A-LevelChemistryAtomic structure9701/43 · Oct/Nov 2021 · Q312 Marks

(a) Define the term electron affinity.

(a)Define the term electron affinity.[2]
(b)Write an equation for the process corresponding to the second ionisation energy of calcium. Include state symbols.[1]
(c)Some data relating to calcium and oxygen are listed. Select relevant data from this list for your answers to parts (c), (d) and (e). [Table of processes and values: first ionisation energy of oxygen +1310, second ionisation energy of oxygen +3390, first electron affinity of oxygen -142, second electron affinity of oxygen +844, enthalpy change for ½O₂(g) + 2e⁻ → O²⁻(g) +951, enthalpy change for Ca(s) → Ca²⁺(g) + 2e⁻ +1933, lattice energy of CaO(s) -3517] Oxygen exists as O₂ molecules. Use the data in this question to calculate a value for the bond energy of the O=O bond. Show all your working.[3]
(d)(i)Suggest why the first electron affinity of oxygen is negative.[1]
(d)(ii)Suggest why the second electron affinity of oxygen is positive.[1]
(e)Calculate the enthalpy of formation of calcium oxide, CaO(s).[2]
(f)The lattice energy of lithium fluoride, LiF(s), is –1022kJ mol⁻¹. Identify the factor that causes the lattice energy of calcium oxide to be more exothermic than that of lithium fluoride. Explain why this factor causes the difference in lattice energies.[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[1] enthalpy/energy change
[1] one mole of electrons gained
[1] by one mole of atoms
[1] gaseous (atoms)
[1] Ca+(g) → Ca²+(g) + e-
[1] M1: selecting correct data 951, 844, 142 only
[1] M2: evaluation to give 249 (ΔHatom) OR 2(951) = BE – 2(142) + 2(844)
[1] M3: evaluation to 498 (2 × 249) ecf M2 951 = ΔHatom -142 + 844 ΔHatom = 249 BE = 498 (kJ mol⁻¹)
[1] attraction between nucleus / protons / nuclear charge and electron
[1] repulsion between 1⁻ ion / electrons of O- and electron
[1] M1: selecting correct data 951, 1933, 3517 only (ignore signs)
[1] M2: evaluation to give –633 (ΔH) ecf ΔHf = 951 + 1933 – 3517 = −633 (kJ mol⁻¹)
[1] ionic charge / charge density (of the ions)
[1] greater (attractive) force between the ions

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