TJC Prelim P1 P2 ANS

Answ ers to H2 Chemistry Preliminary Examinations 2009 Paper 1 and 2 Paper 1 1 B 6 D 11 D 16 A 21 B 26 A 31 B 36 C 2 A 7 C 12 B 17 C 22 C 27 B 32 B 37 D 3 A 8 D 13 C 18 C 23 C 28 D 33 B 38 A 4 A 9 D 14 A 19 C 24 B 29 C 34 A 39 C 5 B 10 C 15 D 20 C 25 C 30 A 35 A 40 C Paper 2 1 This question is about chlorine and its compounds. (a) A teacher instructed a st udent to react some chlorine gas with 250 cm 3 of 0.100 mol dm-3 aqueous sodium hydroxide, and left the lab. After he returned, he realized that he had not specified the reaction temperature. The teacher found that the pH of the solution had become 12.0. After acidifying the solution and adding excess silver nitrate, 2.69 g of silver chloride was obtained. He then concluded that the student had heated the sodium hydroxide. For this question, you may assume that all of the chloride ions are precipitated by the addition of silver ions. (i) Determine how the teacher arrived at his conclusion. ● Number of moles of NaOH used = 0.250 x 0.100 – 0.250 x 10 (14 – 12.0) = 0.0225 mol ● Number of moles of chloride = 2.69 / (107.8 + 35.5) = 0.0188 mol Number of moles of chloride : number of moles of NaOH = 0.0188 : 0.0225 ● = 5 : 6 Which is in line with the equation: 3Cl 2 + 6OH- → 5Cl- + ClO3 - + 3H2O 1

(ii) If the student had not heated the aqueous sodium hydroxide and the same amount of chlorine gas was used, what would be the resulting mass of silver chloride obtained when excess silver ni trate is added after acidifying the solution? [5] Number of moles of chlorine gas used = 0.0188 x 3 / 5 = 0.0113 mol Cl2 + 2OH  C l + ClO + H2O ● Number of moles of chloride formed (cold) = 0.0113 mol ● Mass of silver chloride formed = 0.0113 x (107.8 + 35.5) = 1.62 g (b) The melting point of POCl 3 is 1.25 oC whereas the melting point of PC l5 is 166.8 oC. ) ame and draw diagrams to illustrate the shapes of POCl3 and PCl5.   (ii) , in terms of structure and bonding, why PC l5 has a higher melting point. [4] ● ed-dipole-induced- ● resulting in stronger van der Waals’ forces. (i N Explain POCl 3 and PC l5 both are simple discrete molecules with weak van der Waals’ forces of attraction between the molecules (or POC l3 has permanent-dipole-permanent- dipole interactions while PC l5 has induc dipole interactions between molecules). PCl5 has a much larger M r than POC l3, and therefore a larger electron cloud that is mo re easily distorted, giving rise to larger partial charges, Cl P Cl O C Cl C l POCl3 Tetrahedral Trigonal bipyramidal PCl5 PCl C Cl l l 2

(c) hosp uch as the following: ) ith r