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 orange 490 – 550 green red 550 – 580 yellow violet 580 – 650 orange blue

650 – 700 red green The diagram below shows the UV-Visible spectrum of the complex formed between Ru2+ and bpy. Use the data given to suggest the colour of the complex formed between Ru 2+ and bpy. ………………………………………………………………………………………….. [1] Blue is absorbed hence the complex is orange ; (d) Ru2+ also forms an octahedral aqua complex with the formula [Ru(H2O)6]2+. Typically, the colour of the complex changes when the ligands are different. This is due to different ligands causing the five d-orbitals to be split to different extent. The following diagram shows how the five d-orbitals are split in an octahedral environment. (i) With reference to this diagram, outline why [Ru(H2O)6]2+ is coloured. ………………………………………………………………………………………. ………………………………………………………………………………………. …………………………………………………………………………………… [2] wavelength /nm Absorbance 400 450 500 550 energy gap E degenerate d-orbitals of an isolated Ru2+ ion non-degenerate d-orbitals of Ru 2+ in aqua complex

The electrons in the lower energy d-orbital absorbs radiation from the visible region of the electromagnetic spectrum and get promoted to the higher energy d-orbital. The complementary colour of the light absorbed is shown as the colour of [Ru(H 2O)6]2+. The electrons of transition metal ions in complexes can fill the non-degenerate d-orbitals in two different ways, namely the ‘high spin’ state and the ‘low spin’ state. This is dependent on the magnitude of the energy gap, E, and the pairing energy, P. Electrons usually prefer to occupy orbi tals singly, rather than in pairs. Pairing energy, P, is the energy needed for an electron to fill an orbital that is already occupied by another electron. In the ‘high spin’ state, the electrons occupy all the d-orbitals singly, before starting to pair up in the lower energy d-orbitals. This occurs because the magnitude of the energy gap, E, is smaller than the pairing energy, P. In the ‘low spin’ state, the lower energy d-orbitals are filled first, by pairing up if necessary, before the higher energy d-orbitals are used. This occurs because the pairing energy, P, is smaller than the magnitude of the energy gap, E. For Period 4 d-block elements, the electronic configuration of the 3d electrons can be either ‘high spin’ or ‘low spin’. However, for Period 5 d-block elements, the 4d electrons are always in the ‘low spin’ state. (ii) Suggest why electrons usually prefer to occupy orbitals singly, rather than in pairs. ………………………………………………………………………………………. …………………………………………………………………………………….[1] Electrons are negatively charged and will exert repulsive force against each other. (iii) With reference to the relative sizes of 3d and 4d orbitals, suggest a reason why 4d electrons prefer to pair up in the lower energy d-orbital before filling the higher energy d-orbitals. ………………………………………………………………………………………. …………………………………………………………………………………… [1] 4d is bigger in size compared to 3d orbitals. The bigger space causes the repulsion between electrons to be smaller hence the pairing energy becomes smaller than the magnitude of the energy gap. (iv) In the diagram below, show the electronic distribution of a Ru2+ ion in the ‘low spin’ state, given that the electronic configuration of Ru2+ is [Kr] 4d6.

[1] (e) Two galvanic cells were set up under standard conditions to determine the standard electrode potential of Ru3+/Ru2+. (i) Define the term standard electrode potential. ………………………………………………………………………………………. …………………………………………………………………………………….[1] energy gap E degenerate d-orbitals of an isolated Ru2+ ion non-degenerate d-orbitals of Ru 2+ in aqua complex V e.m.f = 1.16 V Cu 1 mol dm-3 Cu2+ Au 1 mol dm-3 Au3+ - + V e.m.f = 1.25 V Pt 1 mol dm-3 Ru2+ 1 mol dm-3 Ru3+ Au 1 mol dm-3 Au3+ - + cell A cell B energy gap E degenerate d-orbitals of an isolated Ru2+ ion non-degenerate d-orbitals of Ru 2+ in aqua complex

Standard electrode potential is the relative potential of the electrode under standard conditions compared with the standard hydrogen electrode whose electrode potential is assigned as zero. (ii) Using the data given and relevant data from the Data Booklet, determine the standard electrode potentials of Au3+/Au and that of Ru3+/Ru2+ respectively. E⊖(Au3+/Au) = …………………………………. E⊖(Ru3+/Ru2+) = …………………………………. [3] 1.16 = E⊖(Au3+/Au) – (+0.34) E⊖(Au3+/Au) = +1.50 V ; 1.25 = +1.50 - E⊖(Ru3+/Ru2+) E⊖(Ru3+/Ru2+) = +0.25 V ; working ; (iii) Hence using your answer in (e)(ii) and relevant data from the Data Booklet, state and explain whether Ru 3+ is able to act as a homogenous catalyst f