the halogen displacement and reaction chemistry of (i) and (ii) to other halogens and halide systems. In OCR H432 the halogen section covers redox behaviour of the halogens and halide ions, including displacement reactions in which a more reactive halogen oxidises a halide ion of a less reactive halogen. A learner should apply the same pattern to any halogen pair: chlorine displaces bromide and iodide, bromine displaces iodide, and iodine displaces neither. Write the ionic equation, for example Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, and identify the oxidation and reduction changes. The analogous reaction is predicted from the trend in oxidising power down Group 17, which decreases as the halogen gets larger and less able to gain an electron.

This clause requires a balanced evaluation of chlorine in water treatment. The benefit is that chlorine kills bacteria in drinking water and so reduces the spread of waterborne disease. The risks include the toxicity of chlorine gas to workers and the public if it escapes, and the possibility that chlorine reacts with organic matter to form chlorinated hydrocarbons, some of which are suspected carcinogens. A strong answer states both sides and reaches a judgement, for example that the disease-prevention benefit outweighs the risks when chlorine is handled and dosed correctly. Use the equation Cl₂ + H₂O ⇌ HCl + HClO to show how chlorine forms the disinfecting species.

Halide ions show characteristic reactions that let you identify them. With aqueous silver nitrate, chloride gives a white precipitate of silver chloride, bromide a cream precipitate of silver bromide and iodide a pale yellow precipitate of silver iodide. Adding dilute ammonia dissolves the silver chloride; concentrated ammonia dissolves the silver bromide, and silver iodide is essentially insoluble in ammonia. These differences arise because the solubility of the silver halide in ammonia decreases from chloride to iodide. A reducing property also distinguishes the ions: concentrated sulfuric acid oxidises iodide to iodine, giving a purple vapour, while bromide gives bromine vapour and chloride is largely unaffected.

This test identifies halide ions in solution. Add dilute nitric acid to remove carbonate and other interfering ions, then add aqueous silver nitrate. Chloride ions give a white precipitate of silver chloride, bromide ions a cream precipitate of silver bromide and iodide ions a pale yellow precipitate of silver iodide. The ionic equation is Ag⁺(aq) + X⁻(aq) → AgX(s), where X is the halide. Next add aqueous ammonia: dilute ammonia dissolves silver chloride, concentrated ammonia dissolves silver bromide, and silver iodide remains insoluble even in concentrated ammonia. The sequence of precipitate colour followed by ammonia solubility therefore distinguishes the three halide ions. Learn each observation, the ionic equation and the order of the test so you can interpret an unknown result.