2024 MI PU3 H2 Chem EOY P1 answers (final) w FIDI

Class Adm No Candidate Name: This question paper consists of 12 printed pages and 2 blank pages. 2024 Preliminary Examinations Pre-University 3 H2 CHEMISTRY 9729/01 Paper 1 Multiple Choice 18th Sep 2024 1 hour Additional materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class and admission number in the spaces provided at the top of this page and on the Multiple Choice Answer Sheet 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 Multiple Choice Answer Sheet provided. Read the instructions on the Multiple Choice 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 question paper. The use of an approved scientific calculator is expected, where appropriate. FOR EXAMINER’S USE TOTAL (30 marks)

2 DA 1 2 3 4 5 6 7 8 9 10 A 75.6% 88.4% 4.7% 10.5% 2.3% 7.0% 24.4% 18.6% 10.5% 3.5% B 4.7% 2.3% 7.0% 10.5% 20.9% 69.8% 10.5% 37.2% 20.9% 51.2% C 11.6% 2.3% 27.9% 69.8% 53.5% 10.5% 9.3% 11.6% 29.1% 32.6% D 3.5% 2.3% 55.8% 4.7% 18.6% 8.1% 51.2% 27.9% 34.9% 7.0% DA 11 12 13 14 15 16 17 18 19 20 A 60.5% 27.9% 68.6% 29.1% 30.2% 44.2% 32.6% 32.6% 8.1% 7.0% B 25.6% 29.1% 4.7% 50.0% 34.9% 11.6% 25.6% 18.6% 12.8% 31.4% C 2.3% 9.3% 17.4% 3.5% 27.9% 3.5% 17.4% 32.6% 20.9% 9.3% D 7.0% 29.1% 4.7% 12.8% 2.3% 36.0% 19.8% 11.6% 53.5% 47.7% DA 21 22 23 24 25 26 27 28 29 30 A 23.3% 0.0% 10.5% 32.6% 12.8% 16.3% 18.6% 17.4% 73.3% 9.3% B 36.0% 2.3% 32.6% 3.5% 1.2% 18.6% 50.0% 15.1% 4.7% 22.1% C 23.3% 83.7% 18.6% 46.5% 22.1% 15.1% 24.4% 41.9% 4.7% 47.7% D 11.6% 9.3% 32.6% 11.6% 59.3% 45.3% 2.3% 19.8% 12.8% 15.1% Answer Key: 1 2 3 4 5 6 7 8 9 10 A A D C C B D D B B 11 12 13 14 15 16 17 18 19 20 B A A B B A A A D D 21 22 23 24 25 26 27 28 29 30 B C B D D D B A A C

3 [Turn over Section A For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 Which of the following statements are correct about 1 mole of C(s)? 1 It contains 6.02 x 10²³ carbon atoms. 2 It has a mass of 12.0 g. 3 It contains 12 moles of electrons. 4 It occupies a volume of 24.0 dm³ at room temperature and pressure. A 1 and 2 only B 3 and 4 only C 1, 2 and 3 only D 2, 3 and 4 only 1 – True. 1 mole of carbon contains 1 × 6.02 × 1023 atoms (Avogadro’s number). 2 – True. Mass of C = amount × Ar = 1 × 12.0 = 12.0 g 3 – Not true. 1 mole of C contains 6 mole of protons and 6 mole of electrons. 4 – Not true. Carbon is a solid at room temperature and pressure and is not a gas (24.0 dm3 is the molar volume of gas), thus C will occupy a volume depending on the density of the allotrope. (Ans: A – 1 and 2 only) 2 A painkiller pill contains 500 mg of acetaminophen, C₈H₉NO₂. acetaminophen What is the mass of nitrogen present in one pill? [1 g = 1000 mg] A 0.046 g B 0.093 g C 0.151 g D 0.700 g Mr of C8H9NO2 = 8(12.0) + 9(1.0) + 14.0 + 2(16.0) = 151.0 Mass of nitrogen in one pill = 14.0 151.0 × 0.500 g = 0.04636 g = 0.046 g (Ans: A)

4 3 Ammonia reacts with boron trifluoride in the following equation. NH₃ + BF₃ → BF₃.NH₃ What is the role of NH₃ in the reaction? A Brønsted-Lowry acid B Brønsted-Lowry base C Lewis acid D Lewis base Brønsted-Lowry acid is a H+ donor while Brønsted-Lowry base is a H+ acceptor. N on NH3 has a lone pair which can be donated (Lewis base) into the energetically accessible vacant 2p orbital of B in BF3 (Lewis acid). (Ans: D) 4 When excess potassium iodide, KI, is added to potassium iodate(V), iodine is produced. IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O Which of the following statements is incorrect about the reaction? A Iodine in IO₃⁻ was reduced. B Iodine in I⁻ was oxidised. C The reaction is a disproportionation reaction. D The oxidation number of iodine in I₂ is 0. A – True. Oxidation number of I in IO3– is +5 and it decreases to 0 in I2 (reduced). B – True. Oxidation number of I in I– is –1 and it increases to 0 in I2 (oxidised). C – Not true. Disproportionation is when one species simultaneously undergoes reduction and oxidation. In this case, two different species are undergoing reduction and oxidation separately (to form the same product – the reverse of disproportionation –called comproportionation) D – True. The oxidation numbers of both I atoms must be the same (no electronegativity difference between the two atoms) and must add up to 0 (no overall charge). (Ans: C)

5 [Turn over 5 Which of the following statements is not true of an element in Period 3? A It has electrons in the spherically shaped orbitals. B It has three dumb-bell shaped orbitals in its outermost shell. C It has five orbitals with four lobes in each orbital. D It is able to accommodate more than 8 electrons in its outermost shell. A – True. Spherically shaped orbitals refer to s orbitals, of which an element in Period 3 does have electrons in the 1s, 2s and 3s orbitals. B – True. The outermost shell of Period 3 elements contain the 3s, 3p and 3d subshells, of which the dumb -bell shaped orbitals are the three 3p orbitals oriented perpendicularly (or orthogonally) to each other: 3px, 3py, 3pz. C – Not true. The orbitals with four lobes are the 3d orbitals (3d xy, 3dyz, 3dxz, 3dx2-y2) and there are only 4, not 5, of them. The 3dz2 orbital has two lobes and a ring. D – True. The outermost shell can contain 2 + 6 + 10 = 18 electrons (> 8). (1 orbital × 2e– = 2e– in 3s, 3 orbitals × 2e– = 6e– in 3p, 5 orbitals × 2e– = 10e– in 3d) (Ans: C)

6 6 A beam of Na+ 11 23 ions were passed through an electric field and it was found to deflect by an angle of +8.5º. What is the angle of deflection for a beam of P3− 15 31 ions travelling at the same speed through the same electric field? A –34.4º B –18.9º C –18.7º D –3.8º Angle of deflection ∝ 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 Angle of deflection = k ( 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 ) +8.5 = k( +1 23) k = +195.5 Angle of deflection = k ( 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 ) = (+195.5)( −3 31) = –18.9° (Ans: B) 7 Which of the following correctly describe the structure and bonding in carbon? 1 It contains strong C–C covalent bonds. 2 It can contain weak instantaneous dipole-induced dipole forces of attraction. 3 It can have a giant lattice structure. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 1 – True as carbon forms giant covalent structures such as in graphite (each C covalently bonded to 3 other C atoms) or diamond (each C covalently bonded to 4 other C atoms). 2 – True as in graphite, each layer of C atoms (in a hexagonal repeating pattern) is not covalently bonded to the adjacent layers but only has weak instantaneous dipole-induced dipole forces of attraction. 3 – True as in diamond, the carbon atoms are arranged tetrahedrally in a giant repeating pattern (lattice). (Ans: D – 1, 2 and 3)

7 [Turn over 8 An unknown element J exists as a liquid at 35 ˚C and is electrically conductive in both the solid and liquid states. Another unknown element K exists as a gas at 35 ˚C and does not conduct electricity in both the solid and liquid states. Which of the following statements are false about the compound formed between J and K? A It could conduct electricity in the molten state. B It could exist as non-polar molecules. C It could have the formula JK₂. D It could