RI 2025+Post+Prelim+TP+Suggested+solutions

© Raffles Institution 2025 9729/02/O/25 2025 Y6 H2 Chemistry Post Prelim Timed Practice – Suggested Solutions 1(a)(i) Comparing sodium and chlorine, chlorine has a higher nuclear charge due to larger number of protons, S hielding effect remains approximately constant as electrons are added to the same electronic shell. As there is stronger electrostatic attraction between the nucleus and the bonding electrons, chlorine has higher electronegativity, resulting in a larger EN value. 1(a)(ii) ∆EN for AlCl3 = 3.16 – 1.61 = 1.55 ∆EN for PCl5 = 3.16 – 2.19 = 0.970 1(a)(iii) Ionic bonds exist in NaCl due to large ∆EN (or large difference in electronegativities) between Na and Cl. Covalent bonds exist in PCl5 due to small ∆EN (or small difference in electronegativities) between P and Cl. 1(b)(i) SeF4 > BrF3 1(b)(ii) There are 4 bond pairs and 1 lone pair in the valence shell of Se atom. To minimise electrostatic repulsion, the shape of SeF4 is see-saw. Se F F F F AlCl3 PCl5

© Raffles Institution 2025 9729/02/O/25 Since lone pair – bond pair repulsion > bond pair – bond pair repulsion, the bond angles are 88o and 118o (other acceptable bond angles < 90o and < 120o). 2(a) By Le Chatelier’s Principle, the position of equilibrium will shift right to decrease the concentration of Cl2. The equilibrium mixture will contain more PCl5, less PCl3 and more Cl2. 2(b)(i) Kc = [PCl5] [PCl3][Cl2] , mol−1dm3 2(b)(ii) Kc = [PCl5] [PCl3][Cl2] = �1.2 2 � �0.4 2 ��0.25 2 � = 24.0 mol−1dm3 2(b)(iii) PCl3(g) + Cl2(g) ⇌ PCl5(g) initial amount /mol 0.4 0.25 + x 1.2 change in amount /mol –0.08 –0.08 +0.08 equilibrium amount /mol 0.4 – 0.08 = 0.32 0.25 + x – 0.08 = 0.17 + x 1.28 Kc = [PCl5] [PCl3][Cl2] = �1.28 2 � �0.32 2 ��0.17+𝑥𝑥 2 � = 24 x = 0.163 mol 2(c) Na2O reacts vigorously with water to give a strongly alkaline solution with a pH = 13–14. Na2O + H2O → 2NaOH P4O10 reacts readily with water to give an acidic solution with a pH = 2. P4O10 + 6H2O → 4H3PO4 Al2O3 does not dissolve in water (energy needed to break down the ionic lattice is more than the energy released on hydrating the ions). pH = 7 3(a)(i) 3(a)(ii) amount of manganate used = (44.8/1000) x 0.02 = 0.000896 mol amount of iron(II) produced = 0.000896 x 5 = 0.00448 mol amount of Fe2+ = amount of Fe3+ amount of iron(III) reacted with hydroxylamine = 0.00448 mol 3(a)(iii) amount of hydroxylamine used = 0.074 / 33 = 0.00224 mol Reduction: Fe3+ + e–  Fe2+ amount of Fe3+ : amount of NH2OH : amount of e– 2 : 1 : 2

© Raffles Institution 2025 9729/02/O/25 final oxidation state of nitrogen in product = – 1 + 2 = +1 N2O 3(a)(iv) Fe3+ + e–  Fe2+ (x4) 2NH2OH  N2O + H2O + 4H+ + 4e– (x1) 4Fe3+ +2NH2OH  4Fe2+ + N2O + H2O + 4H+ 3(b)(i) The second ionisation energy of element F is the energy required to remove 1 mole of electrons from 1 mole of gaseous F+ ions to form 1 mole of gaseous F2+ ions. F+(g)  F2+(g) + e– 3(b)(ii) Large decrease in 2nd ionisation energy from D to E implies that the 2nd electron in E is removed from the outer electron shell. E belongs to Group 2 and B belongs to group 17. B is fluorine 3(b)(iii) A: ns2np4 A+: ns2np3 B: ns2np5 B+: ns2np4 The 2 nd electron removed in B is a paired electron which experiences greater electron-electron repulsion, so less energy is needed to remove the paired electron than the unpaired electron removed in A. Hence, the 2nd ionization energy of B is lower. 4(a)(i) The product of reduction of H 2O2 is water which is clean / non -pollutant / environmentally friendly. 4(a)(ii) Reduction 4(a)(iii) From C 14H8O2 to C14H10O, gain of 2 H and loss of 1 O = reduction or 4(a)(iv) Add 2,4-DNPH. anthrahydroquinone no orange ppt anthraquinone orange ppt formed

© Raffles Institution 2025 9729/02/O/25 O O + O2N NO2N H H2N2 N N NH NO2 NO2 NH NO2 NO2 + 2H2O 4(b) Sufficient energy is released from the formation of hydrogen bonds between water and H2O2 to overcome the hydrogen bonds between water molecules and hydrogen bonds between H2O2 molecules. Insufficient energy is released from the formation of instantaneous dipole- induced dipole interactions between anthraquinone and water to overcome the hydrogen bonds between water molecules. H2O2 dissolve in water / anthraquinone cannot dissolve in water. 4(c)(i) A heterogeneous catalyst is a catalyst which is not in the same phase as the reactants. 4(c)(ii) The reactant molecules are adsorbed onto the surface of the solid catalyst. The adsorption weakens the covalent bonds within the reactant molecules, thereby lowering the activation energy for the reaction. The adsorption increases the concentration of reactant molecules at the catalyst surface and allows the reactant molecules to come into close contact with proper orientation for reaction. The catalyst provided an alternative reaction pathway of lower activation energy. As such, significantly more reactant molecules have energy greater than or equal to the activation energy for the catalysed reaction. Frequency of effective collisions increases and an increase in the rate of the reaction. Rate constant also increases. After the reaction, desorption of product molecules occurs and the product molecules eventually diffuse away from the catalyst surface.

© Raffles Institution 2025 9729/02/O/25 4(d)(i) Finely dividing the palladium into nanoparticles increases its surface area to volume ratio, providing more active sites for adsorption of reactants. 4(d)(ii) Due to their small size, nanoparticles can enter human body through pores on skin, breathing/respiratory system, or ingestion (by digestive system). These particles can travel around the body via the circulatory system, get deposited in body tissues / cells / organs and cause the latter to be damaged / inflamed / develop cancer / cause immune response. 4(d)(iii) Gold particles are larger and the deposition of the nanoparticles decreases the chances / prevents them from being inhaled or ingested. 5(a) Across the period, nuclear charge increases due to increasing number of protons. As electrons are added to the 3d subshell, the increase in the number of inner 3d electrons provide more shielding between the nucleus and the outer 4s electrons. The increase in shielding effect offsets the increase in nuclear charge. Hence, the electrostatic attraction between the nucleus and outer 4s electrons increases minimally. The first IE remains almost constant/ relatively invariant. 5(b)(i) Atoms with the same proton number but different nucleon numbers (i.e. same number of protons but different number of neutrons) 5(b)(ii) Ar of copper = [5(63)+2(65)]/7 = 63.5714 5(b)(iii) atomic mass unit of 114 79Br35Cl atomic mass unit of 116 79Br37Cl and 81Br35Cl atomic mass unit of 118 81Br37Cl 5(b)(iv) percentage of sample with 79Br35Cl = (1/2)(3/4) = 3/8 = 0.375 = 37.5% percentage of sample with 81Br37Cl = (1/2)(1/4) = 1/8 = 0.125 = 12.5% percentage of sample with 79Br37Cl and 81Br35Cl = (1/2)(1/4) + (1/2)(3/4) = 4/8 = 50% 5(c)(i) Oxidising agent. Oxidation state nitrogen decreases from +5 in HNO3 to +4 in NO2. 5(c)(ii) Sulfur precipitate: 4 – y = 64.2 32.1 ∴y = 2 NO2 gas: 13 – (3 + 2x) = 182 22.7 ∴x = 1 6(a) 6(b)(i) buta-1,3-diene or butadiene

© Raffles Institution 2025 9729/02/O/25 6(b)(ii) OH HO OH OH monomer of polymer X alkaline KMnO4, cold acidified KMnO4, heatCO2 6(c)(i) Cis-trans isomerism. There is restricted rotation about the C=C bond and each carbon of the C=C bond is bonded to two different groups. 6(c)(ii) C C CH2 CH2H H Structure of polymer Y (trans isomer) 6(c)(iii) polymer Z (cis isomer) 6(d) Carbon-carbon bonds are non-polar / to