MI 2018_H2_CHEM_(9729 02)_Answers Prelim

2018 Preliminary Exams Pre-University 3 H2 CHEMISTRY 9729/02 Paper 2 Structured Questions 12th Sept 2018 2 hours Candidates answer on the Question paper. Additional materials: Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Question 1 2 3 4 5 Total Marks 20 15 12 12 16 75

1 Ruthenium, Ru, is a Period 5 d-block element. Its ions have the ability to form complexes with both organic and inorganic ligands. One such organic ligand is 2,2’-bipyridine which can be represented by bpy. (a) Define the term ligand and suggest why bpy can act as a bidentate ligand. ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………[2] Ligand is a neutral molecule or anion which contain at least one atom bearing a lone pair of electrons which can form a dative bond to a central atom/ion, resulting in the formation of a complex. Bpy has 2 nitrogen atoms with a lone pair of electrons each, so it can form two dative bonds with the central atom/ion. (b) In an experiment, varying volumes of solutions of 0.1 mol dm -3 Ru 2+ and 0.1 mol dm-3 bpy are mixed to produce a coloured complex. xRu2+ + ybpy → [Rux(bpy)y]2+ The concentration of the coloured complex formed is proportional to the absorbance of the solution which is measured using a colorimeter. The following graph is plotted using the results of the experiment. 0 10 20 30 40 50 Absorbance Vol of Ru 2+ /cm 3 50 40 30 20 10 Vol of bpy /cm 3

By drawing two best-fit lines on the graph, deduce the formula of the complex ion formed between Ru2+ and bpy and hence draw the structure of the complex ion. Formula of complex: ………………………….. Structure of complex ion: [3] Draw two straight lines that intersect one another. Point of intersection shows VRu2+ = 12.5 cm3 and Vbpy = 37.5 cm3 Ratio of Ru2+ to bpy = 1:3 ; formula = [Ru(bpy)3]2+ (c) The table below shows the colour of the radiation of the electromagnetic spectrum and the corresponding wavelength range. Wavelength range (nm) Colour Complementary colour 400 – 450 violet yellow 450 – 490 blue