EJC 2026 Prelim P2 Ans

© EJC [Turn Over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2026 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME CIVICS GROUP 2 5 – INDEX NUMBER CHEMISTRY Paper 2 Structured Questions 9476/02 27 August 2026 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Answer all questions. Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. Write and shade your index number in the spaces provided. Write your answer to each question in the space provided. Do not use an erasable pen. Do not use correction fluid or tape. The use of an approved scientific calculator is expected, where appropriate. 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 Paper 2 1 / 13 2 / 14 3 / 11 4 / 13 5 / 24 Total / 75 This document consists of 22 printed pages and 2 blank pages.

2 © EJC 9476/02/J2PRELIM/26 1 Carbon can form both discrete molecular compounds like benzene and giant covalent networks like graphite. Neighbouring elements boron and nitrogen can combine to form isostructural analogues. One such analogue is borazine, B 3N3H6, frequently termed as inorganic benzene with the structure shown in Figure 1.1. Figure 1.1 (a) Explain why the first ionisation energy of nitrogen is higher than that of carbon, while the first ionisation energy of boron is lower than that of beryllium. ................................ ................................ ................................ ................................ ............. ................................ ................................ ................................ ................................ ............. ................................ ................................ ................................ ................................ ............. ................................ ................................ ................................ ................................ ............. ................................ ................................ ................................ ................................ ............. ................................ ................................ ................................ ................................ ............. ................................ ................................ ................................ ................................ ......... [3] (b) (i) Borazine is a volatile liquid with a density of 0.810 g cm−3. A 0.400 cm3 sample of liquid borazine is injected into a sealed 0.500 dm3 container. The container is heated to 100 ºC to ensure the sample is completely vapourised. Calculate the final pressure in the container at 100 ºC. [2] Nitrogen has a higher nuclear charge (more protons) than carbon, while experiencing similar shielding (electrons are in the same shell), resulting in increase in effective nuclear charge. This results in stronger electrostatic forces of attraction between the nucleus and the outermost electrons in nitrogen, raising IE. Less energy is required to remove the valence 2p electron in boron that is at a higher energy level than the 2s electron in beryllium, hence the first ionisation energy of boron is lower than that of beryllium. pV = nRT p = mRT MV = (0.810 × 0.400) × 8.31 × (273 + 100) 80.4 × (0.5 × 10-3) = 24982.09 Pa = 25.0 kPa

3 © EJC 9476/02/J2PRELIM/26 [Turn Over Borazine can be synthesised via a gas -phase reaction between diborane, B 2H6, and ammonia, NH3. Their boiling points are shown in Table 1.1. Table 1.1 compound boiling point / ºC borazine, B3N3H6 55 diborane, B2H6 –93 ammonia, NH3 –33 (ii) Using the data in Table 1.1, state and explain which of the gaseous reactants, B2H6 or NH3, will show a greater deviation from ideal gas behaviour at low pressure. ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ . [2] (b) The gas phase reaction between diborane and ammonia to form liquid borazine proceeds via a series of simplified steps as shown. Step 1: B2H6(g) + 2NH3(g) → [H2B(NH3)2]+[BH4]−(s) Step 2: [H2B(NH3)2]+[BH4]−(s) → 2X(g) Step 3: 3X(g) → B3N3H6(l) + 6H2(g) (i) Write an overall chemical equation for the reaction between diborane and ammonia to form liquid borazine. ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ . [1] (ii) Given that X is a molecule, write the molecular formula of X in step 2. ................................ ................................ ................................ ................................ . [1] Ammonia will show a greater deviation. Despite having a lower Mr, it has a significantly higher boiling point than B2H6, indicating that NH3 molecules experience much stronger intermolecular hydrogen bonding compared to the instantaneous dipole – induced dipoles interactions between B2H6 molecules. 3B2H6 + 6NH3 → 2B3N3H6 + 12H2 BNH6

4 © EJC 9476/02/J2PRELIM/26 (iii) Given that the boron atom is bonded to nitrogen atom in X, suggest the structure of X. Hence, identify the type of bond formed between the boron and nitrogen atom. Type of bond: ................................ ................................ ................................ ........... [2] (c) Borazine eliminates hydrogen gas to form a giant covalent molecule called hexagonal boron nitride (h-BN). h-BN is structurally similar to graphite. Figure 1.2 shows a layer of h-BN. Figure 1.2 Despite structural similarities, graphite is a highly efficient electrical conductor along its planar sheets while hexagonal boron nitride is an excellent electrical insulator. (i) Using Figure 1.2, state how many electrons occupy the  system in a single B3N3 ring. ................................ ................................ ................................ ................................ . [1] (ii) Using the difference in electronegativit y between boron and nitrogen , s uggest why graphite is a good electrical conductor while hexagonal boron nitride is an electrical insulator. ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ . [1] [Total: 13] Dative covalent bond / coordinate bond 3 N  2 = 6 electrons Each B and N is bonded to 3 other atoms within the h -BN structure, B has no more valence electrons while N has 2 valence electrons left to contribute to  system. In h-BN, nitrogen is more electronegative than boron , so nitrogen pulls the  electrons strongly toward itself rather than allowing them to move freely across the lattice.

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