Group 17 Revision Notes — 9701 Paper 1

Colours, states and volatility of chlorine, bromine and iodine

Syllabus requirement · §11.1.1, 11.1.3

“

describe the colours and the trend in volatility of chlorine, bromine and iodine; interpret the volatility of the elements in terms of instantaneous dipole–induced dipole forces.

”

Three elements, three states, at room temperature

You need three Group 17 elements: chlorine, bromine and iodine. At room temperature each one is in a different state, and you must know each colour and state — they are asked for directly, and they are also how you recognise a halogen forming in a test tube later in this note.

Element	Formula	Colour and state at room temperature	Colour of the vapour
Chlorine	Cl₂	pale green (yellow-green) gas	pale green
Bromine	Br₂	red-brown (dark brown) liquid	orange-brown
Iodine	I₂	grey-black shiny solid	purple (iodine sublimes on gentle heating)

Mark schemes accept "green gas", "brown liquid" and "grey/black solid". A colour-and-state answer needs both parts.

All three are made of small diatomic molecules, $\text{X}_2$ : two halogen atoms joined by one covalent bond. So there are two kinds of attraction to think about — the X–X bond inside each molecule, and the weak forces between molecules. This section is about the second; the next section is about the first.

Volatility falls down the group

A volatile substance evaporates easily, so it has a low boiling point. Going down Group 17 the boiling point rises steeply ( $\text{Cl}_2$ : $-34\,°\text{C}$ ; $\text{Br}_2$ : $59\,°\text{C}$ ; $\text{I}_2$ : $184\,°\text{C}$ ), so volatility decreases down the group.

To boil $\text{Cl}_2$ , $\text{Br}_2$ or $\text{I}_2$ you never break the covalent X–X bond. You only pull whole molecules away from their neighbours. The forces between these molecules are van der Waals' forces — here, instantaneous dipole–induced dipole (id–id) forces, because a symmetrical $\text{X}_2$ molecule has no permanent dipole. You met these in AS Chemical Bonding ("Intermolecular forces: van der Waals' and hydrogen bonding").

Why the id–id forces get stronger

Electrons are always moving. At any instant a molecule's electron cloud is slightly lopsided, so the molecule has a tiny instantaneous dipole. This dipole induces a dipole in the next molecule, and the two attract.

A molecule with more electrons has a bigger electron cloud that is distorted more easily, so its instantaneous dipoles are larger and the attraction is stronger. Down Group 17 the number of electrons per molecule rises ( $\text{Cl}_2$ : 34; $\text{Br}_2$ : 70; $\text{I}_2$ : 106). So:

more electrons in each molecule;
stronger id–id forces between molecules;
more energy needed to separate the molecules;
higher boiling point, so lower volatility.

Those four links are the whole explanation. A mark scheme wants "more electrons" and "stronger instantaneous dipole–induced dipole forces" in those words.

Down the group the colour deepens, the state changes from gas to liquid to solid, and the boiling point rises steeply: the elements become less volatile.

Why iodine is a solid when chlorine is a gas

3 marks

At room temperature chlorine is a gas but iodine is a solid. Explain this difference.

Show full working

1
Both are simple molecular substances made of $\text{X}_2$ molecules. To melt or boil them you separate molecules; you do not break the X–X bond.
Say first which force is being overcome. Answers that talk about breaking the I–I bond score nothing.
2
An $\text{I}_2$ molecule has 106 electrons; a $\text{Cl}_2$ molecule has 34. So $\text{I}_2$ has the larger, more easily distorted electron cloud.
The number of electrons is the cause. It is the first marking point in almost every volatility answer.
3
So the instantaneous dipole–induced dipole forces between $\text{I}_2$ molecules are much stronger than those between $\text{Cl}_2$ molecules.
Name the force in full. "Van der Waals' forces" is also accepted; "intermolecular bonds" is too vague.
4
More energy is needed to separate $\text{I}_2$ molecules, so iodine has much higher melting and boiling points. It is still a solid at room temperature, while chlorine is already a gas.
Finish by linking the force back to the property the question asked about.

Answer

$\text{I}_2$ has more electrons than $\text{Cl}_2$ , so the instantaneous dipole–induced dipole forces between its molecules are stronger. More energy is needed to separate the molecules, so iodine's melting and boiling points are much higher.

State and explain the volatility trend

9701/21 M/J 2024 Q1(b)3 marks

State the trend in volatility of the halogens chlorine, bromine and iodine. Explain your answer.

Show full working

1
Trend: volatility decreases from chlorine to iodine.
This is the first mark on its own. Many students explain well but never state the trend.
2
Cause: there are more electrons in the molecules going down the group.
Second mark. "Bigger molecules" alone is weaker; say more electrons.
3
Force: so the instantaneous dipole–induced dipole forces between molecules get stronger, and more energy is needed to separate the molecules.
Third mark. It must be the force BETWEEN molecules. Mentioning the X–X bond here loses the mark.

Answer

Volatility decreases down the group. The molecules have more electrons, so the instantaneous dipole–induced dipole forces between them are stronger.

3 marks = trend + more electrons + stronger id–id forces. Write each as its own sentence.

Choosing the correct explanation for the volatility trend

9701/12 M/J 2025 Q171 mark

Why do the halogens become less volatile as Group 17 is descended? A The halogen–halogen bond energy decreases. B The halogen–halogen bond energy increases. C The number of electrons in each molecule increases. D The van der Waals' forces between molecules become weaker.

Show full working

1
A and B are about the wrong thing. Boiling separates molecules; it does not break the X–X bond inside them. So a change in X–X bond energy cannot explain volatility.
Check first whether an option describes a force between molecules or a bond inside one.
2
D has the wrong direction. Boiling point rises down the group, so the van der Waals' forces must get stronger, not weaker.
D uses the right force but the wrong direction — a common trap.
3
C is correct. More electrons per molecule means stronger instantaneous dipole–induced dipole forces, so volatility falls.
C gives the cause of the stronger forces, which is what "why" asks for.

Answer

A question about the melting or boiling point of a simple molecular substance is about the forces between molecules, never the covalent bond inside them.

Completing a table of colours and states

9701/22 F/M 2025 Q3(a)(i)1 mark

The halogens chlorine, bromine and iodine show trends in chemical and physical properties down the group.

Table 3.1 shows some properties of chlorine, bromine and iodine.

Table 3.1

property	chlorine	bromine	iodine
colour and state at room temperature	green gas
bond energy / kJ mol⁻¹	242	193	151
electronegativity	3.0	2.8	2.5
formula of sodium halide	NaCl	NaBr	NaI

Complete Table 3.1.

Show full working

1
Only the empty "colour and state" boxes need filling; the other rows are already complete.
Read which cells are blank before writing anything.
2
Bromine: red-brown liquid. Iodine: grey-black solid.
The single mark needs both elements, each with colour AND state.

Answer

Bromine: (red-)brown liquid. Iodine: grey/black solid.

A colour alone, or a state alone, does not score.

Your turn

1
State the colour and physical state at room temperature of chlorine, bromine and iodine.
Stuck? Show hint
One of each state: gas, liquid, solid.
Show solution
1. 1
  Chlorine: pale green (yellow-green) gas.
  The smallest molecule, weakest id–id forces, so a gas.
2. 2
  Bromine: red-brown liquid.
  Bromine is one of only two elements that are liquids at room temperature.
3. 3
  Iodine: grey-black solid.
  Don't confuse the solid (grey-black) with the vapour (purple).
Answer
Chlorine: pale green gas. Bromine: red-brown liquid. Iodine: grey-black solid.
2
Explain, in terms of intermolecular forces, why bromine has a higher boiling point than chlorine.
Stuck? Show hint
Compare the number of electrons in one $\text{Br}_2$ molecule with one $\text{Cl}_2$ molecule.
Show solution
1. 1
  A $\text{Br}_2$ molecule (70 electrons) has more electrons than a $\text{Cl}_2$ molecule (34 electrons).
  Start with the cause.
2. 2
  So the instantaneous dipole–induced dipole forces between $\text{Br}_2$ molecules are stronger.
  More electrons give larger instantaneous dipoles, so the attraction between molecules is stronger.
3. 3
  More energy is needed to separate $\text{Br}_2$ molecules, so bromine has the higher boiling point.
  Link the stronger force to the property asked about.
Answer
$\text{Br}_2$ has more electrons than $\text{Cl}_2$ , so the instantaneous dipole–induced dipole forces between its molecules are stronger and need more energy to overcome.
3
Astatine is below iodine in Group 17. Predict its state at room temperature and whether it is more or less volatile than iodine. Give a reason.
Stuck? Show hint
Follow the trend one step further down the group.
Show solution
1. 1
  $\text{At}_2$ molecules have even more electrons than $\text{I}_2$ molecules.
  The trend continues down the group.
2. 2
  So the id–id forces between $\text{At}_2$ molecules are stronger still, and astatine is less volatile than iodine.
  Stronger forces between molecules mean a higher boiling point.
3. 3
  Astatine is predicted to be a solid at room temperature (darker in colour than iodine).
  Iodine is already a solid, so a less volatile element must be too.
Answer
A solid, less volatile than iodine, because $\text{At}_2$ has more electrons and so stronger instantaneous dipole–induced dipole forces.

Paper 1 · MCQReal past-paper questions · Colours and volatility of Cl2, Br2, I2 Paper 2 · TheoryReal past-paper questions · Colours and volatility of Cl2, Br2, I2

Checking your access…

Can you do all of these?

State the colour and state of chlorine, bromine and iodine, and explain why volatility falls down the group using instantaneous dipole–induced dipole forces
Explain why the X–X bond gets weaker down the group (atom size, bond length, orbital overlap, attraction), and why this trend runs opposite to volatility
Describe the oxidising power of the halogens, write ionic equations for displacement reactions, and give the colours in water and in hexane
Describe the reactions of the halogens with hydrogen, with observations, and explain why they get less vigorous down the group
Describe the reducing power of the halide ions and explain it using ion size and shielding
Describe and explain the thermal stability of the hydrogen halides using H–X bond strength
Give the precipitate colours with AgNO₃(aq) and the solubility of each in dilute and concentrated NH₃(aq), and use them to identify a halide
Write the acid–base and redox equations for the halides with concentrated H₂SO₄, give the oxidation-number change of sulfur, and describe the observations
Write the equations for chlorine with cold and with hot NaOH(aq), name chlorate(I) and chlorate(V), and use oxidation numbers to show disproportionation
Explain, with an equation, how chlorine purifies water, naming HOCl and ClO⁻ as the species that kill bacteria

Paper 1 · MCQ

Real past-paper questions, sorted by difficulty, with mark schemes

Paper 2 · Theory

Real past-paper questions, sorted by difficulty, with mark schemes

Group 17

Colours, states and volatility of chlorine, bromine and iodine

Three elements, three states, at room temperature

Volatility falls down the group

Why the id–id forces get stronger

The X–X bond strength trend — and why it isn't the volatility trend

The halogens as oxidising agents; the halide ions as reducing agents

Can you do all of these?

Colours, states and volatility of chlorine, bromine and iodine

Three elements, three states, at room temperature

Volatility falls down the group

Why the id–id forces get stronger

The rest of this note

Can you do all of these?