JJC H2 Chem 2013 Prelim P3 Soln

© Jurong Junior College 9647/03/PRELIM/2013 2013 JC2 H2 Chemistry Preliminary Examination Paper 3 Suggested Answers 1 (a) (i) Pb(s) + 4HNO3(aq)  Pb(NO3)2(aq) + 2NO2(g) + 2H2O(l) (ii) Al3+ left in solution X exists as Al(H2O)6 3+. Due to high charge density of A l3+, it hydrolyses in water to give a weakly acidic solution. Al3+ polarises the electron cloud of H2O molecule attached to it which will weaken and break its OH bond, releasing H+. Al(H2O)6 3+  Al(H2O)5(OH)2+ + H+ (iii) Since Al2(SO4)3  2Al(H2O)6 3+, [Al(H2O)6 3+] in solution X =  602 0.10060 + 40 = 0. 120 mol dm3 [H+] = Kacid ac? = ..  50120 14 10 = 1.30  103 mol dm3 (b) (i) Hf(NO2(g)) = 0.5 [(+183) + (116)] = +33.5 kJ mol1 (ii) Hr = 3(+33.5) + 1 2 (256) + 1 2 (183) = 137 kJ mol1 (c) (i) (ii) HNO3 can accept H + due to the availability of the lone pair of electrons on O atom of –OH group. After accepting H + from H 2SO4, the unstable H 2NO3 + intermediate formed quickly breaks down to NO2 + and H2O. O N O O H Hr 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g) 3 2 N2(g) + 3O2(g) + H2O(l) 1 2 (+183) 1 2 (256) 3(+33.5) 630

2 © Jurong Junior College 9647/03/PRELIM/2013 1 (d) OR O + HNO 3 O NO2 +H 2O Sn, conc HCl, heat under reflux, followed by NaOH(aq) (e) (i) A weak acid dissociates partially in water. HCN  H+ + CN (ii) At high [H+], equilibrium position of HCN  H + + CN  shifts left, resulting in low [CN ]which slows down step 1. At low [H +], equilibrium position of HCN  H+ + CN shifts right, resulting in low [HCN] which slows down step 2. (iii) I: condensation II: reduction (iv) 631

3 © Jurong Junior College 9647/03/PRELIM/2013 2 (a) (i) Since O 2– has a smaller radius, it is less polarisable than C l–. Hence, Fe 2O3 is more thermally stable/ decomposes at a higher temperature. (ii) Fe2O3 has giant ionic structure while FeCl3 has simple covalent structure. Much larger amount of energy is required to overcome the strong ionic bonds between Fe 3+ and O 2– as compared to that required to overcome the weak van der Waals’ forces between FeCl3 molecules. (b) (i) D : CuCl E: Cu(H2O)6 2+ (ii) Cu(I) in D has a d 10 configuration/ completely filled d–orbitals and hence, electron transition between d–orbitals is not possible. Thus, CuCl is white in colour. Cu(II) in E has a d 9 configuration/ partially filled d–orbital and hence, electrons transition between d–orbitals is possible. In Cu(II) complex ion, the presence of ligands causes the five 3d orbitals to split into 2 sets of different energies. The difference in energies between the 2 sets of 3d orbitals is relatively small such that radiation from the visible regi on region of the electromagnetic spectrum when an electron moves from a d–orbital of lower energy to another partially–filled d–orbi