ACJC H2 CHEM P1 Ans Prelim

1 © ACJC 2017 9729/01/Prelim/2017 [Turn over ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY 9729/01 Higher 2 Paper 1 Multiple Choice 24 August 2017 1 hour Additional Materials: Optical Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluids. Write your name, index number and tutorial class on the Optical Answer Sheet in the spaces provided unless this has been done for you. 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 Optical Answer Sheet. Read the instructions on the Optical 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 14 printed pages, including this cover page. 9729/01/Prelim/17 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2017 Department of Chemistry [Turn over

2 © ACJC 2017 9729/01/Prelim/2017 [Turn over For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 Carbon disulfide is a colo urless volatile liquid with the formula CS2. The compound is used frequently as a building block in organic chemistry as well as an industrial solvent. It reacts with nitrogen monoxide, NO, to form a yellow solid and two gases. These two gases are formed in equal amounts. What are these two gases? A CO2, NO2 B CO2, N2 C CO, N2 D CO2, N2O The yellow solid is sulfur. You cannot balance the equation (with the two product gases in equal amounts) with the three other options. Balanced equation: CS2 + 2NO  CO2 + N2 + 2S 2 Use of the Data Booklet is relevant to this question. Which of the following ions will be deflected the most in an electric field? A S2- B Br - C F - D O2- You are to refer to the Data Booklet for the ionic radii (although it is not absolutely necessarily to do so; you can use the Group trend for ionic sizes). The charge density of the oxide ion is the highest amongst the four. 3 Use of the Data Booklet is relevant to this question. Nuclear magnetic resonance (NMR) spectroscopy is a n analytical t echnique that uses the magnetic properties of certain atomic nuclei in order to elucidate the structure of an organic molecule. Atomic nuclei with an even number of protons and an odd number of neutrons (or vice versa) are most suitable for NMR spectroscopy. Which of the following nuclei is least suitable for NMR spectroscopy? A 28Si B 31P C 103Rh D 19F A: no. of protons 14 no. of neutrons = 28 – 14 = 14 B: no. of protons 15 no. of neutrons = 31 – 15 = 16 C: no. of protons 45 no. of neutrons = 103 – 45 = 58 D: no. of protons 9 no. of neutrons = 19 – 9 = 10

3 © ACJC 2017 9729/01/Prelim/2017 [Turn over 4 A 25.00 cm 3 sample of a solution of 0.150 mol dm -3 MoOx2- was passed through a Jones redu ctor (a column of zinc powder). It was reduced to Mo 3+. The f iltrate required 22.50 cm3 of 0.100 mol dm-3 acidified KMnO4 (aq) to obtain back the original amount of MoOx2-. What is the value of x? A 4 B 3 C 2 D 1 Amt of electrons involved = (22.50 X 0.100 / 1000) X 5 = 0.01125 Change in oxidation state in Mo = 0.01125 / (25.00 X 0.150 / 1000) = 3 Original oxidation state of Mo = 3 + 3 = 6 Hence x = 4.

4 © ACJC 2017 9729/01/Prelim/2017 [Turn over For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 5 Neoxanthin is a major xanthophyll found in green leafy vegetables such as spinach. Which of the following σ bonds are present in neoxanthin? 1 A σ bond formed by sp2–sp3 overlap. 2 A σ bond formed by s–p overlap. 3 A σ bond formed by sp–sp2 overlap. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 Recall: Linear Carbon: sp (eg. =C= or ─C≡) Trigonal Planar Carbon: sp2 Tetrahedral Carbon: sp3 Bent Oxygen (or Tetrahedral in terms of electron pair geometry) in C-O-C and C-O-H: sp3 Option 2 is wrong. For s-p overlap, it means that the s orbital belongs to hydrogen, whereas the p orbital (unhybridised) belong to carbon. But the overlaps in the following cases are:

5 © ACJC 2017 9729/01/Prelim/2017 [Turn over 6 The enthalpy changes for the following reactions were measured experimentally: CH3CH2CH3(g) + 5O2(g)  3CO2(g) + 4H2O(l) H = 2202 kJ mol1 H2(g) + ½ O2(g)  H2O(l) H = 286 kJ mol1 CH3C≡CH(g) + 2H2(g)  CH3CH2CH3(g) H = 310 kJ mol1 What is the enthalpy change of combustion of propyne, CH3C≡CH, in terms of kJ mol-1? A -2226 B -1940 C -1606 D -1320 CH3─C≡C─H(g) + 4O2(g) + 2H2(g) + O2(g) 3CO2(g) + 2H2O(l) + 2H2(g) +O2(g) CH3CH2CH3(g) + 4O2(g) + O2(g) 3CO2(g) + 4H2O(l) By Hess’ Law, ΔHc(propyne) = ─310 ─2202 ─(2)(─286) = ─1940 kJ mol-1 7 35 cm 3 of 0.001 mol dm -3 nitric acid solution was added to 3 5 cm3 of sulfuric acid solution of the same concentration . What is the resulting pH of the combined solution? A 1.5 B 2.5 C 2.8 D 3.0 Amount of H+ from HNO3 = (0.035)(0.001) = 0.000035 mol Amount of H+ from H2SO4 = (2)(0.035)(0.001) = 0.000070 mol [H+] = (0.000035 + 0.000070) ÷ (0.035 + 0.035) = 0.0015 mol dm-3 pH = ─log [H+] = 2.82 ΔHc(propyne) ─310 ─2202 (2)(─286)

6 © ACJC 2017 9729/01/Prelim/2017 [Turn over 8 In which of the following pairs is the bond angle in the first species smaller than that in the second species? 1 PBr3, PBr4+ 2 H2Se, H2O 3 SF2, SCl2 A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 Recall: VSEPR model dictates that “Lone Pair – Lone Pair repulsion > Lone Pair – Bond Pair repulsion > Bond Pair – Bond Pair repulsion” Option 1 Lone pair on P in PBr3 is closer to nucleus of P, compared to the Bond Pair in PBr4+. There is stronger repulsion between lone pair and bond pair in PBr3. Hence the bond angle will be smaller than that in PBr4+. Option 2 Selenium is larger atom than Oxygen. Selenium is also less electronegative than Oxygen. Electron density in Se─H bond is further away from Se, compared to electron density in O─H bond. There is weaker repulsion between bond pairs in H2Se. Hence the bond angle will be smaller than that in H2O. Option 3 Fluorine is more electronegative than chlorine. Electron density in S─F bond is further away from S, compared to electron density in S─Cl bond. There is weaker repulsion between bond pairs in SF2. Hence the bond angle will be smaller than that in SCl2.

7 © ACJC 2017 9729/01/Prelim/2017 [Turn over 9 X and Y are both ideal gases. X has the smaller molecular mass. Which of the following diagrams correctly describe the behaviour of equal masses of these gases? All temperatures are measured in oC. A B C D For Ideal Gas, PV = nRT PV = (m / Mr) RT At constant Volume, P = (m R / VMr) T where Gradient of Line = (m R / VMr) X has smaller Mr than Y. Hence the line for X has a steeper gradient than Y. temperature P Y X o temperature V X Y o temperature P Y X o temperature