ACJC H2 Chem 2021 Paper 3 (suggested solutions for exchange)

Index No. Name Form Class Tutorial Class 2CH________ Subject Tutor ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY H2 9729/03 Higher 2 Paper 3 Free Response 30 August 2021 Candidates answer on the Question Paper. 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your index number and name, form class and tutorial class 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. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all the questions. Section B Answer one question. The use of an approved 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. Please fill in the question number for the question attempted. Section Question No. For Examiner’s Use Marks A 1 2 3 4 B Presentation of answers TOTAL: 80 m This document consists of 31 printed pages and 1 blank page. 9729/03/Prelim/21 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2021 Department of Chemistry [Turn over

2 © ACJC2021 9729/03/Prelim/2021 [Turn over Section A Answer all the questions in this section. 1 (a) Describe the reactions, if any, when separate samples of the oxides Na2O, Al2O3 and P4O10 are added to water. Write equations where appropriate and suggest the pH of any aqueous solution formed. [4] Na2O hydrolyses/dissolves / is soluble in water to give a colourless solution of NaOH with pH 13. Na2O(s) + H2O(l)  2NaOH(aq) Al2O3 is insoluble in water and hence pH remains at 7. P4O10 hydrolyses / dissolves / is soluble in water / reacts with water to give a colourless solution of H3PO4 with pH 2 (accept 1) P4O10(s) + 6H2O(l)  4H3PO4(aq) (b) Phosphorus pentachloride, PCl5, is a white, moisture-sensitive solid. It is a dangerous substance as it reacts violently with water. (i) PCl5 reacts completely with a large excess of water to form phosphoric acid, H3PO4. Write a balanced equati on for this reaction. State the approximate pH value of the resulting solution. [2] The structural formula of phosphoric acid is shown below. Two molecules of phosphoric acid can undergo a condensation reaction producing diphosphoric acid, H4P2O7, and water. The reaction involves an –OH group from each H3PO4 molecule forming an oxygen bridge between the two phosphoric acid units. (ii) Draw the structure of diphosphoric acid. [1] (iii) This condensation reaction may continue to give triphosphoric acid, H5P3O10, and tetraphosphoric acid. Give the molecular formula of tetraphosphoric acid. [1] (i) PCl5 + 4H2O  5HCl + H3PO4 pH = 2 (ii) (iii) H6P4O13

3 © ACJC2021 9729/03/Prelim/2021 [Turn over (c) Phosphorus pentachloride can be used to convert alcohols into chloroalkanes. For example, Phosphoryl chloride, POCl3, is formed as a side-product. (i) The structure of POCl3 is shown below. As there are four bond pairs and zero lone pairs on the phosphorus atom, VSEPR theory predicts that C l-P-Cl bond angle is 109.5 o. However based on experimental evidence, the actual bond angle is only 103o. Suggest a reason why the actual bond angle is smaller than the predicted one. [1] (ii) Phosphoryl chloride can be synthesised by reacting PCl5 with ethanedioic acid in equimolar amounts. Two moles of HCl is formed per mole of PCl5, together with two other gases, one polar, the other non-polar. Identify the polar and the non - polar gas and write a balanced equation of this reaction. [3] (i) The higher electron density of the double bond repels the bond pairs of the single bonds more and hence the bond angle is smaller than expected. or The P=O double bond occupies more space than single bonds, hence forcing the single bonds to be closer to one another, causing the bond angle to be smaller than the ideal angle. (ii) Non-polar: CO2 or O2 polar: CO PCl5 + H2C2O4  2HCl + POCl3 + CO2 + CO Or PCl5 + H2C2O4  2HCl + POCl3 + 2CO + ½ O2

4 © ACJC2021 9729/03/Prelim/2021 [Turn over (iii) The Vilsmeier–Haack reaction involves the use of phosphoryl chloride and a substituted amide to produce an aryl aldehyde or aryl ketone. An example of the Vilsmeier–Haack reaction is shown below. The mechanism of stage 2 consists of two steps. Using curly arrows and showing all relevant charges, propose the mechanism for stage 2. [3] [Total: 15]

5 © ACJC2021 9729/03/Prelim/2021 [Turn over 2 Magnesium reacts with pure nitrogen to form magnesium nitride, Mg3N2, which can be considered as a possible intermediate in the ‘fixing’ of nitrogen to make ammonia -based fertilisers. 3Mg + N2 Mg3N2 Nitrogen is an essential macronutrient needed by all plants to thrive. It is an important component of many structural, genetic and metabolic compounds in plant cells. (a) When water is added to Mg3N2, a colourless gas which turns moist red litmus paper blue is produced. This gas reacts with chlorate(I) ion, ClO– in a 2:1 mole ratio to form a colourless liquid A with empirical formula NH 2. The reaction of A with sulfuric acid in a 1:1 mole ratio produces a salt B, O4N2SH6, which contains one cation and one anion per formula unit. Explain the role of the nitride ion when water was added to Mg3N2. Deduce the structures of compounds A and B. [3] The nitride ion is functioning as a (Bronsted) base by accepting protons from water. A is hydrazine, H2NNH2 B is (H2NNH3)HSO4 or N2H62+ SO42-

6 © ACJC2021 9729/03/Prelim/2021 [Turn over (b) Another type of ammonia-based fertiliser contains urea, NH 2CONH2, and has a large percentage by mass of nitrogen. The nitrogen content of a urea-containing fertiliser can be determined by boiling a sample of known mass of the fertiliser with an excess of NaOH(aq), absorbing the gas evolved in water, and titrating the resulting aqueous solution with hydrochloric acid of a known concentration. NH2CONH2 + 2OH  2NH3 + CO32– NH3 + HCl  NH4Cl When 0.100 g of this fertiliser was subjected to this procedure, the resulting solution of ammonia required 15.0 cm3 of 0.200 mol dm3 HCl for complete neutr