The halogens

In a nutshell

The halogens (Group 17) are reactive non-metals that exist as diatomic molecules and react mainly by gaining one electron to form 1- ions, so they act as oxidising agents.

This subtopic covers the boiling-point trend, the reactivity trend shown by displacement reactions, the disproportionation of chlorine (water treatment and bleach), and the silver nitrate test for halide ions.

Assumed knowledge: Redox, Electron structure, Bonding and structure.

Core content

Characteristic physical properties

Every halogen exists as diatomic molecules, X₂: two atoms joined by a single covalent bond.

Boiling points increase down the group.

The explanation, as a chain of linked points:

1. Down the group, the molecules have more electrons.
2. So there are stronger induced dipole-dipole interactions (London forces) between the molecules.
3. So more energy is needed to overcome these intermolecular forces, and the boiling point rises.

Boiling a halogen separates whole X₂ molecules. The covalent bond inside each molecule does not break.

Redox reactions and reactivity: electron configuration and oxidising power

Every halogen atom has the outer-shell electron configuration s²p⁵: one electron short of a full p sub-shell.

• Cl: [Ne] 3s²3p⁵
• Br: [Ar] 3d¹⁰4s²4p⁵
• I: [Kr] 4d¹⁰5s²5p⁵

In many redox reactions each halogen atom gains one electron to form a 1- ion (a halide ion):

Cl₂ + 2e^- → 2Cl^-

The oxidation number of chlorine falls from 0 to -1, so the halogen is reduced.

Because the halogen takes electrons from another species, it is an oxidising agent: it oxidises the other species and is itself reduced.

Redox reactions and reactivity: displacement reactions

Reactivity (oxidising power) decreases down the group: Cl₂ > Br₂ > I₂.

You see this when a halogen is added to a solution of another halide ion. A more reactive halogen displaces a less reactive halogen from its halide ions.

The three reactions that happen

In each one, the halogen is reduced (0 → -1) and the halide ion is oxidised (-1 → 0). The K⁺ ions are spectator ions.

Colours: how you tell which halogen is present

The colour at the end shows which halogen molecule (X₂) is now in the tube. The halide ions themselves are colourless.

In water, bromine and iodine can both look yellow-brown, so they are easy to confuse.

Adding a few drops of cyclohexane and shaking fixes this:

• The halogens are non-polar, so they dissolve better in the organic solvent than in water.
• Cyclohexane does not mix with water and is less dense, so it forms the top layer.
• The colour of the top layer is clear-cut: orange for bromine, violet for iodine.

Redox reactions and reactivity: explaining the trend

Reactivity falls down the group because the halogen atoms form 1- ions less easily: they attract the extra electron less strongly.

Going from Cl to Br to I:

1. Atomic radius increases: the outer shell is further from the nucleus.
2. Electron shielding increases: there are more inner shells between the nucleus and the outer shell.
3. So there is less attraction between the nucleus and the electron being gained.
4. So the atom gains an electron less easily to form a 1- ion, and the halogen is a weaker oxidising agent.

Nuclear charge does increase down the group, but the bigger radius and extra shielding outweigh it.

Still don't get it? · why reactivity falls down the group

Imagine a magnet trying to pick up a paperclip. Close up, it grabs it easily. Put a few sheets of cardboard in the way and hold it further off, and the pull is much weaker, even if you use a slightly stronger magnet.

Now rebuild it step by step:

• A halogen reacts by taking an electron in from something else. So "reactive" here means "good at pulling an electron in".
• The pull comes from the positive nucleus. The incoming electron has to land in the outer shell.
• Chlorine's outer shell is the third shell. Bromine's is the fourth, so it is further out (bigger atomic radius).
• Bromine also has more full inner shells in between. These inner electrons repel the incoming electron and cancel out some of the nuclear charge. That is shielding (the cardboard).
• Bromine's nucleus has more protons (the slightly stronger magnet), but the extra distance and shielding win.
• So bromine pulls an electron in less strongly than chlorine, forms its 1- ion less easily, and is less reactive. Iodine is weaker still.

The exam version: down the group, atomic radius increases and there is more shielding, so there is less attraction for the electron being gained; the halogen forms 1- ions less easily, so reactivity (oxidising power) decreases. Always say it comparatively, and always say which electron: the one being gained.

Redox reactions and reactivity: disproportionation

Disproportionation is the oxidation and reduction of the same element in one reaction.

Chlorine starts at oxidation number 0. In these reactions some chlorine atoms go up (oxidised) and others go down (reduced) at the same time.

1. Chlorine with water (water treatment)

Cl₂(aq) + H₂O(l) → HClO(aq) + HCl(aq)

Ionic equation (HCl is a strong acid, so it is fully ionised): Cl₂(aq) + H₂O(l) → HClO(aq) + H⁺(aq) + Cl^-(aq)

Hydrogen (+1) and oxygen (-2) do not change, so water is neither oxidised nor reduced here.

The products make the solution acidic, and HClO is a bleach. So if universal indicator is added to chlorine water, it first turns red and then loses its colour.

2. Chlorine with cold, dilute aqueous sodium hydroxide (making bleach)

Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Ionic equation: Cl₂(aq) + 2OH^-(aq) → Cl^-(aq) + ClO^-(aq) + H₂O(l)

• Cl goes from 0 to +1 in NaClO (oxidised).
• Cl goes from 0 to -1 in NaCl (reduced).
• NaClO is sodium chlorate(I). A solution of it is household bleach.
• The conditions matter: the spec reaction uses cold, dilute NaOH.

3. Analogous reactions

Bromine and iodine are in the same group, so you can be asked to write the matching equations by swapping the halogen. For example, bromine with water:

Br₂(aq) + H₂O(l) → HBrO(aq) + HBr(aq)

Bromine goes from 0 to +1 in HBrO and from 0 to -1 in HBr. Worked example 2 does the NaOH version.

Still don't get it? · spotting disproportionation

Think of a group of friends who all start on the ground floor of a building. Some take the lift up a floor, some take the stairs down a floor. Nobody new arrived: the same group just split in two directions.

Step by step:

• Write the oxidation number of the element before the reaction. In Cl₂ it is 0, because it is an element.
• Find every product that contains that element.
• Work out its oxidation number in each product. In HClO: H is +1, O is -2, so Cl must be +1 for the total to be 0. In HCl: H is +1, so Cl is -1.
• One product is higher (that chlorine was oxidised) and one is lower (that chlorine was reduced).
• The same element went both ways in one reaction, so it is disproportionation.

The exam version: disproportionation is oxidation and reduction of the same element. To earn the marks, quote the actual numbers: "chlorine is oxidised from 0 to +1 and reduced from 0 to -1".

Redox reactions and reactivity: chlorine and water treatment

Chlorine is added to drinking water because the chloric(I) acid (and chlorate(I) ions) it forms kill bacteria. This prevents water-borne diseases.

There are risks too:

• Chlorine gas is toxic, so it must be handled and stored carefully.
• Chlorine can react with organic matter in the water to form chlorinated hydrocarbons, which may be harmful to health.

Whether to chlorinate is a balance of benefits and risks. For drinking water, the benefit of killing bacteria is generally judged to outweigh the risks. There are also ethical questions, such as people's right to choose what is added to their water.

Other ways to purify drinking water include:

• ozone (O₃), a strong oxidising agent that kills microorganisms;
• ultraviolet (UV) light, which kills microorganisms.

Unlike chlorine, ozone and UV leave no disinfectant behind in the water to keep protecting it as it travels through the pipes.

Characteristic reactions of halide ions

Aqueous halide ions react with aqueous silver ions (from silver nitrate) to form silver halide precipitates. The colour tells you which halide is present.

Ionic equations (spectator ions such as Na⁺ and NO₃^- left out):

• Ag⁺(aq) + Cl^-(aq) → AgCl(s)
• Ag⁺(aq) + Br^-(aq) → AgBr(s)
• Ag⁺(aq) + I^-(aq) → AgI(s)

A full equation, for comparison: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

White, cream and yellow are hard to tell apart by eye, so the ammonia step confirms the result. You only need the observations for this step: the equations for the silver-ammonia complexes are not needed.

PAG4: qualitative analysis of ions

This test-tube test is part of PAG4 (qualitative analysis of ions).

Method for a halide test

1. Dissolve the unknown solid in water (distilled or deionised) in a test tube.
2. Add a few drops of dilute nitric acid.
3. Add a few drops of aqueous silver nitrate. Record the colour of the precipitate.
4. Add dilute aqueous ammonia and record whether the precipitate dissolves. If it does not, repeat with concentrated ammonia.

Why add nitric acid first?

• Carbonate ions would also form a precipitate with silver ions (silver carbonate), giving a false result.
• The acid removes carbonate ions (as CO₂) before the silver nitrate goes in.
• Use nitric acid: hydrochloric acid would add Cl^- ions, and sulfuric acid would add SO₄^2- ions (silver sulfate is also insoluble).

Where it fits in a full analysis (3.1.4)

When testing an unknown compound for several anions, the order is carbonate, then sulfate, then halide. Doing the halide test last stops carbonate or sulfate ions giving a false silver precipitate.

Recording observations

Always write the colour and the state: "white precipitate", not just "white" or "goes cloudy".

Worked examples

1. Oxidation numbers in a disproportionation reaction

Show that the reaction of chlorine with cold, dilute sodium hydroxide is disproportionation.

Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

1. In Cl₂, chlorine is an element: oxidation number 0.
2. In NaCl: Na is +1, so Cl is -1.
3. In NaClO: Na is +1 and O is -2. The total must be 0, so Cl = 0 - (+1) - (-2) = +1.
4. So chlorine is oxidised from 0 to +1 and reduced from 0 to -1 in the same reaction: this is disproportionation.

Where marks go: the numbers 0, +1 and -1 must all appear, and you must say which change is oxidation and which is reduction.

2. Writing an analogous equation

Bromine reacts with cold, dilute sodium hydroxide in the same way as chlorine. Write the full and ionic equations and name the bromine-containing products.

1. Swap Cl for Br in the chlorine equation:

   Br₂(aq) + 2NaOH(aq) → NaBr(aq) + NaBrO(aq) + H₂O(l)

2. Remove the spectator Na⁺ ions:

   Br₂(aq) + 2OH^-(aq) → Br^-(aq) + BrO^-(aq) + H₂O(l)

3. Check the ionic equation balances: 2 Br, 2 O and 2 H on each side; charge is 2- on the left and (1-) + (1-) = 2- on the right.

4. Name the products: sodium bromide (Br is -1) and sodium bromate(I) (Br is +1, so the Roman numeral is I).

Where marks go: students often write the products as acids (HBr, HBrO). With NaOH the products are sodium salts.

3. Predicting a displacement reaction

Aqueous bromine is added to aqueous potassium iodide, then cyclohexane is added and the tube is shaken. Predict what you see and write the ionic equation.

1. Bromine is above iodine in Group 17, so it is the stronger oxidising agent. It displaces iodine.
2. Ionic equation: Br₂(aq) + 2I^-(aq) → 2Br^-(aq) + I₂(aq)
3. Oxidation numbers: Br goes 0 → -1 (reduced); I goes -1 → 0 (oxidised). Bromine is the oxidising agent.
4. Observations: the aqueous solution turns from orange to brown; after shaking with cyclohexane, the top organic layer is violet (purple).

4. Model answer: explaining the reactivity trend

Explain why chlorine is more reactive than iodine.

1. Iodine atoms have a larger atomic radius than chlorine atoms.
2. Iodine atoms have more shielding from inner shells.
3. So there is weaker attraction between the iodine nucleus and the electron being gained.
4. So iodine forms a 1- ion less easily, so it is less reactive (a weaker oxidising agent) than chlorine.

Notice that each point is comparative (larger, more, weaker, less easily). "Iodine is big" does not compare anything.

5. Model answer: identifying a halide

A student dissolves a white solid in water, adds dilute nitric acid then silver nitrate solution, and sees a cream precipitate. Describe a further test to confirm the halide ion, and write the ionic equation for the precipitate.

1. A cream precipitate suggests bromide, Br^-.
2. Add dilute ammonia: the precipitate does not dissolve (rules out chloride).
3. Add concentrated ammonia: the precipitate dissolves (rules out iodide).
4. Ionic equation: Ag⁺(aq) + Br^-(aq) → AgBr(s)

Common exam mistakes

• Saying boiling points rise because covalent bonds get stronger or must be broken. Boiling separates molecules: the answer needs stronger London forces between molecules.
• Explaining the boiling-point trend with "bigger M_r" or "bigger atoms" only. The link examiners want is more electrons, so stronger induced dipole-dipole interactions.
• Calling the halide ion the oxidising agent ("chloride displaces bromide"). It is the halogen molecule, Cl₂, that does the oxidising; Cl^- has already gained its electron.
• Giving iodine's colour as purple in water. In water iodine is brown; it is violet only in an organic solvent such as cyclohexane.
• In the reactivity explanation, talking about losing electrons or ionisation energy. Halogens gain an electron, so describe attraction for the electron being gained.
• Forgetting to make the explanation comparative, or leaving out shielding. A full answer needs radius, shielding and the attraction for the incoming electron.
• Defining disproportionation as "a reaction where something is oxidised and something is reduced". That is just redox: the key words are the same element.
• Stating chlorine is "oxidised and reduced" without the oxidation numbers (0 → +1 and 0 → -1).
• Giving chlorine's oxidation number in HClO or NaClO as -1. It is +1 (the -1 is in HCl/NaCl).
• Leaving out cold, dilute for the NaOH reaction, or naming NaClO "sodium chloride" or "sodium chlorate". It is sodium chlorate(I).
• Describing the silver halide result as "white", "white colour" or "white solution". Write white precipitate.
• Mixing up the colours: AgCl white, AgBr cream, AgI yellow.
• Saying AgBr dissolves in ammonia without saying concentrated. Dilute ammonia dissolves only AgCl.
• Acidifying with hydrochloric acid. It adds chloride ions, so every sample would give a white precipitate. Use nitric acid.
• Missing state symbols or leaving spectator ions in the ionic equation. AgCl is a solid: Ag⁺(aq) + Cl^-(aq) → AgCl(s).

Key definitions

• Disproportionation: the oxidation and reduction of the same element in a redox reaction.
• Oxidation: loss of electrons, or an increase in oxidation number.
• Reduction: gain of electrons, or a decrease in oxidation number.
• Oxidising agent: a species that accepts electrons (and is itself reduced). The halogens are oxidising agents.
• Displacement reaction (halogens): a reaction in which a more reactive halogen oxidises the halide ions of a less reactive halogen, releasing that halogen.
• Induced dipole-dipole interactions (London forces): attractions between molecules caused by temporary dipoles that induce dipoles in neighbouring molecules; they are stronger when molecules have more electrons.

Specification

• I can state that the halogens exist as diatomic molecules and explain the trend in the boiling points of Cl₂, Br₂ and I₂ in terms of induced dipole-dipole interactions (London forces). (3.1.3 a)
• I can give the outer-shell s²p⁵ configuration and explain that halogens gain one electron in many redox reactions to form 1- ions. (3.1.3 b)
• I can describe the trend in reactivity of Cl₂, Br₂ and I₂ using their reactions with other halide ions, with full and ionic equations and the colour changes in aqueous and organic solutions. (3.1.3 c)
• I can explain the reactivity trend from the decreasing ease of forming 1- ions, in terms of attraction, atomic radius and electron shielding. (3.1.3 d)
• I can explain disproportionation as oxidation and reduction of the same element, using chlorine with water, chlorine with cold, dilute NaOH, and analogous reactions, with oxidation numbers. (3.1.3 e)
• I can weigh the benefits of chlorinating water (killing bacteria) against the risks (toxic chlorine gas, chlorinated hydrocarbons), and mention other methods of purifying water. (3.1.3 f)
• I can describe the reactions of aqueous Cl^-, Br^- and I^- with silver ions and then ammonia, write the ionic equations, and use them as a test for halide ions (PAG4). (3.1.3 g)

Related notes

• Redox: oxidation numbers, oxidising agents and Roman numerals such as chlorate(I)
• Electron structure: s and p sub-shells and electron configurations
• Bonding and structure: induced dipole-dipole interactions (London forces)
• Periodicity: atomic radius and shielding across and down the table
• Group 2: the other group studied in 3.1, as reducing agents
• Qualitative analysis: the full sequence of anion tests (carbonate, sulfate, halide)
• Haloalkanes: organic compounds of the halogens