2020 YIJC Chem Prelim P2 (Suggested Answers)

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUGGESTED ANWERS CG INDEX NO CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper Additional Materials: Data Booklet 9729/02 1 September 2020 2 hours This document consists of 24 printed pages and 4 blank pages. For Examiner’s Use Paper 1 /30 Paper 2 1 /14 2 /13 3 /15 4 /13 5 /20 Penalty /75 Paper 3 /80 Paper 4 /55 Overall Percentage (%) READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. 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.

2 ©YIJC Answer all the questions in this section in the spaces provided. 1 Sodium chlorate(I), NaClO, found in household bleach produces chlorine upon addition of acidified dilute hydrochloric acid according to the equation below. ClO–(aq) + 2H+(aq) + Cl–(aq)  Cl2(aq) + H2O(l) A 15.0 cm 3 sample of a household bleach is diluted to 100 cm 3. 25.0 cm 3 of this solution is then pipetted out into a conical flask containing dilute hydrochloric acid. Excess potassium iodide solution is added which will react with the chlorine present in the sample according to the equation below. Cl2(aq) + 2I–(aq)  I2(aq) + 2Cl–(aq) This resultant solution is then titrated against 0.150 mol dm –3 sodium thiosulfate solution and required 17.20 cm 3 for complete reaction. Sodium thiosulfate reacts with iodine according to the following equation. I2(aq) + 2S2O32–(aq)  2I–(aq) + S4O62–(aq) (a) (i) Calculate the amount of iodine produced in this reaction. Amount of S2O32 = 0.150 0.0172 = 2.58 103 mol Amount of I2 = 0.5 2.58 103 = 1.29 103 mol [1] (ii) Calculate the amount of sodium chlorate(I) present in the 15.0 cm3 sample. Amount of I2 = 1.29 103 mol = Amount of Cl2 Amount of NaClO in 25.0 cm3 of the diluted solution = 1.29 103 mol Amount of NaClO in 15.0 cm3 of the undiluted solution = (1.29 103) (100/25) = 5.16 103 mol [1] (iii) Calculate the concentration of the sodium chlorate(I) in the household bleach in g dm–3. Mass of NaClO in 15.0 cm3 of the undiluted solution = 0.0516 (23.0 + 35.5 +16.0) = 0.38442 g Concentration of NaClO = 0.38442 ÷ 0.015 = 25.6 g dm3 [2]

3 ©YIJC [Turn over (b) In a thiosulfateacid reaction, a pale yellow precipitate of sulfur is formed slowly. S2O32–(aq) + 2H+(aq)  S(s) + SO2(g) + H2O(l) The rate of the reaction can be studied by following the time taken for the reaction mixture to turn opaque. Outline a suitable experiment for determining the order of reaction with respect to S2O32–. You may assume that you are provided with:  100 cm3 of 0.100 mol dm3 sodium thiosulfate, Na2S2O3 and  100 cm3 of 2.0 mol dm-3 hydrochloric acid, HCl No details regarding the use of specific glassware are required. 1. Prepare a conical flask of 20.0 cm3 of 0.100 mol dm3 Na2S2O3. 2. Prepare a second conical flask of 5.0 cm3 of 2.0 mol dm3 HCl. 3. Pour the hydrochloric acid rapidly into the first conical flask. Start the stopwatch during this addition. 4. Mix the contents t horoughly by swirling the flask and place the conical flask on top of a piece of page with a printed logo. 5. Stop the stopwatch when the solution first becomes opaque and note the time, t1. 6. Repeat Steps 1 to 5 with 10.0 cm3 of Na2S2O3, 10.0 cm3 of deionised H2O and 5.0 cm3 HCl. 7. Compare the volume of Na2S2O3 against time.  If t2 is the same as t1, order with respect to thiosulfate is 0  If t2 is twice that of t1, order with respect to thiosulfate is 1.  If t2 is four times that of t1, order with respect to thiosulfate is 2. [3]

4 ©YIJC (c) Thiosulfate, S2O32 and tetrathionate, S4O62 are oxoanions of sulfur. Peroxodisulfate(VI), S2O82, is another oxonaion of sulfur. S2O82 is able to oxidise the tartrate ion, [CH(OH)(CO2)]2 in the following reaction. (i) With the aid of a Boltzmann distribution diagram, explain how an increase in temperature will lead to an increase in the rate of the reaction. When the temperature is increased from T2 to T1,  The average kinetic energy of the particles increases.  There is an increase in the fraction of particles with energy equal to or greater than the activation energy, Ea.  This result in an increase in the frequency of effective collisions, hence the reaction increases.  A higher temperature also results in a larger rate constant and hence an increase in reaction rate. [4] energy, E fraction of particles Ea fraction of particles with E  Ea at high T2 fraction of particles with E  Ea at low T1 T1 (lower temperature) T2 (higher temperature) 0

5 ©YIJC [Turn over (ii) The reaction between S2O82 and [CH(OH)(CO2)]2 is very slow, even at elevated temperature. Suggest an explanation for this observation. Repulsion between two negatively charged ions caused the activation energy to be high. [1] (iii) Steps 1 and 2 represent a possible mechanism for the reaction between S2O82 and the tartrate ion, [CH(OH)(CO2)]2 involving Mn3+ cations. step 1: 6Mn3+ + [CH(OH)CO2]2 + 2H2O  6Mn2+ + 2CO2 + 2HCO2 + 6H+ step 2: 6Mn2+ + 3S2O82  6Mn3+ + 6SO42 Explain the role of the Mn3+ ions in this reaction with reference to the mechanism given. Mn3+ functions as a catalyst. It is used in step 1 and regenerated in the second step. [2] [Total: 14]

6 ©YIJC 2 Nitrosyl chloride, NOCl, is a yellow gas that is most commonly encountered as a decomposition product of aqua regia, a mixture of hydrochloric acid and nitric acid. At 250 oC, NOCl readily dissociates into NO and Cl2. 2NOCl(g) ⇌ 2NO(g) + Cl2(g) (a) (i) In NOCl, the central atom is nitrogen. Draw a dot-and-cross diagram to show the bonding present within a NOCl molecule. You should distinguish carefully between electrons originating from the central atom and those from the other atoms. [1] (ii) State and explain, with reference to the Valence Shell Electron Pair Repulsion theory, the shape of the nitrosyl chloride molecule. There are 3 electron pairs around the central atom, N. To minimise repulsion , the basic shape for the 3 electron pairs are arranged in a trigonal planar arrangement. The 3 electron pairs consists of 2 bond pairs and 1 lone pair and the molecule has a final shape of bent shape. [2]

7 ©YIJC [Turn over (iii) Table 2.1 shows the electronegativity values of the atoms in nitrosyl chloride. Table 2.1 atom electronegativity / Pauling units nitrogen 3.04 oxygen 3.44 chlorine 3.16 Predict whether nitrosyl chloride is a polar molecule. Explain your reasoning. Nitrosyl chloride is a polar molecule. As the N-O bond and N-Cl bond are polar with different dipole moments and there is a net dipole moment as