AQA A-level Chemistry (7405) · Group 7, the Halogens
Mini-Lesson
Group 7, the Halogens
This mini-lesson covers AQA 3.2.3 Group 7, the halogens: the trends in electronegativity and boiling point, the oxidising ability of the halogens and their displacement reactions, the reducing power of the halide ions with concentrated sulfuric acid, the silver nitrate test, and the reactions of chlorine with water and with alkali.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.
Group 7 · trends
Electronegativity, boiling point and oxidising power
The halogens exist as diatomic molecules and form 1− ions by gaining one electron.
Electronegativity decreases down the group. The bonding pair is further from the nucleus and better shielded, so it is attracted less strongly. Fluorine is the most electronegative element of all.
Boiling point increases down the group. The molecules have more electrons, so the van der Waals forces between them are stronger and more energy is needed to separate them: Cl₂ is a gas, Br₂ a liquid, I₂ a solid.
Oxidising ability decreases down the group. Oxidising means gaining an electron — and the incoming electron is added further from the nucleus, with more shielding, so it is captured less readily.
Displacement reactions prove the trend. A halogen will displace the ions of any halogen below it:
Cl₂ + 2KBr → 2KCl + Br₂the colourless solution turns orange — chlorine is the stronger oxidising agent
Br₂ + 2KI → 2KBr + I₂the solution turns brown. But I₂ + KBr → no reaction
Match it
Match each displacement reaction to its result
Tap a reaction on the left, then its result on the right.
Reaction
Result
Group 7 · halide ions as reducing agents
Halide ions with concentrated sulfuric acid
Turn the trend around: the reducing power of the halide ion increases down the group, because a larger ion loses its outer electron more easily. Concentrated sulfuric acid tests this beautifully — the further down the group, the more the sulfur is reduced.
NaCl: Cl⁻ is too weak a reducing agent to reduce the sulfur at all. Only an acid–base reaction occurs: NaCl + H₂SO₄ → NaHSO₄ + HCl. Observation: misty white fumes of HCl only.
NaBr: Br⁻ reduces S from +6 to +4. Products include SO₂ (a choking gas) and orange/brown Br₂: 2Br⁻ + 2H₂SO₄ → Br₂ + SO₂ + 2H₂O + SO₄²⁻.
NaI: I⁻ is the strongest reducing agent and takes S all the way from +6 to −2. Products include H₂S (rotten-egg smell), solid sulfur (yellow), SO₂, and purple I₂ vapour: 8I⁻ + 9H₂SO₄ → 4I₂ + H₂S + 4H₂O + 8HSO₄⁻.
How to answer this in the exam: quote the oxidation state change of the sulfur (+6 → +4 for Br⁻; +6 → 0 or −2 for I⁻) and say the halide is oxidised to the halogen. The observations follow from the products.
Quick check
Quick check
?Solid NaI is warmed with concentrated H₂SO₄ and a gas smelling of rotten eggs is produced. Which species has been reduced, and by how much?
Group 7 · halide tests
Testing for halide ions with silver nitrate
Add dilute nitric acid, then silver nitrate solution. The nitric acid is essential: it removes carbonate and hydroxide ions, which would otherwise give their own precipitates with Ag⁺ (silver carbonate and silver oxide) and produce a false positive. It must be nitric acid — hydrochloric acid would add chloride ions.
Ag⁺(aq) + X⁻(aq) → AgX(s)
Cl⁻ → AgCl, a white precipitate — dissolves in dilute ammonia.
Br⁻ → AgBr, a cream precipitate — dissolves only in concentrated ammonia.
I⁻ → AgI, a yellow precipitate — insoluble even in concentrated ammonia.
The colours are close enough that people misjudge them, so the ammonia step is the decisive confirmation: solubility in ammonia falls as you go down the group, exactly in step with the falling solubility of the silver halide.
Sort it
Which halide gives each result?
Tap an observation, then tap the halide ion responsible.
🟩 Chloride, Cl⁻
🟪 Bromide, Br⁻
🟦 Iodide, I⁻
Calculate
Your turn
1A solution containing 0.0200 mol of Cl⁻ is treated with excess acidified AgNO₃. Calculate the mass of AgCl precipitated. (Mr(AgCl) = 143.4)
g
Hint: The ratio Cl⁻ : AgCl is 1 : 1. m = 0.0200 × 143.4.
Calculate
Your turn
225.0 cm³ of 0.100 mol dm⁻³ NaBr is treated with excess acidified AgNO₃. Calculate the mass of AgBr precipitated. (Mr(AgBr) = 187.8)
Chlorine disproportionates — it is simultaneously oxidised and reduced in the same reaction. Watch the oxidation states.
Cl₂ + H₂O ⇌ HClO + HClCl goes from 0 → +1 (in HClO) and 0 → −1 (in HCl): disproportionation
HClO (chloric(I) acid) and the ClO⁻ ion are powerful bactericides — this is why chlorine is added to drinking water and swimming pools. In bright sunlight a different reaction dominates:
2Cl₂ + 2H₂O → 4HCl + O₂Cl 4 = 4, H 4 = 4, O 2 = 2 ✓
With cold, dilute sodium hydroxide, chlorine disproportionates again — this is household bleach:
Cl₂ + 2NaOH → NaCl + NaClO + H₂OCl: 0 → −1 and 0 → +1. Check: Na 2 = 2, Cl 2 = 2, O 2 = 2, H 2 = 2 ✓
Evaluating water chlorination — a real "society decides" question. Chlorine is toxic, and it can react with organic matter in water to form chlorinated hydrocarbons that are suspected carcinogens. But chlorination kills the bacteria that cause cholera and typhoid, and its benefits to public health overwhelmingly outweigh the risks. Society, not chemistry, makes that judgement.
Calculate
Your turn
3Calculate the oxidation state of chlorine in NaClO (sodium chlorate(I)).
Hint: Na is +1 and O is −2, and the compound is neutral: +1 + x + (−2) = 0.
Calculate
Your turn
4In Cl₂ + H₂O → HClO + HCl, calculate the oxidation state of chlorine in HCl.
Hint: H is +1 and HCl is neutral. (Note that the other chlorine went UP to +1 — that is what disproportionation means.)
Quick check
Quick check
?What does it mean to say that chlorine disproportionates in water?
Quick check
Quick check
?Why is dilute nitric acid — not hydrochloric acid — added before silver nitrate in the halide test?
Quick check
Quick check
?Why is iodine a solid at room temperature while chlorine is a gas?
Group 7 · exam traps
Two opposite trends, and the acid you must use
Oxidising power of the HALOGEN falls down the group; reducing power of the HALIDE ION rises. They are opposite sides of the same coin, and questions deliberately mix them up.
Conc. H₂SO₄ with a halide: quote the SULFUR oxidation-state change. Cl⁻: no redox at all (+6 stays +6, only HCl fumes). Br⁻: +6 → +4 (SO₂). I⁻: +6 → −2 (H₂S). That is the answer the mark scheme wants.
Acidify with NITRIC acid before adding AgNO₃ — never hydrochloric, which would add chloride ions.
Boiling point rises down the group because the molecules have more electrons, so stronger van der Waals forces — not because the covalent bonds are stronger.
Disproportionation means the same element is simultaneously oxidised and reduced. Both of chlorine's reactions — with water and with cold dilute NaOH — are disproportionations, and you should be able to prove it by assigning oxidation states.
Quick check
Quick check
?Which halide ion is the strongest reducing agent, and what is the evidence?
Calculate
Your turn
5Calculate the oxidation state of chlorine in NaClO₃ (sodium chlorate(V)).
Hint: Na is +1, each O is −2, and the compound is neutral: +1 + x + 3(−2) = 0.
Recap
The big ideas to know
Down the group: electronegativity decreases; boiling point increases (more electrons → stronger van der Waals forces)
Oxidising power:decreases down the group: Cl₂ > Br₂ > I₂ — so a halogen displaces the halide of any element below it
Reducing power of X⁻:increases down the group: with conc. H₂SO₄, Cl⁻ gives only HCl; Br⁻ gives SO₂; I⁻ gives H₂S
Halide test: acidified AgNO₃ → AgCl white (dissolves in dilute NH₃) · AgBr cream (conc. NH₃) · AgI yellow (insoluble in conc. NH₃)