2024 Prelims TJC H2 Chem P2 (Ans)

9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over TEMASEK JUNIOR COLLEGE 2024 JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME H2 MARK SCHEME CENTRE NUMBER S INDEX NUMBER Chemistry 9729/02 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 on all the work you hand in. 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. Answer all questions in the spaces provided on the Question Paper. 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 Paper 1 /30 Paper 2 Q1 Q2 Q3 Q4 Q5 Total /75 Paper 3 /80 TOTAL (%) /100

2 DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN Answer all questions in the spaces provided. 1 (a) Table 1.1 shows the successive ionisation energies of an element A. Table 1.1 ionisation energies / kJ mol−1 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th element A 945 1794 2735 4839 6056 11690 14180 17370 20550 23830 (i) Explain why the successive ionisation energies of an element always increase. [1] [1] As electrons are removed, number of electrons decrease while nuclear charge remains the same . Hence, there are stronger electrostatic forces of attraction between the nucleus and the remaining electrons and more energy is required for subsequent removal of electrons. (ii) Element A is in Period 4. Using information from Table 1.1 and the Data Booklet, identify element A. Explain your answer clearly. [2] [1] There is a significant increase in the values of the 5th and 6th ionisation energies. This means that the 6th electron is removed from an inner principal quantum / electron shell which is closer to the nucleus , and hence experiences stronger electrostatic forces of attraction. [1] This means that there are 5 valence electrons in element A, which puts it in Group 15, A is arsenic. (iii) State and explain how the first ionisation energy of element A will compare to that of the element to its right in the Periodic Table. [2] [1] Element A will have a higher first ionisation energy as the electron to be removed from element A is unpaired (4s2 4p3). [1] While the electron removed from the element to its right in the periodic table is paired (4s2 4p4). Less energy is required to remove this electron due to inter- electronic repulsion between the paired electrons in the same orbital. ECF from (ii) (b) (i) With the aid of a labelled diagram. describe the structure and bonding of barium. [2]

3 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over [1] [1] Ba2+ cations are arranged in a giant metallic structure held by strong electrostatic forces of attraction between the sea of delocalised electrons and the Ba2+metal cations. (ii) Explain the trend in atomic radii down Group 2. [2] Down the group, [✓] number of protons increases, nuclear charge increases. [✓] Number of electron shell increases [✓] valence electrons are further away from the nucleus. [✓] atomic radii increase. 2[✓] = [1] (iii) The density of magnesium and barium is 1.7 g cm–3 and 3.6 g cm–3 respectively. Explain why the density of barium is significantly greater than that of magnesium. [1] [✓] Both atomic mass and atomic radius ® of Ba atom are larger than Mg atom due to higher number of protons, neutrons and electrons in Ba. Atomic volume (V) α r3 Since density = mass/volume, [✓] The increase in mass is more significant / greater than the increase in volume , hence the mass of Ba per unit volume is larger, therefore higher density. 2[✓] = [1] (iv) When magnesium is burned in air, a mixture of the ionic solids magnesium oxide and magnesium nitride, Mg3N2, is formed. Adding water to Mg3N2 produces an alkaline gas and a white insoluble solid. Suggest an equation for the reaction between Mg3N2 and water, and use it to calculate the mass of white insoluble solid that would be formed from 2.0g of Mg3N2. [2] [1] Mg3N2 + 3H2O → 3MgO + 2NH3 OR Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3 Mr of Mg3N2 = 100.9, Mr of Mg(OH)2 = 58.3 Amt of Mg(OH)2 = 3 x (2/100.9) = 0.0595 mol

4 DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN [1] Mass of Mg(OH)2 = 0.0595 x 58.3 = 3.47g (If the white solid is MgO: ans is 2.40 g) (c) The scientific community was shocked at the recent claim of the discovery of an isotope of a new element, unbibium, with a mass number of 292. This is over 50 mass units higher than uranium, the heaviest known naturally -occurring element. There is a possibility that there is an ‘island of stability’ beyond the known Periodic Table at some very high atomic numbers. (i) The scientists suggested that the atomic number of unbibium is 122. How many neutrons are there in this isotope? [1] [1] 292 – 122 = 170 (ii) If unbibium really exists then it will require a new block of the Periodic Table, corresponding to the occupancy of another type of subshell, beyond the s, p, d and f. This would be a g subshell, which is predicted to be found in the 5 th shell of an atom, i.e. the 5g subshell. Based on the sequence of subshells in the Periodic Table, s, p, d, f, predict how many orbitals there are in a g subshell. [1] [1] 9 [Total: 14]

5 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over 2 (a) The halogens and their compounds have many applications. (i) Explain, in terms of structure and bonding, the trend in the volatility of the halogens. [2] ✓ The halogens have simple molecular structures . Down the group, size of electron cloud of X2 increases and is more easily distorted. ✓ The strength of the instantaneous dipole -induced dipole attractions increases. ✓ Hence requiring more energy to overcome the intermolecular forces. ✓ The boiling point increases and volatility decreases down the group. 2√ = 1m Note: Volatility depends on the boiling point and covalent bonds are not broken. (ii) When heated in chlorine, phosphorous form phosphorus pentachloride, PCl5. Explain the reaction of PCl5 with water. Include the pH value of the resulting solution and write equation for the reaction that occurs. [2] ✓ PCl5 undergoes hydrolysis due to energetically accessible vacant 3d orbitals for dative bonding with water. ✓ It forms strongly acidic solution, pH = 1. 2√ = 1m PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl (aq) [1] Note: - Be familiar with the state symbols. - In limited water: PCl5(s) + H2O(l) → POCl3(l) + 2HCl (g) POCl3(l) + 3H2O(l) → H3PO4(aq) + 3HCl(aq) There is said to be a diagonal relationship between elements of the second and third periods of the Periodic Table. (iii) By analogy with the reaction of SiCl4, suggest a balanced equation for the hydrolysis of BCl3. [1] Deduce the equation by inferring from SiCl4(l) + 2H2O(l) →