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Why does oxidising power decrease while reducing power of the halide ions increases down Group 7?

The trends in electronegativity and boiling point of the halogens. The trend in oxidising ability of the halogens down the group, including displacement reactions of halide ions in aqueous solution. The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid. The use of acidified silver nitrate to identify and distinguish halide ions, and the use of chlorine in water treatment.

A focused answer to the AQA A-Level Chemistry 3.2.3 specification points on Group 7. Covers the boiling point and electronegativity trends, the decrease in oxidising power down the group with displacement reactions, the increase in reducing power of the halide ions with concentrated sulfuric acid, the silver nitrate test and the use of chlorine in water treatment.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

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What this dot point is asking
Boiling point and electronegativity trends
Oxidising power and displacement reactions
Reducing power of the halide ions
Identifying halide ions
Chlorine in water treatment
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What this dot point is asking

AQA wants you to describe and explain the trends in electronegativity and boiling point, the decrease in oxidising power of the halogens (with displacement reactions), the increase in reducing power of the halide ions (with concentrated sulfuric acid), the silver nitrate test for halide ions, and the use of chlorine in water treatment.

Boiling point and electronegativity trends

Boiling point increases down the group because the molecules get larger, with more electrons, so the van der Waals forces between molecules are stronger and need more energy to overcome. At room temperature $F_2$ and $Cl_2$ are gases, $Br_2$ is a liquid and $I_2$ is a solid.

Electronegativity decreases down the group: the atoms get larger with more shielding, so the nucleus attracts the bonding pair in a bond less strongly.

Oxidising power and displacement reactions

For example chlorine displaces bromine from potassium bromide:

Cl_2(aq) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_2(aq)

The chlorine is reduced (gains electrons) and the bromide is oxidised. The solution turns orange as bromine forms.

Reducing power of the halide ions

Worked example: sodium halides with concentrated sulfuric acid

Compare the reactions of solid $NaCl$ , $NaBr$ and $NaI$ with concentrated $H_2SO_4$ and explain the trend in terms of reducing power.

Step 1: Sodium chloride, the weakest reducing agent

Chloride ion ( $Cl^-$ ) is the smallest and least polarisable halide ion, so it holds its electrons too tightly to reduce the sulfur in sulfuric acid. The only reaction that occurs is a straightforward acid-base proton transfer, producing misty fumes of hydrogen chloride gas.

NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl

Observable: misty white fumes of $HCl$ only.

Step 2: Sodium bromide, a stronger reducing agent

Bromide ( $Br^-$ ) is larger and loses its outer electrons more easily, so it is strong enough to reduce sulfuric acid partially. Sulfur is reduced from the $+6$ oxidation state in $H_2SO_4$ to $+4$ in $SO_2$ .

Observable: misty $HBr$ fumes, choking $SO_2$ gas and brown $Br_2$ fumes.

Step 3: Sodium iodide, the strongest reducing agent

Iodide ( $I^-$ ) is the largest halide ion and loses electrons most readily. It reduces sulfuric acid all the way down to $-2$ in hydrogen sulfide, passing through intermediate products.

Observable: misty $HI$ fumes, yellow solid sulfur ( $S$ ), choking $SO_2$ , and purple $I_2$ vapour with the distinctive smell of rotten eggs from $H_2S$ .

Step 4: Explain the trend

Down Group 7 the halide ions increase in size and the outer electrons are further from the nucleus with more shielding, so they are lost more easily. This means reducing power increases from $Cl^-$ to $Br^-$ to $I^-$ , reflected in the progressively more vigorous reduction of sulfuric acid.

Final answer: $Cl^-$ gives only $HCl$ ; $Br^-$ also gives $SO_2$ and $Br_2$ ; $I^-$ gives $H_2S$ , $SO_2$ , $S$ and $I_2$ . Reducing power increases down the group because the larger ions lose their outer electrons more easily.

Identifying halide ions

Add dilute nitric acid (removes carbonate or hydroxide ions), then silver nitrate solution:

$Cl^-$ gives a white precipitate ( $AgCl$ ), soluble in dilute ammonia.
$Br^-$ gives a cream precipitate ( $AgBr$ ), soluble only in concentrated ammonia.
$I^-$ gives a yellow precipitate ( $AgI$ ), insoluble in ammonia.

Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)

Chlorine in water treatment

Chlorine is added to water to kill bacteria and make it safe to drink. With water it disproportionates:

Cl_2 + H_2O \rightleftharpoons HClO + HCl

The chlorine here is both oxidised and reduced (disproportionation). The benefit of killing pathogens outweighs the risk of toxic chlorine and possible formation of chlorinated hydrocarbons.

Try this

Q1. Explain why bromine displaces iodine but not chlorine. [2 marks]

Cue. Oxidising power decreases down the group; bromine is a stronger oxidiser than iodine (so displaces it) but weaker than chlorine.

Q2. Give the colours of the silver halide precipitates. [3 marks]

Cue. $AgCl$ white, $AgBr$ cream, $AgI$ yellow.

Q3. Write an equation for chlorine reacting with water and name the type of reaction. [2 marks]

Cue. $Cl_2 + H_2O \rightleftharpoons HClO + HCl$ ; disproportionation.

Exam-style practice questions

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

AQA Paper 1 (style)3 marksExplain why the oxidising ability of the halogens decreases down Group 7.

Show worked answer →

A 3-mark answer links oxidising power to the ease of gaining an electron.

A halogen acts as an oxidising agent by gaining one electron to form a 1- ion. Down the group the atomic radius increases and there is more shielding, so the outer shell is further from the nucleus and the nuclear attraction for an incoming electron is weaker. The atoms therefore gain an electron less easily, so oxidising ability decreases down the group.

Markers reward larger radius and more shielding, weaker attraction for the incoming electron, and electron gained less easily.

AQA Paper 1 (style)2 marksDescribe what you would observe when concentrated sulfuric acid is added to solid sodium iodide.

Show worked answer →

A 2-mark answer needs the products and the observations.

Iodide is a strong reducing agent, so it reduces sulfuric acid all the way to hydrogen sulfide. Observations include a black solid ( $I_2$ ), purple fumes of iodine vapour and a smell of rotten eggs ( $H_2S$ ); misty fumes of $HI$ are also seen.

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Sources & how we know this

AQA A-level Chemistry (7405) specification — AQA (2015)