Group 7: The Halogens

The halogens — fluorine (F₂), chlorine (Cl₂), bromine (Br₂), and iodine (I₂) — are Group 7 elements. They are non-metals that exist as diatomic molecules. Their chemistry is dominated by their high electronegativity and tendency to gain one electron to achieve a noble gas configuration, forming X⁻ halide ions.

Trends in Physical Properties

State at Room Temperature

Halogen	State	Colour	Boiling Point / °C	Electron Count
F₂	Gas	Pale yellow	−188	18
Cl₂	Gas	Green-yellow	−34	34
Br₂	Liquid	Red-brown	59	70
I₂	Solid	Grey solid, purple vapour	184	106

The halogens change from gas to liquid to solid as you descend the group. This is because the number of electrons increases, which increases the strength of London dispersion forces (temporary dipole–induced dipole interactions) between molecules. Stronger intermolecular forces require more energy to overcome, so boiling points increase.

Key detail for exams: When iodine dissolves in organic solvents (such as cyclohexane or hexane), it produces a purple/violet solution. In aqueous solution (with KI), iodine produces a brown solution. This colour difference is frequently tested.

Electronegativity

Electronegativity decreases down Group 7. Fluorine is the most electronegative element in the periodic table (3.98 on the Pauling scale).

Halogen	Electronegativity (Pauling)	Atomic Radius / pm
F	3.98	64
Cl	3.16	99
Br	2.96	114
I	2.66	133

As atomic radius increases down the group, the bonding pair of electrons in a covalent bond is further from the nucleus and experiences more shielding, so the atom's ability to attract the bonding pair decreases.

Oxidising Ability and Standard Electrode Potentials

The halogens are oxidising agents — they gain electrons (are reduced) in reactions:

X₂ + 2e⁻ → 2X⁻

Oxidising ability decreases down the group. This can be quantified using standard electrode potentials (E°):

Half-Reaction	E° / V
F₂ + 2e⁻ → 2F⁻	+2.87
Cl₂ + 2e⁻ → 2Cl⁻	+1.36
Br₂ + 2e⁻ → 2Br⁻	+1.07
I₂ + 2e⁻ → 2I⁻	+0.54

The higher the E° value, the stronger the oxidising agent. Fluorine is the most powerful; iodine is the weakest. This decrease occurs because:

Atomic radius increases down the group
The incoming electron is further from the nucleus
Increased shielding reduces the nuclear attraction on the incoming electron
It becomes harder for the atom to gain an electron

Displacement Reactions

A more reactive halogen can displace a less reactive halide ion from solution. This demonstrates the trend in oxidising ability:

Cl₂(aq) + 2KBr(aq) → 2KCl(aq) + Br₂(aq)

Chlorine oxidises bromide ions to bromine. The solution turns orange/brown.

Cl₂(aq) + 2KI(aq) → 2KCl(aq) + I₂(aq)

Chlorine oxidises iodide ions to iodine. The solution turns brown/dark brown. If an organic solvent (cyclohexane) is added, the iodine dissolves in it to give a purple/violet colour in the organic layer.

Br₂(aq) + 2KI(aq) → 2KBr(aq) + I₂(aq)

Bromine oxidises iodide ions to iodine.

However, bromine cannot displace chloride from KCl, and iodine cannot displace either bromide or chloride.

Complete Displacement Results Table with Observations

	KCl(aq)	KBr(aq)	KI(aq)
Cl₂	No reaction	Orange/brown (Br₂)	Brown (I₂); purple in organic layer
Br₂	No reaction	No reaction	Brown (I₂); purple in organic layer
I₂	No reaction	No reaction	No reaction

flowchart TD
    A["Halogen displacement test"] --> B{"Which halogen<br>is added?"}
    B -->|"Cl₂"| C{"Halide present?"}
    C -->|"Br⁻"| D["Orange/brown → Br₂ formed"]
    C -->|"I⁻"| E["Brown → I₂ formed"]
    C -->|"Cl⁻"| F["No reaction"]
    B -->|"Br₂"| G{"Halide present?"}
    G -->|"I⁻"| H["Brown → I₂ formed"]
    G -->|"Cl⁻ or Br⁻"| I["No reaction"]
    B -->|"I₂"| J["No reaction with any halide"]

Disproportionation of Chlorine

Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced.

Chlorine in Water

Cl₂(g) + H₂O(l) → HCl(aq) + HClO(aq)

One Cl atom is reduced from 0 to −1 (in HCl)
One Cl atom is oxidised from 0 to +1 (in HClO)

This is a reversible reaction. HClO (chloric(I) acid / hypochlorous acid) is a powerful oxidising agent that kills bacteria. This is why chlorine is added to drinking water and swimming pools.

Chlorine in Cold Dilute NaOH

Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Again, chlorine disproportionates from 0 to −1 (in NaCl) and 0 to +1 (in NaClO). Sodium chlorate(I) (NaClO) is the active ingredient in household bleach.

Exam distinction: In cold, dilute NaOH, the product is NaClO (chlorate(I), Cl in +1). In hot, concentrated NaOH, the product is NaClO₃ (chlorate(V), Cl in +5): 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. This distinction is sometimes tested.

Chlorine in Water Treatment

Chlorine is added to water supplies to kill harmful bacteria and make the water safe to drink. The HClO formed when chlorine dissolves in water is responsible for the antibacterial action.

Benefits:

Kills disease-causing bacteria and viruses
Provides residual protection as water travels through pipes
Prevents waterborne diseases like cholera and typhoid

Risks/concerns:

Chlorine can react with organic matter in water to form chlorinated hydrocarbons (trihalomethanes, THMs), some of which may be carcinogenic
Some people object to the addition of a chemical to public water supplies on ethical grounds
THM levels are regulated and monitored to stay below safety thresholds

Worked Example: Predicting Displacement Outcomes

Question: A student adds bromine water to a solution of potassium iodide, then adds cyclohexane and shakes. Describe the observations.

Answer:

The bromine water (initially orange) is added to KI (colourless)
Br₂ is a stronger oxidising agent than I₂ (E° for Br₂ = +1.07 V > E° for I₂ = +0.54 V)
Br₂ oxidises I⁻ to I₂: Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq)
The aqueous layer turns brown (dissolved I₂)
When cyclohexane is added and shaken, I₂ dissolves preferentially in the organic layer
The cyclohexane (top) layer turns purple/violet; the aqueous layer becomes paler

Bond Enthalpies in Halogens

The bond enthalpy of the X–X bond in the diatomic halogen molecules does not follow a simple trend:

Halogen	Bond Enthalpy / kJ mol⁻¹	Bond Length / pm
F–F	158	142
Cl–Cl	242	199
Br–Br	193	228
I–I	151	267

Notice that the F–F bond is unexpectedly weak despite fluorine being the smallest halogen. This anomaly is explained by the very small size of fluorine atoms: the lone pairs on adjacent F atoms are very close together and experience significant repulsion, which destabilises the bond. From Cl to I, bond enthalpy decreases as expected because the larger atoms have less effective orbital overlap.

Exam note: The weak F–F bond is one reason why fluorine is such a powerful oxidising agent — the energy cost of breaking the F–F bond is low, while the electron affinity and hydration enthalpy of F⁻ are very high. This combination makes the overall process of F₂ gaining electrons highly exothermic.

Hydrogen Halides and Acid Strength

All hydrogen halides (HF, HCl, HBr, HI) dissolve in water to form acidic solutions. However, their acid strengths differ:

HF is a weak acid (does not fully dissociate in water)
HCl, HBr, and HI are all strong acids (fully dissociate)

The H–X bond enthalpy decreases down the group (HF: 568, HCl: 432, HBr: 366, HI: 298 kJ mol⁻¹). The weaker the bond, the more readily it dissociates in water. HF has such a strong bond that dissociation is incomplete.

Common Exam Mistakes with Halogens

Saying "reactivity increases down Group 7" — it decreases (halogens become weaker oxidising agents). Reactivity increases down Group 1 and 2 (metals), but decreases down Group 7 (non-metals).
Confusing displacement colours: Br₂ in aqueous solution is orange/brown, not red. I₂ in aqueous solution is brown; only in organic solvents is it purple.
Forgetting the organic solvent layer test: Adding cyclohexane and shaking separates I₂ (purple in organic layer) from Br₂ (orange in organic layer) — this is the definitive way to distinguish them.
Writing products of disproportionation incorrectly: In cold dilute NaOH, the product is NaClO (Cl in +1). Students sometimes write NaClO₃ (Cl in +5), which only forms in hot concentrated NaOH.

Summary

The halogens show clear trends: boiling points and atomic radius increase down the group, while electronegativity and oxidising ability decrease. Displacement reactions confirm the reactivity order — supported by electrode potential data. Chlorine undergoes disproportionation in water and NaOH, and is used in water treatment despite concerns about THM formation. The anomalous weakness of the F–F bond and the trend in hydrogen halide acid strength are important details for exam answers.

A-Level Deep Dive: Group 7: The Halogens

Spec mapping

Edexcel 9CH0 specification Topic 4 — Inorganic Chemistry and the Periodic Table, sub-topic 4.2 covers the trends in physical properties (m.p./b.p., colour/state, polarisability) and chemical reactivity (oxidising power) of the halogens Cl₂, Br₂ and I₂; displacement reactions of halogens with halide ions; and the trend in oxidising ability descending the group (refer to the official specification document for exact wording). Examined in Paper 1 (9CH0/01) and synoptically in Paper 3 (9CH0/03) through CP4 displacement and qualitative analysis. Synoptic dependencies include Topic 8 (Redox I) for half-equations and electrode potentials, Topic 2 (Bonding) for instantaneous-induced dipole forces explaining b.p. trends, and Topic 13 (Energetics II) for hydration vs lattice arguments where ionic halides are concerned.

Worked example with full mark scheme

Question (8 marks):

(a) State the colours and physical states at room temperature of Cl₂, Br₂ and I₂. Explain the trend in melting point. (3)

(b) Predict, with reasoning, what would be observed when bromine water is added to (i) potassium iodide solution and (ii) potassium chloride solution. Write a balanced ionic equation for any reaction. (5)

Solution with mark scheme:

(a) Step 1 — physical data.

Cl₂ — pale green/yellow gas; Br₂ — red-brown liquid (volatile, fuming vapour); I₂ — shiny grey-black solid (sublimes to violet vapour on warming).

B1 — all three colours and states correct.

Step 2 — explain m.p. trend.

m.p. increases F < Cl < Br < I (F₂ −220 °C; Cl₂ −101 °C; Br₂ −7 °C; I₂ +114 °C). The molecules are non-polar diatomic, so the only intermolecular forces are instantaneous-induced (London) dispersion forces. Larger molecules (more electrons, larger electron clouds) have greater polarisability and stronger dispersion forces, so more energy is required to separate them.

M1 — instantaneous-induced/London/dispersion forces named.

A1 — link of polarisability/electron count to dispersion strength to m.p.

(b) (i) Br₂ + 2KI → 2KBr + I₂.

Bromine is a stronger oxidising agent than iodine (oxidising power decreases down Group 7), so Br₂ oxidises I⁻ to I₂. Observation: brown solution forms (I₂ in water); shaking with cyclohexane gives a violet/purple organic layer confirming I₂.

M1 — direction of redox correct (Br₂ oxidises I⁻).

B1 — observation: brown/orange to brown solution + violet/purple cyclohexane layer.

A1 — ionic equation: Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq).

(ii) No reaction. Cl⁻ is a weaker reducing agent than Br⁻; equivalently, Br₂ is a weaker oxidising agent than Cl₂. No oxidation of Cl⁻ occurs.

M1 — no reaction stated.

A1 — justified by E° comparison or oxidising-power trend (Cl₂ > Br₂ > I₂).

Total: 8 marks (M3 A3 B2).

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks): A student adds chlorine water to a colourless solution and observes a yellow-orange colour develop. The student then shakes the mixture with cyclohexane.

(a) Identify the original anion and explain your reasoning, including the redox half-equations. (4)

(b) Predict the colour of the cyclohexane layer and write the overall ionic equation. (2)

Mark scheme decomposition by AO:

Mark	AO	Awarded for
(a) B1	AO1	Anion is Br⁻ (yellow-orange = Br₂ in aqueous)
(a) M1	AO2	Cl₂ is a stronger oxidising agent than Br₂ (oxidising power decreases down Group 7)
(a) M1	AO2	Half-equation oxidation: 2Br⁻ → Br₂ + 2e⁻
(a) A1	AO2	Half-equation reduction: Cl₂ + 2e⁻ → 2Cl⁻
(b) B1	AO1	Cyclohexane layer orange/red-brown (Br₂ in non-polar solvent)
(b) A1	AO2	Overall: Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)