DHS Prelim P2 Suggested Solution updated

This question paper consists of 18 printed pages and 0 blank page. © DHS 2018 [Turn over Suggested solutions DUNMAN HIGH SCHOOL Preliminary Examination 2018 Year 6 H2 CHEMISTRY 9729/02 Paper 2 Structured Questions 13 September 2018 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet INSTRUCTIONS TO CANDIDATES 1 Write your name, index number and class on this cover page. 2 Write in dark blue or black pen. 3 You may use an HB pencil for any diagrams or graphs. 4 Do not use staples, paper clips, glue or correction fluid. 5 Answer all questions in the spaces provided on the Question Paper. The number of marks is given in brackets [ ] at the end of each question or part question. You are advised to show all workings in calculations. You are reminded of the need for good English and clear presentation in your answers. For Examiner’s Use Question No. Marks 1 13 2 12 3 25 4 10 5 15 Total 75

2 © DHS 2018 9729/02 Answer all questions in the spaces provided. 1 Isoprene, E, is an organic compound that could be used to synthesise limonene, which is commonly used in fragrances. (a) E can be synthesised from 3–methylbut–1–ene, A, in a 4–step process as follows. C A E I II III IV HCl (g) B D excess concentrated H2SO4, heat (i) B is a major product of step I. Draw the structures of compounds B, C and D. [3] B : Cl C : D : OH OH (ii) Suggest the reagents and conditions for steps II and III. [2] step II : alcoholic NaOH heat under reflux step III : cold alkaline KMnO4 (iii) Predict, with reasoning, whether the mixture of products formed in step I is optically active. [2] The reaction mixture in B is not optically active. During electrophilic addition, a trigonal planar carbocation is formed. C l– has a 50% chance each of attacking the carbocation from the top or bottom of the plane . Hence, resulting in a racemic mixture formed. CH3 H + :Cl- CH3 H Cl H CH3 H ClH 50% 50% Trigonal Planar carbocation

3 © DHS 2018 9729/02 [Turn Over (b) The following reaction shows an alternative route to form intermediates for the synthesis of isoprene, E. Cl Cl F GA + + other products (i) State the reagents and conditions used to form F and G from A. [1] Limited Cl2, uv (ii) Predict the ratio in which F and G will be formed. [1] F : G 6 : 1 (iii) Z was one of the other products formed in the reaction in (b). It was suggested that the radical, •G, is involved in the formation of Z. The mechanism of •G converting to Z is thought to involve 2 steps. 1. Delocalisation of unpaired electron forming • Z. 2. • Z reacts with Cl2 to form Z and a radical. The structures of Z, •G and •Z are as follows. G Z Z Cl Use the information above to draw out the mechanism for the conversion of •G to Z. You are advised to use skeletal or structural formula for all species, so that it is clear which bonds are broken and which are formed. Indicate any unpaired electrons by a dot (•). Use curly half-arrows to indicate the movement of unpaired electrons. [2] Cl + Cl 1. 2. Cl Cl

4 © DHS 2018 9729/02 (iv) Describe a chemical test that could distinguish between A and F. State reagents, conditions and observations clearly in your test. Add 1 cm3 of A and F each into separate test tubes. Add NaOH (aq) into both test tubes and heat, followed by adding excess HNO3 (aq) then add AgNO3 (aq). F : White ppt of AgCl seen A : no white ppt seen [Total: 13] 2 (a) Fig 2.1 shows a bar chart of the third ionisation energy (3 rd IE) of nine consecutive elements (J to R) in Periods 2 and 3 of the Periodic Table. 3rd IE elements Fig 2.1 J K L M N O P Q R (i) Write an equation for the third ionisation energy of element J. [1] J2+(g) → J3+(g) + e– (ii) Identify element N. [1] Mg (iii) Using your answers in (a)(i), (a)(ii) and the electronic configurations of the species involved, explain the following features of Fig 2.1. 1. The significantly higher 3 rd IE of N compared to O. Electronic configuration: N2+ (1s22s22p6); O2+ ([Ne]3s1) The 3rd ionisation energy of N involves the removal of a 2p electron The 3rd ionisation energy of O involves the removal of a 3s electron A significantly larger amount of energy is required to remove the 2p electron in N2+ which is much more strongly held by the nucleus as it is found in an inner quantum shell compared to 3s electron in O2+. Hence 3rd ionisation energy of N is significantly higher than that of O.

5 © DHS 2018 9729/02 [Turn Over 2. The lower 3 rd IE of Q than P. [4] Electronic configuration: P2+ ([Ne]3s2); Q2+ ([Ne]3s23p1) The 3rd ionisation energy of P involves the removal of a 3s electron The 3rd ionisation energy of Q involves the removal of a 3p electron Smaller amount of energy is required to remove the 3p electron in Q2+ which is of higher energy than 3s electron in P2+. (The 3p electron also experiences increased screening effect provided by the filled 3s subshell.) Hence 3rd ionisation energy of Q is lower than that of P (b) Fig 2.2 shows another bar chart of the logarithm of all the ionisation energies, log (IE), of an element S against the number of electrons removed. log(IE) number of electrons removed Fig 2.2 (i) Explain the general trend shown in Fig 2.2. [2] In an atom of S, Number of protons unchanged  nuclear charge unchanged As electrons are removed from the atom, the increasingly positively charged ion holds the remaining electrons more strongly hence more energy is required to remove the remaining electrons resulting in higher I.E Thus, successive I.E shows an increasing trend. (ii) On the axes provided, draw and label the orbital which the fifth electron is removed from. [1] x y z

6 © DHS 2018 9729/02 2s x y z (c) The Periodic Table shows helium placed at the top of Group 18. (i) Suggest why the element helium could be placed at the top of Group 2. [1] Helium has 2 valence electrons like all other Group 2 elements. (ii) Suggest why the element helium is not placed at the top of Group 2, by comparing one physical property. Explain your answer. [2] Helium atoms are held by weak instantaneous dipole induced dipole interactions while within group 2 elements exist strong electrostatic forces of attraction between cations and a sea of delocalised electrons. Hence, helium has low melting/boiling point and it exist as a gas while group 2 elements has high melting/boiling point and exist as a solid at room temperature. Include energy OR Hence, helium is a non-conductor of electricity due to ( absence of mobile/delocalised electrons) while group 2 elements are good conductor of electricity due to (presence of delocalised electrons). [Total: 12] 3 The ions of transition elements form complexes by reacting with ligands. (a) (i) State what is meant by the terms: Complex Ligand [2] A complex is formed when a metal ion or atom forms dative covalent or coordinate bonds with surrounding ion or molecules. A ligand is a neutral molecule or an anion containing at least one atom with a lone pair of electrons that can be donated into low lying vacant orbital of metal atom/ion to form a coordinate bond.

7 © DHS 2018 9729/02 [Turn Over (ii) Two of the complexes formed by copper are Cu(H 2NCH2CH2NH2)2(OH)2 and CuCl42–. Draw three–dimensional diagrams of their structures in the boxes below and name their shapes. Cu(H 2NCH2CH2NH2)2(OH)2 CuCl42– Shape: Shape: [3] Cu(H 2NCH2CH2NH2)2(OH)2 Shape: octahed