2023 RI Prelim H2 Chem Paper 1 Solutions

2023 H2 Chemistry Preliminary Examinations Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 D D A A B D C D C B 11 12 13 14 15 16 17 18 19 20 C B B C C C C B D C 21 22 23 24 25 26 27 28 29 30 D C D D B B A D B B 1 (D) shape structure polar molecule? CH2F2 tetrahedral (4 bond pairs) H C H F F yes SF4 see-saw (4 bond pairs, 1 lone pair) S F F F F : yes NF3 trigonal pyramidal (3 bond pairs, 1 lone pair) N F F F : yes BeF2 linear (2 bond pairs) F–Be–F no 2 (D) r is a carbon-carbon double bond hence it is stronger than q, which is a carbon-carbon single bond. p has partial double bond character due to continuous side- on overlap of 4 p-orbitals (see diagram). Hence it is stronger than a carbon -carbon single bond but weaker than a carbon-carbon double bond. p q r 3 (A) Substances with giant covalent structure generally do not conduct electricity in all physical states since all its valence electrons are used up for covalent bonding.

4 (A) The largest increase in ionisation energy occurs between the 9th and 10th ionisation energy. The largest ionisation energy is used to remove the 10th electron. Hence, the 10th electron is from an inner shell, which experiences stronger electrostatic forces of attraction with the nucleus. Since there are 9 outer electrons, there must be 8 in n = 2 electron shell and 1 electron in n = 3 valence electron shell. As such, the element must be in Group 1 and it would be Na. 5 (B) Down the group, nuclear charge and shielding effect increase. The increase in the number of protons does not explain the increase in atomic radius of the elements (Option 1 is incorrect). The increase in the number of electronic shells increases the distance between the valence electron and nucleus, which result in a net decrease in strength of electrostatic forces of attraction between the nucleus and valence electrons (option 2 is correct; option 3 is incorrect) 6 (D) For graphs of V against T, pV = nRT ⇒ V =( nR p ) T V is directly proportional to T, i.e. V vs T graph is a straight line with a positive gradient. Since X has a higher mass (i.e. higher n), the gradient ( nR p ) for X will be steeper (than Y). For graphs of (pV/T) against p, pV = nRT ⇒ (pV/T) = nR The graph would show a horizontal line, with y-intercept = nR. A Incorrect. Since X has a higher n, X should have a higher y-intercept value than Y. B Incorrect, because the shape of the graph is incorrect. C Incorrect, because the shape of the graph is incorrect. 7 (C) Since X is a period 3 chloride and forms a solution with a pH of 3, X is likely to be AlCl3 and the solution contains Al3+ ions. When NH3(aq) is added to Al3+ ions, a white ppt of Al(OH)3 is formed, which is insoluble in excess NH3(aq).

8 (D) The stronger the H–X bond strength, the more thermally stable the hydrogen halide. Down Group 17, the H –X bond strength decreases. Thus, the thermal stability of the hydrogen halides decreases. Halogens tend to undergo reduction and act as oxidising agents in redox reactions. Down the group, the tendency of the halogen gaining electrons decreases as can be seen by their less positive E values. Hence, the oxidising power of the halogens decreases down the group. 9 (C) A 9.60 g of ozone: 9.6 48 = 0.2 mol of O3 B 14.2 g of chlorine: 14.2 71 = 0.2 mol of Cl2 C 2.41 × 1023 molecules of methyl ethanoate: 2.41 ×1023 6.02 ×1023 = 0.4 mol of CH3CO2CH3 D 2.40 dm3 of carbon dioxide: 2.4 24 = 0.1 mol of CO2 10 (B) 1 Correct. Since cerium (IV) sulfate is an oxidising agent, it oxidises H 2O2 to O2 as shown in the following half-equation: H2O2 → O2 + 2H+ + 2e−. The gas given off is O2. 2 Correct. Since the mole ratio of the reaction between Ce 4+ and H2O2 is 2:1, the 2 mol of e– released from H2O2 must react with 2 mol of Ce4+. 2Ce4+ + 2e– → ?? 1H2O2 → O2 + 2H+ + 2e− i.e. 2Ce4+ + 2e– → 2Ce3+ The final oxidation state of Ce is +3. 3 Incorrect. The overall reaction is H2O2 + 2Ce4+→ 2Ce3+ + O2 + 2H+. If KMnO4 were used as the titrant, the titration reaction would be 2MnO4− + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O From the two equations, 1 mole of H2O2 requires 2 moles of Ce4+ but only requires 2 5 moles of MnO 4−. Hence, for the same concentration of titrant, the titre volume would be smaller than 25.00 cm3.

11 (C) Number of moles of Ba(OH)2 = 20.0 1000 × 1.00 = 0.02 mol Number of moles of CH3COOH = 20.0 1000 × 2.50 = 0.05 mol 2CH3COOH + Ba(OH)2 → Ba(CH3COO)2 + 2H2O Since Ba(OH)2 is the limiting reagent, amount of H2O = 0.04 mol Total volume = 20.0 + 20.0 = 40.0 cm3 q = mc∆T = 40.0 × 4.18 × (37.5 – 25) = 2090 J ∆Hn = − 2090 0.04 = −52250 J mol−1 = −52.3 kJ mol−1 12 (B) A Incorrect. Hydration is an exothermic process as ion-dipole interactions are formed between water molecules and the ions. B Correct. Energy is required for the formation of gaseous atoms in atomisation. C Incorrect. The enthalpy change of formation can be either exothermic or endothermic. D Incorrect. The enthalpy change of solution can be either exothermic or endothermic. 13 (B) 1 Correct. The increase in temperature favours the backward endothermic reaction, thus decreasing the yield of NH3. 2 Correct. Increasing the volume, decreases the total pressure. Hence, the backward reaction is favoured as it produces more gas particles, thus decreasing the yield of NH3. 3 Incorrect. A catalyst speeds up the rate of both the forward and backward reactions by the same extent. The yield of NH3 is, therefore, unaffected.

14 (C) From the slow step, rate α [N2O2][O2] From step 1, the rate of formation of N2O2 depends on [NO] i.e. [N2O2] α [NO]2 rate α [N2O2][O2] rate α [NO]2[O2] rate = k[NO]2[O2] 1 Correct. The overall order is 2+1 = 3. 2 Incorrect. The rate equation contains both NO and O2. 3 Correct. Summing up both steps gives the overall equation 2NO + O2 → 2NO2. If the reaction occurred via single-step mechanism, then the overall equation is also the equation of the slow step. In that case, the rate equation is rate = k[NO]2[O2]. 15 (C) A Incorrect. During auto-ionisation of water, H2O + H2O ⇌ H3O+ + OH–, equimolar H+ and OH– will be produced, causing [H+] = [OH–], B Incorrect. Since Kw = [H+][OH–], [H+] = �Kw At 30 °C, [H+] = �1.44 × 10−14 = 1.20 × 10–7 mol dm–3 C Correct. Since [H+] = 1.2 × 10–7 mol dm–3, pH = –lg(1.20 × 10–7) = 6.92 < 7. D Incorrect. Since [H+] = 1.2 × 10–7 mol dm–3, pH = –lg(1.20 × 10–7) = 6.92 < 7. 16 (C) As the range of rapid pH change is unknown, the best indicator can only be determined using the pH at equivalence (pH 8.72). Thymol blue is the choice of indicator as the pH of equivalence lies within the pH range for its colour change. 17 (C) A Incorrect. HBr is formed during the propagation step. B Incorrect. Hexane is not a suitable solvent for the reaction because hexane can also undergo free radical substitution with the reactant Br 2. This generates undesirable products. C Correct. The reaction only involves homolytic fission and fusion. D Incorrect. CH3CH2C(CH3)2CH2CH3 is not produced from the reaction. CH3CH2C(CH3)2CH2CH3 refers to

18 (B) The reaction produces 2 possible products (which are structural isomers) and have the following structures. O Br * O Br Next, consider the number of stereoisomers from each of the structural isomers. O Br H Br H O does not exhibit enantiomerism because it has no chiral centres and contains a plane of symmetry This product can exist as two enantiomers This product does not have stereoisomers. Therefore, total number of isomers = 2 + 1 = 3 19 (D) OH HOOC O Mr: 128.0 Mr: 144.0 contains 2 sp2 C contains 2 sp2 C A Incorrect. Octenol has a lower molecular mass than Z. B Incorrect. Octenol and Z cannot be distinguished by addition of PCl5, because bot