2023 ASRJC H2 Chem Prelim P2 MS

ASRJC JC2 PRELIM 2023 9729/02/H2 [Turn over Anderson Serangoon Junior College 2023 JC2 H2 Chemistry Preliminary Examination Paper 2 Suggested Solutions 1 The common chlorides of Period 3 elements are shown in Table 1.1. Table 1.1 Period 3 chloride NaCl AlCl3 SiCl4 PCl5 Bonding C Structure S pH of aqueous solution 3 (a) A 3.30 g sample of a Period 3 chloride is heated to 500 K in a sealed flask of 250 cm3. At this temperature, the chloride is a gas and the pressure in the flask is 323 kPa. Assuming the gas behaves ideally, calculate the Mr of the Period 3 chloride. Hence, use the chlorides given in Table 1.1 to deduce its formula. [2] 563.23 10 250 10 0.01948.31 500n −  ==  Mr = 3.30 0.0194 = 170.1 [1] SiCl4 [1] matches the Mr from the table. 1 (b) (i) Complete Table 1.1 by • identifying the bonding shown by each chloride under standard conditions. Use C = covalent, I = ionic, M = metallic • identifying the structure shown by each chloride under standard conditions. Use G = giant, S = simple • stating the pH of the aqueous solution of each chloride at 298 K. [3] Period 3 chloride NaCl AlCl3 SiCl4 PCl5 Bonding [1] I C C C Structure [1] G S S S pH of aqueous solution [1] 7 3 1 or 2 1 or 2

2 ASRJC JC2 PRELIM 2023 9729/02/H2 (ii) At certain temperature, aluminium chloride exists as A l2Cl6 molecules. Draw the structure of Al2Cl6. State how the shape with respect to each Al atom changes from AlCl3 to Al2Cl6. [2] Al Al Cl Cl Cl Cl Cl Cl [1] Trigonal planar about Al to tetrahedral about Al [1] (iii) SCl2 is also a chloride of Period 3 element. It is a cherry-red liquid that reacts vigorously with water to form a strongly acidic solution. Use this information to deduce the bonding and structure shown by SCl2. Structure: …………………………. Bonding: ………………………….. Explanation:…………………………………………………………………………….…. ……………………………………………………………………………………………… ……………………………………………………………………………………………[2] Structure: simple / molecular, because it has a low melting / boiling point Bonding: covalent, because it undergoes complete hydrolysis. [1] for correct structure and bonding Explanation: it has a low melting point (as it is a liquid) and it undergoes complete hydrolysis (reacts vigorously with water). [1] for correct explanations on m.p. and hydrolysis SCl2 is formed when sulfur, S 8, reacts with an excess of chlorine in a series of steps. step 1 S8(s) + 4Cl2(g) → 4S2Cl2 (l) Hr = –58.2 kJ mol–1 step 2 S2Cl2(l) + Cl2(g) 2SCl2(l) Hr = –40.6 kJ mol–1 (iv) Calculate the enthalpy change of formation, Hf, of SCl2(l). [2] 1 8 S8(s) + Cl2 → SCl2(l) Hf 1/8 S8(s) + Cl2 → SCl2(l) Hf = ? 1/8 S8(s) + 1/2 Cl2(g) → 1/2 S2Cl2 (l) Hr = (–58.2/8) kJ mol–1 1/2 S2Cl2(l) + 1/2 Cl2(g) SCl2(l) Hr = (–40.6/2) kJ mol–1 [M1 for both] 1/8 S8(s) + Cl2 → SCl2(l) –27.6 kJ mol–1 [1]

3 ASRJC JC2 PRELIM 2023 9729/02/H2 (v) State the effect of a decrease in pressure on the position of equilibrium in step 2. Explain your answer. [1] When pressure decrease equilibrium position will shift to the left to increase pressure by producing more gaseous molecules [1] (c) Sulfur also reacts with chlorine to form S2Cl2. Fig. 1.1 shows one of the constitutional isomers of S2Cl2 S Cl S Cl Fig. 1.1 (i) Define the term constitutional isomer. [1] Molecules / isomers with the same molecular formula but different structural formulae [1] (ii) Using the isomer of S2Cl2 shown in Fig 1.1, • State the oxidation state of S. • Suggest a value for the Cl–S–S bond angle. Explain your answer. [3] Oxidation state: +1 [1] -1 +1 +1 -1 104.5o or 105o [1] Sulfur has two bond pairs and two lone pairs of electrons AND lone pair-lone pair repulsion is greater [1]

4 ASRJC JC2 PRELIM 2023 9729/02/H2 (iii) Draw the ‘dot-and-cross’ diagram of another isomer of S2Cl2, given that • one of the S atom has an oxidation state of zero. • the other S atom is the central atom. [1] SCl Cl S X X X X X X XX XX XX or SCl Cl S X X X X XX XX XX X X (d) SO3 can react with SC l2 to fo rm liquid SOCl2 and another gas which turns aqueous acidified potassium manganate(VII) from purple to colourless. (i) Write a balanced chemical equation for the reaction between SO3 and SCl2. [1] SO3 + SCl2 → SOCl2 + SO2 [1] (ii) Draw the organic product formed when compound A is reacted with SOCl2. [1] C HH OH CH2HO COOH Compound A C HH Cl CH2HO COCl [1] [Total: 19]

5 ASRJC JC2 PRELIM 2023 9729/02/H2 2 (a) Phthalic anhydride is an important industrial chemical for the synthesis of phthalic esters, which are used as plasticisers to soften plastics. Phthalic anhydride Phthalic acid (C8H6O4) The reactivity of phthalic anhydride is similar to that of acyl chloride. Hydrolysis of phthalic anhydride produces phthalic acid, which is an aromatic dicarboxylic acid that can ionise in two stages, each with its associated pKa value as shown in Table 2.1. Table 2.1 C8H6O4 C8H5O4– + H+ pKa1 = 2.89 C8H5O4– C8H4O42– + H+ pKa2 = 5.51 (i) Suggest reagents and conditions for hydrolysis of phthalic anhydride to form phthalic acid. [1] H2O(l), room temperature [1] (ii) Suggest two reasons why the pKa2 of phthalic acid is higher than its pKa1. [2] A higher pKa2 suggests that C8H5O4– is a weaker acid than C8H6O4. For pKa1, it involves removal of H + from a neutral C8H6O4/molecule. For pKa2, it is more difficult to r emove H+ from a negatively charged C8H5O4–/conjugate base / results in repulsion between two negatively charged –CO2– in close proximity. [1] (Intramolecular/internal) hydrogen bonding in C8H5O4–/conjugate base which stabilise it would be disrupted when it further dissociates. [1] O O O O H  [1]: idea that pKa2 leads to the formation of a more negative conjugate base which made it harder for the H + to be removed (stronger attraction) / caused repulsion between the 2 negatively charged –CO2– [1]: idea that pKa2 disrupts the (internal) hydrogen bonding within C8H5O4–

6 ASRJC JC2 PRELIM 2023 9729/02/H2 (b) A student pipetted 10 cm 3 of 0.50 mol dm –3 of phthalic acid into a conical flask and 20.00 cm 3 of 0.50 mol dm-3 of potassium hydroxide was required for complete neutralisation. (i) Calculate the pH of 0.50 mol dm –3 solution o