CJC_Prelim H2 Chem 2008 (worked soln for P1 P2 P3) (updated)

CJC H2 Chem Preliminary Exam 2008 Answers Qn No Answer Qn No Answer Qn No Answer Qn No Answer 1 C 11 D 21 B 31 A 2 C 12 D 22 C 32 D 3 C 13 C 23 A 33 C 4 C 14 B 24 D 34 B 5 D 15 B 25 B 35 C 6 A 16 B 26 C 36 B 7 B 17 A 27 C 37 D 8 C 18 C 28 D 38 C 9 A 19 B 29 C 39 D 10 C 20 B 30 B 40 B Paper 1 Paper 2 1 (a) S [Ne] 3s 2 3px 2 3py 1 3pz 1 P [Ne] 3s 2 3px 1 3py 1 3pz 1 S has a lower 1st I.E. than P due to the inter-electron repulsion between paired electrons in a 3p orbital of S. (b) CO 2 exists as simple covalent molecules with weak van der Waals’ forces of attraction between molecules. SiO2 has a giant molecular structure with strong covalent bonds between atoms ∴ SiO2 has a much higher melting point than CO2. (c) 2-nitrophenol is able to form intr a-molecular hydrogen-bonding between the –OH and –NO2 groups. Hence its inter-molecular hydrogen bonding is less extensive than that of 4-nitrophenol. Less energy is required to break the intermolecular forces of attraction and it has a lower melting point. (d) Boiling point is dependent on the intermolecular forces of attraction. Down the group, boiling points of the hydrogen halides generally increase due to stronger van der Waals’ forces of attraction between molecules as the number of electrons increases (with increase in Mr). However, HF has the highest boiling point (an exception) due to the presence of strong hydrogen bonding between molecules. (e) P is more electronegative than As. The bond pair electrons in PCl 3 are closer to the central P atom as compared to that in AsCl3. There is more repulsion between the electrons / greater bond pair-bond pair repulsion and hence the bond angle increases.

2 2 (a) +5 (b) (i) BiO 3 - + 6 H+ + 2 e- → Bi3+ + 3 H2O (ii) 2 Mn 2+ + 5 BiO3 - + 14 H+ → 2 MnO4 - + 5 Bi3+ + 7 H2O (c) Fe 2+ → Fe3+ + e- MnO4 - + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ + 4 H2O ∴MnO4 - ≡ 5 Fe2+ no. of mol of Mn in sample = no. of mol of MnO4 - required = 5 1 × No. of mol of Fe2+ = )1000 00 . 3610 . 0 (5 1 × × = 0.000720 mol ∴ mass of Mn in sample = 0.000720 × Ar of Mn = 0.000720 × 55 = 0.0396 g ∴ % of Mn in sample = 10000 . 1 0396 . 0× = 3.96 % (d) (i) E o cell = E o red - E o oxd Reactants Products Eθ cell / V Can reaction occur? (Y/N) Mn & SO4 2- Mn 2+ & SO2 Eθ cell = +0.17 - (-1.18) = +1.35 Yes Mn2+ & SO4 2- MnO 4 - & SO2 Eθ cell = +0.17 - (+1.52) = -1.35 No Fe2+ & SO4 2- Fe 3+ & SO2 Eθ cell = +0.17 - (+0.77) = -0.60 No (ii) Yes, sulphuric acid is a suitable replacement for nitric acid in this experiment. Reason: H2SO4 can oxidise Mn to Mn2+ but cannot further oxidise Mn2+ to MnO4 -. Also, it does not oxidise Fe2+ to Fe3+. ⇒ H2SO4 does not interfere with the redox reaction. CJC 2008 9746/Answer/prelim/08

3 3 (a) (i) (ii) initial dose = (2 × 500) = 1000 m