2022 RI Prelim P3 (Ans)

1 © Raffles Institution 2022 9729/03/S/22 2022 Y6 H2 Chemistry Preliminary Exam Paper 3 – Suggested Solutions Section A 1(a)(i) [CO32−] = 0.010  20 1000 20 + 20 1000 = 5.00  10–3 mol dm–3 As maximum amount of AgCl have been precipitated without precipitating Ag2CO3, the solution is saturated with Ag2CO3. Hence, ionic product = Ksp(Ag2CO3) [Ag+]2(5.00  10–3) = 8.1  10–12 [Ag+] = 4.025  10–5 mol dm–3 As the solution is also saturated with AgCl, [Ag+][Cl–]remaining = (4.025  10–5)[Cl–]remaining = 2.0  10–10 [Cl–]remaining = 4.97  10–6 mol dm–3 Comments: • Instead of using the concentration of CO32−, s everal students incorrectly substituted the amount/number of moles of CO32− into the Ksp expression. • A handful of students did not consider the dilution of CO32− upon the mixing of the two solution, KCl(aq) and Na2CO3(aq). • Some students attempted to calculate the solubility of Ag2CO3 in water, without realising that there was already CO32− present before the addition of AgNO3. 1(a)(ii) Initial [Cl–] in mixture = 0.010  20 1000 20 + 20 1000 = 5.00  10–3 mol dm–3 Percentage of Cl– precipitated = (5.00  10–3) – (4.97  10–6) 5.00  10–3  100% = 99.9% This is an effective method. Comments: • Several students erroneously subtracted the concentration of Cl– remaining from the amount of Cl– present initially.

2 © Raffles Institution 2022 9729/03/S/22 1(b)(i) Co-ordination number is the number of nearest neighbouring ions that surrounds an ion of opposite charge. Comments: • Students are advised to read the question carefully . The question requires students to define the term co-ordination number when used to refer to a crystal lattice. For the crystal lattices given in this question, the bonds formed are ionic in nature (rather than dative covalent, which are present in trans ition metal complexes). • Weaker responses referred to the number of ionic bonds that an ion forms with other oppositely charged ions. Students should note that ionic bonds are non - directional (i.e. ionic bonds can also be formed between oppositely charged ions that are not in close vicinity). 1(b)(ii) As the Ba2+ ion is larger than the Ca 2+ ion, the Ba2+ ion is able to accommodate a greater number of chloride ions surrounding it. Comments: • Several students have the misconception that the larger Ba2+ ion is less sterically hindered. • A handful of students wrongly concluded that the lower charge density of Ba 2+ allowed for a larger coordination number. 1(b)(iii) The ions in CaCl2 are less tightly packed in the lattice structure. Hence, less energy is required to overcome the weaker electrostatic forces of attraction between Ca 2+ and Cl– ions. OR Due to its smaller size/radius, Ca2+ has a higher charge density and polarising power than Ba2+. As such, CaCl2 has a greater covalent character than BaCl2, resulting in an unexpectedly lower melting point than that of BaCl2. Comments: • Many students focused on explaining why the lattice energy of CaCl2 is expected to be more exothermic than that of BaCl2, rather than why the melting point of CaCl2 is lower. 1(c)(i) As the process occurs readily (or is spontaneous), ∆G is negative. Since there is a decrease in the number of ways the water particles can be arranged, ∆S is negative. ∆G = ∆H − T∆S. Since −T∆S is positive, ∆H1 must be negative. Comments: • Several students simply stated that ∆G is negative and ∆S is negative without any explanation of how they arrived that the conclusions. • In their explanation of why ∆S is negative, students are expected to refer to the decrease in the number of liquid particles (or the number of ways the water particles can be arranged).

3 © Raffles Institution 2022 9729/03/S/22 1(c)(ii) The lattice energy (LE) of calcium chloride is the energy released when 1 mole of solid CaCl2 is formed from Ca2+(g) and Cl−(g) at 1 bar and 298 K. Comments: • Students are advised to learn the definitions well. 1(c)(iii) +177 + 590 + 1150 + 2(–242) + LE(CaCl2) = –796 LE(CaCl2) = –2230 kJ mol–1 Comments: • A number of students erroneously included bond energy of C l-Cl into the calculations, without realising that the enthalpy change of formation of C l−(g) is not the same as the electron affinity of Cl(g). • Some students drew the arrows in the wrong direction (e.g. upwards for exothermic reactions). +177 or ∆Hatom(Ca(s)) Ca2+(g) + Cl2(g) + 2e– Ca2+(g) + 2Cl–(g) + 590 + 1150 or 1st IE + 2nd IE of Ca 2  ∆Hf(Cl–(g)) = 2(–242) LE(CaCl2) Ca(s) + Cl2(g) Ca(g) + Cl2(g) CaCl2(s) –796 or ∆Hf(CaCl2(s)) 0 enthalpy

4 © Raffles Institution 2022 9729/03/S/22 1(d) AlCl3 exists as simple covalent molecules but dissolves in water to form [Al(H2O)6]3+ ions. AlCl3(s) + 6H2O(l) ⎯→ [Al(H2O)6]3+(aq) + 3Cl–(aq) In [Al(H2O)6]3+, the high charge density of Al3+ polarises and weakens the O–H bond of the water molecules, allowing the complex ion to undergo hydrolysis to form H3O+. A weakly acidic solution of pH 3 results. [Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5(OH)]2+(aq) + H3O+(aq) PCl5 exists as simple covalent molecules, which hydrolyses in water as the low-lying d-orbitals of the central P atom accepts a lone pair of electrons from H2O. The reaction produces HCl as a product which dissolves in water to form a strongly acidic solution of pH 2. PCl5(s) + 4H2O(l) ⎯→ H3PO4(aq) + 5HCl(aq) Comments: • Students are advised to learn the reactions well. • A handful of students included the reactions with limited water, without realising that the term “resulting solutions” implies reaction with excess water. 2(a)(i) Cyclopentadienyl anion consists of a ring of five sp2 hybridised carbon atoms. Each carbon atom has an unhybridised p orbital that overlaps continuously. The C atom bearing the negative charge contain s a lone pair of electrons in its unhybridised p orbital. Hence, together with the 4  electrons from the two C=C bonds, the cyclopentadienyl anion has a total of 6 delocalised  electrons. Comments: • Many students did not identify the hybridisation of the carbon atom bearing the negative charge while some identified it wrongly. • Many students simply stated the number of delocalised  electrons without any explanation. 2(a)(ii) When naphthalene undergoes substitution reaction, the product formed retains its aromaticity (of the entire molecule) as there are 10 delocalised  electrons in the continuously overlapping p orbitals. However, i f naphthalene undergoes addition reactio n, the product formed is no longer aromatic (across the entire molecule) due to the absence of continuously overlapping p orbitals (or absence of 4n + 2 delocalised  electrons). Comments: • Many students either did not mention that substitution allows the molecule to retain its aromaticity, or that addition causes the molecule to lose its aromaticity.

5 © Raffles Institution 2022 9729/03/S/22 2(b)(i) SO3 + H2SO4 ⇌ HSO3+ + HSO4− Comments: • SO3 + H+ ⇌ HSO3+ is not accepted because the question clearly stated that the formation of HSO3+ results from the protonation of SO3 by H2SO4. 2(b)(ii) Electrophilic substitution Comments: • The curly arrow should start from inside (not outside) the benzene ring for the first step to show the movement of a pair of delocalised electrons. This arrow should be pointing towards the S atom (not O atom) as a C−S bond (not C-O bond) is formed (as shown in the product). • Some students did not show the regeneration of H2SO4 catalyst in the second step. 2(b)(iii) concentrated H2SO4, concentrated HNO3, heat (under reflux) Comments: • Both acids need to be concentrated for the reaction to work. Students should not use the state sy