Explain Group 17 halogen bond-strength trends
Compare halogen–halogen bond strengths and explain the down-group weakening trend, including the fluorine exception.
Section 11 · Group 17
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. State conditions for halogen reactions, distinguish halogen oxidising strength from halide-ion reducing strength, and balance redox equations. Chlorine, bromine, iodine and concentrated acids are hazardous; examples teach chemistry, not practical instructions. Written practice is self-checked
Before you begin
- Use molecular halogen formulas X₂, oxidation numbers, ionic equations and electron-transfer ideas.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which bond is stronger in the Cl₂/Br₂/I₂ sequence?
What you will learn
- Describe the bond-strength trend for Cl₂, Br₂ and I₂
- Relate bond length to overlap
- Explain why F–F is an anomaly
Halogen–halogen bond strength falls from chlorine to iodine
For chlorine, bromine and iodine molecules, the X–X bond becomes longer and weaker down the group. Larger atoms have valence orbitals farther apart, so their overlap is less effective; less energy is needed to break the bond.
This is distinct from the volatility trend, which is controlled by intermolecular forces. I₂ has weaker covalent X–X bonding than Cl₂ but stronger intermolecular dispersion forces than Cl₂.
Fluorine's bond is unusually weak
The F–F bond is weaker than expected for a small diatomic molecule. Each small fluorine atom has three lone pairs packed near the bonding region; repulsions between these lone pairs weaken the bond.
The expected broad comparison is Cl–Cl > Br–Br > I–I in bond strength, with F–F as an anomalously weak bond. State the comparison requested rather than forcing all four into a simple smooth sequence.
Pause and explain
Why is the F–F bond unusually weak?
Worked example
Explain why the Cl–Cl bond is stronger than the I–I bond, and why F–F is a separate anomaly.
Show the worked solution
- Iodine atoms are larger, so the I–I bond is longer.
- The longer bond has less effective overlap and is weaker.
- F–F is weakened by repulsion between close lone pairs on the small atoms.
Answer Cl–Cl is stronger than I–I because it is shorter with more effective overlap. F–F is unusually weak because lone-pair repulsions are large in small fluorine atoms.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Order Cl₂, Br₂ and I₂ by decreasing X–X bond strength.
Give me a hint
Bond strength decreases as the halogen atoms become larger down the group.
Compare my reasoning
- Chlorine is above bromine and iodine.
- The Cl–Cl bond is shortest and strongest in this set.
- The I–I bond is weakest of the three.
Cl–Cl > Br–Br > I–I in bond strength.
Look for these in your work
- Correct order
- Direction stated
- Bond property named
Explain why an I–I bond is weaker than a Cl–Cl bond.
Give me a hint
Link atomic size to bond length and overlap.
Compare my reasoning
- Iodine atoms are larger than chlorine atoms.
- The I–I bond is longer.
- The longer bond has less effective overlap and is weaker.
The larger iodine atoms form a longer bond with less effective overlap, so I–I is weaker than Cl–Cl.
Look for these in your work
- Size comparison correct
- Bond length linked
- Overlap consequence stated
A data table shows F–F is weaker than Cl–Cl despite fluorine's small atoms. Reconcile the observation with the general down-group trend.
Give me a hint
Identify F–F as an anomaly and use the close lone-pair repulsion model.
Compare my reasoning
- The general Cl₂ to I₂ trend is a decrease in bond strength down the group.
- F–F is an exception to a simple smooth size-only sequence.
- Lone-pair repulsion in small fluorine atoms weakens F–F.
F–F is anomalously weak because close lone pairs repel strongly; for Cl₂, Br₂ and I₂, bond strength decreases down the group as bond length increases.
Look for these in your work
- Exception recognized
- Repulsion explained
- General trend retained
Common mistakes
- Using volatility to rank covalent bond strengths
- Claiming bond strength increases smoothly from F₂ to I₂
- Forgetting the anomalously weak F–F bond
- Saying the entire molecule expands down the group rather than the bond length increasing
What can you explain now?
Which bond is stronger in the Cl₂/Br₂/I₂ sequence?
Compare with the explanation
Cl–Cl
The X–X bond weakens from chlorine to iodine.
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Make it stick
How does X–X bond strength change from Cl₂ to I₂, and why must F₂ be treated separately?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.