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Group 2 and halogens lab answers for A Level chemistry (free PDF download)

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The method and the results are on the group 2 and halogens lab. Do the analysis yourself first, then check the answers below.

Quick answer: Group 2 metals are reactive metals that lose two outer electrons to form 2+ ions, and their reactivity rises down the group. These answers go with the Chemistry 11-12 lab on group 2 and halogens: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.

Group 2 and halogens lab answers for A Level chemistry (free PDF download)

Here are the group 2 and halogens lab answers for chemistry grades 11-12: graph, calculations, uncertainties and conclusion. First do the lab yourself, then compare, and go back to the lesson if needed.

These model answers use the sample results from the table. Your own colours may differ slightly, but the pattern of solubility in ammonia and the order of displacement must match.

Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.2.2-3.2.3

Group 2 and halogens lab answers: the analysis explained

Question 1 Identify the solutions

A is sodium chloride (white, dissolves in dilute ammonia), B is sodium iodide (yellow, insoluble in conc. ammonia) and C is sodium bromide (cream, dissolves only in conc. ammonia). The hexane layers agree: iodine is violet, bromine orange-brown.

Question 2 Ionic equations

Tube A: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Tube C with chlorine: Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq).

Question 3 Role of nitric acid

Nitric acid removes carbonate and hydroxide ions that give false precipitates with Ag⁺. Hydrochloric acid would add chloride ions and give a false white precipitate, ruining the test.

Question 4 Reactivity order

Chlorine displaced both bromine and iodine, while no displacement is possible the other way, so the order is Cl₂ > Br₂ > I₂ in oxidising power. The smaller the atom, the stronger the pull on an incoming electron.

Question 5 Mass of precipitate

n(AgCl) = 0.00500 × 0.200 = 1.00 × 10⁻³ mol; mass = 1.00 × 10⁻³ × 143.4 = 0.143 g.

Question 6 Reliability

Silver bromide only partly dissolves in dilute ammonia, so the change is gradual and hard to see by eye. Compare with a control tube and use a fixed volume of ammonia, or use concentrated ammonia in a fume cupboard.

Question 7 Improvement

Use known standard solutions of each halide alongside the unknowns and compare side by side. Repeat each test to be certain of the colour, using a white card behind the tubes.

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