2021 MI Prelim H2 Chemistry P3 (9729 03) Answers

Class Adm No Candidate Name: This question paper consists of 28 printed pages and 2 blank page. 2021 Preliminary Examinations Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response 17 Sep 2021 2 hours Additional materials: Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. 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. Section A: Answer all questions Section B: Answer any 1 question A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for good English and clear presentation in your answers. 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. Section A Section B Total Question 1 2 3 4 5 Marks 24 20 16 20 20 80

2 Section A Answer all questions from this section. 1 (a) A metallic element, M, exists as three isotopes. The relative isotopic masses and their relative abundances are shown in Table 1.1. Relative Isotopic Mass Relative Abundance 53.94 3 55.93 46 56.94 1 Table 1.1 (i) Define the term relative isotopic mass. [1] Relative isotopic mass is defined as the mass of one atom of an isotope compared to 112 of the mass of one atom of 12C. (ii) Using the information in Table 1.1, calculate the relative atomic mass of M, leaving your answer to 2 decimal places. [1] Relative atomic mass = ("#.%&×#))("".%#×&*))("*.%&×+)(#)&*)+)=55.83 A dilute aqueous solution of the chloride salt of M, MCln, was electrolysed using a current of 0.5 A for 1 hour as shown in the setup in Fig. 1.1. The experiment was carried out at 32 °C and 1 atm. Fig. 1.1 0.3471 g of metal M was deposited at electrode B.

3 [Turn over (iii) Using your answer in (a)(ii), determine n. [3] Amount of M deposited = ,.#&-+""..#=0.006217 𝑚𝑜𝑙; Q = 0.5 × 1 × 60 × 60 = 1800 C Amount of electrons = +.,,%*",,=0.01865 𝑚𝑜𝑙 ; Mn+ + ne- → M n = ,.,+.*",.,,*/+-=3 ; (iv) A gas is produced at electrode A. Write the ion-electron equation for the reaction taking place at electrode A. [1] 2H2O → O2 + 4H+ + 4e– (v) Calculate the volume of gas produced at electrode A. [2] Amount of O2 = 0.01865 ÷ 4 = 0.004663 mol ; V = 0123= ,.,,&**#×..#+×(#/)/-#)+,+#/"=1.166×104& 𝑚#=117 𝑐𝑚# ; (vi) Suggest why electrode A will decrease in mass over time. [1] The oxygen gas produced from the electrolysis reacts with the carbon electrode to form carbon dioxide. (b) Chromium(III) chloride, CrCl3, reacts with water to form a violet complex, [Cr(H2O)6]Cl3. This violet complex is isomeric to two other complexes as shown in Table 1.2. Structural Formula of Complex Colour of Complex [Cr(H2O)6]Cl3 violet [CrCl(H2O)5]Cl2 • H2O pale green [CrCl2(H2O)4]Cl • 2H2O dark green Table 1.2 (i) Give one characteristic properties of transition elements as shown by chromium in the isomers in Table 1.2. [1] Can form coloured compounds

4 (ii) Using [Cr(H2O)6]Cl3, explain what is meant by the term ligand. [1] H2O is a molecule that has lone pairs of electrons on O atom, which can be used to form a dative bond with the central metal ion Cr3+. (iii) When 2.665 g of one of the complexes in Table 2 was reacted with an excess of aqueous AgNO3, 2.868 g of AgCl was obtained. Deduce the structural formula of the complex. [2] Amount of complex = /.**""/.,)+..,×*)#"."×#=0.0100 𝑚𝑜𝑙 Amount of AgCl = /..*.(+,-.%)#".")=0.0200 𝑚𝑜𝑙 both amounts ; 1 mole of complex contains 2 moles of free chloride ions Structural formula = [CrCl(H2O)5]Cl2 • H2O ; (iv) Draw the full structure of the cation in the pale green complex, [CrCl(H2O)5]Cl2 • H2O, showing the shape of the cation. [1] (c) Potassium dichromate(VI) can be produced from chromium(III) chloride via a 3-step reaction. Step 1: Chromium(III) chloride is boiled with hydrogen peroxide in an alkaline medium, forming a bright yellow solution, CrO42-. Step 2: Boiling is continued until excess hydrogen peroxide is destroyed. Step 3: The yellow solution is then cooled and acidified with ethanoic acid, forming the orange dichromate(VI) solution. (i) Write the ion-electron equation for the oxidation of Cr3+ in step 1 and hence the overall balanced equation for the reaction between Cr3+ and H2O2 under alkaline conditions. [2] Cr3+ + 8OH– → CrO42- + 4H2O + 3e– ; 2Cr3+ + 10OH– + 3H2O2 → 2CrO42– + 8H2O ;

5 [Turn over (ii) Suggest how would you know when all the excess hydrogen peroxide has been destroyed in step 2. [1] No more effervescence (iii) The reaction in step 3 has the following equation: 2CrO42-(aq) + 2H+(aq) ⇌ Cr2O72-(aq) + H2O(l) State and explain if the reaction in step 3 is a redox reaction. [1] No. The oxidation state of chromium remains the same at +6. (iv) The reaction in step 3 is an exothermic reaction. State and explain the effect of increasing the temperature of the system on the yield of Cr2O72-(aq). [2] When temperature is increased, position of equilibrium will shift left to remove the excess heat by favouring the endothermic reaction. ; Yield of Cr2O72– will decrease. ; (d) A solution of sodium dichromate(VI) is acidified with dilute sulfuric acid before adding to an organic compound. The solution slowly turns green. Boiling the green solution with more of the organic compound produces a pale blue solution. 0.10 mol of the pale blue ions was found to require 0.020 mol of acidified potassium manganate(VII) solution to oxidise it back to the green ions. (i) State the identity of the green ion. [1] Cr3+ or [Cr(H2O)6]3+ (ii) Deduce the oxidation number of the blue ion. [2] MnO4– + 8H+ + 5e → Mn2+ + 4H2O Crn+ → Cr3+ + (3-n)e Amount of electrons = 5 x 0.020 = 0.100 mol ; Ratio of blue ion : e = 0.10 : 0.100 = 1:1 n = 3 – 1 = +2 ; (iii) State, with a reason, which reagent, acidified H2O2 or aqueous Fe3+ could be used to convert the green ions back to dichromate(VI) ions. [1] Acidified H2O2 Only E°(H2O2/H2O) is more positive than E°(Cr2O72-/Cr3+) OR calculate E°cell [Total: 24]

6 2 (a) Compound P, C8H9NO, can react with Tollen’s reagents but not with Fehling’s solution. P is also soluble in dilute hydrochloric acid. On reacting P with lithium aluminium hydride, Q, C8H11NO, is formed. Q reacts readily with sodium metal. 1 mole of Q reacts with 2 moles of propanoyl chloride to give the following compound. Q also reacts with hot acidified KMnO4 to g