2024 JPJC Prelim P2 answers

© Jurong Pioneer Junior College [Turn Over NAME CLASS 23S JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2024 CHEMISTRY 9729/02 Paper 2 Structured Questions 10 September 2024 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and index number in the spaces at the top of this page. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 7 2 11 3 13 4 14 5 12 6 18 Penalty (delete accordingly) Lack 3sf in final answer –1 / NA Missing/wrong units in final ans –1 / NA Bond linkages –1 / NA Total 75 This document consists of 19 printed pages and 1 blank page.

2 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2024 Answer ALL the questions in the spaces provided. 1 Use of the Data Booklet is relevant to this question. The most recent updates to the Periodic Table occurred in 2016, when IUPAC officially recognised the discovery of four new elements: Nihonium (Nh), Moscovium (Mc), Tennessine (Ts), and Oganesson (Og), completing the seventh row of the table. Table 1.1 lists the number of protons, neutrons and electrons in four particles, each from one of the elements mentioned above. Each particle may be an atom, an anion or a cation. Table 1.1 element particle atomic no. nucleon no. no. of protons no. of neutrons no. of electrons Nihonium (Nh) Nh 113 286 113 173 113 Moscovium (Mc) Mc3− 115 290 175 118 Tennessine (Ts) Ts− 117 177 Oganesson (Og) Og 294 118 118 For Examiner’s Use (a) Based on the positions of the four new elements in the Periodic Table, state the Group that each of the elements belong to. Nh …………………… Mc …………………… Ts …………………… Og …………………… Nh: Group 13 Mc: Group 15 Ts: Group 17 Og: Group 18 All groups correct. Do not accept Group 3, 5, 7, 0. [1] [1]

3 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2024 [Turn Over Examiner’s comments: Students would need to recall that elements in the periodic table are arranged according to their atomic number / proton number. By referring to the Data Booklet, it is clear that the 4 missing elements belong to Group 13, 15, 17 and 18 respectively. Common mistakes that are not accepted: - Group 3, 5, 7, 8, 0 - Group III, V, VII, VIII, VIII (b) Fill in the empty spaces in Table 1.1. element parti -cle atomic no. nucleon no. no. of protons no. of neutrons no. of electrons Nihonium (Nh) Nh 113 286 113 173 113 Moscovium (Mc) Mc3- 115 290 115 175 118 Tennessine (Ts) Ts- 117 294 117 177 118 Oganesson (Og) Og 118 294 118 176 118 [2] Examiner’s comments: Generally, quite well done. A few careless mistakes. Students need to understand the following: - Atomic no. = no. of protons - Nucleon no. = no. of protons + no. of neutrons - To find no. of electrons: o For cations: take away the relevant no. of electrons o For anions: add the relevant no. of electrons

4 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2024 (c) A stream consisting of -particles (He2+) is subjected to an electric field as shown in Fig. 1.1 below. Fig. 1.1 Determine, by calculation, the angle of deflection for a second stream consisting of Mc3- that is passed through the same electric field. Since angle of deflection α charge/mass, angle of deflection of -particles 2 / 4 ------------------------------------------ = ------------ angle of deflection of Mc3- 3 / 290 Solving, angle of deflection of Mc3- = -1.6551 ≈ -1.66o [1] Examiner’s comments: Not so well done. Key concept here is that the angle of deflection α charge/mass. Common mistakes: - did not apply the correct formula; - did not indicate – or state towards the positive plate to show direction (did not penalise this time round); - did not round off to 3 s.f. (d) The simplified electronic configuration of Nihonium is given below. Nh: [Rn] 5f14 6d10 7s2 7p1 (i) Using information from Table 1.1 , g ive the simplified electronic configuration of Moscovium (Mc). Mc: [Rn] 5f14 6d10 7s2 7p3 Examiner’s comments: This was meant to be quite a doable part of the question but there were quite a number of mistakes. Looking at the atomic/proton numbers, Mc should have 2 more electrons than Nh. Common mistakes that are not given credit: - 5f14 6d10 7s2 7p3 - [Rn] 5f14 6d10 7s2 7p6 [1] + − -rays Source 80o

5 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2024 [Turn Over (ii) Predict if N h or Mc would have a higher first ionisation energy. Briefly explain your answer. Mc would have a higher first ionisation energy than Nh. Across the period, nuclear charge increases while shielding effect remains constant, leading to an increase in effective nuclear charge OR strong nuclear attractions on the valence electrons [2] Examiner’s comments: This part was generally well done. A common mistake was to use the explanation of increasing IE down a group, i.e. increase in shielding effect outweighs increase in nuclear charge, rather than across a period. Another point that students tend to miss out is the increase in nuclear charge which is due to an increase in proton number. Please take note that nuclear charge is NOT equivalent to effective nuclear charge. [Total: 7] 2 (a) Catalytic converters containing platinum solid convert over 90% of carbon monoxide and gaseous nitrogen oxides into harmless gases such as CO 2 and N2. Outline the mode of action of the platinum catalyst. The catalyst in the catalytic converter is a heterogeneous catalyst. The reactant molecules adsorb(*) onto the surface of the metal catalyst (Pd & Pt) via formation of weak (temporary) bonds. This weakens the bonds in the reactant molecules and thus provides an alternative reaction mechanism of lower activation energy . The surface concentration of the reactants also increases. The product molecules desorbs from the catalyst surface, making the surface available for adsorption of new reactant molecules. [2] Examiner’s comments: This part was not so well done. Students need to understand and apply the mode of action for heterogenous catalyst. Some wrote as homogenous/heterozygous/heterolytic etc, which were all incorrect. The key mode of actions would include: - Adsorption (not absorption) - Reaction (weaking of bonds and forming of new bonds) - Desorption

6 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2024 0 0.001 0.002 0.003 0.004 0.005 0.006 0 20 40 60 80 100 120 140 160 180 200 One reaction in which air can be polluted with nitrogen oxides is shown below. 2NO(g) + O2(g) → 2NO2(g) 0.05 mol dm -3 of NO and 0.005 mol dm -3 of O2 were mixed and the rate of reaction was followed by measuring the concentration of oxygen remaining after various time intervals. The following graph was obtained. (b) (i) Define the term half-life of a reaction. Half-life, t½, is the time taken for the reactant concentration to decrease to half of