HCI Prelim P1 P2 ANS

HWA CHONG INSTITUTION 2009 C2 Higher 2 CHEMISTRY 9746 PRELIMINARY EXAMINATION PAPER 1 Multiple Choice Questions (ANSWERS) Question Answer Question Answer 1 A 21 D 2 B 22 B 3 C 23 A 4 A 24 A 5 D 25 D 6 A 26 C 7 B 27 D 8 B 28 C 9 C 29 C 10 B 30 B 11 C 31 D 12 B 32 C 13 A 33 B 14 D 34 D 15 D 35 A 16 A 36 B 17 C 37 C 18 C 38 B 19 B 39 A 20 C 40 D 1

PAPER 2 Structured Questions (ANSWERS) 1 (a) The alkaline earth metals exhibit +2 oxidation state in their compounds because they have 2 valence electrons which are lost to achieve stable octet. They do not exhibit +1 oxidation state due to the low lattice energy of the compounds formed. +3 oxidation state is also not favourable due to the very high 3 rd ionization energy required to remove the third electron from the inner quantum shell. (b) (i) [Ba 2+] [F-]2 = 1.84 x 10–7 where [Ba2+] = 0.05 mol dm–3 [F-] =  (1.84 x 10–7 / 0.05) = 1.92 x 10–3 mol dm–3 (ii) [Ca2+]remaining = 3.45 x 10–11 / (1.92 x 10–3)2 = 9.38 x 10–6 mol dm–3 2 (a) (i) no. of moles of glycolic acid = 0.20/76.0 = 2.63 x 10–3 mol volume of NaOH required = 2.63 x 10–3/0.10 x 1000 = 26.3 cm3 (ii) no. of moles of CH2OHCOO- salt formed at equivalence = 2.63 x 10–3 mol conc. of CH2OHCOO- = 2.63 x 10–3 /(26.3 + 20.0) = 0.0568 mol dm–3 CH 2OHCOO- +H 2O = CH2OHCOOH + OH– Initial conc. / mol dm –3 0.0568 – – Eqm conc./ mol dm –3 0.0568 - x x x Kb of glycolate = Kw/Ka of glycolic acid = 1x 10–14/ 1.48 x 10–4 = 6.76 x 10–11 mol dm–3 Kb = x2/(0.0568 – x)  x2/0.0568 (assume x << 0.0568 mol dm–3) x = 1.959 x 10–6 mol dm–3 pOH = 5.708 pH = 14 – 5.708 = 8.29 (iii) Metacresol purple because its working range coincides with the sharp jump of the titration curve which lies in the alkaline pH region (equivalence pH = 8.29). (b) (i) Step 1: LiAlH 4 in dry ether Step 2: PCl5 (s), room temperature Step 3: NaOH in ethanol, heat 2

(ii) Increasing ease of hydrolysis: B, A, C C is an acid chloride which undergoes hydrolysis more easily than A (halogenoalkane), because the carbon in the –COCl group of C is bonded to two electronegative atoms, O and C l, thus the carbon is more electron deficient and more readily attacked by nucleophile (H2O) to undergo nucleophilic substitution. B undergoes hydrolysis less easily than A because in B, the C–C l bond has partial double bond character due to the overlap of the p-orbital of C l with the - electron cloud of the adjacent alkene carbons. Thus the C–C l bond in B is much stronger than in A, and is harder to break. OR The carbon of C–C l in B is less electron deficient than that in A due to the presence of the electron-rich  electron