SAJC H2 Chem 2012 Prelim P2 Soln

Date: 11 September 2012 SAJC 2012 Prelims P2 Solutions 1 No. Experiment Observation Deductions with relevant equation 1. Add ½ spatula of Effervescence OR gas evolved. Al(H2O)6 3+(aq) ↔ solid potassium Al[(OH)(H2O)5]2+(aq) + carbonate to H+(aq) each separate samples (1 cm3) 2H+(aq) + CO3 2-(aq) → of the 4 H2O(l) + CO2 (g) solutions in clean, dry test AlBr3 identified tubes. White precipitate formed ZnCO3, Ag2CO3, CaCO3 2. Add each No precipitate formed ZnSO4 identified unknown sample to 1 cm3 of AlBr3 Off-white precipitate formed Ag+(aq) + Br –(aq) →solution till excess in clean, AgBr(s) dry test tubes. AgNO3 identified White precipitate formed Al3+(aq) + 3 OH–(aq) → Al(OH)3(s) White ppt soluble in excess Al(OH)3(s) + OH–(aq) → Al(OH)4–(aq) Ba(OH)2 identified 3. Add Ba(OH)2 White precipitate formed Ba2+(aq) + SO4 2–(aq) → dropwise till BaSO4(s) excess to 1 cm3 of the last ZnSO4 identified unknown solution

2(a) (i) The iron ion has incompletely filled d-orbitals. In the is olated gas phase atoms, all 5 d orbitals of the iron ion are degenerate In forming complexes, the d orbitals split into 2 groups with a small energy gap between them. When a d-electron from the lower energy group is promoted to the higher energy group, energy from the visible region is absorbed. The light energy not absorbed will be seen as the colour of the complex. (ii) The melting point of iron is significantly higher than that of calcium due to the stronger metallic bonding present in iron and thus more energy needed which is attributed to the following reasons: - more delocalised electrons contributed from 3d and 4s electrons - higher charge density (iii) Fe3+ + e Fe2+ +0.77V MnO2 + 4H+ + 2e Mn2+ + H2O +1.23V Ag metal Student can choose any reducing agent as long as the electrode potential is +0.77V < x < +1.23V. OR student show calculation of Ecell Ecell >0 for Ag/MnO2. Ecell = 1.23- 0.80 = + 0.43V >0 reaction is feasible. Ecell of Ag/Fe3+ < 0V, no reaction, Ag only reduces MnO2. Ecell = 0.77 -0.80 = -0.03V <0 2(b) (i) CO + HCl + FeCl3 à +CHO + FeCl4-- (ii) FeCl3 will hydrolyse with water to yield [Fe(H2O)5(OH)]2+ and Cl- ions. OR dissolve in water to form [Fe(H2O)6]3+. Fe in [Fe(H2O)5(OH)]2+ or Fe in [Fe(H2O)6]3+ has no vacant d-orbitals to accept electron pair from Cl- to generate the electrophile.

2(c) (i) O CO/ HCl FeCl3 Heat with reflux CHO LiAlH4, in dry ether r.t.p CH2OH CH2Cl PCl5, r.t.p OR PCl3, Heat OR SOCl2 r.t.p Heat with reflux O (d) (i) ΔHc C6H5CH3 = ∑nDHformationo(products)∑nDHformationo(reactants) – (-3910) = 7 (-393.5) + 4(-285.8) - ΔHf C6H5CH3 ΔHf C6H5CH3 = +12.3 kJ/mol (ii) 7 C(s) + 4 H2 (g) C6H5CH2• (g) + ½ H2 (g) ΔH f C6H5CH2• (g) = +12.3 + 38 + 410 – 218 = +242.3 kJ/mol = + 242 kJ/mol C6H5CH3 (