2025 H2 Chem Prelim P1 Worked Solution CJC

1 9729/01 CJC JC2 Preliminary Examination 2025 [Turn over CANDIDATE NAME CLASS 2T CHEMISTRY 9729/01 Paper 1 Multiple Choice 18 September 2025 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class and NRIC/FIN number on the Answer Sheet in the spaces provided. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 28 printed pages. WORKED SOLUTIONS Catholic Junior College JC 2 Preliminary Examinations Higher 2

2 9729/01 CJC JC2 Preliminary Examination 2025 1 The relative abundances of all the isotopes present in a sample of zirconium are shown. What is the relative atomic mass of zirconium calculated from these data? A 91.1 B 91.3 C 91.6 D 93.1 2 In the interhalogen compound ICl, there is a single polar covalent bond. Which of the following statement(s) helps to explain the polarity of the I–Cl covalent bond? 1 Cl is more electronegative than I. 2 The outer shell electronic configuration of both elements is s2 p6. 3 The outer shell electrons are more shielded from nuclear charge in I than they are in Cl. A 1, 2 and 3 B 1 and 2 C 1 and 3 D 1 only Topic: Atomic Structure Ar of Zr = 51.5(90)+11.2(91)+17.1(92)+17.4(94)+2.8(96) 100 = 91.3 Answer: B Topic: Chemical Bonding Polarity of a covalent bond depends on the electronegativity difference between two atoms. Electronegativity is a measure of the ability of an atom to attract the bonding electrons. Option 1: True; Cl is more electronegative than I. Option 2: True but does not explain why the bond is polar. Option 3: True; hence the attraction exerted by the nucleus of I on the shared pair of electrons is weaker than that of Cl. Answer: C 90 91 92 93 94 95 96 51.5 11.2 17.1 17.4 2.8 Mass of isotope relative abundance

3 9729/01 CJC JC2 Preliminary Examination 2025 [Turn over 3 In which pairs of compounds does the first molecule have a smaller bond angle than that in the second molecule? 1 NF3 CCl4 2 H2S H2O 3 SF6 CS2 A 1, 2 and 3 B 1 and 2 C 2 and 3 D 3 only 4 The table shows the boiling point of three alcohols. boiling point / °C pentan-1-ol 138 2-methylbutan-2-ol 129 2,2-dimethylpropanol 114 What is responsible for the differences in boiling point? A different relative molecular mass B different number of carbon-carbon bonds C weaker hydrogen bonding between branched chain molecules D more extensive instantaneous dipoles–induced dipoles attractions between straight chain molecules Topic: Chemical Bonding molecule shape bond angle molecule shape Bond angle 1 NF3 trigonal pyramidal 107o CCl4 tetrahedral 109o 2 H2S bent 92 o due to central S atom being less electronegative than O in H2O H2O bent 105o 3 SF6 octahedral 90o CS2 linear 180o Answer: A Topic: Chemical Bonding A False; all three alcohols are isomers with the same relative molecular mass B False; same number of carbon-carbon bonds in the three isomers C False; strength of intermolecular hydrogen bonding in the straight chain and branched chain isomers are similar since all three isomers, on average, will form the same number of hydrogen bonds per molecule (since each molecule consist of 1 -OH group). D True; more extensive instantaneous dipoles-induced dipoles attractions between straight chain molecules (due to larger surface area of contact between the molecules). Answer: D

4 9729/01 CJC JC2 Preliminary Examination 2025 5 Which statements about the behaviour of Group 17 elements from chlorine to iodine are correct? A The elements become stronger oxidising agents. B The volatility of the elements decreases. C The thermal stability of the hydrogen halides increases. D The bond energy of H–X bond increases. 6 0.10 mol of an oxide of nitrogen (NxOy) is mixed with an excess of hydrogen and passed over a catalyst at a suitable temperature. The water produced in this reaction has a mass of 7.2 g. The ammonia produced requires 200 cm3 of 1.0 mol dm‒3 HCl for complete neutralisation. What is the formula of this oxide of nitrogen? A N2O B NO C NO2 D N2O4 Topic: The Periodic Table A False; the elements become weaker oxidising agents from chlorine to iodine as there is a lower tendency for X2 to reduce to X–. B True; The volatility of the elements decreases down the group. As the number of electrons of the halogen molecules increases , the instantaneous dipoles – induced dipoles (id-id) forces of attractions between the halogen molecules increases. Larger amount of energy is required to overcome the stronger id -id forces of attractions, resulting in higher boiling point and hence lower volatility. C False; the thermal stability of the hydrogen halides decreases from H-Cl to H-I as the bond energy of H-X bond decreases, hence the ease of breaking H-X bond increases, resulting in lower thermal stability. D False; the bond energy of H–X bond decreases from H-Cl to H-I due to less effective overlap of orbitals as the valence orbital used in bonding becomes bigger and more diffused from chlorine to iodine. Answer: B

5 9729/01 CJC JC2 Preliminary Examination 2025 [Turn over 7 Sodium thiosulfate is used in the textile industry to remove an excess of chlorine from bleaching processes by reducing it to chloride ions. One mole of thiosulfate ions, S2O32‒, is able to remove 4 moles of chlorine, Cl2, in this process. In this process, S2O32‒ is oxidised. What is the resultant sulfur-containing product in this reaction? A HSO4‒ B S4O62‒ C SO2 D S Topic: The Mole Concept and Stoichiometry Let the oxide of nitrogen be NxOy. Amt of water formed when 0.10 mol of the oxide reacted = 7.2 18.0 = 0.400 mol No of O atoms in NxOy, y = 0.400 0.100 = 4 Amt of HCl required for neutralisation = 200 1000 × 1.0 = 0.200 mol = amt of NH3 formed. No of N atoms in NxOy, x = 0.200 0.100 = 2 Hence oxide of nitrogen is N2O4. Answer: D unbalanced equation: NxOy + H2 → NH3 + H2O Amounts / mol: 0.1 0.2 0.4 Hence: 1NxOy + H2 → 2NH3 + 4H2O where x and y can be derived via inspection Topic: The Mole Concept and Stoichiometry (Redox) Half equation involving chlorine: Cl2 + 2e‒ → 2Cl‒ Hence, 4 moles of Cl2 would gain 8e‒. This means that 1 mole of S2O32‒ loses 8e‒. (one S atom loses 4e‒) Oxidation state of S in S2O32‒ = +2 Final OS of S in the product = +2 ‒ (‒4) = +6 OS of S in (A) HSO4‒ (+6) answer (B) S4O62‒ (+2.5) (C) SO2 (+4) (D) S (0) Answer: A

6 9729/01 CJC JC2 Preliminary Examination 2025 8 Nitric acid is made industrially by the oxidation of ammonia. The overall equation for the process is shown. equation 1 NH3 + 2O2 → HNO3 + H2O The process happens in three stages. The equations and enthalpy changes for these stages are given. stage 1 4NH3 + 5O2 → 4NO + 6H2O ΔH = –904 kJ mol–1 stage 2 2NO + O2 → 2NO2 ΔH = –114 kJ mol–1 stage 3 4NO2 + O2 + 2H2O → 4HNO3 ΔH = –348 kJ mol–1 What is the enthalpy change of the process shown in e