SAJC_H1_CHEM_P2_ANS

SAJC 2011 H1 Chemistry Prelims Paper 2 Solutions Section A (40 marks) Answer all the questions in this section in the spaces provided. 1 (a) (i) BeC l2 is a covalent compound while BeF 2 is an ionic compound. The electron cloud of C l- anion is larger than that of F - anion. Hence, C l- anion is more readily polarised by the Be 2+ cation which has a high charge density, giving rise to its covalent character. (ii) (iii) Let x be the mole fraction of BeC l2 and mole fraction of Be 2Cl4 = 1 – x. x(80) + (1-x)(160) = 126 80x = 34 x = 0.425, 1 – x = 0.575 (b) (i) (ii) % Mass of Be in Beryl = 100 x 3) x 16 (28.1 62 x 27 9 x 3 9 x 3 +++ = 5.022 % 5.022% /barb2right 30.6 kg 100% /barb2right 100 x 5.022 30.6 = 609.3 kg

2 % Mass of Beryl in ore = 11.4% 100 x 5330 609.3 = 1 (c) (i) ∆∆ ∆∆ H1: Enthalpy change of formation of Be 3N2 ∆∆ ∆∆ H3: Bond dissociation energy of N 2 ∆∆ ∆∆ H4: Sum of 1 st and 2 nd IE of Be ∆∆ ∆∆ H6: Lattice Energy (ii) Lattice energy = -588 – (324 + 994 + 3(900 + 1760) + 1404) Lattice energy = - 11290 kJ mol -1 (iii) /uni2502/uni2206HLE /uni2502α/uni2502 q+q -/r+ + r-/uni2502 Product of q+ and q- of Be 3N2 > Product of q+ and q- of Li 2O Since product of charges outweighs the sum of ionic radii, the lattice energy of Li 2O is less exothermic than Be 3N2. 2 (a) First ionisation energy is the energy required to r emove 1 mole of electrons from 1 mole of gaseous atoms to form a mole of singly charged gaseous ions. (b) (i) P: Group VI Q: Group II (ii) Physical properties: Electrical conductor at aqueous/ molten state. High melting and boiling point. Soluble in water/insoluble in organic solvents. 3 (a) SiO 2 is insoluble in water due to the high bond dissociation energy needed to break the strong Si—O covalent bonds. Na 2O(s) + H 2O( l) → 2NaOH(aq) Na 2O is readily soluble in water to form a strongly alkaline solution. SiO 2(s) + 2NaOH(aq) → Na 2SiO 3(aq) + H 2O( l) SiO 2 is an acidic oxide which reacts readily with alkalis to form salt and water.

3 (b) (i) No. of moles of S 2O3 2- = 30/1000 x 0.100 = 0.00300 mol No. of moles of Br 2 = 4 x 0.00300 = 0.01200 mol S2O3 2- ≡ 4Br 2 ≡ 6 SiF 4 No. of moles of SiF 4 = 3 x 0.006 = 0.018 mol Mass of SiF 4 = 0.018 x (28.1 +19.0 x 4) = 1.87 g (ii) Disproportionation 3Br 2 + 6OH - /barb2right 5Br - + BrO 3 - + 3H2O 4 (a) Kc = )2 18 . 0)( 2 24 . 0( )2 80 . 0()2 40 . 2( ]][ [ ][ ][ 3 24 3 2 =OHCH CO H = 64 mol 2 dm -6 (b) When temperature increases (from 550 oC to 850 oC), Percentage of H 2 increases, thus forward reaction is favou