CJC Prelim P1 Worked solutions

1 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/01 Paper 1 Multiple Choice Wednesday 29 August 2018 1 hour Additional Materials: Mult iple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class and NRIC/FIN number on the Answer Sheet in the spaces provided. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 29 printed pages. Catholic Junior College JC 2 Preliminary Examinations Higher 2 WORKED SOLUTIONS

2 9729/01/CJC JC2 Preliminary Examination 2018 1 A sample of tungsten contains four naturally occurring isotopes, 182W, 183W, 184W and 186W. The relative atomic mass of tungsten in this sample is 183.9. What is the percentage of the isotope 182W in this sample? Isotope Relative Abundance (%) 182W ? 183W ? 184W 30.6 186W 28.6 A 10.5 C 26.4 B 14.4 D 40.8 Answer: C Let the percentage of 182W be x. The percentage of 183W would be 100 – 30.6 – 28.6 – x = 40.8 – x Thus, 183.9 = ଵ଼ଶ௫ାଵ଼ଷ(ସ଴.଼ି௫)ାଵ଼ସ(ଷ଴.଺)ାଵ଼଺(ଶ଼.଺) ଵ଴଴ x = 26.4 2 20 cm 3 of 0.100 mol dm −3 of potassium ferrate( VI), K 2FeO4, reacts with sodium ethanedioate, Na2C2O4, in an acidic medium to produce 144 cm 3 of carbon dioxide gas at room temperature and pressure. The half equation of C 2O42– is shown as follows: 2CO2 + 2e– C 2O42– What is the final oxidation state of the iron-containing species after the reaction? A +1 B +2 C +3 D +4 Answer: C Oxidation state of Fe in K 2FeO4 = +6 Amt of FeO 42– reacted = ଶ଴ ଵ଴଴଴ × 0.100 = 0.002 mol Amt of CO 2 formed = 144 ÷ 24 000 = 0.006 mol Amt of C 2O42– reacted = ½ x 0.006 = 0.003 mol

3 9729/01/CJC JC2 Preliminary Examination 2018 [Turn over Ratio of FeO 42– : C2O42– = 0.002 : 0.003 2 FeO 42– Ξ 3 C2O42– 3 mol of C 2O42– will produce 6 mol of e – and 2 mol of FeO42– will accept 6 mol of e–. Therefore, 1 mol of FeO 42– will accept 3 mol of e–. Since FeO 42– is reduced from an oxidation state of +6 to +3. 3 When attracted by a strong magnet, some species are able to exhibit paramagnetism. Such species contain unpaired electrons which are able to spin in a way which aligns parallel to the magnetic field. Which of the following species in the ground state is able to exhibit paramagnetism? 1 O 2 A l+ 3 Ti2+ 4 Cu+ A 1 and 3 only B 2 and 4 only C 1, 3 and 4 only D 2, 3 and 4 only Answer: A ( 1 and 3 only) 1 O : 1s2 2s2 2p4 (2px2, 2py1, 2pz1) there are 2 unpaired electrons in 2p orbital 2 A l+ : 1s2 2s2 2p6 3s2 there are no unpaired electrons 3 Ti2+ : 1s2 2s2 2p6 3s2 3p6 3d2 there are 2 unpaired electrons in 3d orbital 4 Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10 there are no unpaired electrons 4 Phosphorus(V) chloride, PCl5 dissolves in a suitable polar solvent to produce two ions, [PCl4]+ and [PCl6]–. Which of the following shows the correct shape for PCl5, [PCl4]+ and [PCl6]–? PCl5 [PC l4]+ [PC l6]– A trigonal planar square planar square pyramidal B trigonal bipyramidal square planar octahedral C trigonal planar distorted tetrahedral square pyramidal D trigonal bipyramidal tetrahedral octahedral Answer: D

4 9729/01/CJC JC2 Preliminary Examination 2018 PCl5 : 5 bond pairs 0 lone pairs of electrons; shape is trigonal bipyramidal [PC l4]+ : 4 bond pairs 0 lone pairs of electrons; shape is tetrahedral [PC l6]– : 6 bond pairs 0 lone pairs of electrons; shape is octahedral tetrahedral octahedral 5 The table shows the boiling point of some halogenoalkanes. compound boiling point/ °C CH3CH2Cl 12.3 CH3CH2Br 34.8 CH3CH2I 70.0 Which of the following correctly explains the difference in the boiling point? 1 the electronegativity difference between the halogen and carbon increases from C−Cl to C−I 2 the strength of permanent dipole-permanent dipole attraction increases from C −Cl to C−I 3 the strength of instantaneous dipole-induced dipole attraction increases from CH3CH2Cl to CH3CH2I 4 the bond energy of C−X bond decreases from C−Cl to C−I A 1 and 2 only B 2 and 4 only C 3 only D 3 and 4 only Answer: C (3 only) N.B. H-bonding > pd-pd> id-id only if size of electron cloud of molecules are similar. 1 the electronegativity difference between the halogen and carbon should decrease from C−Cl to C−I Statement does not explain for the trend of increasing boiling point from CH3CH2Cl to CH3CH2I. 2 the strength of permanent dipole-permanent dipole attraction decreases from C −Cl to C−I P Cl Cl Cl Cl P Cl Cl Cl Cl Cl Cl -+

5 9729/01/CJC JC2 Preliminary Examination 2018 [Turn over The statement of option 2 is incorrect and does not explain for the trend of increasing boiling point from CH3CH2Cl to CH3CH2I. 3 the strength of instantaneous dipole-induced dipole attraction increases from CH3CH2Cl to CH3CH2I Statement is correct as the total number of electrons increases from CH 3CH2Cl to CH3CH2I and due to the increase in id-id attraction, the boiling point increases from CH3CH2Cl to CH3CH2I. 4 the bond energy of C-X bond decreases from C−Cl to C−I Statement is correct but boiling does not break the C −X bond, so this does not explain for the trend of increasing boiling point from CH3CH2Cl to CH3CH2I. 6 Which of the following changes will result in the greatest decrease in the density of a fixed mass of ideal gas? Pressure Temperature/ K A halves halves B halves doubles C doubles halves D doubles doubles Answer: B Density, ρ = ௠ ௏ Hence, ρ = ௣ெೝ ோ் From the formula above, the greatest decrease in density is brought about when pressure decreases and temperature increases. 7 Consider the following reactions. Reaction 1: CH3+ + Br – → CH3Br Reaction 2: HPO42− + H2BO3− H 2PO4− + HBO32− Which of the following statement is not true about the reactions above? A Both reactions are acid-base reactions. B In reaction 2, HPO42− acts as the Brønsted-Lowry base. C In reaction 2, HBO32− is the conjugate acid of H2BO3−. D In reaction 1, a dative covalent bond is formed between CH3+ and Br – .

6 9729/01/CJC JC2 Preliminary Examination 2018 Answer: C For Reaction 1: CH3+ behaves as the Lewis acid (electron pair acceptor) while Br− behaves as the Lewis base (electron pair donor). Hence it is an acid-base reaction (option A is true) which involves the formation of a dative covalent bond. (option D is true) For Reaction 2: HPO 42− + H2BO3− H 2PO4− + HBO32− H2BO3− behaves as the Br ønsted-Lowry acid (H + donor) while HPO 42− behaves as the Brønsted-Lowry base (H+ acceptor). Hence it is an acid-base reaction. (option B is true) HBO32− is the conjugate base of H2BO3− (option C is not true) 8 Which of the following sketches shows the correct trend in the stated property for the elements in the third period of the Periodic Table? Answer: C Graph A shows the trends of ionic radius across period 3 elements, not atomic radius. Atomic radius across period should be this. A C B D Na Mg Al Si P S Cl Atomic Radius / nm Na Mg Al Si P S Cl First I.E / kJ mol−1 Na Mg Al Si P S Cl Melting Poin