Describe Group 17 colours and volatility
Compare chlorine, bromine and iodine appearance, physical state and volatility using intermolecular forces.
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
What is the room-temperature state of bromine?
What you will learn
- State the characteristic colours and room-temperature states
- Describe the down-group volatility trend
- Explain the trend using instantaneous dipole–induced dipole forces
Colours and states distinguish the three halogens
At room temperature chlorine is a pale green-yellow gas, bromine is a red-brown liquid and iodine is a grey-black solid. Iodine gives a purple vapour when it sublimes. Attach the colour to the correct physical state and observation conditions.
From chlorine to iodine, volatility decreases: boiling points rise and it becomes harder for molecules to escape the liquid or solid. A more volatile substance evaporates more readily; volatility is not the same property as reactivity.
Larger electron clouds strengthen dispersion attractions
Cl₂, Br₂ and I₂ are non-polar molecules. Their electrons move, so a temporary uneven distribution creates an instantaneous dipole that induces a dipole in a neighboring molecule. These London forces are instantaneous dipole–induced dipole attractions.
Down the group the electron clouds become larger and more easily distorted. Instantaneous and induced dipoles become stronger, so intermolecular attractions strengthen, boiling points rise and volatility falls. Do not say that covalent bonds between atoms in one molecule are broken during boiling.
Pause and explain
Why does volatility decrease from chlorine to iodine?
Worked example
Explain why iodine is less volatile than chlorine and state their appearances at room temperature.
Show the worked solution
- Iodine molecules have larger electron clouds than chlorine molecules.
- Stronger instantaneous dipole–induced dipole forces act between iodine molecules.
- More energy is needed for iodine molecules to escape, so iodine is less volatile.
Answer Chlorine is a pale green-yellow gas and iodine is a grey-black solid. I₂ has stronger dispersion attractions, so it is less volatile than Cl₂.
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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.
State the colours and room-temperature states of chlorine, bromine and iodine.
Give me a hint
Match each characteristic colour to the physical state of X₂ at room temperature.
Compare my reasoning
- Chlorine is pale green-yellow and gaseous.
- Bromine is red-brown and liquid.
- Iodine is grey-black and solid.
Cl₂: pale green-yellow gas; Br₂: red-brown liquid; I₂: grey-black solid.
Look for these in your work
- All three colors matched
- All three states matched
- Formulas identify molecules
Explain the trend in volatility from Cl₂ to I₂ using a named intermolecular force.
Give me a hint
Consider molecule size, temporary dipoles and energy needed to separate molecules.
Compare my reasoning
- Electron-cloud size increases down the group.
- Instantaneous dipole–induced dipole attractions become stronger.
- More energy is needed for molecules to escape, so volatility decreases.
Volatility decreases down Cl₂, Br₂, I₂ because larger electron clouds give stronger instantaneous dipole–induced dipole forces.
Look for these in your work
- Force named correctly
- Size linked to force
- Direction of volatility stated
A learner says iodine boils at a higher temperature because the I–I covalent bond is stronger than the Cl–Cl bond. Correct the explanation.
Give me a hint
Separate forces between molecules from the covalent bond within each molecule.
Compare my reasoning
- Boiling separates molecules by overcoming intermolecular attractions.
- I₂ has stronger dispersion attractions because its electron cloud is larger.
- The X–X covalent bond does not need to break for boiling.
Iodine has a higher boiling point because stronger intermolecular dispersion forces must be overcome; the I–I covalent bond within each molecule remains intact.
Look for these in your work
- Intermolecular forces identified
- Trend reason correct
- Intramolecular bond distinguished
Common mistakes
- Calling iodine a purple liquid at room temperature
- Saying volatility increases down the group
- Attributing boiling to breaking the X–X covalent bond
- Mixing volatility with oxidising strength
What can you explain now?
What is the room-temperature state of bromine?
Compare with the explanation
Liquid
Bromine is the red-brown liquid in this three-element comparison.
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Make it stick
How do color, state and intermolecular forces help you describe the Cl₂–Br₂–I₂ volatility trend?
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.