2022 RVHS JC2 H2 CM Prelim P2 (Solutions)

River Valley High School 9729/02/PRELIM/22 [Turn over 2022 Preliminary Examination Suggestion Solutions for 2022 H2 Chemistry Prelim Paper 2 1 (a) The enthalpy change of solution of a substance is the enthalpy change when one mole of the substance is completely dissolved to give an infinitely dilute solution, so that no further enthalpy change takes place on adding more solvent. [1] (b) (i) Assuming no heat loss to surroundings, amount of NH4NO3  26000 = 50  4.18  5 amount of NH4NO3 = 0.04019 mol minimum mass of NH4NO3 = 0.04019  (2(14.0) + 4(1.0) + 3(16.0)) = 3.22 g [1] (ii) [2] (c) By Hess’ Law, Hsoln(NH4Cl) = – (–705) + [(–307) + (–381)] = +17.0 kJ mol–1 [3] (d) Since Cl− has a smaller ionic radius and higher charge density than Br−, it forms stronger ion-dipole interactions with water molecules. Thus, Cl− has a larger magnitude of Hhyd than Br−. [2] [Total: 9]

2 River Valley High School 9729/02/PRELIM/22 2022 Preliminary Examination 2 (a) It is a d-block element that is able to form one or more stable ions with a partially filled d subshell. [1] (b) Both elements have giant metallic lattice structure and exhibit metallic bonding. In Co, both the 4s and 3d electrons can be contributed to form the sea of delocalised electrons as they are very close in energy. The resulting cobalt ion has a higher positive charge and a smaller ionic radius/ higher charge density . This results in stronger electrostatic forces of attraction between the metal cations and the sea of delocalised electrons in Co as compared to Ca, which only contributes 2 valence electrons per Ca atom to form Ca2+. [2] (c) (i) 1s22s22p63s23p63d7 [1] (ii) O2 + 2H2O + 4e– ⇌ 4OH– E = +0.40 V Overall equation: 4[Co(NH3)6]2+ + O2 + 2H2O → 4[Co(NH3)6]3+ + 4OH– Ecell = (+0.40) – (+0.11) = +0.29 V Since E cell > 0, the reaction is feasible. Hence, yellow -brown [Co(NH3)6]2+ is oxidised by oxygen in air to produce red-brown [Co(NH3)6]3+ [2] (iii) (iv) [3] [1] (d) (i) Amount of D reacted with EDTA = 18.75 1000 × 0.0400 = 7.500 × 10–4 mol Amount of D in 3.501 g sample = 500 25.0 × 7.500 × 10−4 = 0.01500 mol Molar mass of D = 3.501 0.01500 = 233.4 g mol–1 [2]

3 River Valley High School 9729/02/PRELIM/22 [Turn over 2022 Preliminary Examination (ii) Amount of D reacted with AgNO3 = 23.34 233.4 = 0.1000 mol Amount of AgCl2 formed = 14.340 (107)+(35.5) = 0.1000 mol Mole ratio of D: AgCl is 1 : 1 [1] (iii) 17a + 35.5(3) = 233.4 – 58.9 Solving: a = 4 Cation: [Co(NH3)4Cl2]+ [2] (iv) [2] [Total: 17] 3 (a) (i) Enantiomerism [2] (ii) An electrophile is an electron pair acceptor and is electron deficient. [1]

4 River Valley High School 9729/02/PRELIM/22 2022 Preliminary Examination (iii) [2]