RVHS Prelim P3_Student_ANS

H2 Chem Prelim 2 Paper 3 Suggested Answers 1 RVHS 2015 1 (a) (i) Across the period, the atomic number increases, increasing nuclear charge. Hence, ionic radii of isoelectronic ions decrease from Na+ to Si4+, and from P3 to Cl. The ions of Na, Mg, Al and Si loses electrons to form cations of one isoelectronic group containing 10 electrons, while the ions of P, S and Cl gains electrons in the outermost valence shell to form anions of another isoelectronic group containing 18 electrons. Between the two isoelectronic groups, the cations have smaller ionic radii than the anions as there is one less quantum shell of electrons. [4] (ii) AlCl3(s) + 6H2O(l)  [Al(H2O)6]3+(aq) + 3Cl (aq) [Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5(OH)]2+(aq) + H3O+(aq) PCl3(l) + 3H2O(l)  H3PO3(aq) + 3HCl(aq) Or PCl5(s) + 4H2O(l)  H3PO4(aq) + 5HCl(aq) [2] (b) (i) Ca(OH)2(aq) + 2HClO(aq)  Ca(ClO)2(aq) + 2H2O(l) [1] Na+ Mg2+ Al3+ Si4+ P3 S2 Cl  11 12 13 14 15 16 17 Radius/ nm Atomic number

H2 Chem Prelim 2 Paper 3 Suggested Answers 2 RVHS 2015 (ii) pH = pKa + log [ClO] [HClO] 7.6 = log(3.5  108) + log [ClO] [HClO] [ClO] [HClO] = 1.39 Amount of Ca(OH)2 added = 56.0 74.1 = 0.7557 mol Amount of OH = 0.7557  2 = 1.511 mol Let the amount of HClO present = x when freshly treated. Ca(OH)2 + 2HClO  Ca(ClO) 2 + 2H2O Initial amt / mol 0.7557 x 0  Change in amt / mol 0.7557 1.511 +0.7557  Eqm amt / mol 0 x1.511 0.7557  Ca(ClO)2  2ClO [ClO] [HClO] = 2×0.7557 V x  1.511 V = 1.39 x = 2.598 mol [HClO] = 2.598 2500×1000 = 1.04  106 mol dm3 [2]

H2 Chem Prelim 2 Paper 3 Suggested Answers 3 RVHS 2015 (c) (i) PbCl2(s) ⇌ Pb2+ + 2Cl Eqm conc / mol dm3 2.45  102 2  2.45  102 Ksp = [Pb2+][Cl]2 = (2.45  102)(2  2.45  102)2 = 5.88  105 mol3 dm9 (ii) Maximum [Cl] in water = 250×103 35.5 = 7.04  103 mol dm3 Precipitation will occur when I.P > Ksp I.P. = [Pb2+][Cl]2 = [Pb2+] (7.04  103)2 = 5.88  105 mol dm3 [Pb2+] = 1.19 mol dm3 = 2.45  102 g dm3 Minimum mass of Pb required for a 50 cm3 sample = 2.45  102  50 1000 = 12.3 g

H2 Chem Prelim 2 Paper 3 Suggested Answers 4 RVHS 2015 (d) C:H is approximately 1:1  M contains a benzene ring M reacts with ferric chloride to give purple solution  M contains a phenol group M undergoes oxidation with hot acidified K2Cr2O7 with no CO2 evolved to give N  M contains an alcohol N undergoes nucleophilic substitution with 1 mol of phosphorus chloride to give P  N contains a carboxylic acid , P is an acyl chloride , M is a primary alcohol . P undergoes acid-metal reaction with sodium, and (intramolecular) condensation  Q contains an ester (OR Q is neutral with 2 O atoms  Q contains an ester ) M: N: P: Q: [8] [Total: 20] Cl

H2 Chem Prelim 2 Paper 3 Suggested Answers 5 RVHS 2015 2. (a) 𝑝𝑉