OxfordAQA A-Level Chemistry: Group 2, the alkaline earth metals (9620) – 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.

Questions 1 to 12 below test section 3.2.2.1 Group 2, the alkaline earth metals of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, together with the Group 2 part of Required practical 3. Everything here is International AS content; nothing is International A-level only. The section belongs to Unit 1 (Inorganic 1 and Physical 1), and Unit 5 can draw on all content.

Learn the content first in the Group 2 study guide, then revise it with the Group 2 revision notes. The OxfordAQA A-Level Chemistry hub and the printable checklist place this topic in the course, and the free 10-minute diagnostics point you to weak areas. For more mole practice, see amount of substance.

Use these Aᵣ values: H 1.0, C 12.0, O 16.0, S 32.1, Cl 35.5, Mg 24.3, Ca 40.1, Ti 47.9, Ba 137.3.

Questions

1. Describe how atomic radius changes from Mg to Ba, and give a reason. [2]

2. Strontium’s first ionisation energy is greater than barium’s. Explain why. [3]

3. (a) Explain why calcium has a higher melting point than strontium. [3] (b) Name the element from Mg to Ba whose melting point does not fit the general trend. [1]

4. (a) Give a balanced equation, including state symbols, for strontium reacting with water. [2] (b) A small piece of calcium and a similar piece of barium are each added to cold water. Give two differences you would observe. [2] (c) Give the equation for magnesium reacting with steam. [1]

5. Titanium is made by reducing titanium(IV) chloride with magnesium.

(a) Write the equation and identify the reducing agent, using oxidation states. [2] (b) A reactor is charged with 0.950 tonnes of TiCl₄. Calculate the minimum mass of magnesium needed and the mass of magnesium chloride formed, both in kg. [3]

6. A student adds 0.687 g of barium to water. When the reaction stops, the solution is made up to 250 cm³ with distilled water. Portions of 25.0 cm³ are then titrated against 0.100 mol dm⁻³ hydrochloric acid.

(a) Give the equation for barium reacting with water. [1] (b) Calculate the expected titre. [3] (c) Suggest why the same method, using the same mass of calcium, would be less reliable. [1]

7. (a) List the hydroxides of Mg, Ca, Sr and Ba in order of increasing solubility. [1] (b) Explain why magnesium hydroxide, rather than barium hydroxide, is used to treat indigestion. [2] (c) A farmer spreads 185 kg of calcium hydroxide on an acidic field. Calculate the amount of H⁺ ions, in mol, that it can neutralise. [2]

8. A power station uses 6.00 tonnes of powdered limestone per day to remove sulfur dioxide from its flue gases. The limestone is 95.0% CaCO₃ by mass.

(a) Write an equation for the reaction of calcium carbonate with sulfur dioxide. [1] (b) Calculate the mass of SO₂ removed per day, in tonnes. [3] (c) Calculate the mass of carbon dioxide released per day, in tonnes. [1] (d) Suggest why the limestone is used as a powder rather than as lumps. [1]

9. A solution contains both sodium carbonate and sodium sulfate.

(a) Describe how you would show that sulfate ions are present, and give the positive result. [2] (b) Explain why barium chloride is the reagent chosen. [1] (c) Explain why the acid is needed for this solution. Include an ionic equation. [2]

10. Excess acidified barium chloride is added to a 500 cm³ sample of spring water. The dried precipitate has a mass of 0.0817 g. Calculate the concentration of sulfate ions in the water, in mg dm⁻³. (Mr SO₄²⁻ = 96.1) [4]

11. Four unlabelled colourless solutions are magnesium nitrate, barium nitrate, sodium sulfate and sodium carbonate. Using only dilute hydrochloric acid, sodium hydroxide solution and the four solutions themselves, describe how you would identify each one. Give ionic equations for the reactions you use. [6]

12. (a) Explain why the reactions of the Group 2 metals with water become more vigorous from Mg to Ba. [3] (b) Explain why barium sulfate can be swallowed safely before an X-ray, even though barium ions are toxic. [2] (c) Barium carbonate is insoluble in water. Suggest, with an equation, why it must not be used in place of barium sulfate. [2]

Answers

1. Atomic radius increases down the group [1], because each element has one more occupied shell, so its outer electrons are further from the nucleus [1]. [2] Examiner insight: A trend with no reason earns only half the credit; “more shells” is the reason expected.

2. Barium’s outer electron is in the 6s sub-shell, further from the nucleus than strontium’s 5s electron [1]. Barium has more inner shells, so there is more shielding [1]. The attraction for the outer electron is weaker, despite barium’s larger nuclear charge, so removing it takes less energy [1]. [3] Examiner insight: Answers that say nuclear charge “decreases” contradict the chemistry; acknowledge that it increases and is outweighed.

3. (a) Both are giant metallic lattices held by attraction between positive ions and delocalised electrons [1]. Both form 2+ ions, but Ca²⁺ is smaller than Sr²⁺ [1]. The delocalised electrons are closer to the calcium ions, so the attraction is stronger and overcoming it takes more energy [1]. (b) Magnesium [1]. [4] Examiner insight: Writing about molecules or intermolecular forces in a metal answer signals the wrong model and costs credit.

4. (a) Sr(s) + 2H₂O(l) → Sr(OH)₂(aq) + H₂(g): balanced equation [1]; correct state symbols [1]. (b) Barium fizzes faster [1]. The barium disappears more quickly (or: calcium leaves a cloudier white mixture) [1]. (c) Mg + H₂O → MgO + H₂ [1]. [5] Examiner insight: “Hydrogen is produced” is a deduction, not an observation; describe what you would see.

5. (a) TiCl₄ + 2Mg → Ti + 2MgCl₂ [1]. Magnesium is the reducing agent: it changes from 0 to +2 while titanium changes from +4 to 0 [1]. (b) n(TiCl₄) = 0.950 × 10⁶ ÷ 189.9 = 5003 mol [1]. Mass of Mg = 2 × 5003 × 24.3 = 2.43 × 10⁵ g = 243 kg [1]. Mass of MgCl₂ = 2 × 5003 × 95.3 = 9.54 × 10⁵ g = 954 kg [1]. [5] Examiner insight: Show the tonnes-to-grams conversion on its own line; a slip there spoils every later value.

6. (a) Ba + 2H₂O → Ba(OH)₂ + H₂ [1]. (b) n(Ba) = 0.687 ÷ 137.3 = 5.004 × 10⁻³ mol, so n(Ba(OH)₂) in 25.0 cm³ = 5.004 × 10⁻⁴ mol [1]. n(HCl) = 2 × 5.004 × 10⁻⁴ = 1.001 × 10⁻³ mol [1]. Titre = 1.001 × 10⁻³ ÷ 0.100 = 1.00 × 10⁻² dm³ = 10.0 cm³ [1]. (c) Calcium hydroxide is only slightly soluble, so some could stay undissolved and the solution would not contain all the hydroxide formed [1]. [5] Examiner insight: The 1 : 10 scaling from 250 cm³ to 25.0 cm³ is easy to miss; write it as a separate step.

7. (a) Mg(OH)₂ < Ca(OH)₂ < Sr(OH)₂ < Ba(OH)₂ [1]. (b) Mg(OH)₂ is sparingly soluble, so it neutralises stomach acid without producing a strongly alkaline solution [1]. Ba(OH)₂ is much more soluble, giving a strongly alkaline solution, and it releases toxic Ba²⁺ ions [1]. (c) n(Ca(OH)₂) = 185 × 10³ ÷ 74.1 = 2497 mol [1]. Each mole neutralises 2 mol of H⁺, so n(H⁺) = 4.99 × 10³ mol [1]. [5] Examiner insight: In (b), one point about each compound is expected; describing only magnesium hydroxide leaves the comparison unfinished.

8. (a) CaCO₃ + SO₂ → CaSO₃ + CO₂ [1]. (b) Mass of CaCO₃ = 0.950 × 6.00 = 5.70 tonnes [1]. n(CaCO₃) = 5.70 × 10⁶ ÷ 100.1 = 5.694 × 10⁴ mol = n(SO₂) [1]. Mass of SO₂ = 5.694 × 10⁴ × 64.1 = 3.65 × 10⁶ g = 3.65 tonnes [1]. (c) n(CO₂) = 5.694 × 10⁴ mol; mass = 5.694 × 10⁴ × 44.0 = 2.51 tonnes [1]. (d) Larger surface area, so SO₂ reacts faster [1]. [6] Examiner insight: Apply the 95.0% before converting to moles; using the full 6.00 tonnes overstates every answer.

9. (a) Acidify a sample with dilute HCl and add a few drops of BaCl₂(aq) [1]. Positive result: a white precipitate of BaSO₄ [1]. (b) Barium sulfate is insoluble, so barium ions precipitate the sulfate ions [1]. (c) Carbonate ions would also form a white precipitate (BaCO₃), giving a false result; the acid removes them [1]. 2H⁺(aq) + CO₃²⁻(aq) → CO₂(g) + H₂O(l) [1]. [5] Examiner insight: In (c) the interfering ion must be named; “to stop other ions reacting” is too vague.

10. Mr(BaSO₄) = 233.4; n(BaSO₄) = 0.0817 ÷ 233.4 = 3.500 × 10⁻⁴ mol [1]. n(SO₄²⁻) = n(BaSO₄), so concentration = 3.500 × 10⁻⁴ ÷ 0.500 = 7.001 × 10⁻⁴ mol dm⁻³ [1]. Mass concentration = 7.001 × 10⁻⁴ × 96.1 = 0.06728 g dm⁻³ [1] = 67.3 mg dm⁻³ [1]. [4] Examiner insight: Divide by 0.500 dm³, not by 500, before converting grams to milligrams.

11.

• Add dilute HCl to a portion of each: only sodium carbonate fizzes [1]. 2H⁺ + CO₃²⁻ → CO₂ + H₂O [1].
• Add NaOH(aq) to fresh portions of the other three: only magnesium nitrate gives a white precipitate [1]. Mg²⁺ + 2OH⁻ → Mg(OH)₂ [1].
• Add the identified sodium carbonate solution to the last two: barium nitrate gives a white precipitate; sodium sulfate gives none [1]. Ba²⁺ + CO₃²⁻ → BaCO₃ [1]. [6]

Examiner insight: Each identification needs both the test and the observation; “add NaOH to find magnesium” without the white precipitate is incomplete.

12. (a) Atoms get larger down the group, so the outer electrons sit further from the nucleus, with more shielding [1]. Ionisation energies decrease [1]. The two outer electrons are lost more easily, so the metal is oxidised to M²⁺ more readily [1]. (b) BaSO₄ absorbs X-rays, so the digestive tract shows up on the image [1]. It is insoluble, so barium ions are not released and absorbed into the blood [1]. (c) Stomach acid would dissolve it, forming soluble barium chloride and releasing toxic Ba²⁺ [1]. BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O [1]. [7] Examiner insight: For (c) the link from acid to a soluble barium compound is the point; “it is toxic” alone does not explain the difference from BaSO₄.

Where marks are usually lost

• Leaving shielding out of ionisation energy answers, or claiming nuclear charge falls.
• Describing metals with words such as “molecules” or “intermolecular forces”.
• Giving Mg(OH)₂ instead of MgO for magnesium with steam.
• Forgetting that one Ba(OH)₂ needs two HCl, or skipping the 250 cm³ to 25.0 cm³ scaling.
• Ignoring the purity of limestone, or the 1 : 2 ratio of TiCl₄ to Mg.
• Not naming carbonate as the ion the acid removes in the sulfate test.
• Identifying an ion without stating the observation that proves it.
• Explaining the safety of BaSO₄ as “barium is harmless” rather than “BaSO₄ does not dissolve”.

Next steps

• Group 2 revision notes
• Group 2 study guide
• OxfordAQA A-Level Chemistry course hub
• Printable OxfordAQA A-Level Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.2.2.1 Group 2, the alkaline earth metals, with the Group 2 part of Required practical 3.