2019 MI Prelim H2 Chem P2 ANS

Class Adm No Candidate Name: This question paper consists of 11 printed pages and 1 blank page. 2019 Preliminary Exams Pre-University 3 H2 CHEMISTRY 9729/02 Paper 2 Structured Questions 16 Sept 2019 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 7 23 16 14 15 75

2 1 (a) Zinc-air batteries worked by oxidising zinc with oxygen from the air. At the cathode, oxygen converts to hydroxide ions. At the anode, zinc reacts with the hydroxide ions to form zincate, [Zn(OH)4]2-. (i) Construct an ion-electron equation for the reaction that take s place at each electrode under alkaline conditions. [2] Cathode: O2 + 2H2O + 4e- → 4OH- Anode: Zn + 4OH- → [Zn(OH)4]2- + 2e- (ii) At 298 K, the standard electrode potential of the [Zn(OH)4]2-(aq)|Zn(s) half-cell is -1.25 V. Calculate the cell voltage using relevant data from the Data Booklet. [1] Ecell = 0.40 – (-1.25) = +1.65 V (iii) Using relevant data from the Data Booklet, deduce if oxygen gas can be replaced with chlorine gas. [2] Ecell = 1.36 – (-1.25) = +2.61 V Since the E cell value is more positive when oxygen gas is replaced with chlorine gas, the reaction is feasible. (b) Another type of battery known as the nickel-cadmium battery is a type of rechargeable battery using nickel oxide hydroxide and metallic cadmium as electrodes. The electrode reaction equations for the discharging process under alkaline conditions is given below. Cathode: Cd + 2OH- → Cd(OH)2 + 2e- Anode: 2NiO(OH) + 2H2O + 2e- → 2Ni(OH)2 + 2OH- (i) Construct the overall equation for the reaction that takes place during charging. [1] 2Ni(OH)2 + Cd(OH)2 → 2NiO(OH) + Cd + 2H2O (ii) Suggest a disadvantage of using nickel-cadmium battery. [1] Cadmium is a toxic heavy metal which can result in pollution if discarded in landfill or incinerated. [Total: 7]

3 [Turn over 2 Paramagnetism and diamagnetism are different forms of magnetism. Paramagnetic materials are weakly attracted by an externally applied magnetic field and form magnetic fields in the direction of the applied magnetic field. In contrast, diamagnetic materials are repelled from magnetic fields and form magnetic fields in the direction opposite to that of the applied magnetic field. Transition metals are mostly paramagnetic or diamagnetic. The magnetic property is characterised by the presence of unpaired electrons in paramagnetic compounds and absence of un paired electrons in diamagnetic compounds. The table below shows the magnetic property of metal complexes. Formula of complex Magnetic property of complex Fe(CO)5 Paramagnetic [Fe(H2O)6]2+ Paramagnetic [Fe(H2O)6]3+ Paramagnetic [ZnCl4]2- Diamagnetic [V(H2O)6]3+ [Sc(H2O)3(OH)3] (a) (i) Write the electronic configurations of V in [V(H2O)6]3+ and Sc in [Sc(H2O)3(OH)3]. [2] Sc3+: 1s2 2s2 2p6 3s2 3p6 V3+: 1s2 2s2 2p6 3s2 3p6 3d2 (ii) State the magnetic property of [V(H2O)6]3+ and [Sc(H2O)3(OH)3] in the above table. [1] [V(H2O)6]3+: paramagnetic [Sc(H2O)3(OH)3]: diamagnetic (iii) Draw labelled diagrams to show the shapes of the d orbitals of the metal ion in [V(H2O)6]3+. Include the relative energy of the d-orbitals in the complex. [3] Draw dxy, dxz and dyz, stating that it is of lower energy and dx2-y2 and dz2 at higher energy (iv) Explain why transition metal can show variable oxidation states. [1] The electrons in the 3d and 4s subshells are similar in energy, thus different number of these electrons are available for use in bond formation / ions formed by using different number of electrons for bonding are of similar stability.

4 (b) Sodium reacts with iron pentacarbonyl to produce a salt known as sodium tetracarbonylferrate, Na2Fe(CO)4. 2Na + Fe(CO)5 ⟶ Na2Fe(CO)4 + CO ------- eqn (1) (i) Explain what is meant by the coordination number of a complex using Fe(CO) 5 as an example. [1] Co–ordination number is the number of dative bonds attached to the central atom or ion in a complex. In the case of Fe(CO)5, there are 5 dative bonds formed between Fe atom and C of CO, thus, the coordination number of Fe(CO)5 is 5. (ii) Draw the structure of Fe(CO)5 and state its shape. [2] Shape: trigonal bipyramidal (iii) Deduce the oxidising and reducing agent in eqn (1). [2] Na is the reducing agent since itself is oxidised from Na to Na + in Na2Fe(CO)4, with an increase in oxidation number from 0 to +1. Fe(CO)5 is the oxidising agent since Fe is reduced with a decrease in oxidation number from 0 in Fe(CO)5 to -2 in Na2Fe(CO)4. (iv) Explain why carbon monoxide, CO, is poisonous. [2] CO competes with O 2 for bonding with haemoglobin as it enters into the bloodstream. Being a stronger ligand than O 2, CO will displace the weaker O 2 ligand in oxyhaemoglobin by forming a stronger dative bond with iron in haemoglobin , resulting in a more stable carboxyhaemoglobin complex. The formation of carboxyhaemoglobin complex is not readily reversible due to its stability, and this reduces the availability of haemoglobin for oxygen transport . Deprivation of oxygen causes damage to living organs and tissues, resulting in death. Hence, Co is poisonous. (v) The tetracarbonylferrate dianion acts as a nucleophile and react with alkyl halide by the SN2 mechanism to form a new C-Fe bond. [Fe(CO)4]2- + CH3Br ⟶ [Fe(CH3)(CO)4]- + Br – ------- eqn (2) Suggest a mechanism between Fe(CO)42- and CH3Br. State clearly any intermediates that may be formed and use curly arrows to indicate the movement of electron pairs. [3]

5 [Turn over SN2 Nucleophilic Substitution (not included in marking point as question already states SN2) (vi) State the rate equation for the reaction in eqn 2. [1] Rate = k [CH3Br][Fe(CO)42-] (vii) I Sketch the graph of [Fe(CH3)(CO)4]- against time. II Sketch the graph of rate against [CH3Br] given that [Fe(CO)4]2- is in excess. [2] [Fe(CH3)(CO)4]- / mol dm-3 Time / s Rate / mol dm-3 s-1 [CH3Br] / mol dm-3

6 I: II: (c) The melting point of sodium tetracarbonylferrate, Na 2Fe(CO)4 is lower than that of sodium oxide. With reference to the structure and bonding, explain the difference in melting point.[3] Both sodium oxide and Na2Fe(CO)4 have giant ionic lattice structure with strong electrost