A-LevelChemistryGroup 29701/41 · Oct/Nov 2020 · Q215 Marks

(a) The lattice energies of three ionic compounds are given. compound lattice energy/kJ mol-¹ LiF(s) -1022 CaO(s) -3513 SrO(s) -3310

(a)(i)Define the term lattice energy.[2]
(a)(ii)Explain why the lattice energy of CaO is more exothermic than the lattice energy of LiF.[1]
(a)(iii)Use the data in the table to estimate approximate values for the lattice energies of magnesium oxide and barium oxide.[1]
(b)(i)Write an equation for the reaction between BaO and H2O. Include state symbols.[1]
(b)(ii)State and explain how the solubilities of the hydroxides of the Group 2 elements vary down the group.[4]
(c)Use the following data and relevant data from the Data Booklet to calculate a value for the lattice energy of magnesium fluoride, MgF2(s). You might find it helpful to construct an energy cycle. Show your working. electron affinity of F(g) = -348 kJ mol-¹ enthalpy change of atomisation of Mg(s) = +147 kJ mol-¹ enthalpy change of formation of MgF2(s) = -1102 kJ mol-¹[3]
(d)(i)Define the term electron affinity.[2]
(d)(ii)The electron affinity of carbon, C(g), is -120 kJ mol-¹. Suggest an explanation for the difference between the electron affinity of fluorine and the electron affinity of carbon.[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

[1] M1 energy released when 1 mole of an ionic compound is formed
[1] M2 from gaseous ions (under standard conditions)
[1] Ca2+ & O2- have a higher charge / charge density (than Li+ and F-)
[1] MgO –3600 or more negative AND BaO –3200 or less negative BOTH
[1] BaO(s) + H2O(l) → Ba(OH)2(aq)
[1] M1 (solubility) increases (down the group)
[1] M2 both ∆Hlatt and ∆Hhyd become less exothermic / less negative
[1] M3 ∆Hlatt changes more / is dominant factor
[1] M4 ∆HSol becomes more negative / more exothermic
[1] M1: Use of 2 × −348 (EA F) and +158 (bond energy of F2)
[1] M2: Use of +147 (at Mg) and +736 and +1450 (IEs of Mg)
[1] M3: evaluation and calculation of their answer (-1102−(147 + 158 + 736 + 1450 – 696)) = −2897 (kJ mol-1) ecf
[1] (energy change) when an / one electron is added to
[1] each atom / ion in one mole of gaseous atoms / ions
[1] F has greater nuclear charge / more protons
[1] AND greater attraction between F atom / nucleus and the electrons

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