2021 DHS H2 Prelim Paper 2 Solutions

© DHS 2021 9729/02 [Turn over Name: Centre/Index Number: Class: DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 15 September 2021 2 hours READ THESE INSTRUCTIONS FIRST Write your centre number, index number, name and class at the top of this page. Write in dark blue or black pen. 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. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 12 2 12 3 13 4 14 5 24 Total 75 This document consists of 30 printed pages.

2 © DHS 2021 9729/02 Answer all the questions in the spaces provided. 1 (a) Water chemistry plays an important role in aquariums. For disinfection purposes, chlorine, Cl2, and chloroamine, NH2Cl, are introduced into our tap water to destroy bacteria and viruses. However, these two chlorine–containing compounds are extremely harmful to fishes as they cause gill burns and blood poisoning. The typical concentration of some chemical species found in our tap water are listed in Table 1.1. Table 1.1 chemical species concentration / mol dm-3 Cl2 3.24 ´ 10-5 NH2Cl 3.88 ´ 10-5 PO43- 1.26 ´ 10-7 Fe (as Fe2+ & Fe3+) 1.08 ´ 10-7 To prepare an aquarium tank, a water conditioner containing sodium thiosulfate, Na2S2O3, is added to the tank to remove both the aqueous chlorine and chloroamine present in tap water. Sodium thiosulfate reacts with chlorine and chloroamine respectively as shown in the reactions below. 4Cl2 + S2O32- + 5H2O ® 8Cl- + 2SO42- + 10H+ NH2Cl + 2S2O32− + H2O ® NH3 + Cl− + S4O62− + OH− (i) Calculate the volume of 0.010 mol dm-3 sodium thiosulfate solution that would be required to prepare a tank containing 360 dm3 of water. [3] moles of Cl2 present = 3.24 x 10–5 x 360 = 1.167 x 10–2 mol moles of NH2Cl present = 3.88 x 10–5 x 360 = 1.397 x 10–2 mol moles of S2O32– to react with Cl2 = ¼ x 1.167 x 10–2 = 0.002916 mol moles of S2O32– to react with NH2Cl = 2 x 1.397 x 10–2 = 0.02794 mol Volume of S2O32– required = !.!!#$%&'!.!#($)!.!%! = 3.09 dm3 (ii) State the full electronic configuration of iron in Fe3(PO4)2. [1] 1s22s22p63s23p63d6

3 © DHS 2021 9729/02 [Turn over (b) The direct reaction of hydrogen with chlorine gives hydrogen chloride. However, hydrogen chloride is produced industrially by treatment of halide salts with sulfuric acid. On the other hand, hydrogen iodide, HI, is produced by the reaction of iodine with hydrogen sulfide or with hydrazine. (i) Explain briefly why it is not industrially viable to produce HCl through the direct reaction of hydrogen with chlorine. [1] Reaction of hydrogen and chlorine is explosive to handle. (ii) Write a balanced equation showing the formation of HI from iodine and hydrogen sulfide. Hence, with reference to the Data Booklet, show that the reaction is spontaneous. S + 2H+ + 2e- ⇌ H2S Eo = +0.14 V [2] I2 + H2S ® 2HI + S S + 2H+ + 2e- ⇌ H2S Eo = +0.14 V I2 + 2e- ⇌ 2 I- Eo = +0.54 V Eocell = Eocathode - Eoanode = 0.54 - 0.14 = +0.40 V > 0 Since Eocell > 0, the redox reaction is spontaneous under standard conditions. (iii) Explain why the pKa values decrease from HCl to HI. [2] Down the group, size of halogen atoms increases. Hence, bond length of H–X increases and H–X bond strength decreases down the group. It is therefore easier to break the H–X bond to dissociate H+ and halide ions in aqueous solution. Acid strength increases from HCl to HI, Ka values therefore increases and pKa decreases. (c) In an experiment, a sample of chloride ions was vapourised and passed through an electric field. Analysis revealed that a beam of 37Cl− gives an angle of deflection of −1.0°. (i) State the angle of deflection for a beam of deuterium nuclei in the same experimental set–up. (D is deuterium,H12) [1] Angle of deflection ∝ !"#$%&'#(( *+,-./0,11 of 37Cl− = −%2( *+,-./0,11of 𝐷'# = + %# Angle of deflection of deuterium nuclei = +1.0 ÷%2(×%# = +18.5°

4 © DHS 2021 9729/02 (ii) Under identical conditions, a beam of particles, Z, each having 32 times the mass of a proton, was deflected by an angle of −2.3°. Given that the proton number of Z is 16, deduce the identity of Z. [2] Since proton number of Z is 16, Z is a sulfur ion. Let charge of Z be q. Z: −2.3° ∝ )*+ 37Cl− : −1.0° ∝ −%2( 3#.23%.!=42#÷−%2( q = –2 (nearest whole number) Hence the identity of Z is S2–. [Total: 12] 2 The electrochemical oxidation of benzylic alcohols to the corresponding aldehydes via a two-phase electrolysis system was reported in 2007. An example is shown in Fig. 2.1. Fig. 2.1 (a) State the reagents and conditions for the conversion of benzylic alcohols to the corresponding aldehydes in a school laboratory. [1] K2Cr2O7 in dilute H2SO4, heat with immediate distillation (b) The two-phase electrolysis system comprises the aqueous NaBr and organic CHCl3 layers. (i) Draw simple diagrams to illustrate how a water molecule can be attached to a sodium cation, and to a bromide anion in the aqueous layer. Label each diagram to show the type of interaction involved. [2] Na+ cation Br- anion 1 OH R H O R R = H or alkyl Pt electrodes, aqueous NaBr/CHC l3 solvent layers room temperature Na + H O H d- d+ d+ ion-dipole interaction H O H d+ d- d+ Br - ion-dipole interaction

5 © DHS 2021 9729/02 [Turn over (ii) Predict and explain which layer the benzylic alcohols would predominantly dissolve in. [2] Layer predominantly dissolved in: CHCl3 / Organic layer Explanation: The less extensive hydrogen bonds formed between benzylic alcohol and water molecules (due to the non–polar benzene ring interfering with the hydrogen bonding between the –OH group of benzylic alcohol and water molecules) will not release sufficient energy to overcome the more extensive hydrogen bonds between water molecules. (c) (i) Describe the change in oxidation state of the benzylic carbon, labelled 1 in Fig. 2.1, when the alcohol is oxidised to the corresponding aldehyde. [1] The oxidation state of the benzylic carbon increases from -1 in benzyl alcohol to +1 in benzaldehyde. (ii) In experiment 1 where R = H in Fig. 2.1, one mole of benzylic alcohol is oxidised to give 96 % yield of aldehyde. Using your answer in (c)(i), determine the charge, in Faraday, required to yield the aldehyde. [1] Since oxidation state of benzylic carbon increases from -1 to +1, 2 moles of electrons are lost per mole of benzylic alcohol. Charge required per mole of benzylic alcohol oxidised = 2F Charge required for yield of benzaldehyde = 2 ´ 0.96 = 1.92 F (iii) Current efficiency describes the efficiency with which charge is transferred in an electrolysis system and is given by the equation below. current efficiency = charge requiredcharge passed ´ 100% In experiment 1, 5.5 F of charge is passed per mole of benzylic alcohol. Using your answer in (c)(ii), calculate the current efficiency of experiment 1. [1] Current efficiency = 1.92 / 5.5 ´ 100% = 34.9 % (iv) Some data obtained from experiments 2 and 3 are shown in Table 2.1. Table 2.1 Experiment No. R group in Fig. 2.1 Current efficiency / % 2 -CH3 42.5 3 -C(CH3)3 64.0

6 © DHS 2021 9729/02 Using the data provided in Table 2.1, ded