CJC 2026 Prelim P1 Ans

1 9476/01 CJC JC2 Preliminary Examination 2026 [Turn over CANDIDATE NAME CLASS 2T CHEMISTRY 9476/01 Paper 1 Multiple Choice 17 September 2026 1 hour You must answer on the multiple choice answer sheet. You will need: Multiple choice answer sheet Data Booklet Soft clean eraser Soft pencil (type B or HB is recommended) INSTRUCTIONS • 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 multiple choice answer sheet. • Follow the instructions on the multiple choice answer sheet. • Write your name, class and NRIC number on the multiple choice answer sheet in the spaces provided unless this has been done for you. • Do not use correction fluid or tape. • You may use an approved calculator. INFORMATION • The total mark for this paper is 30 • Each correct answer will score one mark. • Any rough working should be done on this question paper. This document consists of 28 printed pages. WORKED SOLUTIONS 1 2 3 4 5 6 7 8 9 10 A D A D A A C D D D 11 12 13 14 15 16 17 18 19 20 D B A C C B A D D C 21 22 23 24 25 26 27 28 29 30 C B B C C A C A B B Catholic Junior College JC 2 Preliminary Examinations Higher 2

2 1 Which statement correctly explains the trends in ionisation energies in the Periodic Table? A Oxygen has a lower first ionisation energy than nitrogen because the paired 2p electrons experience repulsion with each other. B Iron and cobalt have first ionisation energies that only differ by 5 kJ moI -1 because they both lose an electron from a 3d orbital. C The first ionisation energies increase from sulfur to argon across Period 3 because each successive element has one more electron in the valence shell. D The second ionisation energy of nitrogen is greater than the first ionisation energy of oxygen because the species involved are isoelectronic but oxygen has the greater nuclear charge. Concept: Atomic Structure – Trends in ionisation energies and exceptions N: 1s22s22px12py12pz1 O: 1s22s22px22py12pz1 Statement A is correct as seen from the electronic configurations of N & O where the 2p electron in O occupy the same orbital hence experiencing inter-electronic repulsion. Thus, less energy is required to remove a paired 2p electron from O as compared from removing an unpaired electron from N. Fe: 1s22s22p63s23p63d64s2 Co: 1s22s22p63s23p63d74s2 Statement B is incorrect as the first electron is removed from 4s orbital in Fe & Co. Statement C is incorrect as increasing IE across the same period is due to increasing nuclear charge across a period and not due to additional electrons across the period. Statement D is incorrect because the ions formed N 2+ (5 electrons) and O + (7 electrons) do not have the same electronic configurations and therefore are not isoelectronic even though oxygen has a greater nuclear charge. Ans: A

3 9476/01 CJC JC2 Preliminary Examination 2026 [Turn over 2 Which of the following statements explains why graphite has a high melting point? 1 The electrons in graphite are delocalised. 2 The intermolecular forces of attractions between graphite molecules are strong. 3 Graphite has an extensive network of strong covalent bonds between carbon atoms. A 1 and 3 B 1 only C 2 and 3 D 3 only Concept: Chemical Bonding – Giant Molecular Structure Statement 1: Although graphite has delocalised electrons (which helps it to conduct electricity due to mobile charge carriers), this does not help to explain the bonding. Statement 2: The intermolecular forces of attractions (instantaneous dipole - induced dipole attractions) between the graphite layers are weak. Statement 3: Graphite has a giant covalent lattice structure and extensive network of strong covalent bonds between the carbon atoms. Hence this requires large amounts of energy to overcome, leading to a high melting point. Answer: D

4 3 Which of the following species is planar? 1 XeF4 2 H2S 3 SO3 4 PCl4+ A 1,2 and 3 only B 1 and 3 only C 1 and 4 only D 3 only Concept: Chemical Bonding – Shapes of Molecules Answer: A 1 2 3 4

5 9476/01 CJC JC2 Preliminary Examination 2026 [Turn over 4 The table shows the boiling points of three alkanes. isomers boiling point / C hexane 68.7 2–methylpentane 60.0 2,2–dimethylbutane 49.7 Which is the factor that causes the difference in boiling point? A different relative molecular masses B different number of carbon–carbon bonds C strength of intermolecular hydrogen bonding D extensiveness of intermolecular instantaneous dipole–induced dipole attractions 5 The successive ionisation energies, in kJ mol–1, of elements A and B are given below, starting with the first ionisation energy. A 580 1820 2740 11600 14800 18400 23300 B 940 2050 2970 4140 6590 7880 14900 Which statements about elements A and B are true? 1 The first ionisation energy of A is lower than that of the element preceding it in the Periodic Table. 2 A and B form a compound with the formula A3B2. 3 A has a giant metallic structure. A 1 and 3 only B 3 only C 2 and 3 only D 1 and 2 only Concept: Chemical Bonding – Intermolecular Forces Boiling points decrease as the degree of branching increases. Elongated straight chain molecules such as hexane have larger surface area of contact between adjacent molecules than their branched chain isomers of 2–methylpentane and 2,2 – dimethylbutane which are spherical in shape. Straight chain molecules have more extensive intermolecular instantaneous dipole– induced dipole (id-id) attractions, hence more energy is required to overcome these attraction resulting in a higher boiling point Answer: D

6 Concept: Atomic Structure/Periodic Table. Identification of Group number from successive ionisation energies. Answer: A A is in Group 13 (Al) and B is in Group 16 (Se). The sharpest increase in ionisation energy occurs between third and fourth ionisation energy of A and sixth and seventh ionisation energy of B. Statement 1: Correct. Valence electron of A is removed from np orbital while valence electron of the element preceding it is removed from ns orbital. Hence, valence electron of A is further away and less strongly attracted to the nucleus. Less energy is required to remove the valence electron from A. Statement 2: Incorrect. The formula of their ions are A3+ and B2- respectively. Thus the formula should be is A2B3 Statement 3: Correct. Since it belong to Group 13, it has a giant metallic structure. 6 Use of the Data Booklet is relevant to this question. Which contains the greatest number of molecules? A 3.00 g of methyl ethanoate B 600 cm3 of helium (measured at r.t.p.) C 3.00 g of liquid bromine D 6.00 x 1021 molecules of water No. of moles of Br2 = (30 / 1000) x 0.200 = 0.006 mol No. of moles of KBr = 1.19 / (39.1 + 79.9) = 0.010 mol Mole ratio of Br2 : KBr is 3: 5 3Br2 + ?KOH → 5KBr + KBrOy + ?H2O 3Br2 + (5+1)KOH → 5KBr + KBrOy + 3H2O 6 O 3 O 3 O Ans: C Concept: MCS – mole calculation A – ➔ Mr = 3(12) + 6 + 2(16) = 74.0 Amt. of methyl ethanoate = 3.00  74.0 = 0.0405 mol No. of molecules = 0.0405 x 6.02 x 1023 = 2.44 x 1022 B – Amt. of He = 600  24000 = 0.025 mol No. of molecules = 0.025 x 6.02 x 1023 = 1.51 x 1022 C – Mr of Br2 = 2(79.9) = 159.8, Amt. of Br2 = 3.00  159.8 = 0.0188 mol No. of mole