2024 Prelims CJC H2 Chem P2 (Ans)

[Turn over 9729/02 CJC JC2 Preliminary Examination 2024 CANDIDATE NAME CLASS 2T INDEX CHEMISTRY 9729/02 Paper 2 Structured Questions 26 August 2024 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class 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. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 16 printed pages. For Examiner’s Use Paper 1 30 Paper 2 Q1 /7 Q2 /11 Q3 /13 Q4 /15 Q5 /15 Q6 /14 75 Paper 3 80 Paper 4 55 OVERALL (100%) GRADE Catholic Junior College JC2 Preliminary Examination Higher 2 WORKED SOLUTIONS WORKED SOLUTIONS

2 9729/02 CJC JC2 Preliminary Examination 2024 Answer all the questions in the space provided. 1 Tetrahalosilanes have the general formula SiX4, where X represents one of the halogens. A sample of SiX4 is atomised and ionised. The ions produced are then analysed. (a) Complete the following energy level diagram to show the arrangement of electrons in the orbitals of Si+ ion. [2] (b) In the first analysis, the second ionisation energy of silicon is recorded. Write an equation for the second ionisation energy of silicon. ……..……………………………………………………………………………………….. [1] (c) Explain why the second ionisation energy of silicon is higher than that of the first. …………………..…………………………………………………………………………...… ……….…….………………………………………………………………………….......…… .…..…………………………………………………………………………………………….. .................................................................................................................................. [1] Energy 1s 2s 3p 3s 2p Si+(g) → Si2+(g) + e– The increase in the second ionisation energy is due to more energy required to remove the second electron from an ion with the same nuclear charge as the atom but attracting fewer electrons due to stronger electrostatic force of attraction between the nucleus and valence electrons.

3 9729/02 CJC JC2 Preliminary Examination 2024 [Turn over (d) In the second analysis, ions of X+ are analysed. A sample each of 28 + 14Si and X+ is passed through an electric field. The angles of deflection of 28 + 14Si and X+ are 5.6o and 2.0o respectively. (i) Deduce, by calculation, the identity of X. [2] (ii) Suggest why there is another beam detected with an angle of deflection of 1.9o. ……..…………………………………………………………………………………. [1] [Total: 7] Let the mass number or nucleon number of X be m. 128 1m = 5.6 2.0 m = 78.4 X is Br. Isotopes of bromine with lower charge/mass ratio

4 9729/02 CJC JC2 Preliminary Examination 2024 2 This question is about phosphorus and its compounds. (a) With reference to relevant electronic configurations where necessary, explain why the first ioni sation energy of phosphorus is higher than the elements that come immediately before and after it in Period 3. …………………..………………………..………………………………………………….... …………………..………………………………………………………………………….... …………………..…………………………………………………………………………..... …………………..………………………………………………………………………….... …………………..…………………………………………………………………………..... …………………..…………………………………………………………………………..... …………….…….…………………………………………………………………………..... …………………..…………………………………………………………………………..... …………………..…………………………………………………………………………..... …………….…….………………………………………………………………………….... ……..……………………………………………………………………………………… [3] (b) With reference to structure and bonding, explain why the melting point of phosphorus is lower than the elements that come immediately before and after it in Period 3. …………………..…………………………………………………………………………..... …………………..…………………………………………………………………………..... …………………..………………………………………………………………………….... . …………………..…………………………………………………………………………..... …………………..…………………………………………………………………………..... …………………..…………………………………………………………………………..... …………….…….…………………………………………………………………………..... …………………..…………………………………………………………………………..... …………….…….…………………………………………………………………………..... ……..……………………………………………………………………………………… [2] As compared to Si, P has a smaller (atomic) radius and greater nuclear charge while shielding effect by same number of inner electrons is similar. Therefore, nuclear attraction in P is larger and hence, 1st ionisation energy of P is higher than that of Si. P 1s2 2s2 2p6 3s2 3px1 3py1 3pz1 S 1s2 2s2 2p6 3s2 3px2 3py1 3pz1 In S, the two electrons occupying the same 3p orbital (i.e. 3p x) give rise to inter- electronic repulsion. Thus, less energy is required to remove a paired 3p electron from S, as compared to the energy required to remove an unpaired 3p electron from P. Si has a giant molecular structure with the Si atoms held together by an extensive network of strong covalent bonds. However, P4 and S8 have simple covalent structures with weak intermolecular instantaneous dipole-induced dipole forces of attraction that require less energy to overcome, thus P4 (and S8) have lower melting points than Si. P4 has fewer electrons than S8, hence its intermolecular instantaneous dipole- induced dipole forces of attraction are weaker and require less energy to overcome, thus P4 has a lower melting point than S8.

5 9729/02 CJC JC2 Preliminary Examination 2024 [Turn over The most important oxide of phosphorus is phosphorus(V) oxide, P 4O10. It is a powerful dessicant and dehydrating agent. (c) Write a balanced equation for the reaction of P4O10 with water and state the pH of the resulting solution. …………….…….…………………………………………………………………………........ ……..………………………………………………………………………………………… [1] The structure of phosphorus(V) sulfide, P4S10, is closely related to that of P4O10. (d) Reaction of P4S10 with water gives two products. One of the products is the same as the product of the reaction in (c), the other product is a gaseous compound. Suggest a balanced equation for this reaction. ……..………………………………………………………………………………………… [1] (e) In vapour form, phosphorus(V) sulfide exists as P2S5 molecules. When P2S5 is heated under a vacuum together with caesium sulfide, Cs 2S, and sulfur, it produces an ionic compound R which has the following composition by mass: Cs, 58.1%; P, 6.78%; S, 35.1%. (i) Calculate the empirical formula of compound R and hence deduce its chemical formula, given that the relative formula mass, Mr, is 914.6. [2] (ii) Compound R contains Cs+ cation and an anion. Given that the cation and anion of compound R are present in a 4:1 ratio, write the formula of the anion. Anion: ………………………………………………………………………………. [1] P4O10 + 6H2O → 4H3PO4 pH = 2 P4S10 + 16H2O → 4H3PO4 + 10H2S Let mass of a sample of the salt be 100g. Cs P S Mass/g 58.1 6.78 35.1 Ar 132.9 31.0 32.1 Moles (% mass/Ar) 58.1/132.9 = 0.437 6.78/31.0 = 0.219 35.1/32.1 = 1.09 Simplest ratio 2 1 5 Empirical formula is Cs2PS5 Let the chemical formula be (Cs2PS5)n. Mr of (Cs2PS5)n = 914.6 (2(132.9) + 31.0 + 5(32.1)) n = 914.6