2013 TPJC H1 Chemistry P2 Suggested Solution

TPJC_2013_8872_02 [Turn over TAMPINES JUNIOR COLLEGE JC2 Preliminary Examination CANDIDATE NAME TUTOR NAME 1 2 CHEMISTRY Paper 2 Candidates answer Section A on the Question Paper. Additional Materials: Answer Paper Data Booklet 8872/02 Thursday, 05 September 2013 2 hours H1 CIVICS GROUP For Examiner’s Use Section A B5 B6 B7 Total READ THESE INSTRUCTIONS FIRST Write your name and civics group on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all the questions. Section B Answer two questions on separate answer paper. 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. This document consists of 18 printed pages.

2 Section A Answer all questions in this section in the spaces provided. For Examiner’s Use 1 Radiotherapy is the medical use of radiation generated from radioactive isotopes to destroy or weaken malfunctioning cells. Examples of radioactive isotopes used for the therapy are those of iodine, phosphorus and lutetium. (a) Lutetium has two naturally occurring isotopes, 175Lu and 176Lu. Their natural abundances are 97.4% and 2.6% respectively. (i) Define the term relative atomic mass. Relative atomic mass is defined as the ratio of the average mass of one atom of an element to 12 1 the mass of an atom of 12C isotope, expressed on the 12C scale. [1] (ii) Calculate, to one decimal place, the relative atomic mass of lutetium. Ar of Lu = 0.175100 1766.21754.97 =×+× [1] [2] (b) 176Lu has a half-life of 101078.3 × years. The half-life of a radioactive isotope is the time taken for half of the atoms in a given mass to decay. Calculate the percentage of a sample of 176Lu isotopes remaining after 1110134.1 × years. number of half-lives = 31078.3 10134.1 10 11 =× × % of sample remaining = % 5.122 1100 3 =⎟ ⎠ ⎞⎜ ⎝ ⎛× [2] [2] Iodine-131 is used to treat the thyroid for cancers and phosphorus-32 is used to control the excess of red blood cells produced in the bone marrow. (c) Complete the table below for the 131I and 32P isotopes. Isotope Number of protons Number of neutrons 131I 53 78 [1] 32P 15 17 [1]

3 Radioactive isotopes are commonly incorporated into compounds to trace the path of biochemical reactions. These compounds are known as radioactive tracers. The structure of fluorodeoxyglucose (18F-FDG), a radioactive tracer widely used in medical imaging, is shown below. O OH OH F18 OH OH 18F-FDG (d) (i) Apart from ether (–O–), circle and name the functional groups that are present in the 18F-FDG shown above. O OH OH F18 OH OH primary alcohol secondary alcohol secondary alcoholsecondary alcoholsecondary fluoroalkane [2] (ii) Calculate the percentage composition by mass of carbon in 18F-FDG. Mr of 18F-FDG = 1810.1111810.1650.126=×+×+×+× % composition by mass of C in 18F-FDG = % 8.39%100181 0.126 =×× [2] (iii) Would you expect 18F-FDG to be soluble in water? Explain your answer. Yes, 18F-FDG is expected to be soluble in water because it has many –OH groups which are capable of forming hydrogen bonds with water. [1] (iv) 18F-FDG is heated under reflux with an excess of the following isotopically labelled carboxylic acid in the presence of concentrated sulfuric acid. CCH3 O18 O H Give the structural formula of the organic product formed and state the type of reaction that has occurred. You may assume that the ether group is inert. [7]

4 O O O F18 O O O O O O Reaction Type: Condensation (or addition-elimination) [2] [Total: 12] 2 In an alkaline fuel cell, the chemical energy from the hydrogen fuel supplied to one electrode is converted into electricity through a chemical reaction with the oxygen supplied to the other electrode. These two electrodes are connected using potassium hydroxide as an electrolyte. A simplified diagram of the fuel cell is shown below. The two half-equations for this cell are 2H2O + 2e− H2 + 2OH− O2 + 2H2O + 4e− 4OH− (a) (i) Combine these two half-equations to show the overall reaction occurring in the cell. Oxidation: H2 + 2OH− → 2H2O + 2e− (×2) Reduction: O2 + 2H2O + 4e− → 4OH− (×1) [2] Overall: 2H2 + O2 → 2H2O (ii) Use oxidation numbers to show which species in your equation is reduced and which is oxidised. H2 is oxidised as the oxidation number of H increases from 0 in H2 to +1 in H2O. [2] O2 is reduced as the oxidation number of O decreases from 0 in O2 to −2 in H2O. [4] Load KOH O2 product H2 electrodes

5 Porous graphite impregnated with suitable catalysts could be used as electrodes for an alkaline fuel cell. (b) (i) Describe the structure of, and the bonding in, the element graphite. Draw a diagram to illustrate your answer. Description: • Graphite has a giant molecular layered structure • Within each layer, each C atom uses three out of its four valence electrons to form covalent bonds with three other C atoms in a trigonal planar arrangement to form hexagonal rings • The 4th valence electron is delocalized over the whole layer • The layers are held together by weak instantaneous dipole – induced dipoles attractions [3] (ii) State a physical property of graphite that metals also possess. Explain, in terms of the bonding present, why it possesses this property. Property Conducts electricity in the solid state. [1] explanation The delocalised electrons along the graphite layers can act as mobile charge carriers to conduct electricity. [4] [Total: 8]

6 3 This question is about period three elements and their compounds. (a) (i) Sketch on the axes provided, the trend in first ionisation energy across period three. [1] (ii) Explain the general trend in first ionisation energy of period three elements. Across the period, nuclear charge increases while screening effect remains relatively constant. Thus effective nuclear charge increases and the valence electrons are more strongly attracted by the nucleus. Hence more energy is required to remove an electron and ionisation energy increases. [1] (iii) Explain the difference between the values of the first ionisation energies of phosphorus and sulfur. The 3p electron to be removed from S is paired and experiences inter-electronic repulsion whereas the 3p electron to be removed from P is unpaired. Thus less energy is required to remove the 3p electron from S and S has a lower ionisation energy than P. [1] [3] (b) Sulfuryl chloride, SO2Cl2, decomposes as follows when heated to 100 °C. first ionisation energy / kJ mol−1 Na Mg Al Si P S Cl Ar

7 SO2Cl2(g) SO2(g) + Cl2(g) (i) Calculate the equilibrium constant, Kc, at 100 °C, given the following values: [SO2Cl2] = 14.6 g dm−3, [SO2] = 3.33 g dm−3, [Cl2] = 11.5 g dm−3. [SO2Cl2] = 3dm mol 108.01.135 6.14 5.3520.1621.32 6.14 −==×+×+ [SO2] = 3dm mol 0520.01.64 33.3 0.1621.32 33.3 −==×+ [Cl2] = 3dm mol 162.071 5.11 5.352 5.11 −==× Kc = ( )( ) ( ) 3 22 22 dm mol 0780.0108.0 162.00520.0 ]Cl[SO ]][Cl[SO −== [3] (ii) Draw a dot-and-cross diagram for sulfuryl chloride. [1] (iii) Complete the electronic configuration of a chlorine atom. 1s2 2s22p63s23p5 [1] Hence describe the bonding in the Cl2 molecule in terms of orbital overlap. Include a diagram in your answer. In