2024 Prelims RI H2 Chem P2 (Ans)

© Raffles Institution 2024 9729/02/S/24 2024 Y6 H2 Chemistry Preliminary Exams Paper 2 – Suggested Solutions 1(a)(i) B3+(g) ⎯→ B4+(g) + e− 1(a)(ii) B3+: 1s2 C3+: 1s2 2s1 C3+ has one more electron shell than B3+; the 2s electron in C3+ is further away from the nucleus than the 1s electron in B3+. Shielding experienced by the 2s electron in C3+ is greater than the 1s electron in B3+. Despite the greater nuclear charge in C 3+, electrostatic attraction between the nucleus and the 2s electron in C3+ is weaker than the 1s electron in B3+. Less energy is required to remove the 2s electron in C3+ compared to the 1s electron in B3+. Thus, B has a higher fourth ionisation energy than C. 1(b)(i) 1(b)(ii) shape: trigonal planar bond angle: 120 1(b)(iii) ∆Hr = energy required to break bonds – energy released from bonds formed = 3BE(B–H) + 3 2 BE(O=O) – BE(B=O) – BE(B–O) – 3BE(O–H) = 3(330) + 3 2 (496) – 837 – 536 – 3(460) = –1019 kJ mol–1 = –1020 kJ mol–1 (3 s.f.) 1(b)(iv) The standard enthalpy change of combustion of borane is more exothermic than ∆Hr as energy is released to condense steam to water and to convert B2O3 from gaseous to solid state. 1(c)(i) Aluminium in AlCl3 has a vacant, low-lying orbital to accept a lone pair of electrons from the chlorine atom of another AlCl3 molecule. 1(c)(ii) sp2 to sp3 1(c)(iii) There is less electron density in each B–Hb bond / fewer shared bonding electrons, hence resulting in weaker attraction to the nuclei. 1(d)(i) B H H H

© Raffles Institution 2024 9729/02/S/24 1(d)(ii) N is more electronegative than B, and this reduces the extent of electron delocalisation (partially delocalised) compared to benzene. 1(d)(iii) 2(a)(i) NO3– + 10H+ + 8e− ⇌ NH4+ + 3H2O E = +0.87 V O2 + 4H+ + 4e− ⇌ 2H2O E = +1.23 V Ecell = +1.23 – (+0.87) = +0.36 V > 0 (reaction is spontaneous) 2(a)(ii) The beneficial bacteria provide enzymes which act as biological catalysts to speed up the nitrification process. 2(b)(i) Day 9 or 10 2(b)(ii) Day 29 2(b)(iii) [NO3−] before water change = 35 ppm [NO3−] after water change = 100-25 100 × 35 = 26.3 ppm 2(b)(iv) Using Henderson-Hasselbalch equation, pH = pKa + lg( [NH3] [NH4 +]), 7.4 = 9.25 + lg( [NH3] [NH4 +]) [NH3] [NH4 +] =107.4-9.25 = 0.014125 [NH3] = 0.014125[NH4 +] z = [NH3] TAN = [NH3] [NH3] + [NH4 +] = 0.014125[NH4 +] 0.014125[NH4 +] + [NH4 +] = 0.014125 0.014125 + 1 = 0.0139 OR Let mole fraction of NH3 be z, hence mole fraction of NH4 + = 1 – z z 1 – z = 0.014125 z = 0.014125 1 + 0.014125 = 0.0139 2(c)(i) When small amount of OH−(aq) is added, HCO3−(aq) + OH−(aq) ⎯→ CO32−(aq) + H2O(l) When small amount of H3O+(aq) is added, CO32−(aq) + H3O+(aq) ⎯→ HCO3−(aq) + H2O(l) 2(c)(ii) Tank water with a higher carbonate hardness has a higher buffer capacity to partially offset increase in [H+] due to the nitrification process.

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