JPJC 2023 Prelim P2 Answers

© Jurong Pioneer Junior College [Turn Over Answers to 2023 JC2 Preliminary Examination H2 Chemistry (9729) Paper 2 1 (a) (i) Since F is more electronegative than Cl, C−F bond is more polar. [1m] (ii) Both are polar and thus have permanent dipole -permanent dipole (pd -pd) attractions and instantaneous dipole -induced dipole attractions between molecules. Since CH3Cl has more electrons per molecule than CH3F, more energy is required to overcome the stronger instantaneous dipole-induced dipole (id- id) attractions between molecules in CH3Cl than id-id attractions between molecules in CH3F. Thus, CH3Cl has higher boiling point. [1m] (iii) CH3Cl and CH3CH2F are isoelectronic so id-id attractions are similar. The additional electron-donating CH3 group attached to C of C−F bond makes C−F bond less polarised than C−Cl bond. Hence, less energy is required to overcome the weaker permanent dipole -permanent dipole attractions between molecules in CH3CH2F than pd-pd attractions between molecules in CH3Cl. Thus, CH3CH2F has lower boiling point. [1m] (b) Lone pairs in axial positions. Lone pairs in equatorial positions. Lone pairs in axial and equatorial positions. number of lone pairs at 90 relative to one another 0 0 1 number of bond pair and lone pair at 90 relative to one another 6 4 3 relative stability most stable [1m] least stable [1m] [1m]

2 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2023 (c) (i) [3m] (ii) E cell = (+2.01) – (+0.54) = +1.47 V [1m] G = −nFE cell = −(2)(96500)(+1.47) = −284 kJ mol−1 [1m] (iii) From the equation, S  0 as the reaction proceeds with no change in number of moles of aqueous particles and hence the reaction is enthalpy driven. [1m] −   −  ? 0 0 ve =TG H S Since G < 0, H must be negative and thus the reaction is exothermic. [1m] 2 (a) (i) Since  (q q ) (r r )LE +− +− + and (ionic) radius increases from Mg2+ to Ba2+ [1m], the magnitude of HLE of Group 2 sulfates decreases from MgSO 4 to BaSO4. [1m] (ii) Since (ionic)) radius increases and thus charge density decreases from Mg2+ to Ba2+ [1m], strength of ion-dipole interactions formed between M2+ with water molecules decreases. Thus Hhyd of M 2+ becomes less negative (i.e. less exothermic) from Mg2+ to Ba2+. (iii) Hsol = − HLE + Hhyd (M2+) + Hhyd (SO42−) Both Hhyd(M2+) and HLE become less exothermic/ less negative. HLE becomes less exothermic/ less negative by a smaller extent than that of Hhyd(M2+). [1m] Thus Hsol becomes more endothermic / less exothermic and h ence solubility decreases from MgSO4 to BaSO4 as shown in Table 2.1. [1m] (iv) Solubility of CaSO4 = − 34.7 10 1001000 = 4.7 × 10−2 mol dm−3 Ksp = [Ca2+][SO42−] = (4.7 × 10−2)2 = 2.209 × 10−3 mol2 dm−6 [1m] Ionic Product =         0.0100 100 0.0200 100 200 200 = 5.00 × 10−5 mol2 dm−6 Since ionic product < K