Edexcel International A Level Chemistry: The elements of Groups 1 and 2 (YCH11) – 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.

Here you practise Topic 8B, The elements of Groups 1 and 2, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11, Issue 1, September 2017), covering outcomes 8.11 to 8.23 and Core Practicals 3 and 4, all within the IAS unit named Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. If you get stuck, go back to the study guide or the condensed revision notes. For the wider course, see the course hub and the printable checklist; our free 10-minute diagnostics show which areas need most work.

Use these values: molar masses (g mol⁻¹) CO₂ 44.0, MgCO₃ 84.3, HCl 36.5, Ca(OH)₂ 74.1, H₂C₂O₄·2H₂O 126.0. At room temperature and pressure, one mole of gas occupies 24.0 dm³.

Questions

1. Explain why the first ionisation energy of strontium is lower than that of calcium. [3]

2. Reactions of Group 1 metals.

(a) Write an equation for the reaction of lithium with chlorine. [1] (b) Sodium is burnt in excess oxygen. Write an equation for the main product and give the oxidation number of oxygen in it. [2] (c) A small piece of potassium is dropped into water. Give two observations and write an equation with state symbols. [3]

3. Cold water attacks magnesium only very slowly, but heated magnesium reacts readily with steam.

(a) Give the equation for magnesium heated in steam, and one thing you would see. [2] (b) Explain why barium reacts much faster than magnesium with cold water. [2]

4. Oxides and hydroxides.

(a) Strontium oxide is added to water containing universal indicator. Write an equation and state the final colour of the indicator. [2] (b) Write an ionic equation for solid magnesium hydroxide reacting with dilute nitric acid. [1] (c) Suggest why lumps of calcium oxide stop reacting with dilute sulfuric acid before dissolving. [1]

5. You are given two unlabelled colourless solutions: magnesium chloride and barium chloride. Name one reagent that distinguishes them and give the result for each. [3]

6. Thermal decomposition.

(a) Write equations for the thermal decomposition of sodium nitrate and of calcium nitrate. [2] (b) Give one observation that would differ between the two reactions in (a). [1] (c) Explain why lithium carbonate decomposes on heating with a Bunsen burner but sodium carbonate does not. [3]

7. A 2.50 g sample of impure magnesium carbonate is heated to constant mass. The mass falls by 1.10 g. The impurity does not decompose.

(a) Calculate the percentage by mass of magnesium carbonate in the sample. [3] (b) Explain why the sample is heated to constant mass. [1]

8. Flame tests.

(a) Explain, in terms of electrons, how a compound produces a flame colour. [3] (b) An unknown metal chloride gives a pale green flame. Identify the metal ion. [1] (c) Why is the wire first dipped in concentrated hydrochloric acid and heated? [1]

9. Two white solids, A and B, are tested.

• A gives a brick-red flame. It fizzes with dilute hydrochloric acid, giving a gas that turns limewater milky.
• B gives no flame colour. Warmed with sodium hydroxide solution, it gives a gas that turns damp red litmus blue. A solution of B gives a white precipitate with dilute hydrochloric acid and barium chloride solution.

(a) Identify A and B. [2] (b) Write ionic equations for: A with the acid; B with hydroxide ions; the formation of the white precipitate. Write a full equation for the limewater reaction. [4] (c) Explain why hydrochloric acid is added before the barium chloride. [1]

10. 10.0 cm³ of concentrated hydrochloric acid is diluted to 250 cm³ with distilled water. The diluted acid is run from a burette into 25.0 cm³ portions of 0.0800 mol dm⁻³ sodium carbonate solution with methyl orange indicator. Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O. Titres (cm³): rough 23.40; then 22.90, 22.80, 22.85.

(a) State the indicator colour change you would see at the end point. [1] (b) Calculate the mean titre. [1] (c) Calculate the concentration of the diluted acid. [3] (d) Calculate the concentration of the concentrated acid in mol dm⁻³ and in g dm⁻³. [2]

11. 1.58 g of ethanedioic acid crystals, H₂C₂O₄·2H₂O, is used to make 250 cm³ of standard solution. 25.0 cm³ portions need a mean titre of 24.10 cm³ of sodium hydroxide, with phenolphthalein. H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. Use these uncertainties: balance ±0.005 g per reading (two readings), burette ±0.05 cm³ per reading, pipette ±0.06 cm³, volumetric flask ±0.3 cm³.

(a) Describe how the standard solution is prepared from the weighed solid. [3] (b) Calculate the concentration of the sodium hydroxide. [3] (c) Estimate the overall percentage uncertainty and the absolute uncertainty in your answer to (b). [2] (d) Suggest one change that would reduce the largest uncertainty. [1]

12. A 25.0 cm³ portion of filtered, saturated calcium hydroxide solution is neutralised by 21.60 cm³ of 0.0500 mol dm⁻³ hydrochloric acid.

(a) Calculate the solubility of calcium hydroxide in mol dm⁻³ and in g dm⁻³. [3] (b) Predict, with a reason, how the titre would change if saturated magnesium hydroxide solution were used instead. [2]

Answers

1. Strontium has an extra occupied shell, so its outermost electrons are more distant from the nucleus [1]. Inner electrons screen them more [1]. Although the nuclear charge is larger, the net attraction falls, so removing an electron takes less energy [1]. Examiner insight: “Bigger atom” alone is often not enough; name distance, shielding and the weaker attraction.

2. (a) 2Li + Cl₂ → 2LiCl [1] (b) 2Na + O₂ → Na₂O₂ [1]; oxygen −1 [1] (c) Any two of: floats, fizzes, moves about, burns with a lilac flame [1][1]. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g) [1] Examiner insight: Naming hydrogen says what forms, not what you see; “fizzing” or “bubbles” is the observation that gains credit.

3. (a) Mg(s) + H₂O(g) → MgO(s) + H₂(g) [1]; bright white glow or white solid forms [1] (b) Barium’s two outer electrons are more distant from the nucleus and better screened, so they are lost more easily [1]; barium is therefore the stronger reducing agent and reacts faster [1]. Examiner insight: Giving Mg(OH)₂ as the product with steam usually costs the equation mark.

4. (a) SrO + H₂O → Sr(OH)₂ [1]; blue or purple (alkaline) [1] (b) Mg(OH)₂(s) + 2H⁺(aq) → Mg²⁺(aq) + 2H₂O(l) [1] (c) Poorly soluble calcium sulfate coats the solid, so the acid cannot reach it [1]. Examiner insight: In (b), keep the solid as Mg(OH)₂(s); OH⁻(aq) implies it had dissolved.

5. Add dilute sulfuric acid (or sodium sulfate solution) [1]. Barium chloride: white precipitate (BaSO₄) [1]. Magnesium chloride: no visible change, as MgSO₄ is soluble [1]. Examiner insight: Describe what you would see for both solutions; “no visible change” is safer than “no reaction”.

6. (a) 2NaNO₃ → 2NaNO₂ + O₂ [1]; 2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂ [1] (b) Brown gas is seen only with calcium nitrate [1]. (c) Li⁺ is smaller than Na⁺ (same charge), so it has a higher charge density [1]. It polarises the carbonate ion’s electron cloud more [1], weakening its C–O bonds so it breaks down at a lower temperature [1]. Examiner insight: Referring to “lithium atoms” rather than ions usually loses the first mark in (c).

7. (a) n(CO₂) = 1.10 ÷ 44.0 = 0.0250 mol [1]. Mass MgCO₃ = 0.0250 × 84.3 = 2.11 g [1]. Percentage = 2.11 ÷ 2.50 × 100 = 84.3% [1] (b) To make sure all the magnesium carbonate has decomposed [1]. Examiner insight: Dividing 1.10 by 2.50 gives the mass lost, not the purity; convert CO₂ back to MgCO₃ first.

8. (a) Heat energy promotes electrons to higher energy levels [1]. They fall back to lower levels, releasing energy as light [1]. The energy gaps are fixed for each element, so specific frequencies (colours) are emitted [1]. (b) Ba²⁺ [1] (c) To remove traces of other compounds that would add their own colour [1]. Examiner insight: The energy comes from the flame; light is given out as electrons return, not absorbed.

9. (a) A: calcium carbonate, CaCO₃ [1]. B: ammonium sulfate, (NH₄)₂SO₄ [1] (b) CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l) [1]. NH₄⁺(aq) + OH⁻(aq) → H₂O(l) + NH₃(g) [1]. Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) [1]. Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l) [1] (c) Carbonate ions would otherwise react with Ba²⁺ to form white BaCO₃, a false positive [1]. Examiner insight: Each identification needs both ions; “a calcium compound” alone is incomplete.

10. (a) Yellow to orange [1] (b) Concordant titres 22.90, 22.80, 22.85; mean = 22.85 cm³ [1] (c) n(Na₂CO₃) = 0.0800 × 0.0250 = 2.00 × 10⁻³ mol [1]. n(HCl) = 4.00 × 10⁻³ mol [1]. c = 4.00 × 10⁻³ ÷ 0.02285 = 0.175 mol dm⁻³ [1] (d) Dilution factor 250 ÷ 10.0 = 25, so c = 0.17505 × 25 = 4.38 mol dm⁻³ [1]; 4.376 × 36.5 = 160 g dm⁻³ [1] Examiner insight: Missing the 1 : 2 ratio or the dilution factor are the usual slips; showing each step keeps method credit available.

11. (a) Weigh the crystals by difference, then dissolve them in a little distilled water in a beaker [1]. Pour into a 250 cm³ volumetric flask, washing the beaker and stirring rod into it [1]. Add water until the bottom of the meniscus touches the graduation mark, then stopper and invert [1]. (b) n(acid) = 1.58 ÷ 126.0 = 0.01254 mol, so 1.254 × 10⁻³ mol in 25.0 cm³ [1]. n(NaOH) = 2 × 1.254 × 10⁻³ = 2.508 × 10⁻³ mol [1]. c = 2.508 × 10⁻³ ÷ 0.02410 = 0.104 mol dm⁻³ [1] (c) Mass 0.63% + titre 0.41% + pipette 0.24% + flask 0.12% = 1.4% [1]; 1.4% of 0.104 = ±0.0015 mol dm⁻³ [1] (d) Weigh a larger mass, or use a three-decimal-place balance [1]. Examiner insight: A titre uses two readings (±0.10 cm³), and the overall value adds percentages, not absolute uncertainties.

12. (a) n(HCl) = 0.0500 × 0.02160 = 1.08 × 10⁻³ mol [1]. n(Ca(OH)₂) = 5.40 × 10⁻⁴ mol; c = 5.40 × 10⁻⁴ ÷ 0.0250 = 0.0216 mol dm⁻³ [1]. × 74.1 = 1.60 g dm⁻³ [1] (b) Much smaller titre [1], because magnesium hydroxide is far less soluble [1]. Examiner insight: The 2 : 1 ratio between HCl and Ca(OH)₂ is easy to miss; write the equation before you calculate.

Where marks are usually lost

• Giving the nuclear charge as the reason for a fall in ionisation energy.
• Writing Na₂O for sodium in excess oxygen, or K₂O for potassium.
• Treating Group 1 and Group 2 nitrates as if they decompose the same way.
• Saying the cation “attracts” the anion, instead of polarising its electron cloud.
• Averaging non-concordant titres or the rough titre.
• Forgetting to scale up for a dilution or for titrating 25.0 cm³ of a 250 cm³ solution.

Next steps

• Go back over the revision notes.
• Revisit weak areas in the study guide.
• Plan the rest of your revision with the course hub and the printable checklist.
• Take all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Scope: Topic 8B, outcomes 8.11 to 8.23.