DHS 2026 Prelim P2 Ans

This document has 22 pages. Any blank pages are indicated. © DHS 2026 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 2 Structured Questions You must answer on the question paper. You will need: Data Booklet 9476/02 14 September 2026 2 hours INSTRUCTIONS • Answer all questions. • Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. • Write your name, centre number and index number in the boxes at the top of the page. • Write your answer to each question in the space provided. • Do not use an erasable pen. Do not use correction fluid or tape. • You may use an approved calculator. INFORMATION • The total mark for this paper is 75. • The number of marks for each question or part question is shown in brackets [ ]. For Examiner’s Use 1 10 2 11 3 15 4 14 5 12 6 13 Total 75

2 © DHS 2026 9476/02 Answer all questions in the spaces provided. 1 (a) (i) Group 2 nitrates decompose when heated strongly according to the following equation. M(NO3)2(s) → MO(s) + 2NO2(g) + 1 2O2(g) The thermal decomposition of Group 2 nitrates is similar to that of carbonates. State how the thermal decomposition temperature of Ca(NO 3)2 will differ from that of Sr(NO3)2. Explain your answer. [3] The decomposition temperature of Ca(NO3)2 is lower than that of Sr(NO3)2. Ca2+ has a smaller ionic radius than Sr 2+ and hence, Ca 2+ has higher charge density and stronger polarising power. Hence, Ca2+ will weaken the N–O bond to a greater extent and a lower amount energy and temperature is required for decomposition to occur. (ii) Be has no reaction with cold water while Ba reacts vigorously with cold water. Explain the difference in observations. [1] Ba is a stronger reducing agent than Be. Hence, Ba reacts more vigorously with cold water. (b) State two basic assumptions of the kinetic theory as applied to an ideal gas. [1] 1. The gas particles have negligible volume compared to that of the container. 2. There are no intermolecular forces of attraction between gas particles. (c) A pure sample of Ca(NO3)2 was completely decomposed in a 15 cm3 sealed test tube at 520 C. The sealed test tube can withstand a maximum pressure of 103421 Pa without cracking. You should ignore the partial pressure of the air in the test tube. (i) Calculate the maximum mass of Ca(NO3)2 that can be decomposed without cracking the sealed test tube. [3] pTVT = nTRT (103421)(15 × 10–6) = nT (8.31)(520 + 273)  nT = 2.3541 × 10–4 mol Since no. of moles of Ca(NO3)2  2.5 moles of gases  no. of moles of Ca(NO3)2 = 2.3541 ×10−4 2.5 = 9.4164 × 10–5 mol Maximum mass of Ca(NO3)2 = 9.4164 × 10–5 × [40.1 + 2(14 + 16 × 3)] = 0.0155 g (3sf) (ii) Suggest why despite using the theoretical maximum mass in (c)(i), the test tube may still crack. You should disregard the effect of temperature on the structural integrity of the test tube. [2] As the decomposition proceeds, any CaO that is produced in the test tube occupies space and this reduces the volume of the test tube that the gases can occupy.

3 © DHS 2026 9476/02 [Turn Over Hence, the pressure exerted by the gases is higher than expected resulting in cracks forming in test tube. [Total: 10] 2 (a) Aluminium chloride is often called a ‘shapeshifter ’ as the hybridisation of the aluminium atom changes when anhydrous AlCl3 acts as a Lewis acid to form [A lCl4]– ion in the presence of chloride ions. (i) Explain what is meant by a Lewis acid. [1] A Lewis acid is an electron pair acceptor. (ii) Draw a ‘dot-and-cross’ diagram to show the bonding in an [AlCl4]– ion. Show outer electrons only. Clx xx xx xx Clx xx x x xx Cl x x xx Cl xxx x x xx Al xx x [1] (iii) Using the Valence Shell Electron Pair Repulsion (VSEPR) theory, predict the shape and bond angle of [AlCl4]–. Explain your reasoning. [2] There are 4 bond pairs and 0 lone pair of electrons around aluminium. According to VSEPR theory, these electron pairs arrange themselves as far apart as possible to minimise repulsion. The shape is tetrahedral and the bond angle is 109.5. (iv) Aluminium undergoes orbital hybridisation, similar to carbon. State the hybridisation of the aluminium atom in AlCl3. Hence explain how AlCl3 acts as a Lewis acid. [1] The hybridisation of the aluminium atom is sp 2 in AlCl3. The aluminium atom in AlCl3 is electron deficient and possesses an energetically accessible vacant p-orbital, allowing it to accept a lone pair of electrons from a Lewis base. (v) Explain, in terms of polarising power of cation and polarisability of anion, why AlCl3 has a greater covalent character than NaF. [2] Al3+ has a smaller ionic radius and greater charge than Na +. Hence, A l3+ has a higher charge density and hence stronger polarising power than Na +. Cl− has a larger ionic radius/electron cloud than F− and hence Cl− is more polarisable than F−. Therefore, AlCl3 has a greater covalent character than NaF. (b) Similar to AlCl3, FeCl3 can also function as a Lewis acid in organic reactions. (i) Write the full electronic configuration of Fe3+. [1] 1s22s22p63s23p63d5

4 © DHS 2026 9476/02 (ii) Sketch, on Fig. 2.1, a diagram of the 2dz orbital present in Fe3+. [1] (iii) Table 2.1 shows the relative abundance of the four main stable isotopes of Fe. Table 2.1 isotope relative abundance (%) 54Fe 5.85 56Fe 91.75 57Fe 2.12 58Fe 0.28 Using the information in Table 2.1, calculate the relative atomic mass of Fe, to one decimal place. Ar of Fe = [(5.85 x 54) + (91.75 x 56) + (2.12 x 57) + (0.28 x 58)] / 100 = 55.90  55.9 (to 1 d.p) [1] (iv) Complete Table 2.2 for 57Fe3+. Table 2.2 species number of protons number of neutrons number of electrons 57Fe3+ 26 31 23 [Total: 11] 3 (a) Methanol is a versatile chemical that can be further utili sed to produce other useful chemicals. It is produced industrially by reacting carbon monoxide with hydrogen gas according to the equation shown below. reaction 1 CO(g) + 2H2(g) ⇌ CH3OH(g) H < 0 (i) An engineer introduced a mixture of carbon monoxide and hydrogen gas into a sealed vessel and it was allowed to reach equilibrium at a fixed temperature. Fig. 3.1 shows the total pressure in the vessel as reaction 1 proceeds. At time , t = 10 min, he then realised that he had used lower amount of carbon monoxide than expected so he introduced more carbon monoxide at t = 10 min and monitored the process till t = 30 min.

5 © DHS 2026 9476/02 [Turn Over Given that equilibrium was established at t = 20 min, complete the sketch in Fig. 3.1. Explain your reasoning. total pressure / Pa time, t / min 10 20 30 Fig. 3.1 total pressure / Pa time, t / min 10 20 30 At t = 10 minutes, there is a sharp increase in the total pressure measured due to the introduction of more carbon monoxide. By Le Chatelier’s Principle, the position of equilibrium will shift to the right to decrease the partial pressure of carbon monoxide. This caused the total pressure to decrease but the new total pressure after the re -establishment of the equilibrium is higher. (ii) Explain, in terms of forward and backward rates of reac