Edexcel International A Level Chemistry: Inorganic chemistry of Group 7 (YCH11) – Revision Notes

These notes condense Topic 8C, Inorganic chemistry of Group 7, from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 8.24 to 8.28 (IAS; Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols). The topic is limited to chlorine, bromine and iodine, with predictions for fluorine and astatine. For full explanations and worked examples, read the Group 7 study guide.

When you are ready to apply it, try the Group 7 practice questions. The course hub links every Edexcel IAL Chemistry resource, the printable checklist has every outcome to tick, and the free 10-minute diagnostic points you to gaps.

8.24 Trends at a glance

Property	Trend from Cl₂ to I₂	One-line reason
Melting and boiling temperature	increases	more electrons per molecule, so stronger London forces between molecules
State at room temperature	gas → liquid → solid	follows from the rising melting and boiling temperatures
Electronegativity	decreases	larger atom, more shielding, bonding pair further from the nucleus
Reactivity (oxidising power)	decreases	an added electron sits further out with more shielding, so it is held less tightly

Method in steps: explaining a boiling-temperature trend

1. State the structure: simple molecular, diatomic molecules.
2. Name the forces broken: London forces between molecules, not covalent bonds.
3. Compare electrons: more electrons means larger instantaneous dipoles and stronger London forces.
4. Conclude: more energy needed, so higher boiling temperature.

Method in steps: explaining the reactivity trend

1. A halogen reacts by gaining an electron to form X⁻.
2. Down the group the atomic radius increases and there are more inner shells.
3. The nucleus attracts an incoming electron less strongly.
4. So the halogen gains electrons less readily and is a weaker oxidising agent.

8.25 Colours and displacement

	Standard conditions	Water	Cyclohexane (non-polar)
Cl₂	pale green gas	very pale green / almost colourless	very pale green / almost colourless
Br₂	red-brown liquid	yellow to orange	orange to red-brown
I₂	grey-black solid	brown	violet

Displacement summary (✓ = reaction):

Halogen added	to Cl⁻	to Br⁻	to I⁻
Cl₂	–	✓ Br₂ formed	✓ I₂ formed
Br₂	✗	–	✓ I₂ formed
I₂	✗	✗	–

General ionic equation: X₂ + 2Y⁻ → 2X⁻ + Y₂, where X is above Y in the group. Halogen reduced, halide oxidised.

Reminder. Iodine water and bromine water can both look orange-brown. Shake with cyclohexane: violet means iodine, orange means bromine.

8.26 Redox equations to know

Reaction	Equation	Oxidation number changes
Group 1 metal	2Na + Br₂ → 2NaBr	Na 0 → +1; Br 0 → −1
Group 2 metal	Mg + Cl₂ → MgCl₂	Mg 0 → +2; Cl 0 → −1
Chlorine with water	Cl₂ + H₂O ⇌ HCl + HClO	Cl 0 → −1 and 0 → +1
Cold, dilute NaOH	Cl₂ + 2NaOH → NaCl + NaClO + H₂O	Cl 0 → −1 and 0 → +1
Hot alkali	3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O	Cl 0 → −1 and 0 → +5

Disproportionation: atoms of one element, all in the same reactant, go both up and down in oxidation number. Chlorine with water, with cold alkali and with hot alkali are all examples.

Names: chloric(I) acid = HClO; sodium chlorate(I) = NaClO, the active substance in bleach; sodium chlorate(V) = NaClO₃.

Water treatment, in three points:

• Chloric(I) acid kills bacteria, reducing waterborne disease.
• Chlorine is toxic, so the dose must be controlled.
• By-products: chlorine may combine with organic material to give chlorinated organics, some possibly harmful.

Analogous reactions: swap the halogen or the metal and keep the pattern. Iodine with sodium gives NaI; bromine with cold potassium hydroxide gives KBr + KBrO + H₂O; bromine with water gives HBr + HBrO.

Worked reminder: checking a hot-alkali equation. In 3Cl₂ + 6NaOH, there are six chlorine atoms. Five gain one electron each (5e⁻ gained); one loses five electrons in going to +5 (5e⁻ lost). Balanced.

8.27 Halides and hydrogen halides

With concentrated sulfuric acid

Solid	Products	Is H₂SO₄ reduced?	What you see
NaCl or KCl	HCl + NaHSO₄ (or KHSO₄)	no	steamy fumes
NaBr or KBr	HBr, then Br₂ + SO₂ + H₂O	yes, S +6 → +4	steamy fumes, orange-brown vapour
NaI or KI	HI, then I₂ with SO₂, S and H₂S	yes, S +6 → +4, 0, −2	purple vapour, black solid, yellow solid, bad-egg smell

Reducing ability: HCl < HBr < HI. The larger the halide, the further its outer electrons are from the nucleus, so the more easily it loses them.

Method in steps: any redox step with sulfuric acid

1. Write the oxidation number of S before (+6) and after (+4 in SO₂, 0 in S, −2 in H₂S).
2. Electrons gained by one S = the change (2, 6 or 8).
3. Each halide ion loses one electron, so you need that many HX per H₂SO₄.
4. Balance H and O with water.

Worked reminder: bromide giving sulfur dioxide. Sulfur goes +6 → +4, gaining 2 electrons. Two bromide ions each lose one electron, so two HBr are needed per H₂SO₄: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O. Check: hydrogen 4 on each side, oxygen 4 on each side. Bromide cannot go further and make sulfur or hydrogen sulfide, because HBr is a weaker reducing agent than HI.

With silver nitrate

Acidify with dilute nitric acid (removes carbonate), then add silver nitrate: Ag⁺(aq) + X⁻(aq) → AgX(s).

Ion	Colour	Dilute NH₃	Concentrated NH₃
Cl⁻	white	dissolves	dissolves
Br⁻	cream	does not dissolve (or only slightly)	dissolves
I⁻	yellow	does not dissolve	does not dissolve

Hydrogen halides

• With ammonia gas: HX + NH₃ → NH₄X, a white smoke (solid).
• With water: HX + H₂O → H₃O⁺ + X⁻, giving an acidic solution; this is why they fume in moist air.

8.28 Prediction rules

	Fluorine	Astatine
State	gas, lowest boiling temperature	solid, highest melting temperature
Electronegativity	highest	lowest
Oxidising power	strongest; displaces all others	weakest; displaced by all others
Silver halide	–	dark precipitate, insoluble in concentrated NH₃
Hydrogen halide as reducing agent	weakest	strongest

Watch fluorine: as the most electronegative element it is −1 in all its compounds, so it cannot form chlorate-type ions with positive oxidation numbers.

Must-know distinctions

• Cold, dilute alkali vs hot alkali: chlorate(I), oxidation number +1, versus chlorate(V), +5.
• Halogen vs halide: the halogen (Cl₂) is the oxidising agent; the halide ion (Cl⁻) is the reducing agent.
• Oxidising power of halogens vs reducing power of halides: the first decreases down the group, the second increases.
• Acid–base step vs redox step: forming HX from a solid halide is acid–base; any later step that reduces sulfur is redox.
• Inside a molecule vs between molecules: boiling breaks London forces; the X–X bond stays intact.

Quick self-test

Relative atomic masses for this test: Na 23.0, K 39.1, Cl 35.5, Br 79.9, Ag 107.9. Gas molar volume: 24.0 dm³ mol⁻¹.

1. State the colour of bromine dissolved in cyclohexane.
2. Write the ionic equation for chlorine reacting with aqueous iodide ions.
3. Give the oxidation number of chlorine in NaClO₃.
4. Why does iodine have a higher boiling temperature than chlorine?
5. Name the precipitate formed when acidified silver nitrate is added to potassium bromide, and give its colour.
6. Which hydrogen halide reduces concentrated sulfuric acid to hydrogen sulfide?
7. Calculate the mass of sodium chloride formed from 0.920 g of sodium and excess chlorine.
8. What volume of chlorine, in cm³ at room temperature and pressure, is needed to make 0.0300 mol of NaClO with cold, dilute sodium hydroxide?
9. The same 0.0300 mol of chlorine is passed into hot sodium hydroxide instead. How many moles of NaClO₃ form?
10. A solution contains 0.595 g of potassium bromide. Calculate the mass of silver bromide precipitated with excess silver nitrate.
11. Predict whether astatine will displace iodine from potassium iodide solution.

Answers

1. Orange (to red-brown).
2. Cl₂ + 2I⁻ → 2Cl⁻ + I₂.
3. +5.
4. I₂ has more electrons, so stronger London forces between molecules need more energy to overcome.
5. Silver bromide, AgBr; cream.
6. Hydrogen iodide, HI.
7. n(Na) = 0.920 ÷ 23.0 = 0.0400 mol = n(NaCl); mass = 0.0400 × 58.5 = 2.34 g.
8. Cl₂ : NaClO is 1 : 1, so 0.0300 mol Cl₂; volume = 0.0300 × 24 000 = 720 cm³.
9. 3Cl₂ gives 1NaClO₃, so 0.0300 ÷ 3 = 0.0100 mol.
10. Mr(KBr) = 119.0, n = 0.595 ÷ 119.0 = 0.00500 mol; Mr(AgBr) = 187.8; mass = 0.00500 × 187.8 = 0.939 g.
11. No. Astatine is a weaker oxidising agent than iodine.

Where marks are usually lost

• Writing “chlorine is reduced” for the reaction with water, without the oxidation to +1.
• Giving +5 for the cold-alkali product or +1 for the hot-alkali product.
• Describing iodine in a non-polar solvent as brown rather than violet.
• Writing Cl⁻ + Br₂ as a reaction that happens; bromine cannot oxidise chloride.
• Explaining boiling temperatures with bond enthalpy instead of London forces.
• Missing the first acid–base step when writing products of halides with concentrated sulfuric acid.
• Listing observations for sodium iodide without the bad-egg smell or the yellow solid.
• Forgetting that nitric acid, not hydrochloric acid, is added before silver nitrate.
• Saying silver bromide dissolves in dilute ammonia.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Scope: Topic 8C, outcomes 8.24 to 8.28.