VJC Prelim P3 Ans

1 2009 H2 Chemistry 9746/3 Suggested Answers 1 Vanadium, a transition metal, is commercially important. (a) (i) Write down the electronic configur ation of V, hence, write down the electronic configuration of V 2+ ion. 23V 1s 22s22p63s23p63d34s2 23V2+ 1s 22s22p63s23p63d3 [1] (ii) Explain why V 2+(aq) is coloured whereas Ca2+(aq) is colourless. In the presence of ligands , the 3d orbitals are split into two groups of slightly different energy levels. V 2+ has a partially filled d orbitals , transition of electrons between the two different levels takes place with absorption of energy in the visible spectrum . The wavelengths that are transmitted correspond to the complementary colour observed. In Ca 2+, there is no d-d transition due to absence of 3d electrons (or absence of partially filled 3d orbitals or electron transition takes place outside the visible region), hence it appears colourless. [ 3 ] (b) Sketch the general trends of the first ionization energy and the third ionization energy of the elements across the first series of tr ansition metals; including Ca to Cu on the same axes. Explain the trends shown for these two properties. ionization energy/ kJ mol -1 M 2+ → M 3+ + e M → M + + e Ca Sc Ti V Cr Mn Fe Co Ni Cu The first ionization energy of the elements from scandium to copper remains relatively invariant as it involves the removal of 4s electrons. The significant screening effect of 3d electrons in the penul timate shell nullifies the increase in nuclear charge.  VJC 2009 9746/03/PRELIM/ANS/09 [Turn over

2 The transition elements have higher 1 st ionization energy than Ca since they have higher effective nuclear charge than Ca due to the fact that d electrons are poorly shielding. The increasing trend in third ioni zation energy is due to the increase in the effective nuclear charge and relatively constant shielding effect as 3d electrons are now in valence shell. Third ionization energy of calcium is much higher than that of the other transition elements as the elec tron to be removed is from inner 3p orbital which is closer to the nucleus and has lower energy , held more strongly by the nucleus (or electron is removed from a full octet to form Ca3+). Therefore, larger amount of energy is needed to remove it. The 3 rd electron from Fe 2+ is removed from an orbi tal containing a pair of electrons, less energy is expected due to inter-electronic repulsion. [ 5 ] (c) Both Ca and Zn play an important role bi ologically in the body system. Vanadium ions are poisonous as they can cause bone porosity by replacing the calcium ions in bones. One treatment for vanadium poisoni ng involves administering a solution of edta, which forms hexacoordinate complexes with many divalent metal ions using the 4 donor oxygen and two donor nitrogen atoms. Comment on the use of edta as a treatment for vanadium poisoning in the light of the following data: Equilibrium K c /mol-1 dm3 Ca2+ + edta 4- [Ca(edta)] 2- Zn2+ + edta 4- [Zn(edta)] 2- V2+ + edta 4- [V(edta)] 2- 5 x 1010 3 x 1016 4 x 1016 What problems might arise during the treatment and how could they be overcome? As K c of [V(edta)]2- is much higher than that of [Ca(edta)] 2- , V2+ will be removed in preference of Ca 2+ when edta is used. Due to the close proximity in K c between [V(edta)]2- and [Zn(edta)]2- , Zn2+, which is essential for health, will also be removed in addition to V 2+. Hence, limited amount of edta should be administered to prevent the loss of Ca 2+, removal of Zn 2+ may be overcome by the administration of Zn ion supplements. [ 2 ] (d) Transition metals are often good hom ogeneous and heterogeneous catalysts. V2O5 is used as catalyst to speed up the conversion of SO 2 into SO 3 in the Contact process for making sulfuric acid.  VJC 2009 9746/03/PRELIM/ANS/09 [Turn over

3 The key reaction in the contact process is as follows: 2SO 2(g) + O2(g) 2SO3(g) ∆H = -197 kJ mol-1 (i) For the reaction above, identify t he type of catalyst involved and explain clearly how the catalyst works. V2O5 acts as a heterogeneous catalyst , it is in a different phase from the reacting mixture. The reactant molecules readily adsorbed onto the catalyst surface. This allows for formation of weak bonds be tw een reactants and the surface catalyst; weakens the intramolecular bonds in the reactants and helps to facilitate the reaction . [2] When a 2:1 ratio of sulfur di oxide and oxygen at a total init ial pressure of 3 atm is passed over V 2O5, the catalyst in a fixed volume steel vessel at 430 οC, the partial pressure of sulfur trioxide at equilibrium is found to be 1.9 atm. (ii) Calculate the percentage conversion of SO 2 to SO 3 and the equilibrium constant, Kp at 430οC . 2SO2(g) + O 2(g) 2SO 3(g) Initial/atm 2 1 0 Change/atm -2x -x +2x Equilibrium/atm 2-2x 1-x 2x Given P sulfur trioxide = 2x = 1.9 x = 0.95atm P total = 0.1 + 0.05 + 1.9 = 2.05 atm % conversion = 1.9/2 x 1/100 = 95% K p = P 2 SO3 / P2 SO2 . PO2 = (1.9) 2 / (0.1)2 (0.05) = 7220 atm -1 [2] (iii) According to Le Chatelier’s Principle, the reaction will be favoured by a low temperature and a high pressure. Give one other reason (other than cost factor) why a higher pressure is not applied? At high pressures, however, sulfur dioxide liquefies and affects the catalytic reaction. [1]  VJC 2009 9746/03/PRELIM/ANS/09 [Turn over

4 (iv) Comment on the effect on t he percentage conversion if another unreactive but toxic gas was accidentally added to V2O5? The catalyst is easily ‘poisoned’ by adsorbing the toxic gas on its surface. These take up the surface of the catalyst and impair its efficiency/slow down the process but no change in the perentage conversion. [1] (e) Given the following information on the co lour of the aqueous vanadium ion of various oxidation states: VO2 +(yellow) VO2+(blue) V3+(green) V 2+(violet) Use relevant E Θ values from the Data Booklet to predict the colour observed upon mixing acidified aqueous ammonium polytri oxovanadate(V) solution with excess aqueous Sn 2+ solution. Calculate the relevant E Θ cell values and write balanced equations. D a t a E Θ/V (yellow)VO2 +/VO2+ + 1 . 0 0 (blue)VO2+/V3+ + 0 . 3 4 Sn4+/Sn2+ + 0 . 1 5 (green)V3+/V2+(violet) -0.26 E cell = +1.00 – 0.15 = +0.85V > 0 feasible E cell = +0.34 – 0.15 = +0.19V > 0 feasible E cell = -0.26 – 0.15 = 0.41V < 0 not feasible Sn2+(aq) is powerful enough to reduce VO 3 - to V 3+. Hence, green solution is expected to form. Sn2+ + 2VO2 + + 4H+  Sn4+ + 2VO2+ + 2H2O [Sn2+ + 2VO3 - + 4H+  Sn4+ + 2VO2+ + 2H2O is acceptable] Sn2+ + 2VO2+ + 4H+  Sn4+ + 2V3+ + 2H2O [ 3 ] [Total: 20] 2(a) The solubility products of some group II compounds at 25oC are shown in the following table. Compound K sp Ba(OH)2 5.0 X 10-3 Ca(OH)2 ? BaCO3 5.1 x 10-9 CaCO3 3.8 x 10-9  VJC 2009 9746/03/PRELIM/ANS/09 [Turn over

5 With reference to the given data, (i) Write an expression for the solubility product of Ca(OH)2. Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq) K sp of Ca(OH)2 = [Ca2+] [OH-]2 [1] (ii) Calculate the pH of a saturated solution of Ca(OH) 2 at 25 oC given that its solubility is 0.830 g dm-3.