Edexcel International A Level Chemistry: Lattice energy (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.

These questions test part 12B, Lattice energy (outcomes 12.12 to 12.19), of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. Part 12B is IA2 content; its home is Unit 4: Rates, Equilibria and Further Organic Chemistry. If you have not met the ideas yet, start with the lattice energy study guide and the lattice energy revision notes. Track each outcome on the Edexcel A Level Chemistry checklist; the course hub links every other topic.

Work only from the data printed in each question. Enthalpy changes are in kJ mol⁻¹ and entropies in J mol⁻¹ K⁻¹.

Questions

1. State what is meant by the standard enthalpy change of atomisation of an element. Write an equation, including state symbols, representing ΔatH for iodine. [3]

2. This question is about the electron affinities of sulfur.

(a) Define the first electron affinity of an element. [2] (b) Write equations, with state symbols, for the first and second electron affinities of sulfur. [2] (c) Explain why the second electron affinity of sulfur is endothermic. [1]

3. (a) Define lattice energy as it is used in this specification. [2] (b) Write an equation for the lattice energy of potassium oxide. [1] (c) Define the enthalpy change of hydration of an ion, and write an equation for the hydration of the aluminium ion. [2]

4. Use these values to construct a Born-Haber cycle for barium iodide, BaI₂, and calculate its lattice energy.

ΔfH(BaI₂) = −602; ΔatH(Ba) = +180; first ionisation energy of Ba = +503; second ionisation energy of Ba = +965; ΔatH(I) = +107; first electron affinity of I = −295. [6]

5. Lithium fluoride has ΔfH(LiF) = −616 and a lattice energy of −1046. Calculate the first electron affinity of fluorine, given ΔatH(Li) = +159, first ionisation energy of Li = +520 and ΔatH(F) = +79. [3]

6. The table compares lattice energies (kJ mol⁻¹) for three iodides.

Compound	Experimental (Born-Haber)	Theoretical (electrostatic)
CsI	−604	−599
AgI	−889	−778
MgI₂	−2320	−1950

(a) Identify the iodide with the greatest degree of covalent bonding. Justify your choice with a calculation. [2] (b) All three compounds contain the same anion. Explain why the extent of covalent character differs so much between CsI and MgI₂. [3]

7. Draw a labelled energy level diagram for dissolving barium chloride, BaCl₂, and use it to calculate the enthalpy change of solution.

Lattice energy of BaCl₂ = −2056; ΔhydH(Ba²⁺) = −1341; ΔhydH(Cl⁻) = −364. [4]

8. Caesium fluoride dissolves exothermically: ΔsolH(CsF) = −37. The lattice energy of CsF is −745 and ΔhydH(F⁻) = −506. Calculate ΔhydH(Cs⁺). [3]

9. (a) Explain why ΔhydH(Ca²⁺) is far more exothermic than ΔhydH(K⁺). [2] (b) Predict, with a reason, which of calcium oxide and calcium sulfide has the more exothermic lattice energy. [2]

10. Data at 298 K for dissolving two salts:

Salt	ΔsolH / kJ mol⁻¹	ΔSsystem / J mol⁻¹ K⁻¹
MgSO₄	−91.2	−209.6
CaF₂	+19.9	−149.6

(a) Calculate ΔSsurroundings and ΔStotal for dissolving each salt at 298 K. [4] (b) Use your answers to predict the solubility of each salt in water. [1] (c) Explain why ΔSsystem is negative for both salts, and why MgSO₄ dissolves even so. [2]

11. In Unit 2 you learned how the solubility of the Group 2 hydroxides and sulfates changes down the group. Explain both trends in terms of lattice energy and enthalpy change of hydration. [6]

12. Calcium oxide, CaO, is an ionic solid. Use these values:

ΔfH(CaO) = −635.1; ΔatH(Ca) = +178.2; first ionisation energy of Ca = +589.8; second ionisation energy of Ca = +1145.4; ΔatH(O) = +249.2; first electron affinity of O = −141.0; lattice energy of CaO = −3401.

(a) Write equations, with state symbols, for the atomisation of oxygen and for the second electron affinity of oxygen. [2] (b) Calculate the second electron affinity of oxygen. [3] (c) Explain the sign of your answer to (b). [1] (d) A theoretical value of −3371 has been calculated for the lattice energy of CaO. Comment on the bonding in CaO. [2]

Answers

1. The enthalpy change for producing 1 mol of gaseous atoms [1] from the element in its standard state, under standard conditions [1]. ½I₂(s) → I(g) [1]. [3] Examiner insight: the equation must start from solid iodine and make one mole of atoms; I₂(s) → 2I(g) represents twice the quantity and would not normally gain the equation mark.

2. (a) The enthalpy change when 1 mol of gaseous atoms [1] each gain one electron to form 1 mol of gaseous 1− ions [1]. (b) S(g) + e⁻ → S⁻(g) [1]; S⁻(g) + e⁻ → S²⁻(g) [1]. (c) S⁻ is already negative, so it repels the incoming electron and energy has to be put in [1]. Examiner insight: in (a), “gaseous” is needed for both the atoms and the ions; in (b), every species needs a state symbol.

3. (a) The enthalpy change when 1 mol of an ionic solid is made [1] from its gaseous ions [1]. (b) 2K⁺(g) + O²⁻(g) → K₂O(s) [1] (c) The enthalpy change as 1 mol of an ion in the gas phase is hydrated, giving 1 mol of that ion in aqueous solution [1]. Al³⁺(g) + aq → Al³⁺(aq) [1]. Examiner insight: describing lattice energy as breaking a solid into ions gives the reverse process, so it does not match this specification’s exothermic definition and loses the definition marks.

4. Cycle: Ba(s) → Ba(g) → Ba⁺(g) → Ba²⁺(g); I₂(s) → 2I(g) → 2I⁻(g); then Ba²⁺(g) + 2I⁻(g) → BaI₂(s), with ΔfH linking Ba(s) + I₂(s) to BaI₂(s) [1]. Every species has a state symbol and every arrow points the correct way [1]. Barium steps: 180 + 503 + 965 = +1648 [1]. Iodine steps: 2(107) + 2(−295) = −376 [1]. −602 = 1648 − 376 + lattice energy, so lattice energy = −602 − 1272 [1] = −1874 kJ mol⁻¹ [1]. [6] Examiner insight: both barium ionisation energies and two moles of iodine atoms and electrons must appear; using a single iodine term is a quick way to lose two marks.

5. −616 = 159 + 520 + 79 + EA + (−1046) [1]; the known steps total −288, so EA = −616 + 288 [1] = −328 kJ mol⁻¹ [1]. [3] Examiner insight: writing the whole Hess equation before rearranging makes a sign slip easy for you, and the examiner, to spot.

6. (a) MgI₂ [1]: its experimental value is 370 kJ mol⁻¹ (19.0%) more exothermic than theory, against 111 (14.3%) for AgI and 5 (0.8%) for CsI [1]. (b) Mg²⁺ is small and doubly charged, so it has a much higher charge density and polarising power than Cs⁺ [1]. I⁻ is large, so its electron cloud is easily distorted [1]. Mg²⁺ draws iodide electron density into the region between the ions, giving partial covalent bonding; large, singly charged Cs⁺ barely does this [1]. Examiner insight: since the anion is the same throughout, the marks in (b) depend on comparing the two cations; a general statement that “MgI₂ is more covalent” earns little.

7. Diagram: Ba²⁺(g) + 2Cl⁻(g) at the top; BaCl₂(s) 2056 below it; Ba²⁺(aq) + 2Cl⁻(aq) 2069 below the gaseous ions [1]. Hydration total: −1341 + 2(−364) = −2069 [1]. ΔsolH = +2056 − 2069 [1] = −13 kJ mol⁻¹ [1]. [4] Examiner insight: the lattice energy arrow points down from the gaseous ions to the solid; drawing ΔsolH from the solid to the aqueous ions makes its sign obvious.

8. ΔsolH = −(lattice energy) + ΔhydH(Cs⁺) + ΔhydH(F⁻) [1]. −37 = 745 + ΔhydH(Cs⁺) − 506 [1]. ΔhydH(Cs⁺) = −276 kJ mol⁻¹ [1]. [3] Examiner insight: hydration is always exothermic, so a positive answer is a signal to recheck your signs before moving on.

9. (a) Ca²⁺ has a 2+ charge, double that of K⁺ [1], and a smaller radius, so it pulls the partially negative oxygen ends of water molecules in more strongly [1]. (b) Calcium oxide [1]: O²⁻ is smaller than S²⁻ with the same charge, so the ions sit closer together and attract more strongly [1]. Examiner insight: in (a) give both charge and radius; naming just one factor usually earns only one of the two marks.

10. (a) MgSO₄: ΔSsurroundings = −(−91 200) / 298 = +306.0 [1]; ΔStotal = −209.6 + 306.0 = +96.4 J mol⁻¹ K⁻¹ [1]. CaF₂: ΔSsurroundings = −19 900 / 298 = −66.8 [1]; ΔStotal = −149.6 − 66.8 = −216.4 J mol⁻¹ K⁻¹ [1]. (b) MgSO₄ is soluble (ΔStotal positive); CaF₂ is insoluble (ΔStotal negative) [1]. (c) Both salts contain small or doubly charged ions that hold many water molecules in ordered shells, which outweighs the disorder gained as the lattice breaks up [1]. Dissolving MgSO₄ is strongly exothermic, so the large positive ΔSsurroundings more than cancels the negative ΔSsystem [1]. Examiner insight: convert ΔsolH to J mol⁻¹ before dividing by T; leaving it in kJ makes ΔSsurroundings a thousand times too small and reverses the MgSO₄ conclusion.

11. Sulfates become less soluble down the group [1]. The sulfate ion is large, so as the cation radius increases the lattice energy becomes only slightly less exothermic [1]. Cation hydration enthalpy becomes much less exothermic, so ΔsolH becomes more endothermic [1]. Hydroxides become more soluble down the group [1]. The hydroxide ion is small, so a larger cation makes the lattice energy much less exothermic [1]. Lattice energy falls faster than hydration enthalpy, so ΔsolH becomes more favourable [1]. [6] Examiner insight: each trend needs the two quantities compared; saying how lattice energy changes without saying what happens to hydration leaves the explanation incomplete.

12. (a) ½O₂(g) → O(g) [1]; O⁻(g) + e⁻ → O²⁻(g) [1]. (b) −635.1 = 178.2 + 589.8 + 1145.4 + 249.2 + (−141.0) + EA₂ + (−3401) [1]. The known steps total −1379.4 [1], so EA₂ = −635.1 + 1379.4 = +744.3 kJ mol⁻¹ [1]. (c) Positive (endothermic): the incoming electron is repelled by the negatively charged O⁻ ion [1]. (d) The experimental value is only 30 kJ mol⁻¹ (0.9%) more exothermic than the theoretical value [1], so CaO is very close to purely ionic, with little polarisation of the oxide ion [1]. Examiner insight: a “comment” needs evidence from the data; quoting the size of the gap and linking it to the ionic model is what earns both marks in (d).

Where marks are usually lost

• Treating lattice energy as an endothermic break-up instead of the formation of 1 mol of solid from gaseous ions.
• Writing I₂(s) → 2I(g) or Cl₂(g) → 2Cl(g) for atomisation, which describes two moles of atoms.
• Leaving out the second ionisation energy of a Group 2 metal.
• Using one electron affinity and one atomisation term for a compound with two anions.
• Missing state symbols, especially (g) on ions in lattice energy and electron affinity equations.
• Dividing a ΔsolH in kJ by T and adding the result straight to an entropy in J mol⁻¹ K⁻¹.
• Assuming that endothermic dissolving means a salt is insoluble.
• Explaining covalent character without mentioning polarisation of the anion by the cation.
• Giving the Group 2 solubility trends without comparing changes in both lattice and hydration enthalpies.

Next steps

• Recap the definitions with the lattice energy revision notes.
• Reread any weak area in the lattice energy study guide.
• See every Edexcel topic on the course hub and track progress on the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source document: Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Section tested here: 12B Lattice energy (outcomes 12.12 to 12.19), within Topic 12, Entropy and Energetics.