CJC 2024 Prelim H1 Chem Paper 1 Solutions

1 CHEMISTRY 8873/01 Paper 1 Multiple Choice 12 September 2024 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 11 printed pages Catholic Junior College JC 2 Preliminary Examinations Higher 1 WORKED SOLUTIONS

2 8873/01/CJC JC2 Preliminary Examination 2024 For each question there are four possible answers, A, B, C and D. Choose the one you consider to be correct. 1 A sample of bromine contains a mixture of two isotopes, 79Br and 81Br. The relative atomic mass of Br in this sample is 79.92. What is the percentage abundance of each isotope? 79Br 81Br A 46.0% 54.0% B 41.2% 58.8% C 54.0% 46.0% D 58.8% 41.2% Concept: Mole Concept and Stoichiometry, percentage abundance Answer: C The relative atomic mass of Br is given as 79.92. As the weighted average is closer to 79 than 81, clearly there is a greater abundance of 79Br. Hence A and B are wrong. Let x be the percentage abundance of 79Br. Hence (100-x) is the abundance of 81Br. ( x 100) 79 + ( 100-x 100 ) 81 = 79.92 x = 54.0% 2 Use of Data Booklet is relevant to this question. How many atoms of silver (proton number 47) are there in a pure silver ring with a mass of 2.88 g? A 1.20 x 1021 C 1.20  1022 B 2.00 x 1021 D 1.61  1022 Concept: Mole Concept and Stoichiometry, Use of Avogadro’s number Answer: D From the Data Booklet, Ar of silver = 107.9 and Avogadro’s number = 6.02 × 1023 Hence number of atoms of silver = 2.88 ÷ 107.9 × 6.02 × 1023 = 1.61 × 1022

3 8873/01/CJC JC2 Preliminary Examination 2024 [Turn over 3 In an experiment, 100 cm3 of a 0.2 mol dm−3 solution of a metallic salt reacts exactly with 50 cm3 of 0.2 mol dm−3 aqueous sodium sulfite, Na2SO3. The half-equation for the oxidation of the sulfite ion is shown below: SO 2– 3 (aq) + H2O(l) → SO 2– 4 (aq) + 2H+(aq) + 2e− If the original oxidation number of the metal ion for the metallic salt is +3, what is the new oxidation number of the metal ion? A +1 B +2 C +4 D +5 Concept: Mole Concept and Stoichiometry, Oxidation numbers Answer: B Amt of SO32− = 𝟓𝟎 𝟏𝟎𝟎𝟎 × 𝟎. 𝟐𝟎 = 1.0× 10−2 mol Amt of e– lost = 2 × 1.0 x 10−2 = 2.00 × 10−2 mol Amt of metallic salt = 𝟏𝟎𝟎 𝟏𝟎𝟎𝟎 × 𝟎. 𝟐𝟎 = 𝟎. 𝟎𝟐𝟎𝟎 mol = 2.0 × 10−2 mol Hence mole ratio of metallic salt: e− gained is 1:1 Since original oxidation number of the metal in the salt is +3, after gaining 1 e−, the new oxidation number will be +2. 4 A and B represent two consecutive elements in the periodic table. A2+ contains n protons while the cation of B contains (n+1) protons and is isoelectronic with A. What is the formula of the chloride formed by B? A BCl B BCl2 C BCl3 D BCl4 Concept: Atomic Structure, Number of protons and charges on ions Answer: C Elements No. of protons No. of electrons Charge A n n-2 +2 B n+1 (n+1)-3 Since A and B are isoelectronic +3 (since there is a loss of 3e-) Hence, chloride of B will have the formula of BCl3.

4 8873/01/CJC JC2 Preliminary Examination 2024 5 Which of the following is the correct electronic configuration of a c opper(II), Cu2+ ion in the ground state? A 1s2 2s2 2p6 3s2 3p6 3d9 4s2 B 1s2 2s2 2p6 3s2 3p6 3d10 4s1 C 1s2 2s2 2p6 3s2 3p6 3d10 D 1s2 2s2 2p6 3s2 3p6 3d9 Concept: Atomic Structure, electronic configuration Answer: D Electronic configuration of Cu element: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Electronic configuration of Cu2+ ion: 1s2 2s2 2p6 3s2 3p6 3d9 (remove one 4s electron first followed by one 3d electron) 6 Which of the following gives the symbols of elements in the order of decreasing first ionisation energy? A Cl, Br, I B F, Ne, Na C Al, Mg, Na D Cl, S, P Concept: Atomic Structure, ionisation energy Answer: A A is correct. Ionisation energy decreases down the group as the distance between the nucleus and the valence electrons increases. B is incorrect. Ionisation energy increases from F to Ne (period 2) but decreases for Na as Na is from period 3. The distance between the nucleus and the valence electrons of Na is further than that of F and Ne, thus smaller electrostatic forces of attraction between the nucleus and valence electrons of Na C is incorrect. There is an anomaly between Al and Mg. Ionisation energy increases from Al to Mg and decreases to Na. Al: 1s2 2s2 2p6 3s2 3p1 Mg: 1s2 2s2 2p6 3s2 The 3p electrons of Al is at a higher energy level than the 3s electrons of Mg hence less energy needed to remove the 3p electron is Al. D is incorrect. There is an anomaly between P and S. Ionisation energy decreases from Cl to S and increases to P P : 1s2 2s2 2p6 3s2 3px13py13pz1 S : 1s2 2s2 2p6 3s2 3px23py13pz1 Due to inter-electronic repulsion between paired electrons in the same orbital of S, less energy is required to remove the 3p electron from S. Students can also check the Data Booklet to confirm their choices.

5 8873/01/CJC JC2 Preliminary Examination 2024 [Turn over 7 Which of the following have giant lattice structures under standard conditions at 298 K? 1 magnesium 2 buckminsterfullerene, C60 3 beryllium chloride 4 sodium fluoride A 1 and 2 only B 1 and 4 only C 3 and 4 only D 1, 2 and 4 only Concept: Chemical Bonding, Giant Lattice Structure Answer: B Option 1: Magnesium is a metal and has a giant metallic lattice structure. Option 2: Buckminsterfullerene exists as simple molecular structure with covalent bonds between the C atoms. Option 3: Beryllium chloride exist as simple molecular structure with covalent bonds between the Be and Cl atoms. Option 4: Sodium fluoride is an ionic compound which has a giant ionic lattice. 8 Which of the following species has the smallest bond angle? A SO3 B BrF2+ C SF6 D IF2− Concept: Chemical Bonding, bond angles Answer: C A trigonal planar 120º B bent 105º C octahedral 90º D linear 180º

6 8873/01/CJC JC2 Preliminary Examination 2024 9 Which one of the following statements about aluminium chloride is incorrect? A AlCl3 has a simple molecular structure. B AlCl3 has a lower melting point than Al2O3. C The Al2Cl6 dimer contains intermolecular hydrogen bonding. D The reaction AlCl3 + Cl ‒ → AlCl4‒ involves the formation of a dative bond. Concept: Chemical Bonding Answer: C A AlCl3 has simple molecular structure due to high charge density of Al3+ which polarizes the Cl- electron cloud to an extent where it forms a covalent bond. B AlCl3 is simple molecular and is held together by intermolecular instantaneous dipole-induced dipole forces and hence will have a lower melting point than Al2O3 which is giant ionic. C The Al2Cl6 dimer contains two co-ordinate bonds, not intermolecular hydrogen bonds. D The Al is electron deficient and hence will accept a lone pair from Cl− to achieve the octet configuration. 10 The table shows the boiling points of some halogenoalkanes. compound boiling point/ °C CH3Cl −24.2 CH3Br 3.6 CH3I 42.4 Which of the following correctl