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How do the trends in Group 17 explain the relative oxidising power of the halogens and the behaviour of their compounds?

Describe the trends down Group 17 in volatility, colour and oxidising power, explain displacement reactions of halogens and halides, describe the reactions of halide ions with silver nitrate and with concentrated sulfuric acid, and the disproportionation of chlorine

A focused answer to the H2 Chemistry learning outcome on Group 17. Trends in volatility, colour and oxidising power down the group, halogen-halide displacement, the silver nitrate and concentrated sulfuric acid tests for halides, and the disproportionation of chlorine in alkali.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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The answer
Examples in context
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What this dot point is asking

SEAB wants you to describe and explain the trends down Group 17 (Cl to I) in volatility, colour and oxidising power, account for halogen-halide displacement reactions, describe the silver nitrate and concentrated sulfuric acid tests for halide ions, and explain the disproportionation of chlorine in alkali. Oxidising-power trends, the two halide tests, and disproportionation are recurring exam content.

The answer

Physical trends down the group

Volatility decreases down the group (boiling points rise) because the molecules have more electrons, so van der Waals forces strengthen. Chlorine is a gas, bromine a liquid, iodine a solid at room temperature.
Colour deepens down the group: chlorine pale green-yellow, bromine red-brown, iodine grey-black solid (purple vapour).

Oxidising power

Halogens act as oxidising agents by gaining an electron ( $\text{X}_2 + 2e^- \rightarrow 2\text{X}^-$ ). Oxidising power decreases down the group: chlorine is the strongest, iodine the weakest. As the atom gets larger with more shielding, its attraction for an incoming electron weakens, so it is reduced less readily.

Displacement reactions

A more reactive (stronger oxidising) halogen displaces a less reactive halide from solution:

\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2 \quad (\text{solution turns orange})

\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2 \quad (\text{solution turns brown})

Bromine displaces iodide but not chloride; iodine displaces neither. These reactions order the oxidising power experimentally.

Test 1: silver nitrate then ammonia

Add dilute nitric acid (to remove carbonate or hydroxide) then silver nitrate. A precipitate of the silver halide forms, identified by colour and solubility in ammonia:

Halide	Precipitate colour	Solubility in ammonia
chloride	white	dissolves in dilute $\text{NH}_3$
bromide	cream	dissolves in concentrated $\text{NH}_3$
iodide	yellow	insoluble even in concentrated $\text{NH}_3$

Test 2: concentrated sulfuric acid

Warming a solid halide with concentrated sulfuric acid distinguishes the halides by the reducing power of the halide ion (which increases down the group):

Chloride: only steamy HCl fumes (chloride too weak to reduce the acid).
Bromide: HBr fumes plus some brown $\text{Br}_2$ and $\text{SO}_2$ (bromide partly reduces the acid).
Iodide: HI plus $\text{I}_2$ (purple vapour) plus reduced sulfur products $\text{SO}_2$ , S and $\text{H}_2\text{S}$ (iodide strongly reduces the acid).

Disproportionation of chlorine

Chlorine disproportionates in alkali (the same element is both oxidised and reduced):

Cold dilute alkali: $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$ (chlorine $0$ to $-1$ and $+1$ ). Sodium chlorate(I) is used as bleach.
Hot concentrated alkali: $3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$ (chlorine $0$ to $-1$ and $+5$ ).

Worked example

A solution of an unknown sodium halide gives a cream precipitate with acidified silver nitrate that dissolves only in concentrated ammonia. Warming the solid halide with concentrated sulfuric acid gives some brown vapour and a choking gas. Identify the halide.

The cream precipitate, soluble only in concentrated ammonia, indicates a bromide (AgBr).

Confirmation: with concentrated sulfuric acid, bromide is oxidised partly to brown $\text{Br}_2$ vapour, and the acid is reduced to the choking gas $\text{SO}_2$ . Chloride would give only steamy HCl; iodide would give purple iodine and the rotten-egg smell of $\text{H}_2\text{S}$ .

So the halide is bromide.

Try it: Halide identification helper - enter the precipitate colour, ammonia solubility and sulfuric-acid observations to identify the halide.

Examples in context

Example 1. Water treatment. Chlorine is added to drinking water and swimming pools because it disproportionates to form chloric(I) acid (HClO), a powerful disinfectant. The equilibrium $\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCl} + \text{HClO}$ is the practical face of disproportionation, and SEAB often frames a question around the benefits and risks of chlorination.

Example 2. Distinguishing halides in an unknown salt. A combined Paper 2 question gives both the silver nitrate result and the concentrated sulfuric acid result for an unknown halide. Using the two tests together (precipitate colour and ammonia solubility, then the reduction products of the acid) uniquely identifies chloride, bromide or iodide, exactly the cross-checking expected in qualitative analysis.

Try this

Q1. Predict and explain what happens when bromine water is added to aqueous potassium iodide. [2 marks]

Cue. Bromine (stronger oxidising agent) displaces iodine: $\text{Br}_2 + 2\text{KI} \rightarrow 2\text{KBr} + \text{I}_2$ ; solution turns brown.

Q2. State the colour of the precipitate and its ammonia solubility for silver iodide. [2 marks]

Cue. Yellow precipitate, insoluble even in concentrated ammonia.

Q3. Write the equation for the disproportionation of chlorine in cold dilute sodium hydroxide and give the oxidation numbers of chlorine in the products. [2 marks]

Cue. $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$ ; Cl is $-1$ in NaCl and $+1$ in NaClO.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Specimen (9729)4 marksDescribe and explain the trend in oxidising power of the halogens down Group 17, and predict what is observed when chlorine water is added to aqueous potassium bromide.

Show worked answer →

Oxidising power decreases down the group: chlorine is the strongest oxidising agent, iodine the weakest.

Down the group the atomic radius increases and there is more shielding, so the attraction for an incoming electron weakens. The halogen is less able to gain an electron (be reduced), so it is a weaker oxidising agent.

Chlorine is a stronger oxidising agent than bromine, so it displaces bromine from bromide: Cl2 + 2KBr -> 2KCl + Br2.

Observation: the colourless solution turns orange (or yellow-brown) as bromine is formed.

Markers reward the trend, the radius and shielding explanation, the displacement equation, and the colour observation.

2022 (style)4 marksSolid sodium chloride and solid sodium iodide are each warmed separately with concentrated sulfuric acid. Describe and explain the different observations.

Show worked answer →

Both initially give the hydrogen halide gas (misty fumes) by acid-base reaction:
NaX + H2SO4 -> NaHSO4 + HX.

With sodium chloride, only steamy fumes of HCl are seen, because chloride is a weak reducing agent and cannot reduce sulfuric acid further.

With sodium iodide, iodide is a strong reducing agent and reduces the sulfuric acid further to a mixture of products including SO2 (choking gas), S (yellow solid), and H2S (rotten-egg smell), while iodide is oxidised to iodine (purple vapour or black solid).

The difference reflects the increasing reducing power of the halide ions down the group.

Markers reward the common HX step, the chloride giving only HCl, the iodide giving reduced sulfur products and iodine, and the reducing-power trend.