OxfordAQA A-Level Chemistry: Group 7(17), the halogens (9620)

This guide works through section 3.2.3, which deals with the Group 7(17) halogens, in the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. You will meet point 3.2.3.1 on trends in properties, point 3.2.3.2 on the uses of chlorine and chlorate(I), and Required practical 3, which the specification lists here. The whole section is International AS; no part of it is International A-level only. Its home is Unit 1 (Inorganic 1 and Physical 1), while Unit 5 (Practical and synoptic) may draw on any content.

For a condensed version, use the halogens revision notes. Then try the practice questions. Every 9620 topic is listed on the course hub, while the printable checklist has a box for each point. Take a free 10-minute diagnostic to find weak spots.

Background you need: van der Waals forces from shapes, polarity and intermolecular forces, oxidation states and half-equations from oxidation, reduction and redox equations, and moles from amount of substance.

What this unit covers

Point (all International AS)	You should be able to
3.2.3.1	Account for electronegativity falling from F to I
3.2.3.1	Link boiling points of F₂ to I₂ to structure and bonding
3.2.3.1	Show how oxidising power changes, using halogen–halide displacements in water
3.2.3.1	Show how reducing power of X⁻ changes, using NaX(s) with concentrated H₂SO₄
3.2.3.1	Identify Cl⁻, Br⁻ and I⁻ with acidified AgNO₃, and justify the AgNO₃, the acid and the NH₃
3.2.3.1	State how AgCl, AgBr and AgI differ in solubility in ammonia
3.2.3.2	Give the two reactions of chlorine with water (to Cl⁻ + ClO⁻; to Cl⁻ + O₂)
3.2.3.2	Weigh chlorination of drinking water, and chemicals in water supplies generally
3.2.3.2	Give the reaction of Cl₂ with cold, dilute NaOH(aq) and what the product is used for
Required practical 3	Simple test-tube tests: Group 2 cations, NH₄⁺, halides, OH⁻, CO₃²⁻, SO₄²⁻

The overview says fluorine is too hazardous for school labs (overviews are not directly assessed).

3.2.3.1 Electronegativity

The specification defines electronegativity as an atom’s power to attract the electron pair in a covalent bond. Down Group 7 it falls. On the Pauling scale the values are roughly F 4.0, Cl 3.2, Br 3.0 and I 2.7.

The reason: each element down the group has an extra occupied shell, so the atom is larger and there is more shielding. The shared pair sits further from a more shielded nucleus, so it is pulled less strongly, even though the proton number rises.

Worked example 1: fluorine against bromine

Explain why fluorine is more electronegative than bromine.

• A fluorine atom has two shells and bromine four, so in a bond to fluorine the shared pair is closer to the nucleus.
• It is shielded only by 1s electrons, so fluorine’s nucleus attracts it more strongly.

3.2.3.1 Boiling points

Element	Formula	Look at room temperature	Electrons per molecule	Boiling point / K
fluorine	F₂	pale yellow gas	18	85
chlorine	Cl₂	pale green gas	34	239
bromine	Br₂	red-brown liquid	70	332
iodine	I₂	grey-black solid	106	457

All four are simple molecular, made of diatomic molecules. Boiling overcomes only the van der Waals (induced dipole–dipole) forces between molecules, which get stronger as the number of electrons per molecule rises. The X–X covalent bond stays intact.

Worked example 2: a gas and a liquid

Bromine boils at 332 K and fluorine at 85 K. Explain the difference.

Structure:  both simple molecular, diatomic
Electrons:  F₂ has 18, Br₂ has 70, so Br₂ has stronger van der Waals forces
Energy:     more energy needed to separate Br₂ molecules: higher boiling point

3.2.3.1 Oxidising ability and displacement

A halogen is an oxidising agent because it takes electrons: X₂ + 2e⁻ → 2X⁻. Its oxidising ability decreases down the group. A larger atom must pull the incoming electron into an outer shell that is both more distant and more shielded, so the pull on it is weaker.

So a halogen oxidises the halide ions of any halogen below it, displacing that halogen from aqueous solution.

Colours in water can be faint, so shake the mixture with a non-polar solvent such as cyclohexane. The halogen moves into the upper organic layer: iodine is violet there (brown in water), and bromine gives an orange layer.

Worked example 3: naming an unknown halide

A colourless solution contains one sodium halide, Y. A student adds chlorine water to one portion, then cyclohexane, and shakes: the organic layer turns orange. With bromine water and cyclohexane, the organic layer just shows the colour of the bromine added, and no new halogen appears. Identify Y and write the ionic equation.

• Chlorine made a new halogen with a bromine-type colour, so Y’s ion was oxidised to bromine.
• Bromine water made no new halogen; an iodide would have turned the layer violet.
• Y is sodium bromide.

Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂
reduction:  Cl₂ + 2e⁻ → 2Cl⁻
oxidation:  2Br⁻ → Br₂ + 2e⁻

3.2.3.1 Reducing ability of halide ions

A halide ion is a reducing agent because it gives electrons away: 2X⁻ → X₂ + 2e⁻. This power increases from F⁻ to I⁻ (F⁻ < Cl⁻ < Br⁻ < I⁻). In a bigger ion the outer electrons are further out and more shielded, so one is lost more easily.

Concentrated sulfuric acid with the solid sodium halides shows this. For all four the first step is acid–base, and the hydrogen halide escapes as steamy fumes:

NaX + H₂SO₄ → NaHSO₄ + HX

What happens next depends on how strong a reducing agent X⁻ is:

Solid	Redox step?	What you see	Sulfur product	S oxidation state
NaF or NaCl	no	steamy fumes only	none	stays +6
NaBr	yes	steamy fumes; orange-brown vapour (Br₂); a choking gas (SO₂)	SO₂	+4
NaI	yes, further	steamy fumes; black solid and purple vapour (I₂); yellow solid (S); bad-egg smell (H₂S)	SO₂, S and H₂S	+4, 0 and −2

The further sulfur falls from +6, the stronger the reducing agent.

Worked example 4: one overall equation from the solid

Write a single equation for solid sodium iodide reacting with concentrated sulfuric acid to give sodium hydrogensulfate, iodine, hydrogen sulfide and water.

Electron count:  S goes +6 → −2, gaining 8e⁻; each I⁻ → ½I₂ loses 1e⁻, so 8 I⁻ are needed
Redox part:      8I⁻ give 4I₂; 1 H₂SO₄ gives H₂S
Sodium:          8 Na⁺ end up in 8 NaHSO₄, which needs 8 more H₂SO₄
Water:           balance O and H with 4H₂O
Overall:         8NaI + 9H₂SO₄ → 8NaHSO₄ + 4I₂ + H₂S + 4H₂O

Check: S 9 = 8 + 1; O 36 = 32 + 4; H 18 = 8 + 2 + 8.

3.2.3.1 Identifying halides with silver nitrate

Acidify the sample solution with dilute nitric acid and add aqueous silver nitrate. Chloride, bromide and iodide give precipitates of different colours; fluoride gives none, because silver fluoride is soluble.

Ag⁺(aq) + X⁻(aq) → AgX(s)

Ion	Precipitate	In dilute NH₃(aq)	In concentrated NH₃(aq)
Cl⁻	white AgCl	dissolves	dissolves
Br⁻	cream AgBr	stays	dissolves
I⁻	yellow AgI	stays	stays

The solubility of the silver halides in ammonia falls from AgCl to AgI. The specification asks you to explain three choices:

• Silver nitrate is used because it gives a precipitate with Cl⁻, Br⁻ and I⁻, and the colours differ.
• It is acidified (with nitric acid) to remove ions such as carbonate or hydroxide, which would otherwise give their own precipitates with silver ions. Hydrochloric acid would add chloride, so it is never used for this.
• Ammonia is added because white, cream and yellow are hard to tell apart. Solubility in ammonia settles it.

Worked example 5: purity of a bromide

A 0.600 g sample of impure calcium bromide, CaBr₂, is dissolved. Excess acidified silver nitrate gives 1.034 g of dry AgBr. The impurity contains no halide. Calculate the percentage purity. (Ar: Ag = 107.9, Br = 79.9, Ca = 40.1)

Mr(AgBr)  = 107.9 + 79.9 = 187.8
n(AgBr)   = 1.034 ÷ 187.8 = 0.005506 mol = n(Br⁻)
n(CaBr₂)  = 0.005506 ÷ 2 = 0.002753 mol      (each CaBr₂ gives two Br⁻)
Mr(CaBr₂) = 40.1 + 2(79.9) = 199.9
mass      = 0.002753 × 199.9 = 0.5503 g
purity    = 0.5503 ÷ 0.600 × 100 = 91.7 %

3.2.3.2 Chlorine and water

Chlorine reacts reversibly with water, forming chloride ions and chlorate(I) ions:

Cl₂ + H₂O ⇌ HClO + HCl      or      Cl₂ + H₂O ⇌ 2H⁺ + ClO⁻ + Cl⁻

Chlorine starts at 0 and ends at −1 (Cl⁻) and +1 (ClO⁻). One element both oxidised and reduced is a disproportionation.

In sunlight, chlorine water gives off oxygen and leaves chloride ions:

2Cl₂ + 2H₂O → 4H⁺ + 4Cl⁻ + O₂

This is not disproportionation: chlorine only goes down (0 to −1), and oxygen goes up (−2 to 0).

Worked example 6: oxygen from chlorine water

A flask of chlorine water left in sunlight releases 0.144 g of oxygen. Calculate the mass of chlorine that reacted. (Mr: O₂ = 32.0, Cl₂ = 71.0)

n(O₂)  = 0.144 ÷ 32.0 = 0.00450 mol
n(Cl₂) = 2 × 0.00450 = 0.00900 mol      (2Cl₂ : 1O₂)
mass   = 0.00900 × 71.0 = 0.639 g

Chlorine in water treatment

Water suppliers dose drinking water with chlorine to kill bacteria and prevent water-borne disease. Chlorine is toxic, and organic matter in the water can turn some of it into chlorinated organic compounds such as chloroform. You should appreciate, as the specification puts it, that the health gain outweighs the toxic effects.

Communities judge the pros and cons of any chemical put into water. Fluoride is one case: added in some places against tooth decay, but too much causes fluorosis, so it is debated.

3.2.3.2 Chlorine and cold, dilute sodium hydroxide

Chlorine bubbled into cold, dilute alkali gives:

ionic:  Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
full:   Cl₂ + 2NaOH → NaClO + NaCl + H₂O

Again chlorine disproportionates, from 0 to −1 and +1. The solution of sodium chlorate(I), NaClO, is used as bleach and as a disinfectant for cleaning.

Worked example 7: making a bleach solution

How much chlorine is needed to make 500 cm³ of a solution that is 0.640 mol dm⁻³ in NaClO, using excess cold, dilute NaOH? (Mr: Cl₂ = 71.0)

n(NaClO) = 0.640 × (500 ÷ 1000) = 0.320 mol
n(Cl₂)   = 0.320 mol            (1 Cl₂ : 1 NaClO)
mass     = 0.320 × 71.0 = 22.7 g

Required practical 3: test-tube tests for ions

Ion	What to do	Positive result
Group 2 cations	add NaOH(aq); separately add dilute H₂SO₄	Mg²⁺ gives a white hydroxide precipitate but no sulfate one; Ba²⁺ gives a white sulfate precipitate, with little or no hydroxide one
NH₄⁺	warm with NaOH(aq); test the gas with damp red litmus	litmus turns blue (ammonia)
halide ions	dilute HNO₃, then AgNO₃(aq), then NH₃(aq)	as in the silver halide table
OH⁻	red litmus or universal indicator	litmus blue; indicator shows alkali
CO₃²⁻	dilute acid; pass the gas into limewater	fizzing; limewater turns milky
SO₄²⁻	acidified BaCl₂(aq)	white precipitate of BaSO₄

Common errors

• Saying the Br–Br or I–I bond breaks when the element boils.
• Treating the NaCl reaction with sulfuric acid as redox; it is acid–base only.
• Giving H₂S as a product of sodium bromide.
• Forgetting the factor of two when a metal halide holds two halide ions.
• Calling the sunlight reaction disproportionation.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards; within it, section 3.2.3, the halogens of Group 7(17), at International AS: point 3.2.3.1, on trends in properties; point 3.2.3.2, on the uses of chlorine and chlorate(I); and Required practical 3.