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Group 17 Chemistry Trends

Learn and apply Group 17 Chemistry Trends in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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H1 Group 17 Chemistry Trends: Orientation

Group 17 combines three different causal models. Volatility depends on intermolecular attraction, oxidising strength on ease of electron gain, and hydride stability on H–X bond energy.

H1 8873 scope
  • H1 explains oxidising strength through ease of electron gain; standard electrode potentials are not required.
  • The required comparisons run from chlorine to iodine.

Definitions (Must Know)

  • A halogen is a Group 17 element; chlorine, bromine and iodine form diatomic molecules, X₂.
  • A halide ion is the 1- ion formed when a halogen gains an electron.
  • An oxidising agent accepts electrons and is itself reduced:

X₂ + 2e⁻ → 2X⁻

  • Volatility describes the tendency to enter the gaseous state.
  • Thermal stability is resistance to decomposition on heating.

Detailed Explanations

A. Volatility

The non-polar halogen molecules are attracted mainly by instantaneous dipole–induced dipole forces. From Cl₂ to I₂, electron number and cloud polarisability increase. Stronger attractions require more energy to overcome, so boiling point increases and volatility decreases.

B. Oxidising strength

Each halogen molecule gains electrons to form halide ions. Down the group, the incoming electron enters a shell farther from the nucleus and experiences more shielding. Attraction for the incoming electron is weaker, so electron gain becomes less favourable and oxidising strength decreases.

Chlorine therefore oxidises bromide ions:

Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂

C. Hydrogen-halide stability

The H–X bond becomes longer down the group because the halogen atom is larger. Orbital overlap is less effective and bond energy decreases. Less energy is needed to break the bond, so thermal stability decreases from HCl to HI.

Worked Examples

Modelled example 1

Explain Why Iodine Is Less Volatile Than Chlorine

Core

Problem

Explain why iodine is less volatile than chlorine.
Study the worked solution

  1. Compare the electron clouds

    Method

    State that an I₂ molecule has more electrons and a larger, more polarisable electron cloud than Cl₂.

    Reason

    Its electron distribution is distorted more readily, producing stronger instantaneous dipoles.

    Working

    Polarisability: I₂ > Cl₂.

  2. Link force to volatility

    Method

    Give iodine stronger London dispersion forces between its molecules.

    Reason

    More energy is needed to separate the molecules, so iodine has a higher boiling point and lower volatility.

    Working

    London forces ↑ ⇒ boiling point ↑ ⇒ volatility ↓.

Guided practice 2

Compare the Thermal Stability of HCl and HI

About 6 min

Problem

Predict which compound, HCl or HI, decomposes more readily on heating. Explain your answer.

Try this before viewing the solution

Decomposes more readily
Weaker bond

Hints

Hint 1: inside the molecule
Thermal decomposition requires a covalent H–X bond to break.
Hint 2: compare atom size
Iodine is larger than chlorine, so compare H–I and H–Cl bond length and orbital overlap.
View solution step by step

  1. Compare the bonds

    Method

    State that the H–I bond is longer than the H–Cl bond.

    Reason

    The larger iodine atom gives poorer orbital overlap with hydrogen.

    Working

    Bond strength: H–I < H–Cl.

  2. Link bond strength to stability

    Method

    Conclude that HI decomposes more readily and is less thermally stable.

    Reason

    Less energy is needed to break the weaker H–I bond.

    Working

    Thermal stability: HI < HCl.

Common misconception 3

Correct a Halogen-Displacement Error

Find and correct the mistake

Learner claim

A learner says Br₂ cannot react with aqueous iodide because a negative iodide ion cannot lose electrons. Correct the claim, identify the oxidising agent and write the ionic equation.

Try this before viewing the solution

Oxidising agent
Does displacement occur?

Hints

Hint 1: oxidising order
Oxidising strength decreases down Group 17, so compare bromine with iodine.
Hint 2: electron balance
Reduce one Br₂ molecule to two bromide ions and oxidise two iodide ions to one I₂ molecule.
View solution step by step

  1. Predict the electron transfer

    Method

    Identify bromine as the stronger oxidising agent.

    Reason

    Bromine is above iodine in Group 17 and accepts electrons from iodide ions.

    Working

    Br₂ is reduced; I⁻ is oxidised.

  2. Balance the ionic equation

    Method

    Use two iodide ions to supply the two electrons accepted by Br₂.

    Reason

    This conserves atoms and charge.

    Working

    Br₂ + 2I⁻ → 2Br⁻ + I₂.

Mind Stretchers

Attempt each unfamiliar application before opening the hint, then compare your reasoning chain with the solution.

Mind stretcher 1: Identifying an unknown halogenExtension

Question. Halogen X oxidises iodide but does not oxidise bromide. Identify X and justify.

Show Hint

Place its oxidising strength between bromine and iodine.

Show Answer

X is bromine. It is strong enough to oxidise iodide ions, but it cannot oxidise bromide ions; chlorine would oxidise both.

Mind stretcher 2: Separating two stability ideasExtension

Question. Iodine has the highest boiling point of Cl₂, Br₂ and I₂, yet HI has the lowest thermal stability of HCl, HBr and HI. Reconcile the observations.

Show Hint

One trend concerns attractions between molecules; the other concerns a covalent bond within a molecule.

Show Answer

I₂ has the strongest intermolecular attractions because its electron cloud is most polarisable, raising boiling point. HI has the weakest H–X covalent bond because iodine is largest and overlap is poorest, so HI decomposes most readily.