2025 H1 Chem JC2 Prelim P1 (Q & A) worked solutions JPJC

© Jurong Pioneer Junior College [Turn over NAME CLASS JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2025 CHEMISTRY 8873/01 Higher 1 Paper 1 Multiple Choice Questions 18 September 2025 1 hour Candidates answer on the Question paper. 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 exam index number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C or 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 14 printed pages and 0 blank page.

2 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2025 1 Use of Data Booklet is relevant to this question. The atoms X and Y have the electronic configurations shown below. X, 1s2 2s2 2p6 3s2 3p6 4s2 ; Y, 1s2 2s2 2p5 Which compound are they likely to form? A X2Y B XY C XY2 D XY4 1 Answer C X, 1s2 2s2 2p6 3s2 3p6 4s2  has 2 valence electrons, Group 2 element (Ca); tendency to lose 2 electrons to give Ca2+ Y, 1s2 2s2 2p5  has 7 valence electrons, group 17 element (F); tendency to accept 1 electron to form F− Hence will form CaF2 which is XY2 2 The trend of first ionisation energy for 10 consecutive elements , in Period 2 and 3, is given below. Which of the following statement is correct? A Element E and N belong to the same group. B L has a giant covalent structure. C The oxide of J is amphoteric. D The chloride of I is insoluble in water. E F G H I J K L M N proton number first ionisation energy

3 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2025 [Turn over 2 Answer B The largest drop in first ionisation energy is between H and I. This indicates H is from 2 period 2 and I is from period 3. H is in group 18 and I is in group 1. Hence, E, nitrogen, and M, phosphorous, belong to the same group, group 15. L is in group 14, silicon, and has a giant covalent structure. The oxide of J, magnesium oxide, is a basic oxide. The chloride of I, sodium chloride, is soluble in water. 3 In which pair of molecules are both non–polar? 1 BeH2 and SO3 2 AlCl3 and CH4 3 PCl5 And SCl6 A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only 3 Answer A 1 BeH2 is linear. SO3 is trigonal planar. Both are non–polar. ✓ 2 AlCl3 is trigonal planar. CH4 is tetrahedral. Both are non–polar. ✓ 3 PCl5 is trigonal bipyramidal. SCl6 is octahedral. Both are non–polar. ✓ The following information is for Questions 4 and 5. Nitrotyrosine has been identified as an indicator for cell damage in the human body. N1 O O OH N2 O3 O H H H nitrotyrosine

4 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2025 4 What are the bond angles around N1, N2 and O3 of nitrotyrosine? Bond angle / ° N1 N2 O3 A 105 107 120 B 105 120 105 C 120 120 120 D 120 107 105 4 Answer D N1 is trigonal planar  120 ° N2 is trigonal pyramidal  107 ° O3 is bent  104.5 ≈ 105 ° 5 Which of the following functional groups can be found in nitrotyrosine? You may refer to the structure above. 1 Amide 2 Amine 3 Carboxylic acid A 1, 2 and 3 only B 1 and 2 only C 2 and 3 only D 3 only 5 Answer C N1 O O OH N2 O3 O H H H 1 Amide  2 Amine ✓ 3 Carboxylic acid ✓ nitro phenol carboxylic acid amine

5 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2025 [Turn over 6 Which row correctly describes the structure and bonding present in the solid lattice of the given substance? substance structure bonding A iodine simple covalent instantaneous dipole–induced dipole only B iron giant metallic ionic bonding only C sodium carbonate giant ionic ionic and covalent bonding D graphite giant molecular covalent bonding only 6 Answer: C A Iodine has both covalent bonding between I atoms and instantaneous dipole – induced dipole forces of attraction between I2 molecules. B Iron has metallic bonding. ✓C Sodium carbonate has ionic bonding between Na + and CO 32– ions. There are covalent bonds between the atoms of CO32–. D There are covalent bonds between the carbon atoms of graphite instantaneous dipole–induced dipole forces of attraction between individual layers. 7 Arrange the following compounds in ascending order of boiling point. Compound Formula Mr P CH3CH2CH2CH3 58 Q CH3COCH3 58 R CH3CH(CH3)2 58 S CH3CH2CH2OH 60 A P → Q → R → S B Q → S → P → R C R → P → Q → S D S → P → R → Q

6 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2025 7 Answer C Since P, Q, R and S all have the same Mr, they would have similar number of electrons to polarise. Hence, in this case hydrogen bonding > permanent dipole –permanent dipole > instantaneous dipole–induced dipole. P and Q have the lowest boiling point as both only have instantaneous dipole–induced dipole forces of attractions between their molecules. R has a lower boiling point than P as it has branching which makes the molecule more spherical and thus reducing the extensiveness of instantaneous dipole–induced dipole forces of attraction. Q is a polar molecule with permanent dipole –permanent dipole forces of attraction between molecules. This attraction is stronger than instantaneous dipole–induced dipole forces of attractions, thus boiling point of Q > P. S has the highest boiling point as it is able to form hydrogen bonding between molecules. Hydrogen bonding is stronger than both permanent dipole –permanent dipole and instantaneous dipole–induced dipole forces of attractions. 8 Which of the following statements can be explained by the presence of intermolecular hydrogen bonding only. 1 Ice is less dense than water. 2 Dimerisation of ethanoic acid in hexane. 3 Dissolution of sodium chloride in water. A 1 and 2 only B 2 and 3 only C 1 only D 3 only 8 Answer: A ✓1 Ice is less dense as upon freezing, water molecules form a rigid, open, hexagonal crystal lattice structure due to hydrogen bonds. ✓2 Ethanoic acid dimerizes in benzene due to strong intermolecular hydrogen bonding between the carbonyl oxygen of one molecule and the hydroxyl hydrogen of another, forming a stable ring-like structure known as a dimer 3 When sodium chloride dissolves in water, the ions form ion –dipole interactions with water molecules.

7 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2025 [Turn over 9 Use of Data Booklet is relevant to this question. The maximum tolerable daily intake of ascorbic acid, also known as vitamin C, for adults is 2000 mg, beyond which there may be adverse effects on the human body. What is the number of molecules in 2000 mg of ascorbic acid, C6H8O6? A 6.84 x 1018 B 1.20 x 1024 C 6.84 x 1021 D 1.20 x 1027 9 Answer: C Mr of C6H8O6 = 6(12) + 8(1) + 6(16) = 176 2000 mg = 2000/1000 = 2 g Amount of C6H8O6 = 0.01136 mol Number of molecules = 0.01136 x 6.02 x 1023 = 6.84 x 1021 10 1 mol of acidified Cr 2O72– reacted with 3 mol of Mn 2+. The products are Cr3+ and a manganese cation, of an unknown oxidation state. What is the oxidation state of the manganese cation? A +3 B