2024 H1 Prelim Answer_HCI

2024 Prelim Exam H1 Chemistry For internal circulation only 2024 Preliminary Examination H1 Chemistry (8873) Suggested Solutions Paper 1 1 C 11 A 21 A 2 B 12 B 22 C 3 C 13 C 23 C 4 D 14 C 24 D 5 B 15 A 25 A 6 C 16 D 26 B 7 A 17 D 27 D 8 A 18 B 28 A 9 D 19 A 29 A 10 B 20 B 30 D Paper 2 Section A 1(a) Number of neutrons 78 [0.5] Number of electrongs 52 [0.5] (b) Te+ ion will be deflected towards the negatively charged plate while the electron will be deflected towards the positively charged plate. Te atom will not be deflected. [1] The angle of deflection of Te+ is much smaller than that for the electron. [1] (c)(i) Te (g) → Te+ (g) + e− [1] (c)(ii) The first ionisation energy increases across Period 3 [0.5] as the nuclear charge increases [0.5] while the shielding effect remains relatively constant or effective nuclear charge increases [0.5] and so more energy is required to remove the first outermost electron across the period. [0.5] (c)(iii) The paired electrons in the p orbital of Te experiences inter-electronic repulsion such that it takes less energy to remove the first outermost electron compared to the preceding element. [1] Note: 4d electrons are poor at shielding the outer electrons from the nuclear charge.

2 (d)(i) [0.5] general increase from 1 to 4 [0.5] general increase from 5 to 6 [0.5] jump from 4 to 5 [0.5] bigger jump from 6 to 7 (d)(ii) As electrons are successively removed from the Te atom, nuclear charge remains the same or number of protons remain the same [0.5] but the remaining electrons experiences stronger attraction by nucleus or shielding effect decreases or increase in effective nuclear charge.[0.5] (e) There are 6 bond pairs and 0 lone pairs of electrons around Te. [1] The 6 electron pairs are arranged as far apart as possible around Te to minimise repulsion and maximise stability. [1] (f)(i) TeF6 + 6H2O → Te(OH)6+ 6HF [1] (f)(ii) hydrolysis [1] 2(a)(i) NaCl [0.5] MgCl2 [0.5] [-0.5] for wrong entry (a)(ii) AlCl3 and SiCl4 and PCl5 [0.5] for any 2 correct ones [1] for all 3 (b)(i) –58.2 = 8 (264) + 4 (244) – 4 [ 264 + 2 (S–Cl) ] [1] quote of the correct values = 261 kJ mol–1 (3 s.f.) [1] multipliers [1] bond break – bond form, ecf (b)(ii) The enthalpy change when 1 mole of a substance is formed from its constituent elements in their standard states at 298 K and 1 bar. [1] (b)(iii) –58.2 S8(s) + 4Cl2(g) 4S2Cl2 (l) +4Cl2(g) 8 x ΔH 4(–40.6) +4Cl2(g) 8SCl2(l)

3 8 × Hf = –58.2 + 4(–40.6) [1] correct values at the arrows and = –27.6 kJ mol–1 [1] multipliers [1] apply the correct Hess’ Law, ecf (c)(i) S2Cl2 is oxidised to SCl2; oxidation number of sulfur increases from +1 to +2. [1] Cl2 is reduced to SCl2; oxidation number of chlorine decreases from 0 to –1. [1] (c)(ii) Cl2 + 2e– → 2Cl– Charge on 1 m