2024 RI Prelim H2 Chem Paper 2 Suggested solutions

© 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− Comments: • State symbols are essential for this equation to avoid ambiguity as the definition of fourth ionisation energy requires the ions to be in gaseous state. Any differences in state will cause the enthalpy change of the reaction to be different. • Some students are unfamiliar with the element boron and wrote Be3+ or Br3+ instead. 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. Comments: • Answers for this question were poorly phrased and lacking in key points. Many students focused only on the difference in electron shell from which the electrons are removed and did not discuss shielding effect and nuclear charge in their answer. • Students are reminded to phrase their answers in terms of B3+ and C3+ instead of ambiguous terms like “boron/B” and “carbon/C” as it is open to interpretation whether the answer is referring to the atom or the ion. For example, “C has one more electron shell than B” would be incorrect for the case of C and B atoms. • Common poor phrasing include: o “1s2 electron”, it should be “1s electron” o “1s shell”, 1s is not an electron shell, it is an orbital/subshell depending on the context used o “inner 1s subshell”, the comparison should be in terms of electron shell and not subshell • Students are reminded that in both cases, the electrons to be removed are from the valence electron shell of B3+ and C3+. 1(b)(i) Comments: • Generally well done. • A number of students erroneously included a lone pair of electrons for B. B H H H

© Raffles Institution 2024 9729/02/S/24 1(b)(ii) shape: trigonal planar bond angle: 120° Comments: • Generally well done. • Students are reminded that the command verb for the question is “state” , thus no explanations are required. 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.) Comments: • Generally well done. • Common mistakes include not considering: o The coefficient of ½ in the equation for B2O3 o The two O−H bonds in H2O 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 B 2O3 from gaseous to solid state. Comments: • Students generally found this question challenging as they did not realise that ∆Hc referred to B2O3 in its standard state (as a solid given in the question) while the calculated ∆Hr involves B2O3 in its gaseous 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. Comments: • Generally well done. • Students should consider that A l in AlCl3 is sp2 hybridised and hence has an empty unhybridi sed p orbital which is energetically accessible (low -lying) to accept a lone pair of electrons from Cl. Some students indicated that 3d orbitals were involved, which is incorrect in this context as A l does not have an expanded octet in Al2Cl6. • A small number of students wrongly considered AlCl3 as an ionic compound and provided explanations in terms of charge density and polarising power. 1(c)(ii) sp2 to sp3 Comments: • Generally well done.

© Raffles Institution 2024 9729/02/S/24 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. Comments: • Students generally found this question challenging as they were unable to account for the factor that leads to a weaker B−H b bond than B−Ha and hence resulting in different bond lengths. • Some students attempted to explain in terms of hybridisation / s−character. However, B is sp 3 hybridised in both B−H a and B−Hb and H does not have a p orbital in the first principal quantum shell for hybridisation. Hence, it is not possible to compare and explain the phenomenon in terms of hybridisation / s−character. 1(d)(i) N B N B N B H H H H H H N B N B N B H H H H H H X Y Comments: • A significant number of students appeared to have overlooked this question and did not attempt it. Students are reminded to take note of the question part labels “(i)”, number of marks for each question “[1]” and to read the question carefully “complete Fig 1.2”. 1(d)(ii) N is more electronegative than B, and this reduces the extent of electron delocalisation (partially delocalised) compared to benzene. Comments: • Students generally found this question challenging due to a lack of understanding of the structure of borazine. The question indicated that “Borazine is similar in structure to benzene” and that its structure “can be represented by two different resonance structures”. Some students contradicted these statement s by suggesting that “delocalisation of electrons in borazine does not occur throughout the ring” or there is “no continuous side- on overlap of orbitals” in borazine. • Another common error suggests that there are less number of delocalised electrons in borazine than benzene. This is incorrect as both borazine and benzene have 6 delocalised electrons. For benzene, each carbon atom contributes only 1 electron for delocalisation. For borazine, each nitrogen atom contributes 2 electrons for delocalisation and each boron atom does not contribute any electrons for delocalisation as its p orbital is empty.

© Raffles Institution 2024 9729/02/S/24 1(d)(iii) N B N B N B H H H H H H H Cl H Cl H Cl Comments: • Students generally found this question challenging as they did not consider the structure of the product formed using the resonance structure X and did not ensure that the atoms are conserved in an addition reaction. • In resonance structure X, the N atom has a lone pair of electron s for coordination with H in HCl. At the same time, the B atom has an empty p orbital to accept electrons from Cl in HCl. • In addition reactions, the total number of atoms must be conserved before and after the reaction. Hence, when “1 mole of X reacts completely with 3 moles of HCl”, the equation should be B3N3H6 + 3HCl → B3N3H7Cl3 where the product should have 3 additional H and Cl respectively. • Several reasonable structures of Z were accepted. 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) Comments: • Generally well done. • A small number of students made the mistake of multiplying +1.23 with a coefficient of 2 when calculating Ecell and/or using −0.87 . Students are reminded that Ecell = Ecathode − Eanode without any coefficients (even if the half- reaction is multiplied by a coefficient) and both Ecathode and Eanode are standard reduction potentials from the Data Booklet without changing the sign. • A positive E cell value is sufficient to show that the reaction is spontaneous without having to calculate ∆ G. If ∆ G is calculated, the correct number of moles of electrons transferred (8 mol in this case) should be used. 2(a)(ii) The beneficial bacteria provide enzymes which act as biological catalysts to