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11.1 Physical properties of the Group 17 elements

Element	Formula	Appearance at room conditions	Relative volatility
chlorine	Cl₂	pale green / yellow-green gas	highest of the three
bromine	Br₂	red-brown liquid with orange-brown vapour	intermediate
iodine	I₂	grey-black solid; purple vapour when heated	lowest of the three

From Cl₂ → Br₂ → I₂, colour becomes darker, melting and boiling points rise, and volatility falls. Volatility is the tendency to enter the gas phase, so a more volatile substance has a lower boiling point under comparable conditions.

State the colour for the specified physical form: iodine solid is grey-black, while iodine vapour is purple. The cause of the volatility trend is taught separately through forces between X₂ molecules, not through the X–X covalent bond.

Molecule	Approximate X–X bond enthalpy / kJ mol⁻¹	Comparison
F₂	158	anomalously weaker than Cl₂
Cl₂	243	strongest of these four
Br₂	193	weaker than Cl₂
I₂	151	weakest

From Cl₂ → Br₂ → I₂, atomic radius and X–X bond length increase. The shared bonding pair is farther from both nuclei and attracted less strongly, so bond enthalpy and bond strength decrease.

F₂ breaks the simple trend. Fluorine atoms are so small that non-bonding electron pairs on the two atoms are very close; strong lone-pair–lone-pair repulsion weakens the F–F bond enough to make it weaker than Cl–Cl.

Bond enthalpy measures the energy needed to break the covalent bond inside X₂. It does not predict boiling point: boiling separates intact molecules and is controlled mainly by forces between them.

Cl₂, Br₂ and I₂ are non-polar simple molecules. Their intermolecular attractions are instantaneous dipole–induced dipole forces: a momentary uneven electron distribution in one molecule induces a dipole in a neighbouring molecule.

Step down the group	Consequence
each X₂ molecule contains more electrons and a larger electron cloud	the cloud is more polarisable and fluctuates more readily
instantaneous and induced dipoles become larger	attractions between neighbouring X₂ molecules become stronger
more energy is needed to separate molecules	melting and boiling points increase
fewer molecules escape into the gas phase at a given temperature	volatility decreases: Cl₂ > Br₂ > I₂

This explains the room-condition sequence gas Cl₂ → liquid Br₂ → solid I₂. The particles remain neutral diatomic molecules throughout; only the strength of attraction between molecules changes.

Do not invoke permanent dipoles: each X₂ molecule contains identical atoms and is non-polar. Do not use increasing X–X bond strength—the intramolecular bond actually weakens from Cl₂ to I₂ while boiling point rises.

11.2 The chemical properties of the halogen elements and the hydrogen halides

A halogen is an oxidising agent when it gains electrons and is reduced: X₂ + 2e⁻ → 2X⁻. Relative oxidising power decreases F₂ > Cl₂ > Br₂ > I₂.

Halogen added to a halide solution	Spontaneous displacement among Cl, Br and I
Cl₂	oxidises Br⁻ and I⁻
Br₂	oxidises I⁻, but not Cl⁻
I₂	oxidises neither Cl⁻ nor Br⁻

$\mathrm{Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2}$

$\mathrm{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$

Down the group, atomic radius and shielding increase. The nucleus attracts an incoming electron less strongly, so formation of X⁻ becomes less favourable and X₂ is a weaker oxidising agent.

In a displacement equation, the more powerful halogen oxidising agent is reduced to X⁻, while the displaced halide is oxidised to its element. Do not reverse this order or confuse halogen oxidising power with halide reducing power, which increases down the group.

$\mathrm{H_2(g) + X_2(g) \rightarrow 2HX(g)}$

Halogen	Reaction with H₂	Relative behaviour
F₂	H₂ + F₂ → 2HF	explosive even in cool, dark conditions
Cl₂	H₂ + Cl₂ → 2HCl	explosive in bright light / sunlight
Br₂	H₂ + Br₂ → 2HBr	slow reaction on heating
I₂	H₂ + I₂ ⇌ 2HI	requires heating and forms an equilibrium mixture

Relative reactivity is F₂ > Cl₂ > Br₂ > I₂. Down the group, greater radius and shielding reduce the ability of X₂ to gain electrons during compound formation, so progressively more demanding conditions are needed and the reaction is less vigorous.

This outcome concerns forming gaseous hydrogen halides directly from the elements. Do not replace it with a comparison of aqueous acid strengths; HF being a weak acid in water does not make F₂ unreactive toward H₂.

Hydrogen halide	Relative H–X bond strength	Relative thermal stability
HF	strongest	highest
HCl	weaker	lower
HBr	weaker again	lower again
HI	weakest	lowest

$\mathrm{2HX(g) \rightleftharpoons H_2(g) + X_2(g)}$

From F to I, the halogen atom is larger, so the H–X bond is longer and overlap between the hydrogen 1s orbital and the halogen orbital becomes less effective. Attraction in the bond weakens, bond enthalpy falls, and less heat is required to decompose HX.

HI therefore decomposes on heating more readily than HBr, which decomposes more readily than HCl; HF is most resistant. A complete explanation names bond length or orbital overlap, then bond strength, then ease of thermal decomposition.

Thermal stability is not boiling point, volatility or aqueous acid strength. It asks how readily the covalent H–X bond breaks when heated.

11.3 Some reactions of the halide ions

A halide ion acts as a reducing agent when it donates an electron and is oxidised: 2X⁻ → X₂ + 2e⁻. Relative reducing power increases F⁻ < Cl⁻ < Br⁻ < I⁻.

Halide	Ease of oxidation	Evidence with concentrated H₂SO₄
F⁻	hardest	acid-base reaction only; no reduction of sulfuric acid
Cl⁻	very difficult	acid-base reaction only; no reduction of sulfuric acid
Br⁻	easier	Br₂ forms while H₂SO₄ is reduced mainly to SO₂
I⁻	easiest	I₂ forms while H₂SO₄ can be reduced to SO₂, S and H₂S

Down the group, ionic radius and shielding increase. The outer electron is farther from the nucleus and less strongly attracted, so it is lost more readily and the ion is a stronger reducing agent.

This is the opposite of halogen oxidising power: X₂ gains electrons, whereas X⁻ loses electrons. Always identify the starting species before choosing the trend.

For an aqueous sample, acidify with dilute nitric acid, add aqueous silver nitrate, then test any precipitate first with dilute aqueous ammonia and, if needed, concentrated aqueous ammonia.

$\mathrm{Ag^+(aq) + X^-(aq) \rightarrow AgX(s)}$

Halide	Silver-ion result	Dilute NH₃	Concentrated NH₃
F⁻	no precipitate because AgF is soluble	—	—
Cl⁻	white AgCl precipitate	dissolves	dissolves
Br⁻	cream AgBr precipitate	does not dissolve	dissolves
I⁻	yellow AgI precipitate	does not dissolve	does not dissolve

With a solid sodium halide, concentrated sulfuric acid first acts as an acid: NaX(s) + H₂SO₄(l) → NaHSO₄(s) + HX(g). NaF and NaCl stop at this acid-base stage, giving steamy acidic HF or HCl fumes; the sulfur oxidation number remains +6.

Reducing halide product	Balanced further reaction with concentrated H₂SO₄	Sulfur product / observation
HBr	2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O	red-brown Br₂ and colourless choking SO₂; S: +6 → +4
HI	2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O	purple I₂ vapour / dark iodine and SO₂
HI	6HI + H₂SO₄ → 3I₂ + S + 4H₂O	yellow sulfur; S: +6 → 0
HI	8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O	H₂S with characteristic rotten-egg odour; S: +6 → −2

Br⁻ and I⁻ are oxidised from −1 to 0 in Br₂ or I₂. I⁻ is the stronger reducing agent, so it can reduce sulfuric acid through more oxidation-number steps than Br⁻.

Nitric acid is used before Ag⁺ because chloride-containing acids would create AgCl. The syllabus does not require the formula or formation equation of the soluble silver-ammonia complex, so identify dissolution by observation only. Toxic gases must be handled in a fume cupboard and never smelled directly.

11.4 The reactions of chlorine

In a disproportionation reaction, the same element is both oxidised and reduced. Chlorine reacts with cold dilute sodium hydroxide to form chloride and chlorate(I), while hot concentrated alkali forms chloride and chlorate(V).

Use oxidation numbers to show the two pathways. In cold solution chlorine goes from 0 to −1 and +1; in hot solution it goes from 0 to −1 and +5. Balance atoms and charge before naming the reaction.

Cold: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Hot: 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O. The conditions determine the product.

Do not use one equation for every sodium-hydroxide condition, and do not call the reaction simple oxidation or reduction.

$\mathrm{Cl_2(aq) + H_2O(l) \rightleftharpoons HCl(aq) + HOCl(aq)}$

$\mathrm{HOCl(aq) \rightleftharpoons H^+(aq) + ClO^-(aq)}$

Hypochlorous acid, HOCl, and hypochlorite ions, ClO⁻, are the active species that kill bacteria. Chloride ions from HCl are not the disinfecting species.

HOCl is a weak acid, so both HOCl and ClO⁻ are present in water. Their relative proportions depend on pH, but both species named in the syllabus contribute to purification.

Do not write ClO₃⁻, the hot-alkali product, in the water-purification equilibrium. This card explains formation of HOCl and ClO⁻; it does not prescribe a chlorine dose or claim that unrestricted chlorine addition is safe.

11. Group 17

Exam analysis

Most tested topics

11.1 Physical properties of the Group 17 elements

Halogen colours deepen and volatility decreases from chlorine to iodine

X–X bond strength generally decreases down Group 17, with F₂ anomalous

Larger X₂ electron clouds create stronger temporary attractions

11.2 The chemical properties of the halogen elements and the hydrogen halides

Halogen oxidising power decreases down Group 17

Reactions with hydrogen become less vigorous down Group 17

Hydrogen halide thermal stability decreases as H–X bonds weaken

11.3 Some reactions of the halide ions

Halide reducing power increases down Group 17

Silver ions and concentrated sulfuric acid distinguish halides

11.4 The reactions of chlorine

Chlorine disproportionates differently in cold and hot alkali

Chlorine forms two active bactericidal species in water