2024 Prelims CJC H2 Chem P1 (Ans)

1 9729/01 CJC JC2 Preliminary Examination 2024 [Turn over CANDIDATE NAME CLASS 2T CHEMISTRY 9729/01 Paper 1 Multiple Choice 12 September 2024 1 hour Additional Materials: Multiple 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 28 printed pages. WORKED SOLUTIONS Catholic Junior College JC 2 Preliminary Examinations Higher 2

2 9729/01 CJC JC2 Preliminary Examination 2024 1 Which of the following statements about the properties associated with ionic and covalent bonds is correct? A The only covalent compounds with high melting points are those in which hydrogen bonds occur. B Any covalent compound that contains both oxygen and hydrogen in its molecule forms hydrogen bonds. C Both ionic bonds and covalent bonds cannot occur in the same compound. D Ionic compounds differ from metals in that ionic compounds do not conduct electricity in the solid state. 2 Which of the following shows the correct shape and bond angle of the molecule given? molecule shape bond angle A OF2 bent 117o B BH3 trigonal pyramidal 120o C XeF4 tetrahedral 109o D SF6 octahedral 90o Topic: Chemical Bonding The shape for each compound is as follows: 105o . ... 120o . 90o . . 90o . bent trigonal planar square planar octahedral Answer: D Topic: Chemical Bonding Option A: Incorrect, diamond and graphite has high melting points and only contain covalent bonds between the carbon atoms. Option B: Incorrect because the hydrogen atom must be directly attached to the oxygen atom for hydrogen bond to be formed. Option C: Incorrect, both ionic and covalent bonds can exist in the same compound, eg NH4+Cl-. Option D: Correct as ions in ionic compounds are in fixed positions in the solid state and hence cannot act as charge carriers in conducting electricity. Answer: D

3 9729/01 CJC JC2 Preliminary Examination 2024 [Turn over 3 Thiophenol and phenol are both acting as weak acids in aqueous solutions. phenol thiophenol pKa = 10.0 pKa = 6.50 Which statements are correct? 1 H+ is less easily removed from thiophenol than phenol because the S–H bond is less polar than O–H bond. 2 The Ka value is smaller for phenol because the O–H bond is stronger than the S–H bond. 3 When pure liquid thiophenol and phenol are mixed together without water, the S –H bond in thiophenol is more likely to dissociate than the O–H bond in phenol. . A 1, 2 and 3 B 1 and 2 C 2 and 3 D 3 only Topic: Chemistry of Aqueous Solutions From pKa values, thiophenol is a stronger acid than phenol. 1: Incorrect. H+ is more easily removed from thiophenol since it is the stronger acid. 2: Correct. The Ka value for phenol is smaller (10–10) compared to thiophenol (10 –6.5). Furthermore O–H bond (460 kJ mol–1) is stronger than S–H bond (347 kJ mol–1), either refer bond energy value from Data Booklet or considering shorter O–H bond length due to smaller atomic radius of oxygen. 3: Correct. Given a larger Ka value of thiophenol, the acid dissociation occurs more likely than phenol, and phenol may act as a base. Answer: C – similar to 2022/P1/Q3

4 9729/01 CJC JC2 Preliminary Examination 2024 4 The value of pV is plotted against p at the same temperature for four gases, where p is the pressure and V is the volume of the gas. What are the possible identities of Gas 1, 2, 3 and 4? Gas 1 Gas 2 Gas 3 Gas 4 A He CH3F HF H2O B He HF CH3F H2O C H2O He CH3F HF D H2O HF CH3F He pV p Gas 4 Gas 3 Gas 2 Gas 1 Topic: The Gaseous State Extent of deviation from ideal gas behaviour: H2O > HF > CH3F > He Both H 2O and HF form intermolecular hydrogen bonding. H 2O has more extensive intermolecular hydrogen bonding than that of HF. Hence. H 2O deviates more from ideality. H2O and HF (with stronger intermolecular hydrogen bonding) deviates more from ideality than CH 3F (with weaker intermolecular permanent dipole - permanent dipole attractions). CH3F (with stronger intermolecular permanent dipole -permanent dipole attractions) deviates more from ideality than He (with weaker intermolecular instantaneous dipole-induced dipole attractions). Answer: D

5 9729/01 CJC JC2 Preliminary Examination 2024 [Turn over 5 The graph below shows the melting points of four consecutive Period 3 elements W to Z. Which of the following graphs represents the correct trend of the stated property for W to Z? A B C D Melting Point (K) Elements Y W Z X Ionic radius Y W Z X Ions Y W Z X Electrical Conductivity Y W Z X 1st Ionisation energy Elements Melting Point Y W Z X Oxides Elements

6 9729/01 CJC JC2 Preliminary Examination 2024 Hence, the graphs showing the correct trend of the stated property for W (Mg) to Z (P) are as follows: A MgO ( W) has the highest melting point. This is due to the weaker ionic bonds within Al2O3 than MgO as A l2O3 is an ionic compound with partial covalent character. P4O10 (Z), is a simple molecular oxide. Only a small amount of energy is required to overcome weak instantaneous dipole - induced dipole attraction between the molecules, hence, explaining its low melting point. B The ionic radius of P3- (Z) is much larger than that of Si 4+(Y)/ A l3+(X)/ Mg 2+(W) due to the extra quantum shell of electrons. From Mg 2+ (W) to Si 4+ (Y), the ionic radii decrease due to an increase in nuclear charge but with approximately constant shielding effect. Ionic radius Y W Z X Ions Melting Point Y W Z X Oxides Topic: The Periodic Table From the graph of melting point against elements, Y has the highest melting point. Y is Si as Si has the highest melting point among the Period 3 elements due to large amount of energy needed to overcome the extensive strong covalent bonds  W – Magnesium, X – Aluminium, Y – Silicon, Z – Phosphorus. Answer: B Melting Point (K) Elements YW ZX

7 9729/01 CJC JC2 Preliminary Examination 2024 [Turn over C Conductivity increases from Mg (W) to A l (X). Si (Y) is a semi –conductor. P (Z) is a non-conductor. D 6 To obtain the magnesium present in a mineral ore filled with impurities of Al2O3 and P4O10, which of the following reagents could be used to dissolve the impurities in the ore? A dilute NH3 B dilute HCl C dilute HNO3 D dilute NaOH Topic: Periodicity Al2O3 (being amphoteric) wil