2022 RVHS JC2 H2 CM Prelim P3 (Solutions)

River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination [Turn over Suggested Solutions for H2 Chemistry Prelim Paper 3 1 (a) (i) S(g) → S+(g) + e– [1] (ii) Less energy is required to remove the higher energy 3p electron from Al compared to the 3s electron from Mg. [1] (b) (i) CuCO3 → CuO + CO2 [1] (ii) Ionic radius of Ca2+ = 0.099 nm Ionic radius of Cu2+ = 0.073 nm Decomposition temperature of CuCO3 is expected to be lower. Charge density of Cu2+ is greater than Ca2+ due to the smaller ionic radius of Cu 2+. Cu 2+ ion is able to polarise (the electron cloud of) CO32– ion to a larger extent , hence weakening the C O bond to a larger extent. [2] (c) (i) Na2O(s) + H2O(l) → 2NaOH(aq) pH = 13 [1] (ii) SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq) pH = 2 [1] (d) (i) Na(s) is an electrical conductor due to the presence of delocalised valence electrons which can migrate freely through the metallic structure when a potential difference is applied. Na2O(s) does not conduct electricity because its ions are held in fixed positions in a giant lattice structure / it has no mobile ions. [2] (ii) [1] (iii) Na2O2 has a giant ionic lattice structure while H2O2 has a simple molecular/ covalent structure. More energy is needed to overcome the stronger electrostatic forces of attraction between Na+ and O22– compared to the hydrogen bonds between H2O2 molecules. Hence, Na2O2 has a higher melting point than H2O2 and is a solid at room temperature. [2] (e) (i) 𝐾𝑐 = [Pb2+][Cr2+]2 [Cr3+]2 [1] (ii) Equilibrium [Pb2+(aq)] = ½(2.96  10−4) [2]

2 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination = 1.48  10−4 mol dm–3 𝐾𝑐 = (1.48 × 10−4)(2.96 × 10−4)2 (0.200)2 = 3.24  10−10 mol dm−3 (iii) The concentration of all aqueous species/ ions will be lowered. Since there are more aqueous species/ ions on the right of the equation, the position of equilibrium shifts right. [2] [Total: 17] 2 (a) (i) [2] (ii) 2H2O(l) → O2(g) + 4H+(aq) + 4e− 2Al(s) + 3/2O2(g) → Al2O3(s) [2] (iii) Volume of Al2O3 = 96.2 × 0.03 = 2.886 cm3 Mass of Al2O3 = 3.95 × 2.886 = 11.40 g Amount of Al2O3 = 11.40 2(27.0)+3(16.0) = 0.1118 mol Amount of O2 = 0.1118 × 3 2 = 0.1677 mol Amount of electrons passed = 0.1677 × 4 = 0.6708 mol Q = nF = It 0.6708 × 96500 = 2.0 × t t = 3.24 × 104 s [3] (b) Step 1: 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq) [2]

3 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination [Turn over Step 2: 2Fe2+(aq) + S2O82−(aq) → 2Fe3+(aq) + 2SO42−(aq) (c) (i) BaSO4 [1] (ii) Upon reaction with Br 2, the oxidation state of sulfur increases from +2 in S2O32− to +6 in SO42−. Upon reaction with I2,