2019 DHS Prelim H2 Chem P2 ANS

© DHS 2019 [Turn over Name: Answers Index Number: Class: DUNMAN HIGH SCHOOL Prelims Examination 2019 Year 6 H2 CHEMISTRY 9729/02 Paper 2 Structured Questions Additional Materials: Data Booklet INSTRUCTIONS TO CANDIDATES 1 Write your name, index number and class on this cover page. 2 Write in dark blue or black pen. 3 You may use an HB pencil for any diagrams or graphs. 4 Do not use staples, paper clips, glue or correction fluid. 5 Write your answers in the spaces provided on this question paper. The number of marks is given in brackets [ ] at the end of each question or part question. You are advised to show all workings in calculations. You are reminded of the need for good English and clear presentation in your answers. For Examiner’s Use Question No. Section A Marks 1 21 2 14 3 12 4 13 5 15 Total 75

2 © DHS 2019 9729/02 Answer all questions in the spaces provided. 1 This question explores the chemistry of transition metal complexes. (a) Cobalt cations readily form complexes with ligands. Some of the cobalt complexes with their absorption frequencies are shown in Table 1.1. The absorption frequency is the frequency of light absorbed for each complex, and is proportional to the energy of light absorbed. Table 1.1 Complex Absorption Frequency / cm–1 [Co(CN)6]4– 15300 [Co(CN)6]3– 33500 [Co(H2O)6]2+ 9300 [Co(H2O)6]3+ 18200 [Co(NH3)6]3+ 22870 (i) State the electronic configuration of the cobalt cation in [Co(CN)6]3–. Electronic configuration: 1s2 2s2 2p6 3s2 3p6 3d6 [1] (ii) Using the information in Table 1.1, state the relationship between the oxidation state of the metal cation and the absorption frequency of its complex. A higher oxidation state of the metal cation will result in a higher absorption frequency of the complex. [1] (iii) Stronger field ligands generate a d orbital splitting pattern with a larger energy gap when bonded to transition metal cations. Using the information in Table 1.1, rank CN –, H2O and NH 3 in order of increasing ligand field strength. Ligand field strength: H2O < NH3 < CN– [1] (iv) Hence, predict a value for the absorption frequency of [Co(NH3)6]2+. Any value between 9300 and 15300 (excluding boundaries) [1] (b) In the presence of an octahedral ligand field, the 3 d orbitals of cobalt are split into two energy levels.

3 © DHS 2019 9729/02 [Turn Over (i) Using the axes below, draw the shape of a 3d orbital of cobalt of  a higher energy level,  a lower energy level, in the presence of an octahedral ligand field. Higher energy level Lower energy level (accept dx2–y2) (accept dxy, dyz) [2] (ii) Hence, explain why the 3d orbitals of cobalt are split into two energy levels in the presence of an octahedral ligand field. The dz2 and dx2-y2 orbitals have lobes that are in the region of the lone pairs of the ligands. There is greater repulsion between these orbitals and the incoming ligands. As a result, these orbitals are more destabilised and higher in energy. The dxy, dyz and dxz orbitals have lobes that project between the lone pairs of the ligands . There is less repulsion between these orbitals and the negative charges (lone pairs of the ligands). As a result, these orbitals are less destabilised and lower in energy. [2] (iii) In the presence of a tetrahedral ligand field, the ligands approach in between the axes. Complete Fig 1.1 to show how the degenerate 3d orbitals of cobalt will split in a tetrahedral ligand field. Label all the orbitals in your diagram. Fig 1.1 [2] y Energy dxy dxz dyz dz2 dx2-y2

4 © DHS 2019 9729/02 (c) Another cobalt complex, [CoCl6]3 undergoes ligand exchange with CN – ligands to form [Co(CN)6]3– according to the following equilibrium. [CoCl6]3(aq) + 6CN–(aq) [Co(CN)6]3–(aq) + 6Cl–(aq) The effect of pH on the concentration of [Co(CN)6]3– formed is shown in the graph below. (i) Write an equation that describes the dissociation of the weak acid, HCN, in water. HCN(aq) H+(aq) + CN–(aq) [1] (ii) Hence, using Le Chatelier’s Principle, explain the shape of the graph above. HCN(aq) H+(aq) + CN–(aq) -----(1) [CoCl6]3(aq) + 6CN–(aq) [Co(CN)6]3–(aq) + 6Cl–(aq) ------(2) When pH increases, the [H+] decreases. This shifts the position of equilibrium (1) to the right to produce more H +. This also increases the concentration of CN–. The increase in CN– causes the position of equilibrium (2) to shift to the right to decrease the concentration of CN–. This results in an increase in concentration of [Co(CN)6]3– as pH increases. [2] (iii) Explain why the Gibbs free energy change and enthalpy change for the forward reaction are approximately equal. Since there is no change of state or number of moles of particles from reactants to products, the entropy change is approximately zero. Hence, G  H. [1] (iv) Hence deduce the sign of the enthalpy change of the forward reaction given that it is spontaneous. Since reaction is spontaneous, G is negative. Hence, given G  H, H must be negative. [1] pH conc. of [Co(CN)6]3–

5 © DHS 2019 9729/02 [Turn Over (d) Similar to organic compounds, transition metal complexes may exhibit stereoisomerism. (i) For square planar complexes, cis-trans isomerism can be exhibited for MA2B2 type complexes (where M is the metal cation, and A and B are different ligands). cis and trans isomers are differentiated in the following way: cis: Both A ligands next to one another trans: Both A ligands are separated by a B ligand [Pt(Cl)2(NH3)2] is a square planar complex. Complete the diagrams below to show the structures of its cis and trans isomers. Pt ClCl NH3H3N Pt NH3Cl ClH3N Cisplatin trans isomer [1] (ii) For any transition metal complex, enantiomerism is exhibited if it has:  Mirror images that are non-superimposable and  No internal plane of symmetry Deduce if the trans isomer of [Pt(Cl)2(NH3)2] can exhibit enantiomerism. It cannot exhibit enantiomerism as it has an internal plane of symmetry and / or has mirror images that are superimposable. [1] (e) The cyclopentadienyl anion is a common ligand in transition metal complexes. This anion can be generated via the following reaction between tert–butoxide anion and cyclopentadiene. H H O+ H OH+ Cyclopentadiene tert-butoxide cyclopentadienyl anion An organic compound is considered aromatic if all of the following three criteria are met:  it is a cyclic planar compound,  it has a delocalised π electron system, and  it has (4n + 2) π electrons, where n is an integer. One such example is benzene.

6 © DHS 2019 9729/02 (i) State the type of reaction that generates the cyclopentadienyl anion from cyclopentadiene. Acid–Base [1] (ii) Given that the cyclopentadienyl anion is aromatic, explain fully how it meets the three criteria stated above. All the carbon atoms in the anion are sp2 hybridised and have trigonal planar geometry. Hence, the anion is a cyclic planar anion. The lone pair of electrons on the carbon is in a p-orbital which is overlapping with the π bonds of the 2 C=C bonds. Hence, there is a delocalised π electron system with 6 π electrons. [3] [Total: 21] 2 Halogenoalkanes are widely used commercially as they are important precursors for many organic synthesis. (a) One method of synthesising halogenoalkanes is the free radical substitution of alkanes. For example, butane can