YIJC 2023 H2 Chem Prelim P2 Worked Solutions (1)

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CG INDEX NO CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 28 August 2023 2 hours READ THESE INSTRUCTIONS FIRST This document consists of xx printed pages and x blank pages. For Examiner’s Use 1 / 9 2 / 8 3 / 8 4 / 8 5 / 12 6 / 20 7 / 10 Penalty units significant figures Overall / 75 Write your name and class in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid/tape. 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.

©YIJC [Turn over 2 Answer all the questions in the spaces provided. 1 Semiconductors are made from a variety of raw materials. Through a complex process, semiconductor grade silicon of high purity can be produced by refining the raw materials. (a) An unknown element R was studied for its semiconductor properties. The table below shows the ionization energies of element R. Table 1 1st 2nd 3rd 4th 5th 6th 7th Ionisation Energy/kJ mol–1 945 2026 2987 4144 6593 7880 14 990 The principal quantum number of Element R is n. Predict, with explanation, the Group where element R will be found in and complete its electronic configuration. [2] Group: . . . . . . . . . . . . . . . . . . . . . . Electronic configuration: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Group 16 R: ns2np6 Largest energy difference: 14900 – 7880 = 7110 kJ mol-1 Based on the I.E data, the largest difference in energy is between the 6th and the 7th I.E. This means that the 7th electron to be removed is found in the inner shel l, and thus element R has 6 valence electrons (b) Silicon is commonly used in semiconductors. One important raw material for semiconductors is silane (SiH4), which is a key precursor in the production of semiconductor-grade silicon. (i) Draw the dot-and-cross diagram of silane. State the molecular shape about the central Si atom in silane. [2]

©YIJC [Turn over 3 Tetrahedral (ii) Deduce the polarity in silane, giving a reason for your answer. [1] Silane is non -polar, as the dipole moments caused by all Si – H bonds in a tetrahedral arrangement cancel each other out, resulting in net zero dipole moment. OR SiH4 has a symmetrical tetrahedral shape. The dipole moments of the polar Si – H bonds cancel each other out, resulting in a non-polar molecule. (c) The chlorides of silicon share similar properties as the chlorides of aluminium and phosphorus when they are dissolved in water. With the aid of equations, explain how chlorides of aluminium, silicon, and phosphorus exhibit acidic properties when dissolved in water. [4] For aluminium: AlCl3 (s) + 6 H2O (l) → [Al(H2O)6]3+ (aq) + Cl–(aq) [Al(H2O)6]3+ (aq) + H2O (l) ⇌ [Al(H2O)5(OH–)]2+(aq) + H3O+(aq) AlCl3 reacts with water to form hydrated ions. Since A l3+ ions have a high charge density, it has a high polarizing power to polarize the electron cloud from the (datively bonded water molecules), weakening the O – H bond, producing H 3O+ ions in water, thereby forming an acidic solution of around pH 3. For SiCl4 and PCl3, SiCl4 (l) + 4H2O(l) → SiO2.2H2O(s) + 4HCl (aq) or SiCl4 (l) + 2H2O(l) → SiO2 (s) + 4HCl (aq)

©YIJC [Turn over 4 PCl3 (s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq) Both chlorides hydrolyse completely in water due to (energetically accessible vacant 3d orbitals) which can accommodate the lone pairs of electrons from water to form very acidic solutions of around pH 2. [Total: 9]

©YIJC [Turn over 5 2 (a) The pKb value of three bases at 25 °C are shown in Table 2.1. Table 2.1 base formula pKb diethylamine (CH3CH2)2NH 2.9 ethylamine CH3CH2NH2 3.3 phenylamine C6H5NH2 9.4 (i) State the relationship between pKb and the strength of a base. [1] Inverse relationship. The larger the pKb, the weaker the base. (ii) Explain the relative magnitudes of the pKb values in Table 2.1. [4]

©YIJC [Turn over 6 Diethylamine, with a pKb value of 2.9, has the lowest pKb among the three compounds. This is due to the presence of two electron donating ethyl groups ( -CH2CH3) attached to the nitrogen atom which increase the electron density on the nitrogen atom. Hence the lone pair of electron on nitrogen atom is more available to accept a proton. Ethylamine, with a pKb value of 3.3, has a slightly higher pKb compared to diethylamine. In ethylamine, there is only one ethyl group attached to the nitrogen atom, which leads to lower electron density on the nitrogen compared to diethylamine. As a result, lone pair of electron on nitrogen atom is less available to accept a proton compared to diethylamine. Phenylamine has the highest pKb value of 9.4, the lone pair of electrons on the nitrogen atoms is delocalised into the benzene ring. Hence lone pair of electron on nitrogen atom is least available to accept a proton. (b) Preparation of N-phenylethanamide can be carried out by acylation reaction of phenylamine with ethanoic anhydride. The reaction is performed in two stages I. An excess of sodium ethanoate is heated with 1.5 g of phenylammonium chloride forming phenylamine, ethanoic acid and sodium chloride. C6H5NH3Cl + CH3COONa ⟶ C6H5NH2 + CH3COOH + NaCl Mr: 129.5 Mr: 93.0 II. Phenylamine then reacts with excess of ethanoic anhydride to form 1.12 g of N-phenylethanamide (as a white solid), together with ethanoic acid. (i) Calculate the percentage yield of N-phenylethanamide. [1] Amount of phenylamine = phenylammonium chloride = 1.5 / 129.5 = 0.01158 mol Theoretical yield of N-phenylethanmide= amount of N-phenylethanmide x molar mass = 0.01158 x 135.0 = 1.563 g Percentage yield = 1.12 /1.563 x 100% = 71.6 %

©YIJC [Turn over 7 (ii) Student A wanted to use sodium hydroxide instead of sodium ethanoate in stage I to produce phenylamine. Suggest an equation for the reaction between phenylammonium chloride and sodium hydroxide to form phenylamine. [1] C6H5NH3Cl + NaOH ⟶ C6H5NH2 + H2O + NaCl (iii) Student B claimed that the percentage yield of N-phenylethanmide will be affected by using sodium hydroxide instead of sodium ethanoate in stage I. Suggest and explain if Student B’s claim is correct. [1] If the stages are done without purifying the products of stage 1: Student B is correct, NaOH is a strong base which can undergo hydrolysis with the product N-phenylethamide and yield obtained will be affected. OR If the stages are done with purification of the products of stage 1 to obtain phenylamine: Student B is incorrect, NaOH is used in excess. Hence, the phenylamine formed is unchanged and yield obtained will not be