At GCSE you learned that the halogens get less reactive down the group and that a more reactive halogen displaces a less reactive one from its salt. A-Level explains why — through electronegativity, oxidising power and the reducing power of the halide ions — and adds the reactions with concentrated sulfuric acid, the silver nitrate test, and the disproportionation chemistry of chlorine.
Electronegativity & boiling point
The halogens are reactive non-metals that exist as diatomic molecules (F2, Cl2, Br2, I2). Two physical trends are examined, and they run in opposite directions.
Electronegativity decreases down the group. The atoms get larger with more shells, so the shared (bonding) pair of electrons is further from the nucleus and more shielded — the nucleus attracts it less strongly.
Boiling point increases down the group. The molecules gain more electrons, so the van der Waals forces between molecules get stronger and more energy is needed to separate them — F2 and Cl2 are gases, Br2 a liquid, I2 a solid.
Interactive — down the group, halogen by halogen
Pick a halogen to see its electronegativity, boiling point, state and colour — and the reasons.
- Electronegativity is about the bonding pair. Explain the fall with larger atoms / more shielding → weaker attraction for the shared pair of electrons, not the “outer electrons”.
- Boiling point rises because of van der Waals forces between molecules — more electrons make them stronger. You are not breaking the covalent bonds inside X2.
🧪 Exam-style questions
Which property increases down Group 7? Tick (✓) one box.
Explain why the electronegativity of the halogens decreases down the group.
Show answer
Down the group the atoms are larger, with more shells / more shielding, so the bonding pair of electrons is further from the nucleus. 1 mark
There is therefore a weaker attraction between the nucleus and the shared (bonding) pair of electrons. 1 mark
Source: AQA A-Level Chemistry past papers.
Oxidising power & displacement
Halogens react by gaining an electron each to become halide ions — that makes them oxidising agents. How readily they do it decreases down the group: for the same reason electronegativity falls, a larger, more shielded atom attracts the incoming electron less strongly.
You see this ranking in displacement reactions: a halogen will displace the halide ions of any halogen below it, but not above. So chlorine displaces bromide and iodide; bromine displaces iodide; iodine displaces neither.
Cl2 + 2Br− → 2Cl− + Br2 (chlorine oxidises bromide; the solution turns orange)
In oxidation-number terms, chlorine is reduced (Cl 0 → −1) and the bromide is oxidised (Br −1 → 0) — the same bookkeeping you apply to the concentrated sulfuric acid and disproportionation reactions below.
Interactive — the displacement lab
Mix a halogen with a halide solution and see whether displacement happens.
add the halogen…
…to this halide solution
- Say what is oxidised. Not “chlorine oxidises” alone — chlorine oxidises the bromide ions (removes electrons from them) and is itself reduced.
- Get the colours right. Displaced Br2 is an orange solution; displaced I2 is a brown solution / grey-black solid. Examiner reports flag wrong colours and products.
- A halogen cannot displace one above it — oxidising power falls down the group, so iodine cannot displace bromide or chloride.
🧪 Exam-style questions
Which equation shows a redox reaction that does not occur? Tick (✓) one box.
Which pair of solutions, when mixed, reacts to form a dark brown solution? Tick (✓) one box.
Which is a correct trend down Group 7 from fluorine to iodine? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Reducing power & conc. sulfuric acid
Turn it around and look at the halide ions. A halide reacts by losing its extra electron, acting as a reducing agent — and this gets easier down the group, because the outer electron of the bigger ion is held less tightly. Iodide is the strongest reducer.
The test bench for this is the reaction of solid sodium halides with concentrated sulfuric acid. All three first make the hydrogen halide, but the more strongly reducing halides then go on to reduce the sulfur in H2SO4 — further and further down the group.
| Sodium halide + conc. H2SO4 | Products | Sulfur change | Observations |
|---|---|---|---|
| NaCl | NaHSO4 + HCl only | no redox | misty white fumes of HCl |
| NaBr | + Br2 + SO2 | S +6 → +4 | orange/brown fumes (Br2), choking SO2 |
| NaI | + I2 + SO2 + S + H2S | S +6 → −2 | grey-black I2, yellow S, rotten-egg H2S |
NaCl — acid–base only; chloride is too weak to reduce the sulfur:
NaCl + H2SO4 → NaHSO4 + HCl
NaBr — the acid–base step first (as for every sodium halide), then the HBr reduces the sulfur to SO2:
NaBr + H2SO4 → NaHSO4 + HBr
2HBr + H2SO4 → Br2 + SO2 + 2H2O (S: +6 → +4)
NaI — the acid–base step first, then the HI reduces the sulfur further and further, giving three different sulfur products:
NaI + H2SO4 → NaHSO4 + HI
2HI + H2SO4 → I2 + SO2 + 2H2O (S: +6 → +4)
6HI + H2SO4 → 3I2 + S + 4H2O (S: +6 → 0)
8HI + H2SO4 → 4I2 + H2S + 4H2O (S: +6 → −2)
Interactive — sodium halides + concentrated sulfuric acid
Pick a solid sodium halide to see how far it reduces the sulfur.
- NaCl is the odd one out — it makes HCl only and produces no sulfur-containing gas, because chloride is too weak to reduce the sulfur.
- H2S (rotten eggs) is the give-away for iodide; SO2 (choking) appears with bromide. Match the gas to the halide.
- Check the equation type. Examiner reports flag full, ionic and half-equations being muddled — give exactly what is asked for.
🧪 Exam-style questions
Some solid sodium halides are reacted with concentrated sulfuric acid. Which solid sodium halide does not produce a sulfur-containing gas as one of the products? Tick (✓) one box.
Which statement is correct? Tick (✓) one box.
Concentrated sulfuric acid oxidises iodide ions, and is itself reduced to H2S. Give the two half-equations and the overall ionic equation for this reaction.
Show answer
Oxidation: 2I− → I2 + 2e− 1 mark
Reduction: H2SO4 + 8H+ + 8e− → H2S + 4H2O 1 mark
Overall: H2SO4 + 8H+ + 8I− → 4I2 + H2S + 4H2O 1 mark
Scale the iodide half-equation by 4 so its 8 electrons match the reduction, then add and cancel.
Source: AQA A-Level Chemistry past papers.
The halide test & Required Practical 4
To identify a halide ion in solution, add dilute nitric acid, then silver nitrate solution. Silver ions give an insoluble silver halide whose colour names the ion:
| Halide | With acidified AgNO3 | In ammonia |
|---|---|---|
| F− | no precipitate (AgF is soluble) | — |
| Cl− | white AgCl precipitate | dissolves in dilute ammonia |
| Br− | cream AgBr precipitate | dissolves in concentrated ammonia |
| I− | yellow AgI precipitate | insoluble, even in concentrated ammonia |
Ag+(aq) + Cl−(aq) → AgCl(s)
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl−(aq)
Three “why” questions are examined together: silver nitrate is used because Ag+ forms an insoluble silver halide; it is acidified with nitric acid to remove carbonate and hydroxide ions that would also give a precipitate (a false positive); and ammonia is added to tell the silver halide precipitates apart by their different solubilities — the surest way to separate white AgCl from cream AgBr.
RP4 is a set of test-tube reactions to identify cations (Group 2 ions and NH4+) and anions (halides, OH−, CO32− and SO42−). The halide test above is its centrepiece; the sulfate test (acidified barium chloride) and the carbonate test (effervescence with acid) sit alongside it. Practical skills are examined on all three papers — see the required practicals page.
Interactive — identify the unknown halide
A solution contains one of F−, Cl−, Br− or I−. Run the tests, then name it.
name the halide
- Acidify with dilute nitric acid — examiner reports flag the missing acid step. Not HCl (adds chloride) and not H2SO4 (adds sulfate), which would ruin the test.
- Learn the colours exactly: AgCl white, AgBr cream, AgI yellow. F− gives no precipitate.
- Ammonia distinguishes them: AgCl dissolves in dilute, AgBr only in concentrated, AgI in neither.
🧪 Exam-style questions
In the test for a halide ion, dilute nitric acid is added before the silver nitrate solution. Why is the nitric acid added? Tick (✓) one box.
State what is observed when silver nitrate solution is added to sodium fluoride solution.
Show answer
No change / colourless solution / no precipitate forms. 1 mark
Silver fluoride is soluble, so fluoride is the one halide that gives no precipitate.
Source: AQA A-Level Chemistry past papers.
Chlorine chemistry
Chlorine’s signature reactions are disproportionation — the same chlorine atoms are both oxidised and reduced, going from 0 to +1 (in the chlorate(I)) and 0 to −1 (in the chloride).
Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced.
With water (the basis of water treatment):
Cl2 + H2O ⇌ HCl + HClO
With cold, dilute sodium hydroxide (this makes bleach):
Cl2 + 2NaOH → NaCl + NaClO + H2O
In both, chlorine is 0 on the left, but ends up as +1 in the chlorate(I) (ClO−) and −1 in the chloride (Cl−) — that split is what “disproportionation” means. In bright sunlight, chlorine instead reacts with water to give chloride and oxygen: 2Cl2 + 2H2O → 4HCl + O2. That reaction is not disproportionation — here chlorine is only reduced (0 → −1), while it is the oxygen in the water that is oxidised (−2 → 0).
Chlorine is added to drinking water because the HClO / ClO− it forms kills bacteria, preventing water-borne diseases such as cholera and typhoid. Chlorine is also toxic, and can form traces of chlorinated organic compounds — so society weighs the risks against the benefits, and judges that the health benefits of safe drinking water outweigh the toxic effects.
Interactive — prove the disproportionation
- Prove disproportionation with oxidation states. Examiner reports flag recall without the redox — show chlorine going 0 → +1 and 0 → −1.
- Know the products of Cl2 + water: chloride and chlorate(I) (HCl + HClO), or chloride + oxygen in sunlight. OH− and O3 are not products.
- Water treatment answers need both sides: the benefit (kills bacteria) and the risk (toxic) — then the judgement that benefits outweigh risks.
🧪 Exam-style questions
Which shows the major product(s) formed when chlorine reacts with cold, dilute, aqueous sodium hydroxide? Tick (✓) one box.
Which species is never formed during the reactions of chlorine with water? Tick (✓) one box.
Chlorine is used to treat drinking water even though it is toxic. Give one reason chlorine is added, and explain why its use is considered acceptable.
Show answer
Chlorine sterilises the water — it kills bacteria / microorganisms. 1 mark
The health benefits (preventing water-borne disease) outweigh the risk from its toxicity, and it is used only in small quantities. 1 mark
Source: AQA A-Level Chemistry past papers.
Capstone: the halogen toolkit
Group 7 is a set of trends that mirror each other, plus a handful of reactions worth knowing cold. Line the trends up first:
| Down F → I | Direction | Why it matters |
|---|---|---|
| Electronegativity | decreases | bonding pair further from the nucleus |
| Boiling point | increases | stronger van der Waals between bigger molecules |
| Oxidising power of the halogen | decreases | displacement: a halogen displaces the halides below it |
| Reducing power of the halide ion | increases | conc. H2SO4: NaCl → HCl; NaBr → SO2; NaI → H2S |
These trends also let you predict a halogen below iodine without ever having met it: it would be an even weaker oxidising agent (unable to displace the halide ions above it), its ion an even stronger reducing agent, and the element itself a darker solid with a higher boiling point — larger molecules mean stronger van der Waals forces.
The one exam skill that ties it together is choosing the right kind of equation. AQA asks for a full equation, an ionic equation or a half-equation, and they are not interchangeable. Take chlorine displacing bromide:
| Type | Chlorine + potassium bromide |
|---|---|
| Full equation | Cl2 + 2KBr → 2KCl + Br2 |
| Ionic equation | Cl2 + 2Br− → 2Cl− + Br2 |
| Half-equations | Cl2 + 2e− → 2Cl− and 2Br− → Br2 + 2e− |
How to tell which is which: a full equation writes every substance out, potassium and all; an ionic equation drops the spectator ions (the K+) and keeps only the species that change; a half-equation shows just the oxidation or just the reduction on its own, with electrons added to balance the charge.
Read the command word, give exactly the equation asked for, and pin every observation to the right colour or gas — and Group 7 becomes one of the most dependable topics on Paper 1.
Interactive — give exactly the equation asked for
- Down F→I: electronegativity decreases (bonding pair further from nucleus/more shielded); boiling point increases (stronger van der Waals between larger molecules).
- Oxidising power of the halogens decreases down the group: a halogen displaces the halide ions below it (Cl2 + 2Br− → 2Cl− + Br2). Colours: Br2 orange solution, I2 brown solution / grey-black solid.
- Reducing power of the halide ions increases down the group (I− strongest). With conc. H2SO4: NaCl → HCl only (no redox); NaBr → Br2 + SO2 (S +6→+4); NaI → I2 + S + H2S (S +6→−2).
- Halide test: dilute HNO3 then AgNO3 → AgCl white, AgBr cream, AgI yellow (F− none). Distinguish with ammonia: AgCl dissolves in dilute NH3, AgBr in conc. NH3, AgI insoluble.
- Acidify with nitric acid (not HCl or H2SO4) to remove carbonate/hydroxide ions that would also precipitate.
- Chlorine disproportionates (Cl 0 → +1 and 0 → −1): Cl2 + H2O ⇌ HCl + HClO (water treatment); Cl2 + 2NaOH → NaCl + NaClO + H2O (bleach). Chlorine’s health benefits in water treatment outweigh its toxicity.