Group 17 Chemistry Trends
Explain Group 17 volatility, oxidising strength and hydrogen-halide stability.
On this page
Group 17 questions connect three different ideas: volatility depends on intermolecular attraction, oxidising strength is deduced from supplied E⦵ values, and hydrogen-halide stability depends on H–X bond energy.
Review standard electrode potentials before using the reduction-potential argument, then return to the Periodic Table hub for the full topic sequence.
Definitions (Must Know)
A. Halogen and halide ion
- A halogen is a Group 17 element; chlorine, bromine and iodine exist as diatomic molecules, X₂.
- A halide ion is the 1- ion formed when a halogen gains one electron, X⁻.
B. Oxidising agent
An oxidising agent accepts electrons and is itself reduced. For a halogen:
X₂ + 2e⁻ ⇌ 2X⁻
A more positive E⦵ means X₂ is reduced more readily and is the stronger oxidising agent.
C. Instantaneous dipole–induced dipole attraction
This intermolecular attraction arises when a temporary uneven electron distribution induces a dipole in a neighbouring particle. It becomes stronger as the electron cloud becomes more polarisable.
D. Thermal stability
Thermal stability is resistance to decomposition on heating. For a hydrogen halide, it depends on the energy required to break the H–X bond.
Key Ideas (What Earns Marks)
- From chlorine to iodine, the outer configuration remains ns²np⁵, while atomic radius increases and first ionisation energy and electronegativity decrease.
- Boiling point increases and volatility decreases because the larger electron clouds are more polarisable, strengthening instantaneous dipole–induced dipole attractions.
- Supplied E⦵(X₂/X⁻) values become less positive down the group, so halogen oxidising strength decreases: Cl₂ > Br₂ > I₂.
- A stronger oxidising halogen displaces the halide ion of a weaker oxidising halogen.
- Hydrogen-halide thermal stability decreases from HCl to HI because H–X bond energy decreases.
- Volatility: electron-cloud size and polarisability → intermolecular attraction. - Oxidising strength: compare supplied reduction potentials; more positive E⦵ → stronger oxidising agent. - Hydride stability: compare H–X covalent bond energies, not intermolecular forces.
Detailed Explanations
A. Electronic configuration and atomic properties
The outer configurations are 3s²3p⁵ for chlorine, 4s²4p⁵ for bromine and 5s²5p⁵ for iodine. Down the group, each element has an additional occupied shell.
The greater distance and shielding outweigh the increased nuclear charge. Therefore atomic radius increases, while first ionisation energy and electronegativity decrease.
B. Volatility
Halogen molecules are non-polar, so their main intermolecular attraction is instantaneous dipole–induced dipole attraction.
Down the group, X₂ has more electrons and a more polarisable electron cloud. Stronger attractions require more energy to overcome, so boiling point increases and volatility decreases.
Data table
| Halogen | Boiling point |
|---|---|
| Cl2 | -34 |
| Br2 | 59 |
| I2 | 184 |
C. Oxidising power from supplied E⦵ values
Representative supplied reduction potentials are:
| Reduction half-equation | E⦵ / V |
|---|---|
| Cl₂ + 2e⁻ ⇌ 2Cl⁻ | + 1.36 |
| Br₂ + 2e⁻ ⇌ 2Br⁻ | + 1.07 |
| I₂ + 2e⁻ ⇌ 2I⁻ | + 0.54 |
The values become less positive down the group, so reduction becomes less favourable and oxidising strength decreases.
For example, chlorine oxidises bromide ions:
Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂
D. Hydrogen-halide thermal stability
Hydrogen halides can decompose on heating:
2HX(g) → H₂(g) + X₂(g)
From HCl to HI, the halogen atom becomes larger, the H–X bond becomes longer and its bond energy decreases. The bond therefore breaks more readily, so thermal stability decreases:
HCl > HBr > HI
Worked Examples
Modelled example 1
Predict Chlorine–Bromide Displacement
Problem
Study the worked solution
Compare reduction tendency
Method
Select the couple with the more positive reduction potential.Reason
Its halogen gains electrons more readily under standard conditions.Working
+ 1.36 V > +1.07 V.Name the oxidising agent
Method
Name chlorine as the stronger oxidising agent.Reason
The oxidising agent is itself reduced while causing bromide to be oxidised.Working
Cl₂ + 2e⁻ → 2Cl⁻.Write the displacement
Method
Combine chlorine reduction with bromide oxidation.Reason
The two electrons cancel in the feasible direction.Working
Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂
Guided practice 2
Test Bromine against Chloride
Problem
Try this before viewing the solution
Hints
Hint 1: desired reduction
Hint 2: cell sign
View solution step by step
Compare strengths
Method
Classify bromine as the weaker oxidising agent.Reason
Its reduction potential is less positive than chlorine’s.Working
E⦵(Br₂/Br⁻) < E⦵(Cl₂/Cl⁻).Reject the direction
Method
State that Br₂ does not oxidise Cl⁻ under standard conditions.Reason
The proposed direction gives a negative standard cell potential.Working
E⦵_cell = 1.07-1.36 = -0.29 V.
Common misconception 3
Explain Iodine’s Lower Volatility
Learner claim
A learner says iodine is less volatile than chlorine because the I–I covalent bond is stronger. Correct the explanation.
Try this before viewing the solution
View solution step by step
Choose the relevant attraction
Method
Compare London forces between halogen molecules.
Reason
Volatility depends on separating intact molecules, not breaking X–X bonds.
Working
Intermolecular attraction.Use polarisability
Method
Give I₂ the more polarisable electron cloud and stronger London forces.
Reason
More energy is required to separate iodine molecules, so boiling point is higher and volatility lower.
Working
T_b(I₂) > T_b(Cl₂); iodine is less volatile.
Challenge 4
Order Hydrogen Halide Thermal Stability
Bond-energy transfer
Try this before viewing the solution
Hints
Hint 1: bond length
Hint 2: bond energy
View solution step by step
Follow the bond trend
Method
Increase H–X bond length and decrease bond energy from H–Cl to H–I.Reason
Larger halogen atoms give longer, weaker bonds.Working
H–Cl is strongest; H–I is weakest.Order stability
Method
Place hydrogen chloride first and hydrogen iodide last.Reason
The weaker bond breaks more readily on heating.Working
HCl > HBr > HI in decreasing thermal stability.
Common Mistakes
- Reading E⦵ values as oxidation potentials; the data-table half-equations are reductions.
- Choosing the less positive X₂/X⁻ value as the stronger oxidising agent.
- Forgetting the coefficient 2 when balancing a halogen displacement equation.
- Explaining hydrogen-halide thermal stability with London forces; decomposition requires breaking the covalent H–X bond.
- Including fluorine when the stated 9476 comparison is chlorine to iodine.
Use the Periodic Table topic check to practise and check your understanding.
Exam Tips
- When E⦵ data are supplied, quote the values and say the more positive reduction potential identifies the stronger oxidising agent.
- For volatility, name instantaneous dipole–induced dipole attraction and connect polarisability to boiling point.
- For hydrides, write the chain: atom size → bond length → bond energy → thermal stability.
Mind Stretchers
Mind stretcher 1Extension
Two supplied reduction potentials are E⦵(X₂/X⁻) = +1.07 V and E⦵(Y₂/Y⁻) = +0.54 V. Predict whether X₂ reacts with Y⁻ and justify the direction.
Show Answer
Mark scheme:
- X₂ has the more positive reduction potential, so it is reduced more readily and is the stronger oxidising agent.
- Y⁻ is oxidised while X₂ is reduced.
- The feasible displacement direction is X₂ + 2Y⁻ → 2X⁻ + Y₂.
Syllabus and review details
- GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus
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