VJC 2022 H2 Chem Prelim P2 ans

1 © VJC 2022 9729/02/PRELIM/22 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/02 Paper 2 Structured Questions Candidates answer on the Question Paper. 15 September 2022 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and CT group in the space at the top of this page. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the space provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 25 2 / 13 3 / 12 4 / 19 5 / 6 Total / 75 This document consists of 18 printed pages.

2 © VJC 2022 9729/02/PRELIM/22 [Turn over 1 (a) The most common oxidation states of iron are +2 and +3. (i) Iron(II) and iron(III) both contain electrons in all five 3d orbitals. Sketch and label the shape of the following two 3d orbitals: • one 3d orbital from the lower energy level in an octahedral complex • one 3d orbital from the higher energy level in an octahedral complex Use the axes below. [2] • Lower energy level (in between axes) 3dxy 3dxz 3dyz • Higher energy level (on the axes) 3dx2-y2 3dz2 (ii) Explain why Fe2+(aq) ions are coloured, whereas Zn2+(aq) ions are colourless. • The degenerate 3d orbitals in Fe 2+ octahedral complex is split into 2 different energy levels due to the presence of water ligands. • d-d transition took place whereby a 3d electron from the lower energy level is promoted to the upper energy level by absorbing energy from the visible region of the electromagnetic spectrum. The colour seen is the complement of the colour absorbed. • Fe2+ has a partially-filled d subshell but Zn2+ has completely filled 3d subshell and so d-d transition whereby an electron is promoted from a lower energy level to a higher energy level is impossible. [3] lower energy level higher energy level z y x

3 © VJC 2022 9729/02/PRELIM/22 [Turn over (iii) Most naturally occurring samples of iron(II) oxide are found as the mineral w üstite. Wüstite has formula Fe20Ox. It contains both Fe2+ and Fe3+ ions. 90% of the iron is present as Fe2+ and 10% is present as Fe3+. Deduce the value of x. 20  [0.9(+2) + 0.1(+3)] – 2x = 0 • x = 21 [1] When aqueous solutions of S 2O82– and tartrate ions, C 4H4O62–, are mixed, the reaction proceeds very slowly. However, this reaction proceeds quickly in the presence of an Fe3+(aq) catalyst. The overall equation for this reaction is as shown. C4H4O62– + 3S2O82– + 2H2O → 2CO2 + 2HCO2– + 6H+ + 6SO42– (iv) State and explain the type of catalyst that Fe3+(aq) functions in the above reaction. • Fe3+(aq) is a homogeneous catalyst in this reaction as it has the same physical state (OR phase) as the reactants. [1] (v) Write two equations to show how Fe3+(aq) functions as a catalyst in this reaction. • 6Fe3+ + C4H4O62– + 2H2O → 6Fe2+ + 2CO2 + 2HCO2– + 6H+ • 2Fe2+ + S2O82– → 2Fe3+ + 2SO42– [2] (b) (i) Complete the ‘dot-and-cross’ diagram below, drawing the outer electrons only, to show the bonding in methanoic acid, HCO 2H. The two oxygen atoms in HCO 2H are labelled O1 and O2 respectively. [1] (ii) The carbon atom in HCO2H is sp2 hybridised. Explain what is meant by sp2 hybridisation with reference to the carbon atom in HCO2H. • In sp2 hybridisation, one 2s and two 2p orbitals of carbon are mixed to form three equivalent sp2 hybrid orbitals that are arranged 120° apart. [1] C O2 H H O1 • • • • • • • • • • • • • x x x x x

4 © VJC 2022 9729/02/PRELIM/22 [Turn over (iii) Similar to carbon, oxygen atom can also undergo hybridisation. By considering the number of electron densities around oxygen atom labelled O 1, suggest the type of hybridised orbitals for the oxygen atom labelled O1. • sp3 [1] (iv) Sketch a diagram to show how two sp2 hybridised orbitals can form a sigma bond. • OR [1] (c) Cyclohexane is immiscible with water. Iodine, I2, can dissolve in both water and cyclohexane. The expression and numerical value for the partition coefficient, Kpc, of iodine between cyclohexane and water are given below. Kpc = concentration of I2 in cyclohexane concentration of I2 in water = 93.8 (i) 15.0 cm3 of C6H12 is shaken with 20.0 cm 3 of an aqueous solution containing I2 until no further change is seen. It is found that 0.390 g of I2 is extracted from water into the C6H12. Calculate the mass of I2 that remains in the aqueous layer. Show your working. Let mass of I2 that remains in the aqueous layer be x g. Kpc = concentration of I2 in cyclohexane concentration of I2 in water • 93.8 = (0.390/15) ÷ (x/20) • x = 5.54  10–3 g [2] (ii) Suggest how the value of Kpc of I2 between hexa-2-one, CH3(CH2)3COCH3, and water would compare to the value given in (c). Explain your answer. • Kpc of I2 between hexan-2-one and water would be lower than 93.8. • Iodine being non-polar would be less soluble in hexa-2-one which is a polar solvent. Hence concentration of I2 in hexa-2-one would be lesser than in cyclohexane in the presence of water. [2]

5 © VJC 2022 9729/02/PRELIM/22 [Turn over (d) Some data relating to calcium and oxygen are listed in table 1.1. Table 1.1 process value / kJ mol–1 first ionisation energy of oxygen +1310 second ionisation energy of oxygen +3390 first electron affinity of oxygen –142 second electron affinity of oxygen +844 enthalpy change for 1 2O2(g) + 2e– → O2–(g) +951 enthalpy change for Ca(s) → Ca2+(g) + 2e– +1933 lattice energy of CaO(s) –3517 (i) Suggest why the first electron affinity of oxygen is negative. • First electron affinity of oxygen is negative (or exothermic) due to energy released from the attraction between the nucleus (or protons) of the gaseous oxygen atom and the incoming electron. [1] (ii) Suggest why the second electron affinity of oxygen is positive. • Second electron affinity of oxygen is positive (or endothermic) due to energy needed to overcome the repulsion between the negatively charged O– ion and the incoming electron. [1] (iii) Oxygen exists as O 2 molecules. Use relevant data from table 1.1 to c alculate the bond energy of the O=O bond. Show your working. • 1 2BE(O=O) –142 + (+844)= +951 • 1 2BE(O=O) = 249 BE(O=O) = 498 kJ mol–1 [2] (iv) Use relevant data from table 1.1 to calculate the enthalpy change of formation of calcium oxide, CaO(s). Show your working. • Hf = +1933 + 951 – 3517 • Hf = –633 kJ mol–1 [2] 1 2O2(g) + 2e– ⎯⎯⎯→ O2–(g) +951 O(g) + 2e– 1 2BE(O=O) –142 + (+844) Ca(s) + 1 2O2(g) ⎯⎯⎯→ CaO(s) Hf Ca2+(g) + O2–(g) +1933 + 951 –3517

6 © VJC 2022