2.3.4Organic Chemistry
Halide Ion Reactions — Edexcel AS Chemistry Revision Notes & Topical Questions
5 min read 5 parts Lesson 13 of 42
Section 1 of 5With Concentrated Sulfuric Acid
Carry out these tests in a fume cupboard because the gases formed are toxic.
Step 1
- Every solid halide reacts to form the hydrogen halide: NaX(s) + H₂SO₄(l) → HX(g) + NaHSO₄(s).
Chloride
- Only this reaction occurs, giving misty white fumes of HCl and no further redox change.
Bromide
- HBr is oxidised by the acid2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O, giving orange-brown Br₂ and choking SO₂.
Iodide
- HI is oxidised more strongly, giving I₂ (purple vapour), SO₂, yellow solid S and rotten-egg smelling H₂S.
Equations
- 16HI + H₂SO₄23I₂ + S + 4H₂O and 8HI + H₂SO₄34I₂ + H₂S + 4H₂O
Section 2 of 5Why Reducing Power Increases
1
Going down the group the halide ions become larger, so the outer electrons sit further from the nucleus.
2
Shielding also increases, so the nucleus holds the outer electrons less tightly.
3
Electrons are lost more easily from the larger ions, so the halide ion is a better reducing agent.
4
Cl⁻ cannot reduce sulfuric acid, Br⁻ reduces it to SO₂, and I⁻ reduces it as far as S and H₂S.
5
The order of reducing power is therefore Cl⁻ < Br⁻ < I⁻.
Section 3 of 5Silver Nitrate and Ammonia Test
Silver Nitrate and Ammonia Test
Method
- Acidify the sample with dilute nitric acid, then add aqueous silver nitrate to form a silver halide precipitate.
Ionic equation
- Ag⁺(aq) + X⁻(aq) → AgX(s), where X is Cl, Br or I.
Why acidify
- Nitric acid removes carbonate or hydroxide ions that would otherwise give a misleading precipitate with silver ions.
Ammonia test
- Add dilute then concentrated ammonia to the precipitate, and note whether it dissolves as a soluble complex.
Complex ion
- Silver chloride dissolves to form the diamminesilver(I) ion, [Ag(NH₃)₂]⁺.
| Halide | Precipitate colour | Dilute NH₃ | Concentrated NH₃ |
|---|---|---|---|
| Cl⁻ | White | Dissolves | Dissolves |
| Br⁻ | Cream | Insoluble | Dissolves |
| I⁻ | Pale yellow | Insoluble | Insoluble |
Section 4 of 5Worked Example
Question
- A 0.0200 mol sample of sodium chloride in solution is treated with excess acidified silver nitrate.
- Calculate the mass of silver chloride formed.
- (Ar: Ag 107.9, Cl 35.5.)
- EquationAg⁺(aq) + Cl⁻(aq) → AgCl(s), so 1 mol Cl⁻ gives 1 mol AgCl.
- Moles of AgCl = 0.0200 mol.
- Mr of AgCl = 107.9 + 35.5 = 143.4.
- Mass = 0.0200 × 143.4 = 2.868 g.
Answer 2.87 g of AgCl
Section 5 of 5Exam Checklist & Tips
2.3.4 – Key Points for the Exam
- Reducing agent A substance that gives electrons to another species and is itself oxidised.
- Precipitate An insoluble solid formed when two solutions react.
- Hydrogen halide A compound HX formed in the reaction of a halide with an acid, such as HCl or HBr.
- Complex ion An ion in which a central metal ion is bonded to surrounding molecules or ions.
- Acidified silver nitrate The reagent used to detect halide ions.
Common Mistakes
- Suggesting concentrated H₂SO₄ oxidises Cl⁻ in this test: it only produces HCl.
- Naming the precipitate colours wrongly, especially cream for AgBr and pale yellow for AgI.
- Forgetting to acidify with nitric acid and using hydrochloric acid, which would add chloride ions.
- Giving the product of sulfuric acid reduction as H₂S for all halides when only iodide reaches that stage.
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