OCR A Level Chemistry A: Group 2 (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers section 3.1.2, Group 2, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions test every learning outcome, 3.1.2(a) to (e), from Module 3, Periodic table and energy, all full A level content. Questions 10 and 11 combine several outcomes.

Revise first with the Group 2 study guide. Browse H432 from the OCR chemistry hub, and see each outcome listed on the printable checklist.

Use 24.0 dm³ mol⁻¹ for the molar gas volume at RTP.

Questions

1. This question is about strontium.

(a) State the outer shell electron configuration of a strontium atom. [1] (b) Write the full electron configuration of the Sr²⁺ ion. [1] (c) Explain, in terms of electrons, why strontium is oxidised when it forms a compound. [1]

2. Write equations, with state symbols, for:

(a) the first ionisation energy of barium [1] (b) the second ionisation energy of barium [1]

3. Heated magnesium ribbon reacts with steam.

(a) Give a balanced equation, with state symbols. [2] (b) Using oxidation numbers, name the element that loses electrons and the element that gains them. [2] (c) State how the metal-containing product differs when magnesium reacts with cold water instead. [1]

4. Use these values, in kJ mol⁻¹.

	First ionisation energy	Second ionisation energy
Mg	738	1451
Ba	503	965

(a) Calculate the total energy needed to turn one mole of gaseous atoms into gaseous 2+ ions, for magnesium and for barium. [2] (b) Explain why barium reacts with cold water much faster than magnesium does. [4]

5. A 1.752 g sample of strontium burns completely in oxygen. Ar(Sr) = 87.6, Ar(O) = 16.0.

(a) Write the equation for the reaction. [1] (b) Find the mass of oxide that forms. [2] (c) Calculate the volume of oxygen used, in cm³, at RTP. [2]

6. A student shakes small samples of MgO, CaO, SrO and BaO with water in four test tubes, then adds universal indicator.

(a) Write the equation for the reaction of strontium oxide with water. [1] (b) The approximate pH values recorded are 10, 12, 13 and above 13. Match each value to an oxide. [1] (c) Explain the trend in alkalinity shown. [2] (d) State, with a reason, whether the reaction in (a) is a redox reaction. [1]

7. 0.613 g of strontium is added to an excess of 0.250 mol dm⁻³ hydrochloric acid. Ar(Sr) = 87.6.

(a) Write the equation for the reaction, including state symbols. [2] (b) How many cm³ of hydrogen, measured at RTP, are released? [2] (c) Calculate the minimum volume of the acid needed to react with all the strontium. [2] (d) The experiment is repeated with 0.00700 mol of calcium in place of the strontium. Predict how the rate of fizzing and the final volume of hydrogen compare with the first experiment. [2]

8. Magnesium hydroxide (Mr 58.3) and calcium carbonate (Mr 100.1) are both used as antacids.

(a) Write an equation for the reaction of each compound with hydrochloric acid. [2] (b) Calculate the amount of HCl, in mol, neutralised by 1.00 g of each compound, and state which is more effective per gram. [3] (c) Suggest one disadvantage of calcium carbonate as an antacid. [1]

9. A soil survey finds that a field contains acid equivalent to 25.2 kg of nitric acid, HNO₃. Ar: H 1.0, N 14.0, O 16.0, Ca 40.1.

(a) Write the equation for calcium hydroxide neutralising nitric acid. [1] (b) Calculate the mass of calcium hydroxide, in kg, needed to neutralise this acid. [3] (c) Calcium carbonate can be used instead. Write the equation for its reaction with nitric acid. [1]

10. A Group 2 metal, X, reacts with excess dilute hydrochloric acid to form XCl₂ and hydrogen. 0.876 g of X gives 1.586 g of anhydrous XCl₂ after the solution is evaporated. Ar(Cl) = 35.5.

(a) Calculate the relative atomic mass of X and identify X. [3] (b) State the volume of hydrogen released, in cm³ at RTP. [1] (c) Write the equation, with state symbols, for X reacting with cold water. [2] (d) The same mass of a second Group 2 metal, Y, reacts with the acid more slowly than X but releases a larger volume of hydrogen. Suggest an identity for Y and explain both observations. [3]

11. 0.4812 g of calcium (Ar 40.1) is burned completely in oxygen. The solid is stirred with water, giving a cloudy mixture, which is then neutralised with 0.200 mol dm⁻³ hydrochloric acid.

(a) Write the equation for calcium burning, and give the oxidation numbers of calcium and oxygen before and after the reaction. [2] (b) Calculate the volume of oxygen used, in cm³, at RTP. [2] (c) Write the equation for the reaction of the solid with water, and state the approximate pH of the liquid above the settled solid. [2] (d) Calculate the volume of hydrochloric acid needed to neutralise the whole mixture. [2]

Answers

1. (a) 5s² [1] (b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ [1] (c) It loses its two outer 5s electrons, so its oxidation number rises from 0 to +2 [1] Examiner insight: In (b) remove the 5s electrons, not 4p ones.

2. (a) Ba(g) → Ba⁺(g) + e⁻ [1] (b) Ba⁺(g) → Ba²⁺(g) + e⁻ [1] Examiner insight: Each equation needs (g) on both ions or atoms; Ba(g) → Ba²⁺(g) + 2e⁻ in (b) answers a different question.

3. (a) Mg + H₂O → MgO + H₂, correct formulae and balanced [1]; state symbols Mg(s), H₂O(g), MgO(s), H₂(g) [1] (b) Mg: 0 → +2, so magnesium loses electrons (oxidised) [1]; H: +1 → 0, so hydrogen gains electrons (reduced) [1] (c) With cold water the product is magnesium hydroxide, Mg(OH)₂, not the oxide [1] Examiner insight: Steam is H₂O(g); writing H₂O(l) in (a) makes the state symbols wrong even when the equation is balanced.

4. (a) Mg: 738 + 1451 = 2189 kJ mol⁻¹ [1]; Ba: 503 + 965 = 1468 kJ mol⁻¹ [1] (b) Barium’s 6s electrons are at a greater distance from the nucleus than magnesium’s 3s electrons [1]; more shielding from inner shells [1]; weaker nuclear attraction on the outer electrons, despite greater nuclear charge, so less energy is needed to remove them [1]; barium is oxidised more readily, so it reacts faster [1] Examiner insight: Four marks need four separate links; nuclear charge rises down the group, so never say it falls.

5. (a) 2Sr + O₂ → 2SrO [1] (b) n(Sr) = 1.752 ÷ 87.6 = 0.0200 mol = n(SrO) [1]; mass = 0.0200 × 103.6 = 2.07 g [1] (c) n(O₂) = 0.0200 ÷ 2 = 0.0100 mol [1]; V = 0.0100 × 24.0 = 0.240 dm³ = 240 cm³ [1] Examiner insight: In (c) write the 2 : 1 ratio of Sr to O₂ before multiplying by 24.0.

6. (a) SrO(s) + H₂O(l) → Sr(OH)₂(aq) [1] (b) MgO about 10, CaO about 12, SrO about 13, BaO above 13 [1] (c) The hydroxides become more soluble down the group [1]; so the concentration of OH⁻ ions in solution is higher and the solution more alkaline [1] (d) Not redox: Sr stays +2, O stays −2 and H stays +1 [1] Examiner insight: In (c), “more reactive” is not an explanation of alkalinity; link solubility to the OH⁻ concentration.

7. (a) Sr + 2HCl → SrCl₂ + H₂, correct and balanced [1]; state symbols (s), (aq), (aq), (g) [1] (b) n(Sr) = 0.613 ÷ 87.6 = 0.00700 mol = n(H₂) [1]; V = 0.00700 × 24.0 = 0.168 dm³ = 168 cm³ [1] (c) n(HCl) = 2 × 0.00700 = 0.0140 mol [1]; V = 0.0140 ÷ 0.250 = 0.0560 dm³ = 56.0 cm³ [1] (d) Calcium fizzes more slowly, as it is higher in the group [1]; the same volume of hydrogen, 168 cm³, forms because the amount of metal is the same [1] Examiner insight: In (d) treat rate and amount separately; slower does not mean less gas.

8. (a) Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O [1]; CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ [1] (b) Mg(OH)₂: 2 × (1.00 ÷ 58.3) = 0.0343 mol [1]; CaCO₃: 2 × (1.00 ÷ 100.1) = 0.0200 mol [1]; so magnesium hydroxide neutralises more acid per gram [1] (c) It releases carbon dioxide, which can cause wind [1] Examiner insight: Both compounds react 1 : 2 with HCl; a 1 : 1 ratio halves both answers even though the comparison still looks right.

9. (a) Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O [1] (b) n(HNO₃) = 25 200 ÷ 63.0 = 400 mol [1]; n(Ca(OH)₂) = 400 ÷ 2 = 200 mol [1]; mass = 200 × 74.1 = 14 820 g = 14.8 kg [1] (c) CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂ [1] Examiner insight: Change 25.2 kg into grams first; treating it as 25.2 g makes the answer a thousand times too small.

10. (a) mass of Cl = 1.586 − 0.876 = 0.710 g, so n(Cl) = 0.710 ÷ 35.5 = 0.0200 mol [1]; n(X) = 0.0200 ÷ 2 = 0.0100 mol [1]; Ar = 0.876 ÷ 0.0100 = 87.6, so X is strontium [1] (b) n(H₂) = n(X) = 0.0100 mol; V = 0.0100 × 24.0 = 0.240 dm³ = 240 cm³ [1] (c) Sr + 2H₂O → Sr(OH)₂ + H₂, balanced [1]; state symbols (s), (l), (aq), (g) [1] (d) More hydrogen from the same mass means more moles, so Y has a lower Ar than strontium [1]; a slower reaction means Y is higher in the group, with higher first and second ionisation energies [1]; Y is calcium (or magnesium) [1] Examiner insight: In (a) find the chlorine first; dividing the chloride mass by n(X) gives the Mr of XCl₂, not the Ar of X.

11. (a) 2Ca + O₂ → 2CaO [1]; Ca 0 → +2 and O 0 → −2 [1] (b) n(Ca) = 0.4812 ÷ 40.1 = 0.0120 mol, so n(O₂) = 0.00600 mol [1]; V = 0.00600 × 24.0 = 0.144 dm³ = 144 cm³ [1] (c) CaO + H₂O → Ca(OH)₂ [1]; pH about 12 [1] (d) n(Ca(OH)₂) = 0.0120 mol, so n(HCl) = 2 × 0.0120 = 0.0240 mol [1]; V = 0.0240 ÷ 0.200 = 0.120 dm³ = 120 cm³ [1] Examiner insight: In (d) all the calcium ends up as Ca(OH)₂, dissolved or not, so the acid depends only on the 0.0120 mol of calcium; show that chain of moles.

Where marks are usually lost

• Writing the second ionisation energy from the atom, as in question 2.
• Using H₂O(l) for steam, or giving Mg(OH)₂ as the steam product, as in question 3.
• Explaining reactivity without both distance and shielding, or saying nuclear charge falls.
• Missing the 2 : 1 mole ratio for O₂ or HCl in questions 5, 7 and 11.
• Calling the oxide-water reaction redox in question 6.
• Explaining alkalinity with reactivity rather than hydroxide solubility.
• Forgetting the CO₂ in carbonate equations, as in questions 8 and 9.
• Mixing up rate and amount of gas (questions 7 and 10).

Next steps

• Find weak spots across the course: all free 10-minute diagnostics
• Recap the tables and methods: Group 2 revision notes
• Full explanations and worked examples: Group 2 study guide
• Moles, masses and gas volumes: Amount of substance practice
• Every H432 section: OCR chemistry hub
• Outcome tracker: H432 printable checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.1.2 Group 2, learning outcomes (a) to (e). Published by OCR.