# Group 17 Properties and Reactions

> Chemistry · CIE A-Level 9701
> Source: https://www.owlsprep.com/study/cie-9701-u10-group-17-properties-and-reactions/

This subtopic covers key physical and chemical trends of Group 17 halogens, their displacement reactions, and disproportionation reactions, all core frequently tested content for CIE A-Level Chemistry.

**Prerequisites:** [Redox reactions and oxidation numbers](https://www.owlsprep.com/study/cie-9701-u7-redox-reactions-oxidation-numbers/); [Intermolecular forces](https://www.owlsprep.com/study/cie-9701-u3-intermolecular-forces/)

## Learning objectives

- Describe key physical trends of Group 17 elements down the group
- Explain the trend in oxidising power of halogens
- Write balanced equations for halogen displacement and disproportionation reactions
- Identify observation trends in halogen displacement experiments

## Physical Trends Down Group 17

Group 17 elements (the halogens) are all reactive non-metals that exist as diatomic molecules ($X_2$) at room temperature. Clear, predictable trends in physical properties are observed down the group, driven by increasing atomic and molecular size.

**Halogens** — A group of non-metals with 7 valence electrons that typically form -1 anions (halides) in ionic compounds.

*Notation:* Group 17 (Group VII)

*Example:* Chlorine ($Cl_2$) and iodine ($I_2$) are common halogens.

Key physical trends down the group are: atomic radius increases, electronegativity decreases, melting/boiling point increases, and element colour darkens from pale yellow (fluorine) to black solid (iodine).

Melting and boiling point increase because the only intermolecular forces between non-polar diatomic halogen molecules are London dispersion (instantaneous dipole-induced dipole) forces. As the number of electrons increases down the group, the electron cloud becomes more polarisable, so intermolecular attractions get stronger.

**Worked example:** Explain why fluorine has a lower boiling point than bromine, with reference to intermolecular forces.

1. 1. Both fluorine ($F_2$) and bromine ($Br_2$) are non-polar diatomic molecules, held together by London dispersion forces.
2. 2. Bromine atoms have more electrons than fluorine atoms (35 vs 9 per atom).
3. 3. This means bromine's electron cloud is more polarisable than fluorine's, so intermolecular forces are stronger.
4. 4. More energy is required to overcome stronger intermolecular forces, so bromine has a higher boiling point.

> **Exam tip:** Always mention that intermolecular forces (not covalent bonds) are broken when halogens boil/melt. CIE examiners regularly penalise incorrect references to breaking covalent bonds.

## Trend in Oxidising Power

Halogens almost always act as oxidising agents in reactions, by accepting an electron to form a halide ion, per the half-equation: $X_2 + 2e^- \rightarrow 2X^-$. The strength of oxidising power decreases consistently down Group 17.

**Oxidising power** — The ability of a halogen to act as an oxidising agent (accept electrons from other species), linked to the attraction of the nucleus for an extra outer electron.

Down the group, atomic radius increases and shielding from inner electron shells increases. This reduces the attraction of the nucleus for an extra electron needed to form a halide ion, so oxidising strength decreases.

**Worked example:** Predict whether bromine will displace iodide ions from aqueous potassium iodide, explain your answer and write a balanced equation.

1. 1. Oxidising power decreases down Group 17, so bromine is a stronger oxidising agent than iodine.
2. 2. A stronger oxidising halogen will always displace a halide ion of a weaker oxidising halogen from solution.
3. 3. Therefore, bromine will displace iodide ions from potassium iodide.
4. 4. The balanced overall equation is:
5. $$Br_2(aq) + 2KI(aq) \rightarrow 2KBr(aq) + I_2(aq)$$

**Check your understanding**

Check your understanding of the trend:

1. Which of the following is the correct order of oxidising power (strongest first)?

   - F > Cl > Br > I
   - I > Br > Cl > F
   - Cl > F > Br > I
   - Br > Cl > I > F

   *Why:* Correct! Oxidising power decreases down Group 17, so fluorine is the strongest oxidising agent.

## Displacement Reactions of Halogens

Displacement reactions are the standard practical experiment to demonstrate the trend in oxidising power. When a halogen solution is added to a halide salt solution, a colour change occurs if the halogen is a stronger oxidising agent than the halide.

To clearly identify the product, an non-polar organic solvent like cyclohexane is often added after the reaction, and the mixture is shaken. The halogen dissolves in the organic layer, giving a distinct characteristic colour.

**Worked example:** A student adds chlorine water to potassium bromide solution, then adds cyclohexane and shakes. Describe the observations and write the balanced ionic equation.

1. 1. Chlorine is a stronger oxidising agent than bromine, so it will displace bromide ions from solution.
2. 2. The aqueous layer first turns pale orange-yellow as bromine forms.
3. 3. After adding and shaking cyclohexane, the less dense organic (upper) layer turns orange-brown, confirming bromine is present.
4. 4. The balanced ionic equation for the reaction is:
5. $$Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)$$

> **Exam tip:** CIE frequently asks for the colour of the organic solvent layer, not the aqueous layer. Remember: iodine is purple in organic solvent, but brown in aqueous solution.

## Disproportionation Reactions

Halogens undergo disproportionation when they react with cold or hot alkalis, and with water. In these reactions, the same halogen atom is both oxidised and reduced.

**Disproportionation** — A redox reaction where the same element in a single starting species is simultaneously oxidised (oxidation number increases) and reduced (oxidation number decreases).

**Worked example:** Write the balanced equation for chlorine reacting with cold dilute sodium hydroxide, and show it is disproportionation by calculating oxidation numbers.

1. 1. Chlorine reacts with cold dilute NaOH to form sodium chloride, sodium chlorate(I) and water.
2. 2. The balanced molecular equation is:
3. $$Cl_2(aq) + 2NaOH(aq) \rightarrow NaCl(aq) + NaClO(aq) + H_2O(l)$$
4. 3. Oxidation number of Cl in $Cl_2$ is 0. In NaCl it is -1 (reduced), in NaClO it is +1 (oxidised).
5. 4. Since Cl is both oxidised and reduced, this is a disproportionation reaction.

With hot concentrated sodium hydroxide, chlorine disproportionates to form sodium chloride and sodium chlorate(V) ($NaClO_3$), where chlorine has oxidation numbers of -1 and +5 respectively.

## Common pitfalls

- **Wrong:** Stating covalent bonds are broken when halogens melt or boil.
  - Why it fails: Only intermolecular forces between separate $X_2$ molecules are broken; the covalent bond inside each molecule stays intact.
  - Correct: State that London dispersion intermolecular forces are overcome when halogens change state.
- **Wrong:** Claiming oxidising power increases down Group 17.
  - Why it fails: Atomic radius increases down the group, so nucleus attraction for an extra electron decreases, reducing oxidising strength.
  - Correct: Remember fluorine is the strongest oxidising agent, iodine the weakest, so oxidising power decreases down the group.
- **Wrong:** Stating iodine is purple in aqueous solution.
  - Why it fails: Iodine is only purple in non-polar organic solvents; aqueous iodine is brown.
  - Correct: Memorise: purple = iodine in organic solvent, brown = aqueous iodine for CIE exams.
- **Wrong:** Writing $NaClO_3$ as the product for chlorine reacting with cold dilute NaOH.
  - Why it fails: $NaClO_3$ (sodium chlorate(V)) only forms with hot concentrated sodium hydroxide, not cold dilute.
  - Correct: Cold dilute NaOH gives sodium chlorate(I) ($NaClO$), hot concentrated gives sodium chlorate(V) ($NaClO_3$).
- **Wrong:** Confusing displacement and disproportionation reactions.
  - Why it fails: Displacement involves two different halogen species changing oxidation state, disproportionation involves the same halogen species changing into two different oxidation states.
  - Correct: Check oxidation numbers: if the same starting element has two different oxidation numbers in products, it is disproportionation.

## Cheatsheet

| Property | Trend down Group 17 | Key Notes |
| --- | --- | --- |
| Atomic radius | Increases | More electron shells, increased shielding |
| Electronegativity | Decreases | Less attraction for bonding electrons |
| Boiling point | Increases | Stronger London dispersion forces |
| Oxidising power | Decreases | Less attraction for an extra electron |
| Displacement order | F > Cl > Br > I | Stronger oxidiser displaces weaker halide |
| Organic layer colours | Cl = pale green, Br = orange, I = purple | Aqueous iodine is brown |
| $Cl_2$ + cold NaOH | $NaCl + NaClO + H_2O$ | Disproportionation, Cl = -1, +1 |
| $Cl_2$ + hot NaOH | $NaCl + NaClO_3 + H_2O$ | Disproportionation, Cl = -1, +5 |

## What's next

Mastering Group 17 properties and reactions builds a foundation for learning halide ion qualitative testing, a common topic in both multiple-choice and practical exam questions. You will apply the periodic trends you learned here to analyse other groups in the periodic table, and extend your knowledge of redox and disproportionation reactions to more complex problems involving transition metals and organic chemistry. Recognising the patterns in halogen reactivity will help you quickly answer common exam questions and avoid common pitfalls.

- [Nitrogen and sulfur](https://www.owlsprep.com/study/cie-9701-u11-overview/)
- [Nitrogen and its compounds](https://www.owlsprep.com/study/cie-9701-u11-nitrogen-and-its-compounds/)
- [Sulfur and its compounds](https://www.owlsprep.com/study/cie-9701-u11-sulfur-and-its-compounds/)

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