PJC_H2 Chemistry_2008 JC2 H2 Chemistry Prelim Paper 2 Suggested Answers

1 Pioneer Junior College JC2 Preliminary Examination 2008 H2 Chemistry Paper 2 Suggested Answers 1 (a) (i) Both N2 and P 4 have simple molecular structure . They are non polar molecules, held together by weak van der waals forces due to induced dipole – induced dipole attractions. A s P 4 has a greater number of electrons / larger electron cloud to be polarised leading to stronger van der waals forces between P 4 molecules. This results in higher melting point in P4, hence P4 exists as solid. (ii) Phosphorus is a relatively big at om with diffused orbitals, side-on overlap of its p orbitals to form π bonds is much less effective than head-on overlap to form sigma bond. (b) (i) N O O O - 120o 3- 109.5o P O O O O Shape of NO3 - is trigonal planar. Shape of PO 4 3- is tetrahedral. (ii) To form NO 4 3-, N must be able to accommodate 10 electrons in its valence shell. Since N is in Period 2, it has no energetically accessible/low lying d orbitals to expand its octet. 2 (a) SrF2(s) + aq ⇌ Sr2+(aq) + 2F-(aq) Solubility of strontium fluoride = 0.073 / 125.6 = 5.81 x 10-4 mol dm-3

2 K sp = [Sr2+]eqm[F-]eqm 2 = (5.81 x 10-4) (2 x 5.81 x 10-4)2 = 7.85 x 10-10 mol3 dm-9 (b) [Sr2+] = (0.0100 x 25.0/1000) ÷ (45.0/1000) = 0.00556 mol dm-3 [F -] = (0.0150 x 20.0/1000) ÷ (45.0/1000) = 0.00667 mol dm-3 Ionic product of SrF 2 = (0.00556) (0.00667)2 = 2.47 x 10-7 mol3 dm-9 Since ionic product is greater than solubility product, precipitate of SrF 2 will form. 3 (a) CH 2C O O - CH 2C O O - NC H 2 CH 2 N CH 2 CH 2 C O O - C O O - edta (b) (i) On dropwise addition of NH 3(aq), blue precipitate of Cu(OH) 2(s) is observed. Equilibrium (1): Cu2+(aq) + 2OH-(aq) ⇌ Cu(OH)2(s) On adding excess NH 3(aq), blue precipitate dissolves to give a deep blue solution due to formation of [Cu(NH3)4(H2O)2]2+ complex. [ Equilibrium (2): Cu(OH) 2(s) + 4NH3(aq) + 2H2O ⇌ [Cu(NH3)4(H2O)2]2+ + 2OH-(aq) By forming the complex, [Cu2+(aq)] decreases, leading to decrease in ionic product of Cu(OH) 2 to the extent of ionic product < K sp. ∴ Precipitate dissolves.

3 O r Equilibrium (2): [ C u ( H 2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O By forming the complex, [Cu2+(aq)] decreases, position of equilibrium (1) shifts to the left accordi ng to Le Chatelier’s Principle, ∴ precipitate dissolves. (ii) Type of reaction: Ligand exchange [ C u ( N H 3)4(H2O)2]2+ + edta4- → [Cu(edta)]2- + 4NH3 + 2H2O (c) (i) Due to high K c (position of equilibrium lie s more to