2024 Prelims HCI H2 Chem P1 (Ans)

2024 HCI C2 H2 Chemistry Prelims / Paper 3 HWA CHONG INSTITUTION 2024 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 A D C B C B A A C D 11 12 13 14 15 16 17 18 19 20 A D B B D C C D C A 21 22 23 24 25 26 27 28 29 30 D D C A B B B A B C Comments 1 A 2H has 1 proton, 1 electron and 1 neutron. 16O has 8 protons, 8 electrons and 8 neutrons. For D3O+, Number of neutrons = 1 + 1 + 1 + 8 = 11 Since D3O+ is cationic, it is short of 1 electron overall. Number of electrons = 1 + 1 + 1 + 8 – 1 = 10 2 D Here are the electron-in-box diagrams of the valence orbitals in each species. Only sulfur has 2 unpaired electrons in its electronic configuration. 3 C Based on the number of bond pairs and lone pairs listed in the question, these are the corresponding geometries and bond angles: A: Bent, 118° B: Bent, 104.5° C: Trigonal pyramidal, 107° D: T-shaped: 88°

4 B Since electrons are delocalised in the carbonate ion, there is partial double bond character, and the bond length is likely to be between a C –O bond and a C=O bond. From Data Booklet , the average C –O bond energy is 360 kJ mol –1 while the average C=O bond energy is 740 kJ mol –1 thus it is likely that the C –O bond energy in carbonate is 485 kJ mol–1. 5 C bonding type physical properties 1 giant covalent high melting point, conducts electricity when in solution but not when solid does not conduct electricity in any state. 2 simple covalent low melting point, does not conduct electricity in any state 3 metallic variety of melting points, conducts electricity when solid and when molten 4 ionic low melting point high melting point, conducts does not conduct electricity in any state. 6 B The ordinate (i.e. y-coordinate) is PV = nRT = mRT/Mr = constant / Mr Hence the y-value of the graph is inversely proportional to the molar mass of the gas. In order of increasing molar mass, the gases are helium (Mr = 4.0), methane (Mr = 16.0), nitrogen ( Mr = 28.0) and chlorine ( Mr = 71.0). Hence the graph for methane should be the second highest horizontal line (i.e. second largest y - value). 7 A The gas is compressed to a pressure 4 times its original. If the gas were an ideal gas, its new volume would be expected to be one -quarter of its original, i.e. 19.0 cm 3. However, its new volume was 20.5 cm 3, so the gas did not behave ideally. The gas could not have dimerised, as the number of moles of gas would have decreased and the final volume would have been smaller than 19.0 cm3. Option 3 also does not explain the observation, as the presence of significant intermolecular forces of attraction would have also caused the actual volume to be smaller than 19.0 cm3. 8 A First, use the Data Booklet to find the first ionisation energies of the elements. 1st IE / KJ mol–1 Mg 736 Al 577 Si 786 P 1060 From this data, A l will be the leftmost on the x -axis while P will be the rightmost on the x -axis. Option C will be eliminated at this point. Option A, B and D are possible.

2024 HCI C2 H2 Chemistry Prelims / Paper 3 The melting point of Si will be the highest among the 4 while P has the lowest. Hence on the y-axis Si will be the highest point while P is the lowest point. Only Option A fits both sets of data and is the correct answer. FYI: Melting points of the following Period 3 elements are as follows. mp / oC Mg 659 Al 703 Si 1410 P 44 9 C No. of moles of FeC2O4 = 0.020 × 0.020 = 0.000400 mol No. of moles of KMnO4 = 0.015 × 0.020 = 0.000300 mol No. of moles electrons transferred = 0.000400 × 3 = 0.00120 mol No. of moles of MnO4− : No. of moles of electrons = 0.000300 : 0.00120 = 1:4 Each mole of MnO4− takes in 4 moles of electrons. Hence the oxidation number of Mn changes from +7 to +3. 10 D Options 1 and 2 allow us to determine the total number of moles of carbonates in the sample. Since the mass of the sample is known, we can then solve for the mole fraction of magnesium carbonate using the molar masses of the two carbonates. For option 3, BaSO 4 is collected as a precipitate, allowing us to determine the number of moles of barium carbonate. We can then calculate the mass of barium carbonate, mass of magnesium carbonate and number of moles of magnesium carbonate. 11 A The only enthalpy change that is always exothermic in the list provided is the lattice energy as it is a process that forms ionic bonds. Bond breaking and ionisation energy are both endothermic processes. As for electron affinity, 1st electron affinity to form monoanions are exothermic, but 2nd electron affinity and beyond are endothermic processes. 12 D [H+] will remain constant as it is the catalyst. Hence methods that measure changes in [H+] (options A, B and C) are not suitable for monitoring the reaction. Experiments that use different [H +] have to be set up in order to determine the order for H+ (option D). 13 B From the given data, the maximum volume of oxygen that can be collected at the end of the experiment = 2.38  280 1000  1 2  24 = 8.0 dm3. Option 1 is correct.

From the graph, first t1/2 is found from 0 to 4.0 dm3 (50% of 8.0), which is 42 min. The second t 1/2 is found from 4.0 dm 3 to 6.0 dm 3 (75% of 8.0), which is also 42 min. Hence the reaction is first order with respect to hydrogen peroxide and is also overall first order. The rate constant of an overall first order reaction is k = 𝑙𝑛 2 𝑡1/2. Hence the value of rate constant, k = 𝑙𝑛 2 42 = 0.017. For a first order reaction, rate = k[A]. The units of rate constant = 𝑚𝑜𝑙 𝑑𝑚–3 𝑚𝑖𝑛–1 𝑚𝑜𝑙 𝑑𝑚–3 = min–1 Option 2 is correct. For a first order reaction, half -life is independent of initial concentration of hydrogen peroxide. Hence option C is wrong as it should have remained at 42 min. 14 B The Haber process is used to manufacture ammonia from nitrogen and hydrogen. N2(g) + 3H2(g) ⇌ 2NH3(g) Hr < 0 When temperature increases, the backward endothermic reaction will be favoured, leading to a decrease in yield. Hence A is wrong. High pressure favours the forward reaction as it reduces the number of moles of gases. In addition, rate increases as the frequency of effective collisions increases. Thus B is correct. The presence of a catalyst only affects the rate of reaction but not the position of equilibrium, so C is wrong. The introduction of more nitrogen will favour the forward reaction, causing an increase in the yield. Hence D is wrong. 15 D A negative value for Go represents a driving force in the forward direction, and position of equilibrium lies to the right. A positive value for Go represents a driving force in the reverse direction, and position of equilibrium lies to the left. For point 4, the position of equilibrium lies to the right, which means that the concentration of H2 should be higher than the concentration of CO. 16 C Option 1 is incorrect. If lead( II) nitrate is limiting, the mass of precipitate should remain constant beyond V. Option 2 is correct. When more KI is added, a soluble lead iodide complex is formed. This decreases the concentration of Pb 2+(aq) in solution, causing the position of equilibrium of PbI2(s) ⇌ Pb2+(aq) + 2I−(aq) to shift right, and mass of the precipitate will decrease. Option 3 is correct. The formula of the soluble lead complex is PbI42–.

2024 HCI C2 H2 Chemistry Prelims / Paper 3 17 C Total = 7 constitutional isomers 18 D 19 C A H• radicals are not produced in the propagation steps, so no H 2 can form in the termination step. Incorrect. B Incorrect. There are only five monosubstituted products as shown below. C Only two of the monosubstituted products above are chiral. Correct. D Z only has 9 carbon atoms. If two radicals combine,