CJC H1 CHEM P2 SOL

[Turn over CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 1 CANDIDATE NAME CLASS 2T CHEMISTRY 8872/02 Paper 2 Thursday 27 August 2015 2 hours Candidates answer Section A on the Question Paper Additional Materials: Answer Paper Data Booklet Graph Paper (2 sheets) READ THESE INSTRUCTIONS FIRST Write your name and HT group on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all the questions. Section B Answer two questions on separate answer paper. You are advised to spend not more than 1 hour for Section B. You are advised to show all working in calculations. You may use a calculator. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 15 printed pages and 1 blank page. For Examiner’s Use Section A A1 /7 A2 /14 A3 /7 A4 /3 A5 /9 Section A 40 Section B B6 /20 B7 /20 B8 /20 Section B 40 P2 Total 80 P1 Total 30 Overall % Grade ANSWERS

2 8872/02/CJC JC2 Preliminary Exam 2015 Section A Answer all questions in this section in the spaces provided. 1 (a) Potassium manganate (VII), KMnO 4, can be reduced by iron ( II) sulfate, FeSO 4. The relevant half-equations for the reaction are given below: MnO4 – + 8H+ + 5e– ⇌ Mn2+ + 4H2O Fe3+ + e– ⇌ Fe2+ Write the balanced equation for the redox reaction which can occur. ……………………………………………………………………………………..…………….... ………………………………………………………………………………………………….... (b) Ketones can be reduced by lithium aluminium hydride, LiA lH4, in dry ether to form secondary alcohols. State another set of reagents and conditions that can also be used for the reduction of ketones. ………………………………………………………………………………………………..….... (c) Ketones are not easily oxidised . However, methyl ketones are a specific class of ketones with the acetyl functional group, RCOCH3, and these can be easily oxidised in the triiodomethane test. State the conditions and reagents necessary for this test and write an equation for the reaction, using RCOCH3. [2] …………………………………………………………………………………………………….. ………………………………………………………………………………………………..….... ………………………………………………………………………………………………..….... (d) Many organic compounds undergo oxidation, albeit to varying degrees. Compare the relative ease of oxidation of benzene and methylbenzene. Write appropriate equations and state reagents and conditions to illustrate your answer. [3] ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ [Total: 7] 8H+ + MnO4 – + 5Fe2+  5Fe3+ + Mn2+ + 4H2O NaBH4 in methanol I2 with aq. NaOH; heat under reflux RCOCH3 + 3I2 + 4NaOH  RCOO-Na+ + CHI3 + 3NaI + 3H2O Benzene does not undergo oxidation due to the stable ring structure but methyl benzene undergoes side-chain oxidation ; Reagents and conditions for side-chain oxidation of methylbenzene: KMnO4(aq), dilute H2SO4 (or H2SO4(aq),) heat under reflux ; at room temperature, in the dark.

3 8872/02/CJC JC2 Preliminary Exam 2015 [Turn over 2 (a) Define the term relative atomic mass of chlorine. ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ (b) Write the spdf electronic configuration of a chloride ion, Cl–. ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ (c) (i) Draw a dot-and-cross diagram to show the bonding in calcium chloride. (ii) Explain how and why the lattice energies of calcium chloride , CaC l2, and potassium chloride, KCl, have different numerical values. [4] ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ ……………………………………………………………………………………………............ 1s2 2s2 2p6 3s2 3p6 Lattice energy,    rr qq Hlatt The charge on the ion Ca2+ in CaCl2 is higher than K+ of KCl. Also, the ionic radius of Ca2+ is smaller than the ionic radius of K+. Thus, CaCl2 has a more exothermic lattice energy than KCl. Relative atomic mass is ratio of the average mass of one atom of chlorine to 1/12 of the mass of one 12C atom / isotope.

4 8872/02/CJC JC2 Preliminary Exam 2015 The first ionisation energies of the elements lithium to fluorine is shown below. (d) (i) Describe and explain the general trend in first ionisation energies for the elements lithium to fluorine. …………………………………………………………………………………….... …………………………………………………………………………………….... …………………………………………………………………………………….... …………………………………………………………………………………….... …………………………………………………………………………………….... (ii) Describe and explain how the first ionisation energy for chlorine compares to that of fluorine. [4] …………………………………………………………………………………….... …………………………………………………………………………………….... …………………………………………………………………………………….... …………………………………………………………………………………….... The first ionisation energy for chlorine is lower than that of fluorine. The nuclear charge of chlorine is greater than that of fluorine, but the atomic radius of chlorine is larger than fluorine. Thus, the outermost electron is further from the nucleus and is better shielded by the inner shell of electrons. As a result, less energy is required to remove the outermost electron from chlorine than fluorine. Note: Cambridge insists on the mention of the distance of the valence electron from the nucleus is progressively greater. The first ionisation energy generally increases across the period. This general increase is due to the increase in the nuclear charge (or number of protons), the decrease in atomic radius across the period, while the screening effect remains almost the same. Therefore a greater attraction between nucleus and valence electron results in more energy required to remove the outermost electrons.

5 8872/02/CJC JC2 Preliminary Exam 2015 [Turn over (e) (i) Solid iodine burns in fluorine gas to produce iodine pe ntafluoride which has the chemical formula of IF5. Draw and state the shape of the IF5 molecule. Shape: ................................................. When liquid IF5 reacts with fluorine gas, a compound known as iodine ( VII) fluoride is formed. This has the formula of IF7 where all the fluorine atoms are evenly distributed around the central iodine atom as shown in the diagram below: IF7 IF5 and IF7 differ in their boiling points as shown in this table: Compound Boiling point /oC IF5 97.9 IF7 4.8 (ii) With reference to bonding, suggest an explanation for the difference in boiling points. [4] …………………………………………………………………………………………….... …………………………………………………………………………………………….... …………………………………………………………………………………………….... …………………………………………………………………………………………….... [Total: 14] Both compounds have simple molecular structures, but IF5 is polar while IF7 is non-polar (because IF7 is perfectly symmetrical). The permanent dipole – permanent dipole attractions between IF5 is stronger than the van der Waals’ forces of attraction between IF7 molecules. Hence more energy is required to overcome the stronger permanent dipole – permanent dipole attractions, leading to a higher boiling point for IF5. The first ionisation energy for sodium is lesser than that of lithium The nuclear charge of sodium is greater than that of lithium, but the atomi