Halogen Trends: Electronegativity, Boiling Point & Oxidising Ability Revision Notes | A Level AQA Chemistry

1. Why Group 7 Trends Matter

Fluorine, chlorine, bromine, and iodine sit in the same group, yet their physical and chemical behaviour changes dramatically as you move down. Understanding why those changes happen — and being able to justify them with atomic-level reasoning — is exactly what AQA examiners test at A Level.

Three trends run through this topic: electronegativity decreases, boiling point increases, and oxidising ability decreases. Each trend has a different explanation rooted in atomic structure or intermolecular forces. Keeping those explanations separate is the key to scoring full marks.

2. Electronegativity Trend Down Group 7

Electronegativity measures how strongly an atom pulls a shared pair of electrons towards itself in a covalent bond. Fluorine is the most electronegative element on the periodic table (Pauling value 4.0), and the value falls steadily as you descend the group: Cl (3.2), Br (3.0), I (2.7).

Two factors drive this decrease. First, atomic radius increases down the group because each successive halogen has an extra filled electron shell. Second, the number of inner electron shells also increases, so the bonding electrons experience greater shielding from the nuclear charge. The combined effect is that the nucleus exerts a weaker effective pull on any shared pair of electrons.

A useful way to picture this: imagine the nucleus trying to grip a bonding pair of electrons. In fluorine, the nucleus is close and the grip is tight. In iodine, the nucleus is far away and surrounded by layers of inner electrons that push back against the outer electrons — the grip is much weaker. That weakening grip is the electronegativity trend in one image.

3. Boiling Point Trend Down Group 7

Now that we have the electronegativity trend established, we can turn to a physically observable consequence of moving down the group: the dramatic rise in boiling point. At room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. That progression is not a coincidence — it follows directly from molecular structure and intermolecular forces.

All halogens exist as diatomic molecules ($\text{X}_2$) with non-polar covalent bonds (the two identical atoms share electrons equally). The only intermolecular forces between these molecules are London dispersion forces — temporary induced dipole–induced dipole attractions. These arise when the electron cloud of one molecule momentarily shifts, creating an instantaneous dipole that induces a dipole in a neighbouring molecule.

The strength of London dispersion forces depends on the number of electrons in the molecule. More electrons means a larger, more polarisable electron cloud — the cloud distorts more easily, creating stronger and more frequent instantaneous dipoles. As you descend Group 7, each halogen molecule has more electrons ($\text{F}_2$: 18, $\text{Cl}_2$: 34, $\text{Br}_2$: 70, $\text{I}_2$: 106), so London dispersion forces become progressively stronger.

The chart makes the trend unmistakable. More energy is required to overcome the stronger London dispersion forces in $\text{I}_2$ than in $\text{F}_2$, so iodine has a far higher boiling point. The explanation must always link: more electrons → stronger London dispersion forces → more energy needed to separate molecules → higher boiling point.

4. Oxidising Ability Trend Down Group 7

With the physical trends covered, we can now address the most chemically significant trend in Group 7: the decrease in oxidising ability from fluorine to iodine. This is where the electronegativity reasoning from Section 2 pays off directly.

Halogens act as oxidising agents by gaining electrons — each halogen atom accepts one electron to form a halide ion ($\text{X} + e^- \rightarrow \text{X}^-$). The ease with which this happens determines oxidising ability. Fluorine accepts electrons most readily; iodine does so least readily.

The explanation mirrors the electronegativity argument. As atomic radius increases down the group and shielding increases, the incoming electron is attracted less strongly by the nucleus. The atom has a weaker tendency to gain that electron. Fluorine, with its small radius and minimal shielding, pulls an electron in powerfully. Iodine, with its large radius and extensive shielding, does so far less effectively.

There is a clear link here to standard electrode potentials, which you may encounter in later study. For now, the AQA specification requires you to explain the trend in terms of atomic radius and shielding, and to use displacement reactions as experimental evidence. That evidence is what we examine next.

5. Displacement Reactions as Evidence

A displacement reaction occurs when a more reactive halogen oxidises the halide ion of a less reactive halogen, displacing it from aqueous solution. These reactions are the experimental proof of the oxidising ability trend — and they are a favourite source of AQA exam questions.

The general principle: if halogen A can displace halide ion $\text{B}^-$ from solution, then A is a stronger oxidising agent than B. Chlorine displaces bromide and iodide; bromine displaces iodide only; iodine displaces neither chloride nor bromide. This hierarchy maps directly onto the decreasing oxidising ability down the group.

The ionic equations for these reactions show the electron transfer explicitly. For chlorine displacing bromide:

For bromine displacing iodide, the same pattern applies: $\text{Br}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Br}^-\text{(aq)} + \text{I}_2\text{(aq)}$. In both cases, the halogen with the smaller atomic radius and stronger oxidising ability wins the electrons.

5a. Using Cyclohexane to Confirm the Identity of the Displaced Halogen

In practice, the colour changes in aqueous solution can be ambiguous — bromine and iodine both produce brown colours. Adding a small volume of cyclohexane (a non-polar organic solvent) and shaking the mixture separates the layers. The halogen, being non-polar, preferentially dissolves in the organic layer, giving a much clearer colour: bromine gives an orange layer; iodine gives a violet layer. This technique is commonly tested in AQA questions on identifying the products of displacement reactions.

6. Connecting the Three Trends: A Unified Picture

It is worth pausing to see how the three trends connect. Electronegativity and oxidising ability both decrease down the group for the same underlying reason: increasing atomic radius and shielding reduce the effective nuclear pull on electrons, whether those electrons are in a covalent bond or being gained from a halide ion. The boiling point trend has a completely different cause — it is driven by London dispersion forces between molecules, not by atomic-level electron attraction.

Examiners frequently set questions that require you to distinguish these explanations. A common trap is to use the London dispersion force argument for oxidising ability, or to use the atomic radius argument for boiling point. Keep the two explanations in separate mental compartments.

7. Exam Technique and Mark-Winning Guidance

AQA questions on halogen trends tend to use the command words 'explain', 'describe', and 'compare'. Each demands a different response structure. 'Describe' asks what happens — state the observation without giving a reason. 'Explain' asks why — you must include the scientific mechanism. 'Compare' requires you to address both similarities and differences explicitly.

Worked example: Answering a 4-Mark 'Explain' Question on Oxidising Ability

Explain why chlorine is a better oxidising agent than iodine. [4 marks]

1. An oxidising agent accepts electrons from another species. — [B1] — this anchors the answer in the correct chemical definition and shows you know what oxidising ability means.
2. Chlorine has a smaller atomic radius than iodine because it has fewer electron shells. — [M1] — you must state the comparison explicitly, not just say 'chlorine is smaller'.
3. Iodine has more inner electron shells, so the outer electrons (and any incoming electron) experience greater shielding from the nuclear charge. — [M1] — shielding is the second required factor; omitting it typically costs a mark.
4. Therefore, the nucleus of a chlorine atom exerts a stronger effective pull on an incoming electron than the nucleus of an iodine atom. Chlorine accepts electrons more readily and is the stronger oxidising agent. — [A1] — the conclusion must explicitly connect the structural reasoning to oxidising ability. Saying 'chlorine is more electronegative' alone is insufficient without the structural justification.

Answer: Chlorine is a stronger oxidising agent than iodine because it has a smaller atomic radius and less electron shielding, so its nucleus attracts an incoming electron more strongly. Iodine's larger radius and greater shielding mean it accepts electrons less readily.

Notice that the model answer above earns marks at each stage: definition, structural comparison, shielding, and conclusion. Answers that state only 'chlorine is more electronegative' without explaining why typically score 1 out of 4. The examiner wants the chain of reasoning, not just the conclusion.

When writing ionic equations for displacement reactions, always include state symbols and check that the equation is balanced — both in terms of atoms and charges. In the bromine/iodide equation, two iodide ions ($2\text{I}^-$) each lose one electron, and bromine ($\text{Br}_2$) gains two electrons in total. Charge and atom balance must both be confirmed before writing the final equation.

To consolidate the full picture: electronegativity and oxidising ability both decrease down Group 7 because increasing atomic radius and shielding reduce the effective nuclear attraction for electrons; boiling point increases because larger molecules have more electrons and therefore stronger London dispersion forces between them. These three trends, their explanations, and the displacement reaction evidence together cover every AQA specification requirement for this topic.