2019 RVHS Prelim P2 ANS

River Valley High School 9729/02/PRELIM/19 [Turn over 2019 Preliminary Examination Suggested Answers for 9729 H2 CM 2019 JC 2 Prelim Paper 2 1 (a) Barium has a larger nuclear charge and larger shielding effect than beryllium. Although there is an increase in nuclear charge in barium, it is cancelled out by the simultaneous increase in shielding effect by inner shells of electrons. Barium has a bigger atomic radius than beryllium. Hence the valence electrons become increasingly less attracted by the positive nucleus and less energy is required to remove the valence electrons. Therefore 1st IE of Ba is lower than the 1st IE of beryllium. [3] (b) (i) Ca(NO3)2 (s) → CaO (s) + 2NO2 (g) + 1 2O2 (g) [1] (ii) cations Ionic radius/nm Pb2+ 0.120 Zn2+ 0.074 Ca2+ 0.099 The ionic radius of the metal cation increases from Zn 2+ to Ca2+ to Pb2+ and its charge density decreases. As a result, the ability of M 2+ to polarise the electron cloud of the large NO3– anion decreases and the NO bonds are weakened to a smaller extent. Hence Zn(NO3)2 decomposes at the lowest temperature followed by Ca(NO3)2 and lastly Pb(NO3)2. [3] (c) [2] 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 for Pb2+ = +2 207.2= 0.00965 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 for Pb4+ = +4 207.2 = 0.01930 Angle of deflection= 0.01930 0.00965 ×8 = 16C [Total: 9]  + Beam of 207Pb4+ ions 16

2 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination 2 (a) Ka1 = X2/ (0.300 - X) ; assuming x is very small and 0.300 - X = 0.300 X = 0.0207mol dm–3 pH = -lg 0.0207 = 1.68 [1] (b) [Na3C6H5O7] = 46/ (23.0 x 3 + 12.0 x 6 + 16.0 x 7 + 1.0 x 5) = 0.178 mol dm–3 Kb = Kw / Ka = 10–14 / (3.98 x 10–7) = 2.51 x 10–8 mol dm–3 Kb = [OH–]2 / 0.178 = 2.51 x 10–8 [OH-] = 6.69 x 10–5 mol dm-3 pH = 14 – pOH pH = 9.82 [3] (c) (i) [salt] = [acid] / Amount of acid = Amount of salt / Larger volume of buffer Amounts of salt and acid are relatively higher than the amount of acid or alkali added. [2] (ii) pH = 6.40 = pKa3 buffer consist of HC6H5O72- and C6H5O73- Amt of of citric acid = 0.300 x 0.05 = 0.0150 mol Amt of NaOH needed = 0.0150 x 2.5 = 0.0375 mol mass = 0.0375 x (23.0 + 16.0 + 1.0) = 1.50 g [3] [Total: 9] 3 (a) (i) [1] (ii) Ion-dipole interactions Hydrogen bonding Instantaneous dipole-induced dipole forces [2] (iii) Neutral FeC l3. Violet colouration will be observed if tyrosol is present. There will be no violet colouration if tyrosol is not present. [2]

3 River Valley High School 9729/PRELIM/19 [Turn over 2019 Preliminary Examination (iv) When the NaOH solution is changed, [T]NaOH=0/ lowered. Equilibirum position shifts right to dissolve more tyrosol and counteract the change. More tyrosol will be removed from the olives. [2] (b) (i) D: E: [2] (ii) The brominated tyrosol is a stronger acid than tyrosol. In the brominated phenoxide ion, the electron-withdrawing bromine atoms increas