MJC H2 CHEM P2 ANS Prelim

Reg NumberClass Candidate Name _____________________________ Meridian Junior College 2009 JC 2 Preliminary Examination Answers H2 Chemistry 9746 22 September 2009 1 hr 30 min No part of this paper should be repro duced for commercial use without the prior written permission of Meridian Junior College. Paper 2 STRUCTURED QUESTIONS Additional Materials Data Booklet INSTRUCTION TO CANDIDATES Write your name, class and register number in the spaces provided at the top of this page. Write in dark blue or black ink. You may use a soft pencil for any diagrams or graphs. You are reminded on the need for good English and clear presentation of your answers. INFORMATION FOR CANDIDATES FOR EXAMINER’S USE The number of marks is given in brackets [ ] at the Q1 / 8end of each question or part of question. Q2 / 8 Q3 / 11 Q4 / 11 Q5 / 12 Q6 / 10 Total / 60 This question paper consists of 17 printed pages 1

Answer all the questions in the space pro vided 1 Early Periodic Tables, such as that devised by Mendeleev, listed the then known elements in order of their relative atomic mass. (a) (i) When Mendeleev created the table, t here were uncertainties regarding the relative atomic mass of telluri um. It is now known that there are eight isotopes of tellurium. Comple te the following table and use it to calculate the relative at omic mass of tellurium. isotope percentage abundance isotopic mass x percentage abundance tellurium-120 0.09 11 tellurium-122 2.46 300 tellurium-123 0.87 107 tellurium-124 4.61 572 tellurium-125 6.99 874 tellurium-126 18.71 2357 tellurium-128 31.79 4069.12 tellurium-130 34.48 4482.40 Complete the above table and give your answers to two decimal places. Hence, calculate the relative atomic mass of tellurium to five significant figures. Ar of Te = (11 + 300 + 107 + 572 + 874 + 2357 + 4069.12 + 4482.40) 100 = 127.73 2

(ii) TeO2 resembles aluminium oxide in its acid-base properties. TeO 2, SiO2 and P 4O10 cannot be distinguished based on their physical appearance. If a sample of one of the oxides was provided as a fine powder, describe the reactions you could ca rry out on the powder to determine which of the three oxides it was. In clude all relevant equations in your answer. Dissolve a portion of the sample in water. If the oxide is soluble in water, then the oxide is P4O10 P 4O10 + 6H2O 4H3PO4 Dissolve another sample of the oxide in HCl(aq). If the oxide is soluble in HC l, then the oxide is TeO 2 The oxide that is insoluble in HC l is then SiO2 TeO2 + 4 HCl  TeCl4 + 2H2O . [5] (b) Another chemist looking for patterns in the properties of the elements around the time of Mendeleev was Julius Lothar Meyer. Lothar Meyer looked at how the atomic volume of an element varies with its relative atomic mass. A simplified version of the graph he plotted is shown below. There are 2 circled regions on the graph. Explain why there is only a slight decrease in atomic volume bet ween the elements in region II compared to region I. I II 3

Across the transition elements series, –Inner 3d electrons provide effective shielding of the 4s electrons. - Increase in shielding effect almosts cancels the increase in nuclear charge - Effective nuclear charge increases slightly. - Electrostatic forces of attracti on between outer electrons and nucleus increases slightly hence atomic radius decreases slightly. However, across the main group elem ents, increase in nuclear charge outweighs the negligible increase in screening effect. - Effective nuclear charge increases dr astically and atomic radius decreases drastically. [3] [Total: 8] 2 ‘Hard’ water is water that has high mi neral content such as calcium ions. A common washing-up liquid contains sodium dodecylbenzenesulphonate, C18H29SO3Na [M r = 348]. In ‘hard’ water, it is ineffective as a detergent because it reacts with calcium ions to gi ve a precipitate. The solubility product, Ksp, for (C18H29SO3)2Ca (s) is 1.20  10-17 mol3 dm-9. The washing-up liquid contains 17.4% of sodium dodecylbenzenesulphonate by mass. In a factory production, 0.500 dm3 of 0.102 g dm-3 of the washing-up liquid was accidentally mixed with 0.200 dm3 of ‘hard’ water. A typical sample of ‘hard’ water has a conc entration of calcium ions of 2.50  10-4 mol dm-3. (a) Determine whether a prec ipitate is formed when the common washing-up liquid was accidentally mixed with 0.200 dm3 of ‘hard’ water. Ionic product of (C18H19SO3)2Ca = (7.143  10-5) (3.643  10-5)2 = 9.48 x 10-14 mol3 dm-9 Ionic product > Ksp  precipitate of (C18H19SO3)2Ca is formed. [2] (b) In order for the detergent to be used in ‘hard’ water, sodium tripolyphosphate, Na5P3O10, is added as a water softening agent. The sodium tripolyphosphate ‘softens’ water by complexing with calcium ions. The complexation reaction is as follows: 4

Ca2+ (aq) + P3O10 5- (aq) CaP3O10 3- (aq) For this reaction with calcium ions, the equilibrium cons tant is 7.7  108 mol-1 dm3. (i) Write the Kc expression for the reaction. K c = 3- 31 0 2+ 3- 31 0 [CaP O ] [Ca ][P O ] (ii) Hence, calculate the concentration of tripolyphosphate ions required to reduce the calcium ion concentration in a typical sample of “hard water” to 1.0  10-6 mol dm-3. Let x be the initial concentration of P3O10 5- required At equilibrium, 7.7  108 = 4 64 2.49 10 (1.0 10 )( 2.49 10 )xx     x = 2.493 10-4 mol dm-3 [3] (c) ‘Hard’ water also contains magnesium ions which can form a precipitate with the detergent. For example, magnesiu m ions form magnesium iodide, Mg I2, in the presence of aqueous potassium iodide. (i) The lattice energy of Mg I2 is –2327 kJ mol -1 while the values of the enthalpy changes of hydration are listed below: Ions Hhyd / kJ mol-1 Mg2+ -1920 I- -295 Calculate the enthalpy change of solution of magnesium iodide. Enthalpy change of solution of magnesium iodide = -(-2327) + (-1920) + 2(-295) = -183 kJ mol -1 5

(ii) By using y our answer from (c)(i), and given the entropy change of solution of magnesium iodide is posit ive, predict whether magnesium iodide is soluble in water at room temperature and pressure. Give your reasoning. At lower temperatures, and is both negative hence G = is also negative . Therefore the dissolution pr ocess is feasible and hence MgI oΔS T oΔH 2 is soluble in water. [3] [Total: 8] 3 Cobalt is a bluish-white element that can form various cobalt compounds with the ligands ethylenediamine, H 2NCH2CH2NH2, and ethanedioate, C 2O4 2-. Ethylenediamine is commonly expressed with the short form “en”. (a) Passing air through an aqueous soluti on containing CoCl 3, en and HC l produces a green complex cation, [Co( en)2Cl2]+. Another complex, [Co(C2O4)3]3- is obtained on passing air through an aqueous solution of CoCl3 and C 2- 2O4 . (i) State the coordination number of Co in [Co(en)2Cl2]+. Coordination number of Co = 6 (ii) Explain why transition element complexes such as [Co( en)2Cl2]+ are coloured. In Co2+ ions, the d orbitals are split into two groups due