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 r