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 increase the delocalisation of the negative charge into the benzene ring, making the brominated phenoxide ion more stable than the phenoxide ion in tyrosol. [1] [Total: 10] 4 (a) Bond angle is 104.50 [2] (b) All two compounds have simple covalent structures/ are simple covalent molecules. For SOCl2 and S2Cl2, there exists permanent dipole-permanent dipole interactions and instantaneous dipole-induced dipole interactions between molecules. [3]

4 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination As S2Cl2 has a larger number of electrons/electron cloud and is more polar, more energy is required to overcome the stronger instantaneous dipole-induced dipole interactions and stronger permanent dipole- permanent dipole interactions between S2Cl2 molecules than SOCl2 molecules. Therefore, S2Cl2 has a higher boiling point than SOCl2. (c) (i) Step 2: NH3, rtp Step 3: LiAlH4 in dry ether.. [2] (ii) angelic acid T U [3] (iii) angelic acid cis  trans isomerism / enantiomerism T enantiomerism [2] [Total: 12] 5 (a) (i) The difference in energies (E) between these 2 sets of 3d orbitals is small and radiation from the visible region of the electromagnetic spectrum is absorbed when an electron is promoted from a lower energy d orbital to another unfilled/partially -filled d orbital of higher energy. The colour observed (violet) corresponds to the complement of the absorbed colours (yellow). [2]

5 River Valley High School 9729/PRELIM/19 [Turn over 2019 Preliminary Examination (ii) Vanadium(II) / V 2+(aq). Because V 2+(aq) is violet and V 3+(aq) is green, which means that for V 2+ and V3+, the complement colours, yellow and red respectively, are absorbed. Since the energy of light is inversely proportional to its wavelength , the energy gap between the 2 sets of d orbitals in V 2+ will be larger as the wavelength of yellow light is shorter than that of red light. [2] (b) (i) V3+ : ( 3 , −2.7 ) VO2+: ( 4 , −2.3 ) [1] (ii) E(VO2+/V2+) = gradient of line joining VO2+ and V2+ points = 2.40 ( 1.32) 25     = +0.36 V [1] (iii) −ΔG/F is the most negative here, hence ΔG for V 3+ is the most positive. Thus V3+ is the most (thermodynamically) stable species of vanadium / +3 is the most stable oxidation state of vanadium. Or V3+ will not undergo redox (i.e. oxidation (to VO 2+) and reduction (to V2+)) easily [1]

6 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination (c) (i) Catholyte chamber: VO2+ / VO2+ Anolyte chamber: V3+ / V2+ [2] (ii) V2+ + VO2+ + 2H+ → VO2+ + V3+ + H2O [1] (iii) From anolyte chamber towards catholyte chamber / Right to left [1] (iv) Increase the concentration of VO2+ in the VO2+/VO2+ redox couple (i.e. catholyte) and/or increase the concentration of V 2+ in the V3+/V2+ redox couple (i.e. anolyte) or Attach a number of VRFB cells in series [1] [Total: 12] 6 (a) (i) Energy is needed to overcome the repulsion to add an electron to a negatively charged O2‒, hence the ΔHf(O22‒) is more positive than ΔHf(O2‒). ΔHf (O2‒) is highly endothermic/ requires a lot of energy as it involves the breaking of a double bond and the addition of 2 electrons to a single atom/ single ‒OO‒ bond requires the addition of an electron to a negatively charged O‒. [2] (ii) Enthalpy change when 1 mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditions. [1] (iiI) 12 12 .qqLE rr  . As the charges of O2‒ and O22‒ are higher than O 2‒, the ionic size of O 22‒ is larger than that of O 2‒, the magnitudes of LE of decreases in the order K2O > K2O2 > KO2. [2]

7 River Valley High School 9729/PRELIM/19 [Turn over 2019 Preliminary Examination (b) 2KO2(s) + H2O(l) 2KOH(aq) + 3 2 O2(g) K2CO3(aq) + 3 2 O2(g) + H2O(l) 2KHCO3(aq) + 3 2 O2(g) By Hess’ Law, 2(ΔHr) = (‒113) + (‒219) +(‒100) ΔHr = ‒103 kJ mol‒1 [3] (c) (i) ∆Hppt ‒T∆Sppt = ‒ RT ln(Ksp) ln( ) sp HSK RT R  From graph, gradient = 5.3 ( 6.3) 0.0027 0.0035     =‒1250 K At (0.0027, ‒5.3), ‒5.3 = ‒1250(0.0027) + S R  1.9258.31 SS R   1.925 8.31 16.0S    J mol‒1 K‒1 [1] (ii) Entropy/ Disorder decreases because there are less ways to distribute the energy/ arrange the ions/ particles as they have to take up fixed positions in the lattice structure. [1] [Total: 10] 7 (a) (i) NaCN Na+ + CN [3] R is +CO2(g) ‒100 +CO2(g) ‒219 +2CO2(g) 2(ΔHr) R 6 R R H H δ– δ+

8 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination (ii) step 1 Anhydrous AlCl3, step 3 H2SO4 (aq), heat (under reflux) step 5 (excess) alcoholic NH3, heat in a sealed tube [3] (b) (i) [1] (ii) Use a longer gel plate / Apply a higher voltage / Change the p