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AQA GCSE Chemistry Smart practice A mixed round aimed at your weak spots
Unit 1
Atomic structure and the periodic table
Unit 2
Bonding, structure, and the properties of matter
Unit 3
Quantitative chemistry
Unit 4
Chemical changes
Unit 5
Energy changes
Unit 6
The rate and extent of chemical change
Unit 7
Organic chemistry
Unit 8
Chemical analysis
Unit 9
Chemistry of the atmosphere
Unit 10
Using resources
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Group 7

Practise 27 exam-style questions on Group 7 for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

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  • 1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Describe the appearance of chlorine and of bromine at room temperature.

Model answer

Chlorine is a green gas. Bromine is a red-brown liquid.

Explanation

Chlorine is one of only a few coloured gases. Bromine is one of only two elements that are liquid at room temperature (the other is mercury); it gives off an orange-brown vapour.

Mark scheme
  1. 1 mark

    chlorine: green gas

    allow yellow-green / pale green; both colour and state needed

  2. 1 mark

    bromine: red-brown liquid

    allow brown / dark red / orange-brown; both colour and state needed

Question 2

2 marks

A student adds iodine solution to potassium bromide solution.

Predict what the student would observe. Give a reason for your answer.

Model answer

There would be no change; the mixture stays the colour of the iodine solution. Iodine is less reactive than bromine, so it cannot displace bromine from potassium bromide.

Explanation

Displacement only happens when the halogen added is more reactive than the halogen in the salt. Iodine is below bromine in Group 7, so no reaction takes place.

Mark scheme
  1. 1 mark

    no (colour) change / no reaction

    allow solution stays brown / the colour of the iodine solution

  2. 1 mark

    iodine is less reactive than bromine (so cannot displace it)

Question 3

3 marks

Hydrogen reacts with bromine to form hydrogen bromide, HBr. Hydrogen bromide is a gas at room temperature.

Predict the type of bonding in hydrogen bromide and give a reason. Predict what you would find if universal indicator were added to a solution of hydrogen bromide in water.

Model answer

Hydrogen bromide has covalent bonding, because hydrogen and bromine are both non-metals, so their atoms share a pair of electrons to form HBr molecules. Hydrogen bromide dissolves in water to form an acidic solution, so the indicator would turn red.

Explanation

Halogens form covalent compounds with non-metals. The hydrogen halides (HCl, HBr, HI) are gases made of small molecules that dissolve in water to form acids, just as hydrogen chloride forms hydrochloric acid.

Mark scheme
  1. 1 mark

    covalent (bonding)

  2. 1 mark

    hydrogen and bromine are both non-metals / atoms share a pair of electrons

  3. 1 mark

    the indicator turns red / the solution is acidic

    allow orange / pH less than 7

Question 4

4 marks

Excess chlorine is bubbled through a solution containing 8.3 g of potassium iodide.
Cl₂ + 2KI → 2KCl + I₂

Calculate the maximum mass of iodine that could be produced.

Relative atomic masses (Ar): K = 39, I = 127

Model answer

Mr of KI = 39 + 127 = 166
Mr of I₂ = 2 × 127 = 254
moles of KI = 8.3 ÷ 166 = 0.05 mol
2 mol KI produce 1 mol I₂, so moles of I₂ = 0.05 ÷ 2 = 0.025 mol
mass of I₂ = 0.025 × 254 = 6.35 g

Explanation

Work out moles of the substance you know the mass of, use the balanced equation to find moles of the product (here 2 : 1), then convert back to mass with the product's Mr. Remember iodine is I₂, so its Mr is 254, not 127.

Mark scheme
  1. 1 mark

    Mr of KI = 166 and Mr of I₂ = 254

    both needed

  2. 1 mark

    moles of KI = 8.3 ÷ 166 = 0.05 (mol)

    allow ecf from incorrect Mr

  3. 1 mark

    moles of I₂ = 0.05 ÷ 2 = 0.025 (mol)

    allow ecf

  4. 1 mark

    mass of I₂ = 0.025 × 254 = 6.35 (g)

    allow 6.4 g; allow ecf; correct answer with no working scores 4 marks; 12.7 g (ratio not used) scores 3 marks

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