2025 H1 Chem 8873 Prelim P1 WORKED SOLUTIONS CJC

1 CHEMISTRY 8873/01 Paper 1 Multiple Choice 18 September 2025 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, HT group and NRIC/FIN number on the Answer Sheet in the spaces provided. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. 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 10 printed pages Catholic Junior College JC 2 Preliminary Examinations Higher 1 WORKED SOLUTIONS

2 8873/01/CJC JC2 Preliminary Examinations 2025 For each question there are four possible answers, A, B, C and D. Choose the one you consider to be correct. 1 Use of the Data Booklet is relevant to this question. Which of the following contains 1 mole of the stated particle? A gold atoms in 79 g of solid gold B aqueous H+ ions in 10 dm3 of 0.10 mol dm-3 of H2SO4 solution C hydrogen atoms in 6 dm3 of methane gas at r.t.p. D chlorine molecules in 22.4 dm3 of chlorine gas at r.t.p. 2 Carbon disulphide vapour was burnt in oxygen according to the equation: CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g) A fixed amount of carbon disulphide was completely burnt in 100 cm3 of oxygen. Upon cooling, the mixture of gases was treated with an excess of aqueous sodium hydroxide and the volume of gas reduced by 30 cm3. All measurements were made at the same temperature and pressure. What was the final volume of gas left? A 10 cm3 B 30 cm3 C 40 cm3 D 70 cm3 Concept: Mole Concept and Stoichiometry Answer: C A Amount of Au atoms = mass/Mr = 79/197 = 0.401 mol B Amount of H+ ions = concentration × volume = 2 × 10 × 0.10 = 2.0 mol C At r.t.p., 1 mol of any gas occupies 24 dm3. Therefore, number of moles of gas in 6 dm3 = 0.25 mol In 1 mol of methane, there are 4 mol of H atoms. Amount of H atoms in 0.25 mol of methane = 4 × 0.25 = 1 mol D At r.t.p., 1 mol of any gas occupies 24 dm3. Therefore, number of moles of gas in 2.4dm3 = 22.4/24 = 0.933 mol

3 8873/01/CJC JC2 Preliminary Examinations 2025 [Turn over 3 Element X can exist in a few oxidation states. It was found that 0.01 mol of X2+ requires 0.004 mol of acidified potassium manganate (VII), KMnO4 for complete reaction. What is the final oxidation state of X in the above reaction? A +4 B +3 C 0 D −1 4 Use of the Data Booklet is relevant to this question. Which of the following has more neutrons than electrons and more electrons than protons? A 37Cl– B 48Ti4+ C 79Br+ D 32S2‒ Concept: Mole Concept and Stoichiometry Answer: D CO2 and SO 2 are both acidic gases, so both of them will react with NaOH in a neutralisation reaction. Vol. of CO2 (g) + Vol. of SO2 (g) = 30 cm3 Since CO2 ≡ 2 SO2 Vol. of CO2 (g) = 10 cm3, Vol. of SO2 (g) = 20 cm3, Vol. of CS2 (g) = 10cm3 Vol. of O2 (g) reacted = 10 × 3 = 30 cm3 Vol. of O2 (g) left = 100 – 30 = 70 cm3 Concept: Mole Concept and Stoichiometry (Redox) Answer: A 0.01 mol of X2+ reacts with 0.004 mol of MnO4– Thus 5X2+ ≡ 2MnO4– From data booklet: MnO4− + 8 H+ + 5 e → Mn2+ + 4 H2O Since electrons transferred in a redox reaction is equal (i.e. the no. of e accepted = no. of electron donated), thus 2 MnO4− gains 10 e– while 5 X2+ donates 10 e–  each X2+ loses 2 e–  X is oxidised from +2 to +4 oxidation state.

4 8873/01/CJC JC2 Preliminary Examinations 2025 5 What is the electronic configuration of Fe3+ ion? A 1s2 2s2 2p6 3s2 3p6 3d5 B 1s2 2s2 2p6 3s2 3p6 3d3 4s2 C 1s2 2s2 2p6 3s2 3p6 3d6 4s2 D 1s2 2s2 2p6 3s2 3p6 3d10 4s1 6 The successive ionisation energies (IE) of element X are given below: IE/ kJ mol–1 1st 2nd 3rd 4th 5th 6th 7th 8th X 550 1065 4138 5500 6910 8760 10230 11800 What is the likely formula of the compound that is formed when X reacts with fluorine? A XF B XF2 C X2F D X2F3 Concept: Atomic Structure Answer: A Iron has 26 protons, Fe has electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d6 4s2 Thus Fe3+ e.c. is 1s2 2s2 2p6 3s2 3p6 3d5 since electrons will be lost from 4s subshell first followed by electron from 3d subshell. Concept: Atomic Structure Answer: B There is a sharp increase from the 2 nd to 3 rd ionisation energy, indicating that the 3 rd electron is from the inner quantum shell. Hence, X has 2 valence electrons and will form X2+ ion. Concept: Atomic Structure Answer: A 37Cl– 48Ti4+ 79Br+ 32S2- Protons: 7 22 35 16 Electrons: 18 18 34 18 Neutrons: 20 26 44 16 Note: Students should eliminate option s B and C since the qns stated that there are “more electrons than protons”. Thus the only options that required calculation of no. of neutrons and electrons are options A and D.

5 8873/01/CJC JC2 Preliminary Examinations 2025 [Turn over 7 In which pair of molecules is the bond angle in the first molecule smaller than that in the second molecule? A BF3 and PH3 B BeCl2 and H2S C SO3 and BCl3 D F2O and SO2 8 Which of the following correctly describes the dominant forces of attraction between two of the same compounds shown below? compound dominant forces of attraction A covalent bonds B hydrogen bonding C ionic interaction D permanent dipole – permanent dipole Concept: Chemical Bonding Answer: D A BF3 120o PH3 107o B BeCl2 180o H2S 105o C SO3 120o BCl3 120o D F2O 105o SO2 119o

6 8873/01/CJC JC2 Preliminary Examinations 2025 9 Which statements help to explain the fact that magnesium has a higher melting point than sodium? 1 Magnesium ions have more protons than sodium ions. 2 Magnesium ions have a greater charge than sodium ions. 3 Magnesium ions have a smaller ionic radius than sodium ions. A 1, 2 and 3 B 2 and 3 C 2 only D 1 only Concept: Chemical Bonding Answer: C Correct type of attraction is as follows: Compounds Dominant forces of attraction A Instantaneous dipole – induced dipole B Permanent dipole-permanent dipole C Ionic interaction D Hydrogen bonding Concept: Chemical Bonding; Metallic bonding [2016 A level] Answer: A Magnesium ions have more protons than sodium ions and this results in magnesium ions having smaller ionic radii. Since magnesium ions have a greater charge (and smaller radii) than sodium ions, the charge density of magnesium ions are higher. This, along with the sea of delocalised electrons in magnesium is greater, will give rise to greater attraction between magnesium ions and the sea of delocalised electrons and hence higher melting point.

7 8873/01/CJC JC2 Preliminary Examinations 2025 [Turn over 10 Which of the following sets of substances consists of simple molecular structure, giant ionic structure and giant covalent structure? A SiCl4, MgCl2, C(graphite) B AlF3, MgCl2, C(graphite) C MgCl2, SO2, SiCl4 D Al2O3, Cu, PCl5 11 Which of the following shows the correct trend for Period 3 elements? A B C D Concept: Chemical Bonding; structure and bonding. Answer: A simple molecular structure giant ionic structure giant covalent structure metallic A SiCl4 MgCl2 C(graphite) B AlF3, MgCl2 C(graphite) C SO2, SiCl4 MgCl2 Cu D PCl5 Al2O3 For students who do not recall that BeCl2 is simple covalent, refer to Chemical Bonding Notes on pages 3-32 to 3-33. electronega