Group 17

Inorganic Chemistry Weightage: AS · Papers 1 and 2 (assumed in Paper 4) Topic 11

“Welcome! Group 17 is a topic where every answer is a trend plus a reason. The halogens get less reactive as oxidising agents down the group, while the halide ions get better reducing agents. Learn that one idea, then attach the observations: colours, precipitates, fumes, and the three chlorine disproportionation reactions.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in Group 17

Trends with reasons: volatility (instantaneous dipole–induced dipole forces), oxidising power (atomic size and shielding), thermal stability of hydrides (bond energy).
Observations: colours of the halogens, colours of the silver halide precipitates, the effect of dilute and concentrated ammonia, and the gases from halides with concentrated sulfuric acid.
Equations: displacement, precipitation (ionic), the reactions with concentrated H₂SO₄, and the three chlorine disproportionation reactions, with state symbols when asked.
Oxidation numbers: show the change for each element (0, −1, +1, +5) and say which species is oxidised or reduced.

Where this topic appears. Recent AS papers have asked for: the products when solid halides react with concentrated sulfuric acid, the silver nitrate and ammonia test on chloride ions, the equation for chlorine with hot sodium hydroxide, the ion in bleach that kills bacteria, and the idea of disproportionation in multiple-choice items. The topic is assumed knowledge in Paper 4.

1. Physical Properties

Group 17 (the halogens) has seven electrons in the outer shell. The elements are non-metals that exist as diatomic molecules (F₂, Cl₂, Br₂, I₂) with a single covalent bond between the atoms.

Element	Atomic radius / nm	Melting point / °C	Boiling point / °C	Colour and state at 20 °C
fluorine, F₂	0.072	−220	−188	pale yellow gas
chlorine, Cl₂	0.099	−101	−35	green-yellow gas
bromine, Br₂	0.114	−7	59	red-brown (orange-brown) liquid
iodine, I₂	0.133	114	184	grey-black solid, purple vapour

Going down Group 17 the halogens get darker and less volatile. The molecules are simple (X₂); the instantaneous dipole–induced dipole forces between them get stronger as the number of electrons increases, so the boiling point rises (−35 °C, 59 °C, 184 °C).

Trend in volatility. The halogens get less volatile down the group (the boiling point rises). The molecules are simple and the only forces between them are weak instantaneous dipole–induced dipole forces. Larger molecules have more electrons, so there are more instantaneous dipoles and the forces are stronger. More energy is needed to separate the molecules.

Trend in colour. The colour gets darker down the group: pale yellow, green-yellow, red-brown, grey-black (purple vapour).

⚠️ Examiner Trap: Which bonds break? When a halogen melts or boils, only the weak forces between molecules are overcome. The covalent bond inside Cl₂ or I₂ does not break, and the bond strength does not explain the trend in boiling point.

2. Halogens as Oxidising Agents

A halogen atom gains one electron to form an X⁻ ion, so each halogen is an oxidising agent (electron acceptor). Its oxidation number falls from 0 to −1.

Ca(s) + Cl₂(g) → CaCl₂(s)
Cl₂ + 2e⁻ → 2Cl⁻ (0 to −1)

Trend: the oxidising power decreases from fluorine to iodine. This matches the reactivity (fluorine is the most reactive) and the electronegativity (F 4.0, Cl 3.0, Br 2.8, I 2.5).

Half-reaction (Data Booklet)	E^⊖ / V
F₂ + 2e⁻ ⇌ 2F⁻	+2.87
Cl₂ + 2e⁻ ⇌ 2Cl⁻	+1.36
Br₂ + 2e⁻ ⇌ 2Br⁻	+1.07
I₂ + 2e⁻ ⇌ 2I⁻	+0.54

Explanation. Down the group the atoms are larger. The incoming electron enters a shell that is further from the nucleus and more shielded, so the attraction is weaker and an electron is accepted less readily. These effects outweigh the larger nuclear charge.

Displacement reactions. A more reactive halogen displaces a less reactive halide from solution.

Cl₂(aq) + 2NaBr(aq) → 2NaCl(aq) + Br₂(aq) (ionic: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂): solution turns orange or yellow-brown
Cl₂(aq) + 2KI(aq) → 2KCl(aq) + I₂(aq) (ionic: Cl₂ + 2I⁻ → 2Cl⁻ + I₂): solution turns brown
Br₂(aq) + 2NaI(aq) → 2NaBr(aq) + I₂(aq) (ionic: Br₂ + 2I⁻ → 2Br⁻ + I₂)
Bromine with chloride, iodine with bromide or chloride: no reaction.

A more reactive halogen displaces a less reactive halide from solution, for example Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq). In cyclohexane the colours are clear: chlorine very pale green, bromine orange, iodine purple.

To identify the halogen formed, shake the mixture with cyclohexane (it forms an upper organic layer): chlorine gives a very pale green layer, bromine an orange layer and iodine a purple layer.

In each displacement the halogen is reduced (0 to −1) and the halide ion is oxidised (−1 to 0). The halide ions are therefore reducing agents whose power increases down the group: I⁻ is the best electron donor because the large ion has its outer electrons far from the nucleus and well shielded.

⚠️ Examiner Trap: Two opposite trends. Oxidising power of the halogens decreases down the group; reducing power of the halide ions increases. Write “chlorine is a stronger oxidising agent than bromine because the smaller atom attracts the incoming electron more strongly (less shielding, closer to the nucleus)”, not “chlorine is more reactive”.

3. Hydrogen Halides

The halogens react with hydrogen to form hydrogen halides, H₂(g) + X₂(g) → 2HX(g). The reactions get less vigorous down the group:

Reaction	Conditions
H₂ + F₂ → 2HF	explosive, even in the cold and dark
H₂ + Cl₂ → 2HCl	explosive in sunlight
H₂ + Br₂ → 2HBr	slow on heating
H₂ + I₂ ⇌ 2HI	an equilibrium mixture on heating

The hydrogen halides become less thermally stable down the group. HI decomposes when heated (2HI → H₂ + I₂, purple fumes of iodine); HF and HCl do not decompose up to 1500 °C; HBr is in between.

Bond	H–F	H–Cl	H–Br	H–I
Bond energy / kJ mol⁻¹	562	431	366	299

The H–X bond gets longer and weaker down the group (the outer shell of a larger halogen atom overlaps less with hydrogen), so the hydrogen halides become less thermally stable. HI decomposes on heating to H₂ and purple I₂ vapour; HF and HCl do not decompose up to 1500 °C.

Explanation. Down the group the halogen atom is larger, so the overlap of its outer orbital with the hydrogen 1s orbital is less effective. The bond is longer and weaker (lower bond energy), so less energy is needed to break it and the hydride decomposes more easily.

Predicting astatine. At is below iodine: a darker solid, the least reactive of the halogens (reaction with H₂ is the least vigorous), and HAt would be the least thermally stable hydride (the weakest bond).

4. Halide Ions and Silver Nitrate

Test for halide ions. Add dilute nitric acid (to remove ions such as carbonate), then aqueous silver nitrate. A precipitate of the silver halide forms: Ag⁺(aq) + X⁻(aq) → AgX(s). Then add aqueous ammonia to tell the precipitates apart.

Halide ion	Precipitate with AgNO₃(aq)	Dilute NH₃(aq)	Concentrated NH₃(aq)
Cl⁻	white (AgCl)	dissolves	dissolves
Br⁻	cream (AgBr)	insoluble	dissolves
I⁻	pale yellow (AgI)	insoluble	insoluble

Acidify with dilute nitric acid, add silver nitrate, then ammonia. Ag⁺(aq) + X⁻(aq) → AgX(s). The precipitates look similar, so the ammonia test separates them: AgCl dissolves in dilute NH₃, AgBr only in concentrated NH₃, and AgI does not dissolve.

Full equations for the precipitation:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
AgNO₃(aq) + NaBr(aq) → AgBr(s) + NaNO₃(aq)
AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq)

The precipitate dissolves in ammonia because a soluble complex ion forms, for example AgCl(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq) (the equation is extra detail at AS; the observations are what is tested).

⚠️ Examiner Trap: Colours and the ammonia step. AgBr is cream (not “white”), and it does not dissolve in dilute ammonia. Always give the colour of the precipitate and the result with both dilute and concentrated ammonia. Acidify with nitric acid, not hydrochloric acid (which would add Cl⁻).

5. Halide Ions and Concentrated Sulfuric Acid

Solid halides react with concentrated sulfuric acid in a fume cupboard. The first step is the same for all three: the acid protonates the halide and the hydrogen halide gas forms.

NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g)
NaBr(s) + H₂SO₄(l) → NaHSO₄(s) + HBr(g)
NaI(s) + H₂SO₄(l) → NaHSO₄(s) + HI(g)

Chloride: HCl is not oxidised by concentrated sulfuric acid. The observation is white fumes of HCl (steamy fumes in moist air) and nothing else.

Bromide: the HBr formed is oxidised by the acid.

2HBr(g) + H₂SO₄(l) → Br₂(g) + SO₂(g) + 2H₂O(l)
Observations: red-brown fumes of bromine and colourless SO₂ (which has a choking smell). Sulfur: +6 to +4.

Iodide: HI is a stronger reducing agent, so the sulfuric acid is reduced further.

2HI(g) + H₂SO₄(l) → I₂(g) + SO₂(g) + 2H₂O(l) (S: +6 to +4)
6HI(g) + H₂SO₄(l) → 3I₂(g) + S(s) + 4H₂O(l) (S: +6 to 0)
8HI(g) + H₂SO₄(l) → 4I₂(g) + H₂S(g) + 4H₂O(l) (S: +6 to −2)
Observations: purple fumes of iodine, a yellow solid (sulfur), and a bad-egg smell (hydrogen sulfide).

All three first make the hydrogen halide: NaX + H₂SO₄ → NaHSO₄ + HX. Concentrated sulfuric acid is a moderately strong oxidising agent: it oxidises HBr and HI, but not HCl. The more easily the halide is oxidised, the further the sulfur is reduced.

The halide ions are oxidised from −1 to 0 (HBr and HI act as reducing agents), and it gets easier to oxidise the hydrogen halides down the group. This is why pure HBr and HI cannot be made by this method.

⚠️ Examiner Trap: Do not mix up the two sets of tests. Silver nitrate tests are done on aqueous halide ions; the concentrated sulfuric acid tests are done on the solid halide. For iodide give all of purple I₂, yellow S and H₂S (bad egg) when the question asks for observations. For chloride there is no redox, so there is no Cl₂ and no SO₂.

6. Disproportionation and Water Treatment

Disproportionation is a redox reaction in which the same element is both oxidised and reduced. Chlorine (oxidation number 0) disproportionates in alkali, and the products depend on the temperature.

Conditions	Equation	Oxidation numbers of Cl
cold (about 15 °C), dilute NaOH	Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)	0 → −1 (reduced) and +1 (oxidised)
hot (about 70 °C), concentrated NaOH	3Cl₂(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l)	0 → −1 (reduced) and +5 (oxidised)
water	Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HClO(aq)	0 → −1 (reduced) and +1 (oxidised)

In disproportionation the same element is oxidised and reduced. Chlorine (0) goes down to chloride (−1) and up to chlorate(I) (+1) in cold alkali or to chlorate(V) (+5) in hot alkali. In water it forms HCl and HClO, the sterilising agent.

Ionic equations and half-equations (cold alkali).

Overall: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
Reduction: Cl₂ + 2e⁻ → 2Cl⁻
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻

Hot alkali: 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O. The names are chlorate(I) for ClO⁻ and chlorate(V) for ClO₃⁻.

Chlorination of water. Adding a little chlorine to drinking water kills bacteria (for example the cholera bacterium). Chlorine disproportionates in water to give HCl and HClO (chloric(I) acid). HClO and its ion ClO⁻ are the sterilising agents. Household bleach is a solution of sodium chlorate(I), NaClO, which contains the ClO⁻ ion.

⚠️ Examiner Trap: Disproportionation needs two changes. Show that the oxidation number of the same element goes both up and down, for example “chlorine goes from 0 to −1 (reduced) and from 0 to +1 (oxidised)”. Check the balance of the hot-alkali equation: 6 Cl atoms on the left (3Cl₂), 5 + 1 = 6 on the right; 6 O and 6 H on the left, 3 + 3 = 6 O and 6 H on the right.

7. Quick Sheet and Checklist

Property (down the group)	Trend	Reason
Colour	darker	more electrons (larger molecules)
Volatility	decreases (boiling point rises)	stronger instantaneous dipole–induced dipole forces
Oxidising power of X₂	decreases	larger atom, more shielding, weaker attraction for an electron
Reducing power of X⁻	increases	outer electron further away and more shielded
Thermal stability of HX	decreases	longer, weaker H–X bond
Reaction with H₂	less vigorous	decreasing oxidising power

Before you leave the question, check:

Each trend says the direction and the reason (forces, shielding, bond energy).
Precipitate colours (white, cream, pale yellow) and the dilute and concentrated ammonia results are both given.
With concentrated sulfuric acid: chloride gives only HCl; bromide gives Br₂ and SO₂; iodide gives I₂ with S and H₂S as well.
Disproportionation answers show oxidation numbers rising and falling for the same element, and the equations are balanced.
ClO⁻ is the species in bleach and HClO is the sterilising acid in chlorinated water.