MI Prelim P2 QP

Class Adm No Candidate Name: This question paper consists of 20 printed pages and 2 blank pages. 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

2 Answer all the questions in the spaces provided. 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. For Examiners’ Use (a) Define the term ligand and suggest why bpy can act as a bidentate ligand. ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………[2] (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. Fig 1.1 shows the results experiment. Fig 1.1 0 10 20 30 40 50 Absorbance Vol of Ru 2+ 3 50 40 30 20 10 Vol of bpy

3 [Turn Over By drawing two best-fit lines in Fig 1.1, deduce the formula of the complex ion formed between Ru2+ and bpy and hence draw the structure of the complex ion. Formula of complex ion: ………………………….. Structure of complex ion: [3] For Examiners’ Use

4 (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 Ru 2+ and bpy. Using the data provided, suggest the colour of the complex formed between Ru 2+ and bpy. ………………………………………………………………………………………….. [1] For Examiners’ Use wavelength /nm Absorbance 400 450 500 550

5 [Turn Over (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. Fig 1.2 shows how the five d-orbitals are split in an octahedral environment. Fig 1.2 For Examiners’ Use (i) With reference Fig 1.2, outline why [Ru(H2O)6]2+ is coloured. ………………………………………………………………………………………. ………………………………………………………………………………………. …………………………………………………………………………………… [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 orbita ls 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. energy gap E degenerate d-orbitals of an isolated Ru2+ ion non-degenerate d-orbitals of Ru 2+ in aqua complex

6 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. For Examiners’ Use (ii) Suggest why electrons usually prefer to occupy orbitals singly, rather than in pairs. ………………………………………………………………………………………. …………………………………………………………………………………….[1] (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] (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] energy gap E degenerate d-orbitals of an isolated Ru2+ ion non-degenerate d-orbitals of Ru 2+ in aqua complex

7 [Turn Over (e) Two galvanic cells were set up under standard conditions to determine the standard electrode potential of Ru3+/Ru2+. Fig 1.3 For Examiners’ Use (i) Define the term standard electrode potential. ………………………………………………………………………………………. ………………………………………………………………………………………. …………………………………………………………………………………….[1] (ii) Using the data given in Fig 1.3 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] 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

8 (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 for the reaction between S2O82- and I-. S2O82-(aq) + 2I-(aq) → 2SO42-(aq) + I2(aq) ………………………………………………………………………………………. …………………………………………………………………………………...….. ……………………….…………………………………………………………...…. ……………………………………………………………………………………. [3] For Examiners’ Use (f) Pyridine and phenylamine are two nitrogen-containing compounds. Pyridine has a resonance structure with six p electrons delocalised over the ring. The molecule is planar, with all atoms forming the ring being sp 2 hybridised. The lone pair of electrons on nitrogen occupies one of its sp2 hybrid orbitals. With reference to the shape and orientation of the orbitals about the nitrogen atom in both compounds, suggest why pyridine has a lower pK b value. ………………………………………………………………………………………………. ………………………………………………………………………………………………. ………………………………………………………………………………………………. …………………………………………………………………………………………... [2] [Total: 20]

9 [Turn Over 2 (a) Upon heating at 160 °C, magnesium eth