VJC 2020 H2 Chem Prelim P2_Ans

 VJC 2020 9729/02/PRELIM/20 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/02 Paper 2 Structured Candidates answer on the Question Paper. 15 September 2020 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 20 2 / 11 3 / 8 4 / 19 5 / 17 Total / 75 This document consists of 16 printed pages.

 VJC 2020 9729/02/PRELIM/20 2 Answer all the questions in the spaces provided. 1 (a) (i) Complete the diagram to show the relative energies of all the electrons in an iron atom. [1] (ii) Write an equation for the second ionisation energy of iron.  Fe+(g)  Fe2+(g) + e [1] (iii) Explain why the second ionisation energy of iron is lower than that of chromium.  For the second ionisation energy, 4s electron is removed from iron while 3d electron is removed from chromium . As 4s electron is further away from the nucleus than 3d electron, less energy is needed to remove it, resulting in a lower second ionisation energy. [1] (b) A piece of rusted iron was analysed to find out how much of the iron had been oxidised to rust (also known as hydrated iron(III) oxide). A small sample of the rusted iron was dissolved in excess dilute sulfuric acid to give 250 cm3 of solution. The resultant solution contains Fe2+ and Fe3+ from iron and rust respectively. (i) 25.0 cm 3 of this solution required 16.90 cm 3 of 0.0200 mol dm 3 KMnO4 for complete oxidation of Fe2+. Calculate the amount of Fe2+ in 25.0 cm3 of the solution. 5Fe2+ + MnO4 + 8H+  5Fe3+ + Mn2+ + 4H2O Amount of Fe2+ in 25.0 cm3 of the solution = 5  0.0200  16.90  103  = 0.00169 mol [1] (ii) To another 25.0 cm3 of the solution, an oxidising agent was added to convert all the Fe2+ ions present to Fe3+ ions. The Fe3+ ions were then titrated with 0.100 mol dm 3 EDTA4 solution and 17.60 cm3 was required. Assuming 1 mol of EDTA4 reacts with 1 mol of Fe3+,