2022 TJC Prelims H2 Paper 2 Structured Qns and Ans

9729 / TJC Prelims / 2022 DO NOT WRITE IN THIS MARGIN [Turn over TEMASEK JUNIOR COLLEGE 2022 JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME WORKED SOLUTIONS CENTRE NUMBER S INDEX NUMBER Chemistry 9729/02 Paper 2 Structured Questions 25 August 2022 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name in the spaces 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 the questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 23 printed pages and 1 blank page. For Examiner’s Use Paper 1 /30 Paper 2 Q1 /9 Q2 /14 Q3 /10 Q4 /8 Q5 /20 Q6 /14 Total /75 Paper 3 /80 TOTAL (%) /100

2 DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2022 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN Answer all the questions in the space provided. 1 (a) Element W is from Period 5 of the Periodic Table. The first eight ionisation energies of element W, in kJ mol-1, are 869 1800 2690 3610 5720 6670 12000 13800 (i) Identify element W and explain your answer. [2] ✓ Element W is Tellurium/ Te. Electronic configuration: [Kr]4d105s25p4. ✓ There is a large difference between the 6th and 7th ionisation energies of W. ✓ The 7th electron is from an inner electronic shell. ✓ W has 6 valence electrons, hence it is from Group 16. 2 ticks – 1 mark (ii) Explain the difference between the first ionisation energy of element W compared to the element to its left on the Periodic Table. [2] ✓ 1st ionisation energy of Te involves the removal of paired 5p electrons ✓ which experiences inter-electronic repulsion. ✓ Less energy is needed to remove the electron from W/ Te than the valence electron/ unpaired 5p electron in Sb/ element to its left. ✓ IE of element W is lower. 2 ticks – 1 mark (b) The chlorides of elements in Period 3 of the Periodic Table show different behaviours on addition to water. (i) Describe and explain the reactions of aluminium chloride, A lCl3, and phosphorus pentachloride, PCl5, with excess water. Write equations for any reactions that occur. [3] AlCl3 undergoes hydration and hydrolysis. ✓ Hydrolysis take place as Al3+ has a high charge density hence a high polarising power. ✓ Al3+ draws electrons from its surrounding water molecules and weakens the O-H bond. It is easier for a H+ ion to leave the water molecule. ✓ AlCl3 + 6H2O → [Al(H2O)6]3+ + 3Cl– ✓ [Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+ ✓ PCl5 undergoes hydrolysis in water due to the presence of energetically accessible vacant 3d orbitals in P for dative bonding with water molecules. ✓ PCl5 + 4H2O → H3PO4 + 5HCl 2 ticks – 1 mark (ii) The reaction of PC l5 with limited amount of water involves step -wise substitution of –Cl with –OH.

3 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2022 DO NOT WRITE IN THIS MARGIN [Turn over Write the overall balanced equation for the reaction of PC l5 with limited amount of water. Hence, suggest a three-step reaction sequence for this reaction. ✓ Balanced equation: PCl5 + H2O → POCl3 + 2HCl ✓ Step 1: PCl5 + H2O → PCl4OH + HCl ✓ Step 2: PCl4OH + H2O → PCl3(OH)2 + HCl ✓ Step 3: PCl3(OH)2 → POCl3 + H2O [Total: 9] 2 Boron trifluoride, BF3, and aluminium fluoride, AlF3, differ markedly in their physical properties. compound melting point / C BF3 –127 AlF3 1291 (a) (i) State the type of bonding present in each of these compounds and draw ‘dot-and-cross’ diagrams in the boxes below to illustrate this bonding. ✓ Type of bonding: ✓ covalent bonding ✓ Type of bonding: ✓ ionic bonding BF3 AlF3 4 ticks – [2] 2 to 3 ticks – [1] [2] (ii) Outline the principles of the Valence Shell Electron Pair Repulsion theory. Electron pairs in the valence shell of the central atom arrange themselves as far as possible to minimise repulsion and maximise stability. [1] Strength of repulsion between electron pairs decreases in the order: Lone pair – lone pair repulsion > lone pair – bond pair repulsion > bond pair – bond pair repulsion [1] [2] (iii) Boron trifluoride forms a compound with ammonia. Use the Valence Shell Electron Pair Repulsion theory to predict and draw the likely shape of the product formed from this reaction. F B F F xx x Al 3+ F x3

4 DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2022 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN [1] Shape of product: tetrahedral around B OR N atom [1] (Dative covalent bond must be clearly shown with an arrow from N to B atom.) [2] Cyclobutane is an example of a cyclic alkane. While itself has no commercial or biological significance, more complex derivatives of cyclobutane are important in biology and biotechnology. Cyclobutane decomposes to ethene as shown by the following equation. (g) → 2CH2=CH2(g) Hdecomp  A student planned to determine Hdecomp  by using the standard enthalpy changes of combustion of cyclobutane and ethene, represented by Hc (cyclobutane) and Hc (ethene) respectively. Both substances are gases at 298 K. Hc (ethene) was previously determined to be –1411 kJ mol–1 in a flame calorimetric experiment. To determine Hc  (cyclobutane), the student decided to conduct another flame calorimetric experiment under the same conditions. There are two stages to this experiment. Stage I : Calibration of calorimet er (i.e. copper can, water and other components in the calorimeter) In the experiment, the calorimet er must first be calibrated by determining its heat capacity, C, which is the amount of heat required to raise the temperature of the calorimeter by 1 K. Stage II : Determination of Hc (cyclobutane) using the calibrated calorimeter. The student carried out the flame calorimetric experiment using the calorimeter. Part of his results is shown below. Stage I, using ethene as the fuel, • change in temperature of water = 33.8 C • mass of gaseous ethene burnt = 1.00 g Stage II, using cyclobutane as the fuel, N B H HH F F F

5 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2022 DO NOT WRITE IN THIS MARGIN [Turn over • change in temperature of water = 32.9 C • mass of gaseous cyclobutane burnt = 1.00 g (b) (i) Use the above information and experimental results to calculate the heat capacity, C, of the calorimeter. heat produced from combustion of ethene = heat gained by calorimeter amount of ethene burnt  |Hc (ethene)| = C  |T | 1.00 28.0  1411 = C  33.8 C = 1.4909  1.49 kJ K–1 Working – [1] Answer – [1] [2] (ii) Use your answer to (b)(i) to determine Hc (cyclobutane). heat produced from combustion of cyclobutane = heat gained by calorimeter amount of cyclobutane burnt  |Hc (cyclobutane)| = C  |T | 1.00 56.0  |Hc (cyclobutane)| = 1.4909  32.9  Hc (cyclobutane) = –2746.9  –2750 kJ mol–1 Working – [1] Answer (with correct sign) – [1] [2] (iii) Hence, calculate the standard enthalpy change of decomposition of cyclobutane to ethene, Hdecomp  . Hdecomp  = Hc (cyclobutane) – 2 