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Group 7(17), the halogens

The halogens are the most reactive non-metals, and their chemistry shows clear trends down the group. This chapter develops the trends in electronegativity and boiling point, the halogens as oxidising agents in displacement reactions, the halide ions as reducing agents with concentrated sulfuric acid, the silver-nitrate test for halide ions, and the disproportionation of chlorine used to treat water.

4 sections·~12 min reading time·3 competencies·Level Standard 3 · Advanced 1

T·101010 / 18
Exam profile
AO1 · Describe the Group 7 trends and the displacement, silver-nitrate and disproportionation reactionsAO2 · Explain the trends in oxidising and reducing power and construct the relevant equationsAO3 · Interpret test results to identify anions and evaluate the benefits and risks of chlorinating water
Operators:statedescribeexplaindeduceconstructevaluate

basic level

AS-Level requires the Group 7 trends, the displacement reactions, the silver-nitrate halide test and the disproportionation of chlorine.

higher level

The full A-Level expects the reactions of solid halides with concentrated sulfuric acid, the reducing-power trend, and the anion tests in the correct order, applied to unfamiliar problems.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 4 sections▾
  1. Group 7(17), the halogens
    • 01Trends in electronegativity and boiling point◐
    • 02Halogens as oxidising agents and displacement reactions◐
    • 03Halides as reducing agents with concentrated sulfuric acid●
    • 04Testing for halide ions and disproportionation of chlorine◐
§ 01

Trends in electronegativity and boiling point#

●●○StandardLPAQA 7405 3.2.3.1LPDfE GCE Chemistry - Group 7

Boiling point down Group 7

Boiling point down Group 7Line chart: boiling point / K by halogen, Data: boiling point / K · F2: 85; boiling point / K · Cl2: 239; boiling point / K · Br2: 332; boiling point / K · I2: 4570100200300400F2Cl2Br2I2boiling point / Khalogen
Fig. 1Boiling point rises down the group as the larger molecules have more electrons and stronger van der Waals forces.

Key points

The halogens are the Group 7 non-metals - fluorine, chlorine, bromine and iodine - each with seven outer-shell electrons and existing as diatomic molecules (F2\text{F}_2F2​, Cl2\text{Cl}_2Cl2​, Br2\text{Br}_2Br2​, I2\text{I}_2I2​). Their electronegativity decreases down the group: fluorine is the most electronegative element of all. This is because down the group the atoms get larger and the bonding electrons are further from the nucleus and more shielded, so the atom's power to attract a bonding pair falls.
The boiling points increase down the group, so at room temperature fluorine and chlorine are gases, bromine is a liquid and iodine is a solid. The molecules are held to one another only by van der Waals (London) forces. Going down the group the molecules have more electrons, so the instantaneous and induced dipoles are larger and the van der Waals forces stronger; more energy is therefore needed to separate the molecules, and the boiling point rises.
It is important to be clear about what is being overcome when a halogen boils: the weak van der Waals forces between the molecules, not the strong covalent bond within each X2\text{X}_2X2​ molecule. This is why the boiling points, though rising down the group, remain relatively low - characteristic of simple molecular substances. The colour also deepens down the group, from pale yellow fluorine to dark grey-black iodine.
These two trends have opposite underlying causes and should not be confused. Electronegativity falls because the larger atom attracts a bonding pair less strongly; boiling point rises because the larger molecule has more electrons and stronger van der Waals forces. When you explain a Group 7 trend, name the specific cause - atomic size and shielding for electronegativity, number of electrons and van der Waals forces for boiling point.
Worked example

Explaining the boiling-point trend

Explain why the boiling point increases from chlorine to iodine.

  1. 01Identify the forces

    The X2 molecules are held together by weak van der Waals (London) forces.

  2. 02Effect of molecular size

    Iodine molecules have more electrons than chlorine, so they form larger instantaneous and induced dipoles and stronger van der Waals forces.

  3. 03Effect on boiling point

    More energy is needed to separate the iodine molecules, so iodine has the higher boiling point.

Result: Iodine boils at a higher temperature than chlorine because its larger molecules have more electrons and stronger van der Waals forces.

Exam focus

  • Explain the decrease in electronegativity down Group 7 in terms of atomic size and shielding.
  • Explain the increase in boiling point down the group in terms of van der Waals forces and the number of electrons.

Typical mistakes

  • Saying covalent bonds break when a halogen boils - only the van der Waals forces between molecules do.
  • Confusing the causes of the two trends (size/shielding for electronegativity, electron count for boiling point).

Active revision

Explain why iodine is a solid at room temperature whereas chlorine is a gas, in terms of intermolecular forces.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

§ 02

Halogens as oxidising agents and displacement reactions#

●●○StandardLPAQA 7405 3.2.3.2LPDfE GCE Chemistry - oxidising ability of halogens

Displacement reactions of the halogens

Halogen displacement of halidesTable with 4 columns and 3 rows, Data: Halogen added · KCl · KBr · KI; chlorine · no reaction · orange (Br2) · brown (I2); bromine · no reaction · no reaction · brown (I2); iodine · no reaction · no reaction · no reactionHALOGEN ADDEDKCLKBRKIchlorineno reactionorange (Br2)brown (I2)bromineno reactionno reactionbrown (I2)iodineno reactionno reactionno reaction
Fig. 2A more reactive halogen displaces a less reactive one; the colour identifies the halogen formed.

Key points

The halogens are oxidising agents: they gain an electron to form halide ions, X2+2e−→2X−\text{X}_2 + 2e^- \rightarrow 2\text{X}^-X2​+2e−→2X−, oxidising whatever supplies the electron. Their oxidising power decreases down the group, because down the group the atom is larger and its outer shell more shielded, so it attracts an incoming electron less strongly - it is less able to gain an electron. Fluorine is the strongest oxidising agent and iodine the weakest.
This trend is shown by displacement reactions: a more reactive (more strongly oxidising) halogen will displace a less reactive one from a solution of its halide. Chlorine displaces both bromine and iodine (from bromide and iodide solutions), and bromine displaces iodine, but iodine cannot displace either. For example, Cl2+2KBr→2KCl+Br2\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2Cl2​+2KBr→2KCl+Br2​ - the chlorine is reduced and the bromide oxidised.
The observations identify which reaction has happened. Adding chlorine water to potassium bromide turns the solution orange (bromine formed); to potassium iodide it turns brown, and a brown-black solid (iodine) may appear. Adding an organic solvent such as cyclohexane and shaking makes the colours clearer: bromine gives an orange organic layer and iodine a violet (purple) one, distinguishing the two products.
The displacement reactions are redox reactions in which the halogen is reduced (gains electrons, oxidation state 0→−10 \rightarrow -10→−1) and the displaced halide is oxidised (loses electrons, −1→0-1 \rightarrow 0−1→0). A common exam task is to predict whether a displacement will occur and to write the ionic equation, remembering that only the more reactive halogen higher in the group can displace one below it.
Cl2+2KBr→2KCl+Br2\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2Cl2​+2KBr→2KCl+Br2​

Displacement of bromide by chlorine

Chlorine (the stronger oxidising agent) is reduced to chloride and bromide is oxidised to bromine.

Worked example

Predicting a displacement reaction

Bromine water is added to (a) potassium chloride solution and (b) potassium iodide solution. Predict what happens in each case and give any equation.

  1. 01With potassium chloride

    Chlorine is more reactive (a stronger oxidising agent) than bromine, so bromine cannot displace chloride - no reaction.

  2. 02With potassium iodide

    Bromine is more reactive than iodine, so it displaces iodide: Br2 + 2KI -> 2KBr + I2, and the solution turns brown.

  3. 03Identify the redox

    Bromine is reduced (0 to -1) and iodide is oxidised (-1 to 0).

Result: No reaction with KCl; with KI, bromine displaces iodine (brown solution), Br2 + 2KI -> 2KBr + I2.

Exam focus

  • Predict whether a displacement reaction occurs and write the ionic equation, using the oxidising-power trend.
  • State the colour observations (with and without an organic solvent) that identify the halogen formed.

Typical mistakes

  • Claiming iodine can displace chlorine or bromine - only a higher, more reactive halogen displaces a lower one.
  • Confusing the colours of bromine (orange) and iodine (brown/violet in an organic layer).

Active revision

Chlorine water is added to a solution of potassium iodide. State what you would observe and write the ionic equation for the reaction.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 03

Halides as reducing agents with concentrated sulfuric acid#

●●●AdvancedLPAQA 7405 3.2.3.2LPDfE GCE Chemistry - reducing ability of halides

Halides with concentrated sulfuric acid

NaX with concentrated H2SO4Table with 4 columns and 3 rows, Data: Halide · Products · S oxidation state · Reducing power; NaF / NaCl · HF / HCl (no redox) · +6 (unchanged) · too weak; NaBr · HBr, Br2, SO2 · +6 to +4 · moderate; NaI · HI, I2, SO2, S, H2S · +6 to -2 · strongHALIDEPRODUCTSS OXIDATION STATEREDUCING POWERNaF / NaClHF / HCl (no redox)+6 (unchanged)too weakNaBrHBr, Br2, SO2+6 to +4moderateNaIHI, I2, SO2, S, H2S+6 to -2strong
Fig. 3Down the group the halide reduces sulfur further: no redox for Cl-, SO2 for Br-, and down to H2S for I-.

Key points

While the halogens are oxidising agents, the halide ions are reducing agents, and their reducing power increases down the group - the opposite trend. A larger halide ion (such as iodide) holds its outer electrons less tightly, because they are further from the nucleus and more shielded, so it loses an electron more easily and is a stronger reducing agent. Fluoride and chloride are poor reducing agents; iodide is a good one.
This trend is revealed by the reactions of solid sodium halides with concentrated sulfuric acid. With sodium fluoride and sodium chloride, the only reaction is an acid-base displacement giving the hydrogen halide (HF\text{HF}HF or HCl\text{HCl}HCl) as steamy fumes - the halide is too weak a reducing agent to reduce the sulfur in sulfuric acid, so no redox occurs, for example NaCl+H2SO4→NaHSO4+HCl\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}NaCl+H2​SO4​→NaHSO4​+HCl.
With sodium bromide, HBr\text{HBr}HBr forms first, but bromide is a strong enough reducing agent to reduce some of the concentrated sulfuric acid: sulfur is reduced from +6+6+6 in H2SO4\text{H}_2\text{SO}_4H2​SO4​ to +4+4+4 in sulfur dioxide, and brown bromine fumes and choking SO2\text{SO}_2SO2​ are seen. With sodium iodide the reducing power is greatest of all: iodide reduces the sulfuric acid all the way down through SO2\text{SO}_2SO2​ (+4+4+4) and sulfur (000) to hydrogen sulfide, H2S\text{H}_2\text{S}H2​S (−2-2−2), so the products include purple iodine, sulfur, and rotten-egg-smelling H2S\text{H}_2\text{S}H2​S.
The set of reactions is a clear illustration of the reducing-power trend: the further down the group, the more strongly the sulfur is reduced. When answering, identify the changing oxidation state of sulfur (+6→+4→0→−2+6 \rightarrow +4 \rightarrow 0 \rightarrow -2+6→+4→0→−2) as evidence of increasing reducing power, and use the observations (steamy fumes, then coloured halogen, then the distinctive smells) to distinguish the halides.
NaCl+H2SO4→NaHSO4+HCl\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}NaCl+H2​SO4​→NaHSO4​+HCl

Chloride with concentrated sulfuric acid

Only an acid-base reaction; chloride is too weak a reducing agent to reduce the sulfur (no redox).

Worked example

Comparing bromide and iodide with sulfuric acid

Explain, using oxidation states, why sodium iodide reduces concentrated sulfuric acid further than sodium bromide does.

  1. 01Bromide

    Bromide reduces sulfur from +6 in H2SO4 to +4 in SO2; brown bromine and SO2 are seen.

  2. 02Iodide

    Iodide is a stronger reducing agent and reduces sulfur further, from +6 down to -2 in H2S (via SO2 and S); purple iodine and rotten-egg H2S are seen.

  3. 03Explain the difference

    The iodide ion is larger and less tightly held, so it loses electrons more readily and reduces sulfur to a lower oxidation state.

Result: Iodide, being a stronger reducing agent than bromide, drives the sulfur all the way to -2 (H2S), whereas bromide only reaches +4 (SO2).

Exam focus

  • Describe the products and observations for each sodium halide with concentrated sulfuric acid, and use the oxidation state of sulfur as evidence.
  • Explain why reducing power increases down the group in terms of ionic size and shielding.

Typical mistakes

  • Saying chloride reduces sulfuric acid - it only gives HCl with no redox.
  • Not tracking the oxidation state of sulfur (+6 to +4 to 0 to -2) to show the increasing reduction.

Active revision

Describe what is observed when concentrated sulfuric acid is added to (a) solid sodium chloride and (b) solid sodium iodide, and use the oxidation state of sulfur to compare the reducing power of the two halides.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 04

Testing for halide ions and disproportionation of chlorine#

●●○StandardLPAQA 7405 3.2.3.2LPAQA 7405 3.2.3.3LPDfE GCE Chemistry - halide tests and disproportionation

Silver-nitrate test for halide ions

Silver-nitrate halide testTable with 4 columns and 3 rows, Data: Halide · Precipitate colour · Dilute NH3 · Concentrated NH3; Cl- · white (AgCl) · dissolves · dissolves; Br- · cream (AgBr) · insoluble · dissolves; I- · yellow (AgI) · insoluble · insolubleHALIDEPRECIPITATE COLOURDILUTE NH3CONCENTRATED NH3Cl-white (AgCl)dissolvesdissolvesBr-cream (AgBr)insolubledissolvesI-yellow (AgI)insolubleinsoluble
Fig. 4Colour then ammonia solubility together identify the halide: chloride, bromide or iodide.

Key points

Halide ions are identified using acidified silver nitrate solution, which forms a coloured precipitate of the silver halide: silver chloride is white, silver bromide is cream, and silver iodide is yellow. The silver nitrate is acidified with dilute nitric acid first, to remove carbonate ions that would otherwise also precipitate with silver and give a false result. The colours are then confirmed by adding ammonia solution.
The solubility of the silver halides in ammonia distinguishes them, because it decreases down the group. Silver chloride dissolves in dilute ammonia; silver bromide is insoluble in dilute but dissolves in concentrated ammonia; silver iodide is insoluble even in concentrated ammonia. So the sequence of colour then ammonia solubility gives an unambiguous identification of the halide present.
Chlorine undergoes disproportionation - a reaction in which the same element is simultaneously oxidised and reduced. With cold water, chlorine forms a mixture of hydrochloric acid and chloric(I) acid: Cl2+H2O⇌HCl+HClO\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCl} + \text{HClO}Cl2​+H2​O⇌HCl+HClO, in which chlorine goes from 000 to −1-1−1 (in HCl\text{HCl}HCl, reduced) and to +1+1+1 (in HClO\text{HClO}HClO, oxidised). The chloric(I) acid (and the chlorate(I) ion) is what kills bacteria, which is why small amounts of chlorine are added to drinking water and swimming pools.
With cold dilute sodium hydroxide, chlorine again disproportionates: Cl2+2NaOH→NaCl+NaClO+H2O\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}Cl2​+2NaOH→NaCl+NaClO+H2​O, forming sodium chlorate(I), the active ingredient of household bleach. The use of chlorine in water treatment is a balance of benefit and risk: it kills harmful microorganisms and prevents diseases such as cholera, but chlorine is toxic and can react with organic matter to form chlorinated hydrocarbons; the public-health benefit is generally judged to outweigh the risks.
Cl2+H2O⇌HCl+HClO\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCl} + \text{HClO}Cl2​+H2​O⇌HCl+HClO

Disproportionation of chlorine in water

Chlorine is both reduced (0 to -1 in HCl) and oxidised (0 to +1 in HClO); HClO kills bacteria.

Cl2+2NaOH→NaCl+NaClO+H2O\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}Cl2​+2NaOH→NaCl+NaClO+H2​O

Chlorine with cold dilute NaOH

Disproportionation forming sodium chlorate(I), the active ingredient of bleach.

Worked example

Disproportionation of chlorine in water

Chlorine reacts with water: Cl2 + H2O -> HCl + HClO. Show that this is a disproportionation and state why chlorine is added to drinking water.

  1. 01Assign oxidation states

    Chlorine starts at 0 in Cl2; in HCl it is -1 (reduced) and in HClO it is +1 (oxidised).

  2. 02Identify disproportionation

    The same element (chlorine) is simultaneously oxidised and reduced, which is disproportionation.

  3. 03Use in water

    The chloric(I) acid (HClO) formed kills bacteria, so small amounts of chlorine make water safe to drink.

Result: Chlorine is oxidised to +1 and reduced to -1 in the same reaction (disproportionation); the HClO produced kills bacteria in drinking water.

Exam focus

  • Describe the silver-nitrate test in full, including acidifying the reagent and using ammonia to distinguish the halides.
  • Explain disproportionation using oxidation states, and evaluate the benefits and risks of chlorinating water.

Typical mistakes

  • Forgetting to acidify the silver nitrate with nitric acid, allowing carbonate to give a false precipitate.
  • Not showing that chlorine is both oxidised and reduced when explaining disproportionation.

Active revision

Explain, using oxidation states, why the reaction of chlorine with cold dilute sodium hydroxide is described as a disproportionation, and name the useful product.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Contents

Section -- / 04

    • 01Trends in electronegativity and boiling point◐
    • 02Halogens as oxidising agents and displacement reactions◐
    • 03Halides as reducing agents with concentrated sulfuric acid●
    • 04Testing for halide ions and disproportionation of chlorine◐

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Group 7(17), the halogens

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References & sources

Sources

Department for Education

  • GCE AS and A level subject content for the sciences

AQA

  • AQA A-level Chemistry 7405 specification

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