SRJC H1 CHEM P2 ANS

1 SRJC 8872 / 02 / JC2 Prelims/ 2015 SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1 CANDIDATE NAME CLASS CHEMISTRY 8872/02 JC 2 Preliminary examination 20 August 2015 Paper 2 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Candidates answer on the question paper. Write your name and class on all the work you hand in. Write in dark or blue pen. Do not use paper clips, glue or correction fluid. The number of marks is given in bracket [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. This document consists of 13 printed pages and 3 blank page. FOR EXAMINER’S USE P1 (MCQ) 30 P2 A1 8 A2 9 A3 16 A4 7 B5 20 B6 20 B7 20 Total 110

2 SRJC 8872 / 02 / JC2 Prelims/ 2015 Section A Answer all the questions in the spaces provided. 1 Elements A, B, C, D and E are five consecutive elements from Period 3 and 4 of the Periodic Table. The following shows the successive ionisation energies of element C. No. of electrons removed 1 2 3 4 5 6 7 8 Ionisation energy / kJ mol–1 1260 2300 3850 5150 6542 9362 11018 33604 (a) (i) Deduce and explain which group element C belongs to. [2] Group VII. Largest jump in ionisation energy from 7 th to 8 th electron. Thus, there are 7 valence electrons. The 8th electron is located in an inner principal quantum shell nearer the nucleus, thus experience stronger electrostatic forces of attraction, requiring much more energy to remove. (ii) Hence, state the identity of element C and write down its electronic configuration. [2] Chlorine . Electronic configuration: 1s22s22p63s23p5. (b) Explain the following observations, giving equations where appropriate. (i) The first ionisation energy of element B is lower than the first ionisation energy of element A. [2] B is sulfur: [Ne]3s23p4, A is phosphorus: [Ne]3s23p3 In B, there is interelectronic repulsion between the pair of electrons in the doubly -filled 3p orbital of B . Less energy is required to overcome the weaker electrostatic forces of attraction between the nucleus and the paired valence 3 p electron in B compared to the unpaired valence 3p electron in A. (ii) A strongly acidic solution is formed when the chloride of A reacts with water. [2] Chloride of A undergoes hydrolysis to give white fumes of HCl gas PCl3 (l) + 3 H2O (l) → H3PO3 (aq) + 3 HCl (aq) Or PCl5 (l) + 4 H2O (l) → H3PO4 (aq) + 5 HCl (aq) pH of solution ≈ 2 [Total: 8] For Examiner’s use

3 SRJC 8872 / 02 / JC2 Prelims/ 2015 2 A sequence of reactions is shown below. (a) In the appropriate boxes draw the structures of compound F, G, H, J and K. (b) For the reaction in the scheme shown above state - the reagents and condition for reaction II and III, Reaction II: Cl2 (g), AlCl3 Reaction III: dilute HCl, heat - the type of reaction for reactions I and III. Reaction I: Condensation Reaction III: Acidic Hydrolysis Total: [9] F G, C6H10O2 H, C6H10O J KMnO4, H+, heat II I2 (aq) + NaOH(aq) heat I HCN + NaCN, cold K III

4 SRJC 8872 / 02 / JC2 Prelims/ 2015 3 Aldehydes are commonly used to produce resins to make plastics and adhesives. It exists in equilibrium with its isomer, enol through a process called enolization. The reaction involves the transfer of one proton and the shift of the double bond. An example of enolization of acetaldehyde is shown below. ∆H ɵ rxn > 0 acetylacetaldehyde enol (a) (i) Write an expression for the equilibrium constant, Kc, for this reaction. Kc = [enol] [acetylacetaldehyde] (ii) It was found that 76% of acetylacetaldehyde exist as an enol when dissolved in water. Calculate the equilibrium constant, Kc, for this reaction. acetylacetaldehyde enol Initial concentration / moldm-3 1 0 Change in concentration / moldm-3 -0.76 +0.76 Final concentration / moldm-3 0.24 0.76 Kc = 0.76 0.24 = 3.17 (iii) Explain the significance of this value on the equilibrium position and the relative concentrations of the two species present at equilibrium. The large equilibrium constant indicates that the equilibrium position lies to the right. Hence at equilibrium, most of the species present is the enol. (iv) Suggest what will happen to the composition of the equilibrium mixture when the system is heated. [6] By Le Chatelier’s Principle , the position of equilibrium shifts to the right to absorb the excess heat, favouring the endothermic reaction. [enol] will increase while [acetylacetaldehyde] will decrease until a new equilibrium is reached.

5 SRJC 8872 / 02 / JC2 Prelims/ 2015 (b) Acetylacetaldehyde can undergo oxidation with potassium dichromate( VI) under heat to form a carboxylic acid. (i) Write two redox half-equations to represent the reaction between acetylacetaldehyde acetaldehyde and Cr2O7 2– ions and prove that the overall equation is as follows: 3CH3COCH2CHO + Cr2O7 2– + 8H+  3CH3COCH2COOH + 4H2O + 2Cr3+ Cr2O7 2– + 14H+ + 6e  2 Cr3+ + 7H2O ___(1) CH3COCH2CHO + H2O  CH3COCH2COOH + 2H+ + 2e ___(2) To balance the no of e, 3 x (2) 3 CH3COCH2CHO + 3H2O  3 CH3COCH2COOH + 6H+ + 6e Overall equation: 3CH3COCH2CHO + Cr2O7 2– + 8 H+  3CH3COCH2COOH + 4H2O + 2Cr3+ (shown) (ii) 8 cm3 of liquid acetylacetaldehyde was dissolved in water and made up to 250 cm 3. 25.0 cm3 of this solution was titrated with 0.10 mol dm –3 acidified potassium dichromate(VI). Given that the density of acetylacetaldehyde is 0.956 g cm–3, calculate the volume of potassium dichromate(VI) required to complete the titration. 𝑛𝑎𝑐𝑒𝑡𝑦𝑙𝑐𝑒𝑡𝑎𝑙𝑑𝑦ℎ𝑦𝑑𝑒 in 250 cm3 = 8 × 0.956 86.0 = 8.893 x 10–2 mol 𝑛𝑎𝑐𝑒𝑡𝑦𝑙𝑎𝑐𝑒𝑡𝑎𝑙𝑑𝑦ℎ𝑦𝑑𝑒 in 25 cm3 = 8.893 x 10–3 mol 𝑛𝐶𝑟2𝑂72− 8.898 ×10−3 3 = = 2.964 x 10–3 mol 𝑛𝐶𝑟2𝑂72−= 2.964 ×10−3 0.1 = 2.96 x 10 -2 dm 3 = 29.6 cm 3 (iii) Propose a simple chemical test that allows you to confir m the presence of acetylacetaldehyde and state any observations clearly. [6] Test: Add a few drops of Tollen’s reagent/Fehling’s reagent and warm. Observations: A silver mirror/reddish-brown ppt will be observed. (c) (i) Deduce which of the two acids, CH3COCH2COOH or CH3COCHClCOOH, has a higher pKa value. The electron-donating R group ( -CH3 ) intensifies the negative charge on the carboxylate anion hence destabilising the carboxylate anion relative to the acid, hence, CH3COCH2COOH is the weaker acid and has a higher pKa. The electron -withdrawing chloro group disperses the negative charge on the carboxylate anion hence stabilising the carboxylate anion relative to the acid.

6 SRJC 8872 / 02 / JC2 Prelims/ 2015 (ii) Propose a simple test -tube test which can be achieved in the school laboratory to differentiate CH3COCH2COOH from CH3COCHClCOOH. [4] Test: Add NaOH (aq), heat, followed by aqueous HNO3 then aqueous AgNO3 Observation: White ppt of AgCl will be observed for CH2ClCOOH but not for CH3COOH [Total: 16] 4 Aluminium reacts with fluorine, chlorine and oxygen to for m aluminium fluoride, aluminium chloride and aluminium oxide respectively. (a) Using chemical equations only, explain the action of water on aluminium chloride and suggest a pH value of the solutio