11. Group 17
- Syllabus
- 9701–2028–2029
- Section
- 11
- Level
- AS
11.1 Physical properties of the Group 17 elements
- Syllabus
- 9701–2028–2029
- Topic
- 11.1
- Level
- AS
Halogen colours deepen and volatility decreases from chlorine to iodine
| 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.
X–X bond strength generally decreases down Group 17, with F₂ anomalous
| 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.
Larger X₂ electron clouds create stronger temporary attractions
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
- Syllabus
- 9701–2028–2029
- Topic
- 11.2
- Level
- AS
Halogen oxidising power decreases down Group 17
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⁻ |
Cl2+2Br−→2Cl−+Br2
Br2+2I−→2Br−+I2
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.
Reactions with hydrogen become less vigorous down Group 17
H2(g)+X2(g)→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 thermal stability decreases as H–X bonds weaken
| Hydrogen halide | Relative H–X bond strength | Relative thermal stability |
|---|---|---|
| HF | strongest | highest |
| HCl | weaker | lower |
| HBr | weaker again | lower again |
| HI | weakest | lowest |
2HX(g)⇌H2(g)+X2(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
- Syllabus
- 9701–2028–2029
- Topic
- 11.3
- Level
- AS
Halide reducing power increases down Group 17
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.
Silver ions and concentrated sulfuric acid distinguish halides
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.
Ag+(aq)+X−(aq)→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
- Syllabus
- 9701–2028–2029
- Topic
- 11.4
- Level
- AS
Chlorine disproportionates differently in cold and hot alkali
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.
Chlorine forms two active bactericidal species in water
Cl2(aq)+H2O(l)⇌HCl(aq)+HOCl(aq)
HOCl(aq)⇌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.