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.

© Raffles Institution 2024 9729/02/S/24 2(d)(i) n(S2O32−) = 13.40 / 1000  0.0100 = 0.000134 mol mole ratio of O2 : I2 : S2O32− = 0.5 : 1 : 2 = 1 : 2 : 4 n(O2) = 0.000134 / 4 = 0.0000335 mol [O2] = 0.0000335 / 100  1000 = 3.35  10−4 mol dm−3 (within recommended range) 2(d)(ii) Oxidation: N3−(aq) ⎯→ 3 2 N2(g) + e− Reduction: 8H+(aq) + 2NO2−(aq) + 6e− ⎯→ N2(g) + 4H2O(l) Overall: 4H+(aq) + 3N3−(aq) + NO2−(aq) ⎯→ 5N2(g) + 2H2O(l) 2(e) Decreasing basicity: N(1) > N(2) > N(3) N(1) is a primary amine. The electron-donating –CH2– group increases the electron density on N( 1), making the lone pair of electrons on N more readily available to form a dative covalent bond with a proton. N(2) is an aromatic amine. The orbital containing the lone pair of electrons on the nitrogen atom overlaps with the π electron cloud of the benzene ring and the lone pair of electrons is delocalised and is less available to form a dative covalent bond with a proton. N(3) is a sulfonamide which has similar basicity as amides. It has the lowest basicity among the three nitrogen-containing groups because the orbital containing the lone pair of electrons on the N atom overlaps with the  electron cloud of the adjacent S=O group and the lone pair of electrons is delocalised to a greater extent, and hence least/not available to form a dative covalent bond with a proton. 3(a)(i) 3(a)(ii) The carbocation that is produced in the reaction is trigonal planar with respect to the positively charged carbon. Cl– can attack the positively charged carbon from either side of the trigonal plane with equal likelihood . This results in the formation of a racemic mixture. The optical activity of each enantiomer cancels out each other. The product mixture is unable to rotate plane-polarised light.

© Raffles Institution 2024 9729/02/S/24 3(a)(iii) For visualization, 3(b)(i) In this reaction, a secondary carbocation, F’, rearranges to form a tertiary carbocation, G’ which is more stable because it has more electron donating alkyl groups bonded to the positively charged carbon. This helps to disperse the positive charge. G will be formed in a larger proportion. 3(b)(ii) 3(b)(iii) Although both carbocations are tertiary carbocations, L’ is more stable as there is less repulsion between bond pairs of electrons in the cyclopentane ring, compared to K’. L’ is formed faster and hence, L is the major product. 3(c)(i) Step 2: PCl5 / PCl3 / SOCl2, or dry HCl, ZnCl2, heat 3(c)(ii) 1-chloro-3-methylbutane

© Raffles Institution 2024 9729/02/S/24 3(c)(iii) Electrophilic substitution 3(c)(iv) 3(c)(v) SN2 3(c)(vi) To 1 cm 3 of each compound, add 1 cm 3 NaOH(aq), followed by 2 to 3 drops of KMnO4. Heat the solution. S: Purple KMnO4 is decolourised. Black solid MnO2 is formed. T: Purple KMnO4 remains. 3(d)(i) CFCs contain C–Cl bonds which undergo homolytic fission in the presence of UV light to form chlorine radicals which catalyse the decomposition of ozone. 3(d)(ii) HFCs have higher GWP than CFCs. Hence, HFCs trap more heat and cause global warming to a larger extent than CFCs. 4(a)(i) The order of reaction with respect to a particular reactant is the power to which the concentration of that reactant is raised in the rate equation. The rate constant, k, is a constant of proportionality in the rate equation. It is constant for a particular reaction at a given temperature.

© Raffles Institution 2024 9729/02/S/24 4(a)(ii) From Fig. 3.1, first t1/2 = 2nd t1/2 = 2.3  10−4 s reaction is first order with respect to •OH. Fig. 3.1 4(a)(iii) k’ = k[CH3CHO]n A straight line graph with positive gradient passing through the origin is obtained for the graph of k’ vs [CH3CHO], OR k’  [CH3CHO], OR k’ is directly proportional to [CH3CHO]. The reaction is first order with respect to CH3CHO. k = gradient = (4.0-0.0)  103 (4.5-0.0) × 10-7 = 8.89  109 4(a)(iv) rate = k[•OH] [CH3CHO] 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 [•OH]/[•OH]0 time/ 10-3 s1st t½ 2nd t1/2

© Raffles Institution 2024 9729/02/S/24 4(a)(v) Yes I agree, since equation 1 shows a bimolecular reaction involving 1 •OH reacting with 1 ethanal which agrees with the rate equation in (a)(iv) showing that the reaction is first order with respect to both •OH and CH3CHO. 4(a)(vi) A catalyst will increase the magnitude of the rate constant, k, and decrease the magnitude of the activation energy, Ea. 4(b)(i) 4(b)(ii) 4(c)(i) 2H2O2 ⎯→ •OH + HOO• + H2O 4(c)(ii) Fe2+ is a homogeneous catalyst as it is in the same phase as H2O2. It is used in equation 2 and regenerated in equation 3. 4(c)(iii) Add aq, NaOH/aq. Na2CO3 to precipitate out Fe(OH)2 / Fe(OH)3. OR Add aq. NaS2O3 (or other known reducing agent of H2O2) t