RVHS H2 Chem 2012 Prelim P3 Soln

H2 Chemistry (9647) Prelims 2012 Paper 3 1 (a) (i) C O (ii) Nickel carbonyl has a simple covalent structure with weak intermolecular forces which require little energy to overcome. (iii) M r = 2.00´8.31´(273 + 50) 1.01´105 ´314´10-6 =169 (iv) 58.7 + (12.0 +16.0)x =169 x = 3.93 = 4 (nearest whole number) Coordination number is 4 Tetrahedral (accept “square planar”) [7] (b) (i) In transition metal compounds, the presence of ligands causes the 3d orbitals to split into 2 sets of non-degenerate orbitals. The difference in energies (DE) between these 2 sets of 3d orbitals is small and radiation from the visible region of the electromagnetic spectrum is absorbed when an electron moves from a lower energy d-orbital to another unfilled/partially-filled d orbital of higher energy. The colour observed correspond to the complement of the absorbed colours. Hence, transition metal compounds are often coloured. [3] (ii) Blue solution [Cu(H2O)6]2+ Yellow solution CuCl4 2- Blue solution was formed from the oxidation of Cu (to Cu2+(aq)). Blue solution formed a white solid on heating to due to loss of water ligands. / Splitting of 3d orbitals into 2 sets of different energy no longer occurs. CuCl42- + 6H2O = [Cu(H2O)6]2+ + 4Cl - On addition of water, position of equilibrium of CuCl4 2- + 6H2O = [6][Cu(H2O)6]2+ + 4Cl - shifts to the right, forming blue [Cu(H2O)6]2+ 1

2 (c) Limonene: aqueous bromine cold alkaline KMnO4 Menthone: 2,4-DNPH Menthol: PCl 5 / SOCl2 Acidified potassium dichromate, heat Acidified KMnO 4, heat; followed by test with limewater [4] 2 (a) (i) Oxidation number = 0 – (-2) – 4(+1) = -2 (ii) Zero H C H O OR C H HO OH H +2 C HHO O OR C%O (iii) CH3CH2CH2CH2OH(l) + 6O2(g) ® 4CO2(g) + 5H2O(l) (iv) Temperature increase = 58 - 23 = 35 °C Heat evolved = 35 ´ 515 ´ 4.18 = 7.53 ´ 104 J Amount of butan-1-ol = 2.30 ¸ (4 ´ 12.0 + 10 ´ 1.0 + 16.0) = 3.11 ´ 10-2 mol DHc, = 4 2 7.53 10 3.11 10 - æö ´-ç÷ ´èø = -2.42 ´ 106 J mol-1

3 = -2.42 ´ 103 kJ mol-1 (v) The calculated value is less negative (less exothermic ) than the true value as there is heat loss to the surroundings and copper can calorimeter. [7] (b) (i) Propanedioic acid is a stronger acid than butanedioic acid due to the stabilisation of the anion by hydrogen bonding with the unionised –CO2H group. HO2C CO2H + H2O O H2C O- O O H + H3O+ A (ii) Removal of H + from an anion ( A) that already carries a negative charge is electrostatically unfavourable. OR The stabilising hydrogen bonding would be destroyed by the ionisation of the second –CO2H group. (iii) Let the acid be HA and the concentration of H+(aq) be x. 2 a 4.20 [H ][A ] [HA] 10 0.25 K +- - = = - x x Since x is small, 2 4.2010 0.25 - = x x = 3.97 ´ 10-3 mol dm-3 pH = -lg (3.97 ´ 10-3)