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

This guide teaches Topic 8C, Inorganic chemistry of Group 7, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers outcomes 8.24 to 8.28. The specification limits this part to chlorine, bromine and iodine, with fluorine and astatine met only as predictions. This is IAS content, placed in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. Topic 8C has no core practical.

For fast recall, use the Group 7 revision notes, then test yourself with the Group 7 practice questions. Every Edexcel IAL Chemistry page is listed on the course hub. Mark off each outcome on the printable checklist. Our free 10-minute diagnostic can point out weak spots. Oxidation numbers and half-equations come from Topic 8A, Redox chemistry.

Data used on this page (use these values): H 1.0, N 14.0, K 39.1, Ca 40.1, Br 79.9, Ag 107.9, I 126.9. Gas molar volume (room temperature and pressure) = 24.0 dm³ mol⁻¹.

Scope of Topic 8C

Outcome	What you must be able to do
8.24	Explain trends in melting and boiling temperatures, physical state, electronegativity and reactivity
8.25	Explain the reactivity trend using displacement reactions of Cl₂, Br₂ and I₂ with halide ions; know the colours
8.26	Use oxidation numbers for halogen reactions with Group 1 and 2 metals, with water, and with cold and hot alkali, plus analogous reactions
8.27	Explain halides with concentrated sulfuric acid, silver nitrate tests with ammonia, and hydrogen halides with ammonia and water
8.28	Predict the properties of fluorine, astatine and their compounds from the trends

8.24 Physical trends, electronegativity and reactivity

Use these values:

Halogen	Melting temperature / °C	Boiling temperature / °C	State at room temperature
Cl₂	−101	−34	gas
Br₂	−7	59	liquid
I₂	114	184	solid

All three are simple molecular, made of diatomic molecules. Melting or boiling overcomes the London forces between molecules; it does not break the X–X covalent bond.

Down the group, each molecule has more electrons, so instantaneous dipoles are larger and London forces are stronger. More energy is needed to separate the molecules, so melting and boiling temperatures rise. Room temperature (about 20 °C) is above both values for chlorine, between them for bromine and below both for iodine.

Electronegativity measures how strongly an atom pulls the shared electron pair of a covalent bond towards itself. Its value falls down the group. The atoms get bigger and have more inner shells. The extra nuclear charge is largely cancelled by extra shielding, so the more distant bonding pair is attracted less strongly.

Reactivity decreases down the group. A halogen reacts by gaining electrons, so it acts as an oxidising agent: X₂ + 2e⁻ → 2X⁻. An electron arriving at a larger atom enters a shell further from the nucleus, with more shielding between them, so it is attracted less. Iodine therefore gains electrons least readily and is the weakest oxidising agent of the three.

8.25 Displacement reactions and colours

Any halogen higher in the group will oxidise the halide ions of one lower down. Learn the colours first.

	Standard conditions	In aqueous solution	In a non-polar solvent such as cyclohexane
Chlorine	pale green gas	very pale green, almost colourless	very pale green, almost colourless
Bromine	red-brown liquid	yellow to orange	orange to red-brown
Iodine	shiny grey-black solid	brown (yellow-brown when dilute)	violet

The reactions that happen:

• Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)
• Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq)
• Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq)

Bromine does not react with chloride ions; iodine reacts with neither. The halogen is reduced (0 to −1) and the halide ion oxidised (−1 to 0).

Worked example 1: reading a displacement grid

A student adds a few drops of each aqueous halogen to potassium halide solutions, then shakes each tube with cyclohexane and lets the layers separate. Predict what she sees in the top (organic) layer when bromine water is added to potassium iodide.

1. Bromine is above iodine, so it oxidises iodide: Br₂ + 2I⁻ → 2Br⁻ + I₂.
2. The product halogen is iodine. Iodine is non-polar, so it moves into the cyclohexane.
3. The organic layer turns violet. In water, brown iodine is easily confused with orange bromine; the solvent makes the difference clear.

8.26 Redox reactions in terms of oxidation number

Reactions with Group 1 and 2 metals

Halogens oxidise these metals to their ions. For example, 2K + Cl₂ → 2KCl and Ca + Br₂ → CaBr₂. In the calcium reaction calcium goes from 0 to +2 and each bromine atom goes from 0 to −1. The reactions become less vigorous from chlorine to iodine, matching the fall in oxidising power.

Worked example 2: calcium and bromine

What mass of calcium bromide forms if 2.00 g of calcium is heated in excess bromine vapour?

1. n(Ca) = 2.00 ÷ 40.1 = 0.04988 mol.
2. Ca + Br₂ → CaBr₂ is 1 : 1, so n(CaBr₂) = 0.04988 mol.
3. Mr(CaBr₂) = 40.1 + 2(79.9) = 199.9.
4. Mass = 0.04988 × 199.9 = 9.97 g.

Chlorine with water

Cl₂ + H₂O ⇌ HCl + HClO

Chlorine goes from 0 to −1 in HCl and from 0 to +1 in chloric(I) acid, HClO. One element in a single species both rises and falls in oxidation number, which makes this disproportionation.

Chloric(I) acid kills bacteria, so chlorine is used in water treatment to cut the risk of waterborne disease such as cholera. Against this, chlorine is toxic and must be dosed carefully, and it may combine with organic material in the water, giving chlorinated organic by-products, some possibly harmful.

Making bleach: chlorine and cold, dilute alkali

Cl₂ + 2NaOH → NaCl + NaClO + H₂O

Chlorine goes from 0 to −1 in NaCl and to +1 in sodium chlorate(I), NaClO. The resulting solution is household bleach.

Chlorine with hot alkali

3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O

With hot alkali, chlorine goes from 0 to −1 in chloride and to +5 in sodium chlorate(V), NaClO₃. Check the ratio by counting electrons.

Worked example 3: an analogous reaction with bromine

Bromine reacts with hot potassium hydroxide in the same way as chlorine does with hot alkali. Write the equation and show that the electrons balance.

1. Products by analogy: KBr, KBrO₃ (potassium bromate(V)) and water.
2. Equation: 3Br₂ + 6KOH → 5KBr + KBrO₃ + 3H₂O.
3. Oxidation numbers: Br goes 0 → −1 in KBr (five atoms, each gaining one electron: 5e⁻ gained).
4. One Br atom goes 0 → +5 in KBrO₃ (5e⁻ lost).
5. Electrons gained = electrons lost = 5, so the equation balances in redox terms.

Analogous reactions use the chlorine pattern for bromine or iodine, or potassium in place of sodium.

8.27 Reactions of halides and hydrogen halides

Solid halides with concentrated sulfuric acid

All three solid sodium or potassium halides first give the hydrogen halide, seen as steamy fumes in moist air. For example: KCl + H₂SO₄ → KHSO₄ + HCl. This step is acid–base, not redox.

What happens next depends on how strongly the hydrogen halide reduces sulfuric acid.

Halide	Further reaction	Sulfur’s oxidation number	Observations
Chloride	none: HCl cannot reduce H₂SO₄	stays +6	steamy fumes only
Bromide	2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O	+6 → +4	steamy fumes, orange-brown vapour, choking gas
Iodide	2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O; 6HI + H₂SO₄ → 3I₂ + S + 4H₂O; 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O	+6 → +4, 0 and −2	some steamy fumes, purple vapour, black solid, yellow solid, smell of bad eggs

So the reducing ability of the hydrogen halides increases HCl < HBr < HI. In a larger halide ion the outermost electrons are more distant from the nucleus and better shielded, so they are lost more easily.

Worked example 4: building the hydrogen sulfide equation

Show how 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O balances.

1. Sulfur goes from +6 in H₂SO₄ to −2 in H₂S, so one sulfur atom gains 8 electrons.
2. Each iodine goes from −1 to 0, losing one electron. Eight iodide ions lose 8 electrons, forming 4I₂.
3. Hydrogen: 8 from HI plus 2 from H₂SO₄ = 10; on the right 2 in H₂S plus 8 in 4H₂O = 10.
4. Oxygen: 4 on each side. The equation balances.

Halide ions with silver nitrate

Add dilute nitric acid, then aqueous silver nitrate. The nitric acid removes ions such as carbonate that would otherwise give a precipitate with silver ions. Do not use hydrochloric acid: it adds chloride ions.

Ag⁺(aq) + X⁻(aq) → AgX(s)

Ion	Precipitate	Dilute aqueous ammonia	Concentrated aqueous ammonia
Cl⁻	white precipitate, AgCl	soluble	soluble
Br⁻	cream precipitate, AgBr	insoluble (at most slightly soluble)	soluble
I⁻	yellow precipitate, AgI	insoluble	insoluble

The ammonia step matters because white and cream look alike.

Worked example 5: precipitate mass to concentration

A 30.0 cm³ portion of potassium iodide solution is acidified and treated with excess silver nitrate. The dried yellow precipitate weighs 0.646 g. Find the concentration of the potassium iodide solution.

1. Mr(AgI) = 107.9 + 126.9 = 234.8.
2. n(AgI) = 0.646 ÷ 234.8 = 2.751 × 10⁻³ mol = n(I⁻), since the ratio is 1 : 1.
3. Concentration = 2.751 × 10⁻³ ÷ 0.0300 = 0.0917 mol dm⁻³.

Hydrogen halides with ammonia and with water

With ammonia gas, each hydrogen halide gives a white smoke of the ammonium halide: HBr(g) + NH₃(g) → NH₄Br(s). The same reaction with HCl is used as a test for either gas.

With water, hydrogen halides dissolve readily and ionise to give acids: HCl(g) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq). This is also why they fume in moist air.

Worked example 6: mixing two gases

60.0 cm³ of hydrogen bromide is mixed with 45.0 cm³ of ammonia at room temperature and pressure. Find the gas left over and the mass of ammonium bromide formed.

1. The ratio is 1 : 1, so ammonia (45.0 cm³) is limiting.
2. Gas left: 60.0 − 45.0 = 15.0 cm³ of hydrogen bromide.
3. n(NH₄Br) = 45.0 ÷ 24 000 = 1.875 × 10⁻³ mol; Mr = 14.0 + 4(1.0) + 79.9 = 97.9.
4. Mass = 1.875 × 10⁻³ × 97.9 = 0.184 g.

The specification’s suggested practicals are solid potassium halides with concentrated sulfuric acid (small amounts, in a fume cupboard, as the gases are toxic) and precipitation reactions of anions.

8.28 Predicting fluorine and astatine

Extend each trend one step up or down.

• Fluorine: a pale yellow gas with the lowest boiling temperature, the highest electronegativity and the strongest oxidising power. It would displace chlorine, bromine and iodine from their halides.
• Astatine: a dark solid with the highest melting temperature and the weakest oxidising power. It would be displaced by every halogen above it. Silver astatide would be a dark precipitate, insoluble in concentrated ammonia. Hydrogen astatide would be the strongest reducing agent of the hydrogen halides.

Worked example 7: a prediction question

Predict the products when solid sodium astatide is warmed with concentrated sulfuric acid.

1. First, acid–base: NaAt + H₂SO₄ → NaHSO₄ + HAt.
2. HAt should be a stronger reducing agent than HI, so expect astatine plus reduced sulfur products such as SO₂, S and H₂S.

Be careful with fluorine. It is always −1 in compounds because it is the most electronegative element, so it cannot take the positive oxidation numbers that chlorine has in chlorate(I) and chlorate(V). Predictions based on chlorine’s disproportionation do not carry over to fluorine.

Common errors

• Saying boiling breaks the X–X bond. It overcomes London forces between molecules.
• Describing iodine in cyclohexane as brown. It is violet; brown is its colour in water.
• Calling the chlorine–water reaction plain reduction. Chlorine is reduced and oxidised, so it is disproportionation.
• Writing NaClO₃ for cold alkali or NaClO for hot alkali.
• Acidifying a silver nitrate test with hydrochloric acid.
• Stating that sulfuric acid is oxidised. It is reduced.

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.