2024 Prelims TJC H2 Chem P3 (Ans)

9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over TEMASEK JUNIOR COLLEGE 2024 JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CENTRE NUMBER S INDEX NUMBER Chemistry 9729/03 Paper 3 Free Response Questions 11 September 2024 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name 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 questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. 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 This document consists of 23 printed pages and 1 blank page. For Examiner’s Use Paper 3 Q1 /18 Q2 /20 Q3 /22 Q4 /20 Q5 /20 Total /80

2 DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN Section A Answer all the questions in this section. 1 (a) In Periodic Table, Group 14 consists of 5 elements. The elements at the top of the group, carbon to germanium, have very different properties from those at the bottom, tin and lead. For instance , carbon to germanium tend to adopt oxidation states of +4, whereas the heavier elements, such as tin and lead, exhibit the +2 oxidation state due to the inert pair effect. (i) Table 1.1 shows the melting points of the elements of Group 14. Table 1.1 element C Si Ge Sn Pb melting point / C >3550 1410 937 232 327 Carbon, silicon and germanium each forms a solid with the same type of structure. Using structure and bonding and the information from the Data Booklet, suggest why the melting points of these elements decrease from carbon to germanium. Carbon, silicon and germanium have √ giant molecular structure. Melting involves breaking √ numerous strong covalent bonds between atoms. Atomic radius nm C 0.077 Si 0.117 Ge 0.122 √ The atomic radius increases from C to Ge, leading to less effective overlap of the orbitals. √Bond strength decreases and hence less energy is required for melting. 2 √ : 1 mark [2] Carbon forms many allotropes such as graphite and diamond . Recent scientific research has found that replacing the graphite electrodes with graphene in lithium-ion batteries can extend battery life. Graphene is a single, one atom thick layer of graphite. (ii) Explain why graphene can conduct electricity along the plane. The remaining 2p electron on each carbon atom can delocalise along the plane when a potential difference is applied. [1] [1] (iii) State and explain how you would expect the bond strength of carbon-carbon bonds in graphene to differ when compared with that in diamond.

3 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over The carbon-carbon bond in graphene is stronger than that in diamond. [1] • This is due to the √ presence of both  and  bonds / partial double bond character in the C–C bond in graphene, √while there are only  bonds present in diamond OR greater s character / smaller p character of the sp2−sp2 overlap in graphene as compared to the sp3−sp3 overlap in diamond. [2] (b) Carbon forms carbide anion, C 2– 2 in calcium carbide while silicon mostly forms Si 4+ ions. Calcium carbide is a crystalline solid with a melting point of approximately 2160 °C. (i) Draw the dot-and-cross diagram of C 2– 2 . Anion: Accept additional electrons of opposite symbol/ third symbol do not accept dative bonds [1] (ii) When some calcium carbide is added to excess water with a drop of Universal Indicator, ethyne gas, C2H2, is produced. The solution changes from green to blue after the reaction is complete. Write an equation, with state symbols, for this reaction. CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g) [1] [1] state symbols [2] (iii) In a particular experimental set-up, a beam of 12C 2– 2 ions was deflected by an angle of -2.45°. Assuming an identical set of conditions, calculate the angle of deflection for 28Si4+ ions. 12C 2– 2 whose −= =+ 21 gives an angle of deflection, , of -2.4 5º24 12 q m    = − = +  =       12.45º 29.4 12 qk k km For 28Si4+ whose += =+41 ,28 7 q m =  + =+ 1angle of deflection, 29.4 .2 7 4º [2]

4 DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN (c) Using the data from Data Booklet , describe and explain the reducing strength of Group 2 elements from magnesium to strontium. Group 2 element Standard electrode potential / V Magnesium -2.38 Calcium -2.87 [1] √ The standard electrode potential values become more negative down the group. This is due to√ increasing atomic radius down the group, √the attraction of nucleus for the valence electrons decreases. Hence the electrons can be removed more easily through ionisation. √Reducing strength increases down the group. 2√, 1 mark [3] (d) Aluminium chloride is used as a catalyst for the reaction between methylbenzene and 2-chloropropane to form two constitutional isomers L and M with the formula C10H14. (i) Suggest the mechanism for this reaction to form the major product, L. Show all charges and relevant lone pairs and show the movement of electron pairs using curly arrows. Type of reaction: Electrophilic Substitution [1] Step 1: Generating the electrophile, + Step 2: Electron-rich methylbenzene attacks the electrophile [1] Step 3: Abstraction of proton followed by regeneration of the catalyst [1] [3]

5 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelims / 2024 DO NOT WRITE IN THIS MARGIN [Turn over 2 Many chemical processes release waste products into the atmosphere. Scientists are developing new catalysts to convert these emissions into useful products, such as methanol which can be used as a fuel. An equation for the synthesis of methanol is given below. CO2(g) + 3H2(g) ⇌ CH3OH(g) + H2O(g) ΔH = – 49 kJ mol-1 (a) (i) State and explain the effect on the position of equilibrium when the temperature is increased. [2] By LCP, the position of equilibrium shifts left [✓] to favour the endothermic [✓] reaction to absorb heat [✓] and decrease temperature of the system. [✓] (ii) State and explain the effect on the position of equilibrium when the pressure is increased. [1] By Le Chatelier’s Principle, the position of equilibrium shifts to the right to decrease the number of moles of gases to decrease the pressure. [1] (iii) Write an expression for Kp of the reaction. [1] Kp = (PCH3OH)(PH2O) (PCO2)(PH2)3 [1] (ii) L and M are formed in a 60:40 ratio. Suggest why L is the major product. The ortho-positions are adjacent to the substituent which cause steric effects and block the path of incoming electrophile . The para-position is more accessible for the incoming electrophile to attack. [1] (e) (i) Write an equation for the reaction of with acidified KMnO4. [1] [1] (ii) Name the oxidised product formed in (e)(i).