2024 TJC Prelims H1 Paper 2 Structured Qns and Ans

8873 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over TEMASEK JUNIOR COLLEGE 2024 JC2 PRELIMINARY EXAMINATION Higher 1 CANDIDATE NAME MARK SCHEME CENTRE NUMBER S INDEX NUMBER Chemistry 8873/02 Paper 2 Structured Questions 21 August 2024 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name in the spaces provided at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all the questions. Section B Answer one question. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in bracket [ ] at the end of each question or part question. This document consists of 23 printed pages and 1 blank page. For Examiner’s Use MCQ / 30 Section A Q1 / 8 Q2 / 8 Q3 / 12 Q4 / 10 Q5 / 10 Q6 / 12 Section B Q7 / 20 Q8 / 20

2 DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN Section A Answer all the questions in this section in the spaces provided. 1 This question is about copper-containing species. (a) Copper can exist as a number of isotopes. Some copper isotopes are unstable but are potentially useful for nuclear medicine. (i) Complete Table 1.1 for two isotopes of copper. [2] Table 1.1 isotopes protons neutrons electrons 63 29Cu 29 34 [✓] 29 67 29Cu [✓] 29 38 29 Cu isotope and neutrons 2✓ = [1] All protons, electrons (29) [1] (b) Tumbaga is an alloy of copper and gold. A sample of tumbaga was analysed and the composition of the isotopes present in the sample are shown in Table 1.2. Table 1.2 Mass number % abundance 63 56.4 65 x 197 18.5 (i) Determine the percentage abundance, x, of the species of mass number 65. [1] x = 100 – 56.4 – 18.5 = 25.1 [1] (ii) Hence, calculate the relative atomic mass, Ar, of the copper present in this sample of tumbaga. [1] Ar of Cu = (56.4 63) (25.1 65) 56.4 25.1  +  + = 63.6 [1] Common mistake: included gold in the calculation. (c) Brass is an alloy of copper and zinc. The percentage of mass of copper varies from 50% to 85%, depending on the properties needed in the alloy. A series of reactions is carried out in the laboratory to determine the percentage of copper in a sample of brass. 0.30 g of brass is dissolved in acid to give 100.0 cm3 of solution in which all copper is present as Cu 2+ ions. 25.0 cm 3 of the solution is transferred to a conical flask using a pipette and reacted with excess potassium iodide, K I. All the copper is precipitated as Cu I solid and iodine is formed. The iodine requires 21.50 cm3 of 0.0400 mol dm-3 S2O32- to reach the end- point of titration.

3 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over 2Cu2+(aq) + 4I–(aq) → 2CuI(s) + I2(aq) I2(aq) + 2S2O32-(aq) → 2I–(aq) + S4O62-(aq) (i) Determine the number of moles of iodine reacted with S2O32- in the titration. [1] I2  2S2O32- Number of moles of S2O32- = 421.50 0.0400 8.60 101000 − =  mol Number of moles of I2 = ½ x 8.60 x 10–4 = 4.30 x 10–4 mol [1] (ii) Use your result in (c)(i) to calculate the number of moles of Cu 2+(aq) in 100.0 cm3 of the solution. [1] 2Cu2+  I2 Number of moles of Cu2+ in 25.0 cm3 solution = 2 x 4.30 x 10–4 = 8.60 x 10–4 mol Number of moles of Cu2+ in 100.0 cm3 solution = (100/25) x 8.60 x 10–4 = 3.44 x 10–3 mol [1] (iii) Hence, determine the percentage by mass of copper in this sample of brass. [2] Mass of Cu = 3.44 x 10–3 x 63.5 = 0.218 g [1] Percentage by mass of Cu = (0.218/0.30) x 100 = 72.7% [1] [Total: 8]

4 DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 2 Phosgene, Cl2C=O, is a colourless, toxic gas. It is used in the production of pesticides, rubbers and adhesives. (a) (i) State the shape of a phosgene molecule. [1] [1] trigonal planar Note: the structure is (ii) The molecule of phosgene contains both σ (sigma) and π (pi) bonds. Draw labelled diagrams to show how orbitals overlap to form • a σ (sigma) bond • a π (pi) bond. [2] [1] [1] Head-on overlap between p orbital of C and px orbital of O Sideway overlap between p orbitals of C and py or pz orbital of O (b) Table 2.2 shows the electronegativity values of the atoms in phosgene. Table 2.2 atom electronegativity / Pauling units carbon 2.5 chlorine 3.0 oxygen 3.5 (i) Explain what is meant by the term electronegativity. [1] A measure of an atom’s ability to attract shared electrons / electrons in a covalent bond. [1] The greater the electronegativity of an atom, the greater the electron-attracting ability If two atoms of a covalent bond have different electronegativity, the bonding electrons are not equally shared between them and hence a polar covalent bond is formed.

5 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over (ii) Using the information in Table 2.2, deduce whether phosgene is a polar molecule. Explain your answer. [2] The dipole moment [✓] on C=O and C-Cl do not cancel out each other in a trigonal planar arrangement due to different electronegativity difference [✓] between C=O (1 Pauling unit) and C-Cl (0.5 Pauling unit). So, phosgene is a polar molecule [1] which has a net dipole moment. (c) Phosgene reacts with methylamine to form methyl isocyanate. Cl2C=O + CH3NH2 CH3NCO + 2HCl methylamine methyl isocyanate (i) Draw a dot -and-cross diagram to show the bonding present within a methylamine molecule. [1] [1] Common mistake: missing lone pair electron on N (ii) In the presence of a catalyst, methyl isocyanate reacts with itself to form compound A, which is a non-polar molecule. 3CH3NCO(g) C6H9N3O3(s) A Suggest why, at room temperature, methyl isocyanate is a gas but A is a solid. A has larger electron cloud size, leading to greater extent of distortion of electron cloud. [✓] More energy is needed to overcome the stronger id-id interactions between molecules of A, hence A has a higher melting point. [✓] OR At r.t.p, the energy provided is insufficient to overcome the stronger id -id between A molecules hence A exists as a solid. [1] δ+ δ- δ- δ-

6 DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN [Total: 8] 3 (a) Magnesium, aluminium and phosphorus are elements in Period 3 of the periodic table. (i) State the electronic configuration of aluminium. [1] 1s22s22p63s23p1 [1] (ii) State and e xplain the difference in first ionisation energies of aluminium and magnesium. [2] Ionisation energy of Al is lower .[✓] Less energy is required to remove the 3p electron in Al [✓] as it is further away from the nucleus [✓] compared to the 3s electron in Mg. [✓] (iii) Write a balanced equation for the reaction of phosphorus(V) oxide with excess water. [1] P4O10(s) + 6H2O(l) → 4H3PO4(aq) [1]