RVHS Prelim P3_Soln

1 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations Section A 1 (a) (i) Element E⦵ / V ½F2 + e− ⇌ F− +2.87 ½Cl2 + e− ⇌ Cl− +1.36 ½Br2 + e− ⇌ Br− +1.07 ½ I2 + e− ⇌ I− +0.54 The order of oxidising strength of the halogens, as observed from the E⦵ values above, decreases down the Group. Due to the strong oxidising power of F 2, the oxidation number of Phosphorus +5 in PF5, +5 and +3 in PC l5 and PCl3, +3 in PBr3 and PI3 respectively. [3] (ii) The smaller the p Ka, the stronger the acid, indicating H I is the strongest acid, followed by HBr, HC l and HF. Down the group from F to C l to Br to I, atomic radius increases, effectiveness of orbital overlap between H and X decr eases. The H–X bond becomes increasingly weaker, making it easier to lose the H+. [2] (b) (i) (Acidic) hydrolysis [1] (ii) O and CH3MgBr [2]

2 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations (iii) [3] (c) (i) Ksp = [Ca2+]3[PO43–]2 [1] (ii) [Ca2+] = 1.14 × 10–7 × 3 = 3.42 × 10–7 mol dm–3 [PO43–] = 1.14 × 10–7 × 2 = 2.28 × 10–7 mol dm–3 Ksp = (3.42 × 10–7)3(2.28 × 10–7)2 = 2.08 × 10–33 mol5 dm–15 [2] (iii) Let the solubility of calciu m phosphate in the presence of potassium phosphate be y. 2.08 × 10–33 = (3y)3(0.15 + y)2 Assume that y is small such that 0.15 + y ≈ 0.15 2.08 × 10–33 = (3y)3(0.15)2 y = 1.51 × 10–11 mol dm–3 [2] (iv) An increase in [SO 42–] causes the equilibrium position of CaSO4 ⇌ Ca 2+ + SO 42– to shift left, decreasing the solubility of CaSO4. Since the solubility of CaSO 4 is lowered, there will be less Ca 2+ from dissolution of CaSO 4. Equilibrium position of Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43– shifts right, increasing the solubility of Ca3(PO4)2. [2] [Total: 18] δ+ δ–

3 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations 2 (a) (i) NaOH(aq), heat under reflux [1] ( i i ) The C–Cl bond has a partial double bond character as the lone pair of electrons in the p orbital of the C l atom can delocalise into the π electron cloud of the benzene ring. This strengthens the C–C l bond, making it more difficult to break. [1] ( i i i ) [1] ( i v ) Add PCl5 / SOCl2 2,4-dichlorobenzyl alcohol: white fumes (of HCl) observed amylmetacresol: no white fumes observed or Add Br2(aq) 2,4-dichlorobenzyl alcohol: Br2(aq) remains orange amylmetacresol: orange Br2(aq) decolourises or Add neutral FeCl3 2,4-dichlorobenzyl alcohol: no violet complex formed amylmetacresol: violet complex formed [2]

4 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations (b) Menthol < amylmetacresol < tartaric acid Menthol is the weakest acid. is the least stable anion as the electron-donating alkyl gr oup intensifies the negative charge on the oxygen atom in the anion. Amylmetacresol is more acidic than menthol as the anion is more stable. The negative charge on the oxy gen atom can be delocalised into the ben