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 (l) ΔHf = +12.3 C6H5CH3 (g) ΔHvapouration = + 38.00 C6H5CH2• (g) + H• (g) + 410 ½ B.E. (H-H) = +218 ΔHformation of C6H5CH2•

(iii) Benzyl radical C6H5CH2• is less stable than its reactant, ΔHf C6H5CH2• (g) is endothermic. The radical is very reactive as it requires only 1 more electron to form a stable configuration. [Total marks : 21] 3(a) (i) 120 0 (ii) Coordination no: 6 3(b) (i) Cl2 + 2e 2Cl- +1.36V MnO4 - + 8H+ + 5e Mn2+ + 4H2O +1.52V MnO4– is able to oxidise Cl– to Cl2. Ecell = +0.16V, reaction is feasible. Hence more MnO4– will be used for the titration. (ii) When temperature is increased, the molecules gain kinetic energy and move about faster. This increases the number of molecules having energy E ≥ EA. Thus, the frequency of effective collisions increases. Reaction rate increases. 3(b) (iii) Calculation of % weight of C2O42-ion Al complex H+ + 2 CO2 + 2 e ⇌ H2C2O4 MnO4- + 8H+ + 5e ⇌ Mn2+ +4H2O 5H2C2O4 ≡ 2MnO4– Amt of MnO4– = 27.50/1000 x 0.0213 = 5.86 x 10-4 mol Amt of H2C2O4 in 25 ml = 5.86 x 10-4 x 5/2 = 1.466 x 10-3 mol Amt of H2C2O4 in 250 ml = 1.466 x 10-3 x 10 = 1.466 x 10-2 mol Mass C2O42- = 1.466 x 10-2 x 88.0 = 1.29 g % of C2O42- in complex = 1.29/1.77 x 100% = 72.9% (iv) If n =2, K[Al(C2O4)2] % of C2O42- in complex = 176/242.1 x 100% = 72.7% If n =3, K3[Al (C2O4)3] % of C2O42- in complex = 264/408.3 x 100% = 64.7% Since the value obtained is closer to n = 2, the complex is K[Al(C2O4)2] 3(c) (i) Due to the presence of different ligand, the d orbitals are split to different extent/ energy gap, DE. Rank of the d-orbital splitting NH3 > H2O > F-

(ii) Since EDTA is able to displace H2O, hence it will form a stronger dative bond to Co2+. Hence, ∆H is negative. ∆S is positive as there is an increase in entropy due to more molecules at the product side, 7 molecules vs 2 molecules. Using ∆G =∆H -T∆S, Since ∆S >0 and ∆H is <0, -T∆S would always be <0 ∆G would be always <0 at all temperature. [Total marks:16] 4(a) (i) Nucleophilic addition (ii) OH- will react with ethanol to generate a stronger nucleophile CH3CH2O-. (iii) · Generation of Nucleophile · Slow · Intermediate · Arrow · Partial charges + lone pair (iv) OCH2CH3 OH OCH2CH3 OCH2CH3 + H2O CH3CH2OH, H+ catalyst Step II An Acetal Calcium chloride is a drying agent, which help to remove water. Hence the position equilibrium will shift right to increase the yield of the product. 4(b) (i) C H O S Percentage by mass / % 51.2 7.7 13.7 27.4 Mole ratio 4.3 7.7 0.86 0.85 Simplest ratio 5 9 1 1 ` Empirical formula = C5H9OS

(ii) Using the ideal gas equation, pV = nRT = (m/Mr)RT Mr = (mRT)/pV = [0.219 X 8.31 X (95 + 273)]/(150 X 103 X 38.2 X 10-6) = 116.9 [Total: 10 marks] 5(a) (i) 0.160 (ii) Experiment 2: 18 seconds. Since r = k’ [Cl2], hence t1/2 = ln 2 / k’. Where k’ = k [ O ]2, thus when O halves, the time x 4 . (b) (i) HCl + BaSO4 or HCl + Ba(HSO4)2 (ii) O Cl O Cl OR (c) Step Reagents and conditions I X2 (aq) II Ethanolic KCN heat with reflux III Aqueous H2SO4, heat with reflux IV PCl5 or SOCl2 (with pyridine) or PCl3 (heat) A O Br OH B O Br C O HOOC D O C O Cl [Total: 13 marks]