2021 CJC H2 CHEM Prelim P1 WORKED solutions to be uploaded

1 9729/01 CJC JC2 Preliminary Examination 2021 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/01 Paper 1 Multiple Choice September 2021 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 12 printed pages. Catholic Junior College JC 2 Preliminary Examinations Higher 2 WORKED SOLUTIONS General comments Students found questions 2, 3, 14, 18, 19, 20, 22, 23, 24, 25, and 30 to be the most challenging. (You should revisit and fully understand these even you got it correct)

2 9729/01 CJC JC2 Preliminary Examination 2021 1 Nitrogen exists as a diatomic molecule, N 2. Hydrazine, N2H4, and dinitrogen difluoride, N2F2, are compounds of nitrogen. Which of the following gives the correct number of  bonds in N2, N2H4 and N2F2? number of  bonds in N2 number of  bonds in N2H4 number of  bonds in N2F2 A 2 0 1 B 2 1 1 C 1 1 2 D 3 0 2 (Inspired by H1 2020 MCQ6) Concept: Chemical Bonding,  and  bonds Answer: A number of  bonds in N2 number of  bonds in N2H4 number of  bonds in N2F2 (either cis or trans) A (1), 2 (5), no  (3), 1 2 Aluminium chloride is a covalent compound that forms a dimer with the formula Al2Cl6. A compound of gold and chlorine has a similar molecular formula of Au2Cl6 and has the following structure: The three statements below are properties of the gold compound, Au2Cl6. 1 The oxidation state of the metal is +3. 2 The dimer exists in the vapour phase. 3 The Cl-Au-Cl bond angle is 90o. Which property described is different from that of the aluminium compound, Al2Cl6? A 1 and 2 only B 1, 2 and 3 C 3 only D 2 and 3 only

3 9729 CJC JC2 Preliminary Exam 2021 [Turn over (Inspired by H2 2018 MCQ2) Concept: Chemical Bonding, Al2Cl6. Answer: C 1. This is a similar property to Al2Cl6. Chlorine is more electronegative than Al. Each Al also has an oxidation state of +3 as each Cl has an oxidation state of -1. 2. This is a similar property to Al2Cl6. Al2Cl6 dimer exists in the gas phase. 3. Not similar. The Cl-Al-Cl bond angle is 109.5o. 3 The table below lists three compounds: compound boiling point / oC CH3CH2‒S‒H 35 CH3‒S‒CH3 37 CH3CH2‒O‒H 78 Which of the following statements about the compounds is true? A The C‒S‒H bond angle is larger than the C‒O‒H bond angle because S is larger than O. B CH3CH2‒S‒H has weaker intermolecular hydrogen bonding than CH3CH2‒O‒H. C CH3CH2‒S‒H and CH 3‒S‒CH3 have similar boiling points because they have intermolecular permanent dipole – permanent dipole forces of attraction of similar strengths. D CH3CH2‒O‒H has the highest boiling point because the O‒H bond energy is higher than the S‒H bond energy. (Inspired from 2019 H2 A level P3 Q2(f)(i)) Concept: Chemical Bonding, Bond angle and IMF Answer: C A The C‒S‒H bond angles are smaller than the C‒O‒H bond angle because S is less electronegative than O. The valence electrons around S will be further away compared to the valence electrons in O. As a result, there is weaker bond pair – bond pair repulsion around S than around O, giving rise to a smaller bond angle. B CH3CH2‒S‒H does not have intermolecula r hydrogen bonding as the H present in the molecule is not bonded to F, O or N. C True. Both are isomers of each other (same number of electrons) and both are polar molecules. D CH3CH2‒O‒H has the highest boiling point due to stronger intermolecular hydrogen bonding which is absent in the other two compounds. The strength of covalent bonds is not relevant to the boiling point for simple molecules.

4 9729/01 CJC JC2 Preliminary Examination 2021 4 Use of the Data Booklet is relevant to this question. An isotope of a metal, Z, undergoes radioactive decay to form helium and an element hafnium, Hf, according to the following equation. Z He2 4 + Hf Given that Hf has a nucleon number of 176, which row correctly shows the identity and composition of Z? identity of Z number of nucleons in Z A tungsten 180 B tungsten 178 C osmium 180 D osmium 178 Concept: Atomic Structure, sub-atomic particles. Answer: A From Data Booklet, the atomic number of hafnium is 72. Given that the nucleon number is 176, this will be a more detailed equation of the decay: Z He2 4 + Hf72 176 Hence, proton number of Z is 72 + 2 = 74. This correlates to tungsten. The nucleon number of Z is 176 + 4 = 180 5 Use of the Data Booklet is relevant to this question. The first six ionisation energies of an element, Y, in kJ mol‒1 are shown. 738; 1451; 7733; 10543; 13630; 18020 Y forms an oxide by heating Y with oxygen gas. What is the spdf electronic configuration of Y in its oxide form? A 1s2 2s2 2p6 3s2 3p6 B 1s2 2s2 2p6 C 1s2 2s2 2p6 3s2 D 1s2 2s2 2p6 3s2 3p6 4s2 Concept: Atomic Structure, ionisation energy. Answer: B The significant increase from the second to the third ionisation energy indicates that Y is from Group 2. From Data Booklet, the ionisation energies match Mg (1s2 2s2 2p6 3s2) most closely.

5 9729 CJC JC2 Preliminary Exam 2021 [Turn over When Mg forms MgO, it loses its two valence electrons and hence its electronic configuration is 1s2 2s2 2p6. 6 What is the element that has a second ionisation energy lower than that of each of the elements either side of it in the Periodic Table? A boron B nitrogen C oxygen D fluorine (Inspired from 2010 H1 A level MCQ3) Concept: Atomic Structure. Second Ionisation energy Answer: D Second IE is F+  F2+ + e‒ Element before F: O Element after F: Ne O+: 1s22s2 2px12py12pz1 F+: 1s2 2s2 2px22py12pz1 Ne+: 1s22s2 2px22py22pz1 Comparing F and O: Less energy is required to remove a paired 2px electron from F + due to interelectronic repulsion. Hence second IE of F is less than O. Comparing F and Ne: Less energy is required to remove the electron from F + because it has a lower nuclear charge than Ne and a smaller ionic size than Ne+. Alternatively, check the Data Booklet: Second IE values in kJ mol-1: O F Ne 3390 3370 3950 7 Analysis of a mixture of two sulf ur-containing gases show that hydrogen sulf ide, H2S, and carbon sulfide, CS2, are present in a 3 : 1 mole ratio. This mixture is burned in excess oxygen. What will be the CO2 : SO2 mole ratio in the mixture obtained after complete combustion? A 1 : 2 B 1 : 3 C 1 : 4 D 1 : 5 2016 P1 Q2 modified Concept: Mole concept & stoichiometry Ans: D Let amount of H2S be 3x mol and CS2 be x mol, H2S + 3 2O2 → SO2 + H2O

6 9729/01 CJC JC2 Preliminary Examination 2021 Amount of SO2 produced by H2S = 3x mol CS2 + 3O2 → 2SO2 + CO2 Amount of SO2 produced by CS2 = 2x mol, Amount of CO2 produced = x mol Hence, mole ratio of CO 2 : SO2 in the mixture after complete combustion will be x : (3x + 2x) = 1 : 5 8 Use of the Data Booklet is relevant to this question. A mordant is a soluble salt which forms an acidic aqueous solution and improves the binding of the molecules of a dyestuff to a material.