NJC 2022 H1 Chem Promo P2 Ans

NJC SH1 Promotional Examination 8873/02/2022 [Turn over NATIONAL JUNIOR COLLEGE SH1 Promotional Examination Higher 1 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 8873/02 29 September 2022 1 hour 30 minutes READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name 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 a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all questions on the Question Paper. The use of an approved 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. Appropriate significant figures and units are expected for final numerical answers. For Examiner’s Use 1 /17 2 /9 3 /8 4 /15 5 /11 Penalty Presentation Paper 2 /60 Paper 1 /20 Total /80 Promo Percentage /100 This document consists of 21 printed pages and 3 blank pages.

2 NJC SH1 Promotional Examination 8873/02/2022 For Examiner’s Use 1 Some properties of SCl2, S2Cl2, S8 and Cl2 are shown in the table. property SCl2 S2Cl2 S8 Cl2 density / g cm−3 1.62 1.69 2.07 0.00290 melting point / C −78 −80 115 −102 boiling point / C decomposes at 59 C 137 445 −34 ∆Hfꝋ / kJ mol−1 −49.8 −59.4 (a) S2Cl2 can be made by direct combination of sulfur in its standard state, S8, and chlorine gas, Cl2. (i) Write a thermochemical equation that represents the standard enthalpy change of formation of S2Cl2. [2] ¼ S8 (s) + Cl2 (g) ⎯→ S2Cl2 (l) ∆Hfꝋ = −59.4 kJ mol−1 Correct species (in particular S8) and balanced eqn, with ∆Hfꝋ value Correct state symbols of all species (ii) A chemist wishes to prepare 10.0 cm3 of S2Cl2 from its elements. Use the data in the table to calculate the mass of sulfur required to produce this volume of S2Cl2. Give your answer to 4 significant figures. Mass of S2Cl2 = 1.69 g cm−3 × 10.0 cm3 = 16.9 g Amount of S2Cl2 = 16.9 / 135.2 = 0.125 mol Amt of sulfur, S8 = ¼ × 0.125 = 0.03125 mol Mass of sulfur, S8 = 0.03125 × (8 × 32.1) = 8.025 g (4 s.f.) [2]

3 NJC SH1 Promotional Examination 8873/02/2022 [Turn over For Examiner’s Use (iii) Suggest a reason for the difference in physical state of S 8 and C l2 at room temperature and pressure. [2] Both are non-polar simple covalent molecules. S8 has a higher Mr, hence larger electron cloud which is more easily distorted than that of C l2, forming stronger intermolecular instantaneous dipole -induced dipole (id-id) interactions. Energy at room temperature and pressure is sufficient to overcome the weaker id-id between C l2 but insufficient to overcome the stronger id -id between S 8, hence, Cl2 exists as a gas while S8 exists as a solid at r.t.p. (b) S2Cl2 can also be formed when SCl2 decomposes, as shown in the equation below. 2SCl2 ⎯→ S2Cl2 + Cl2 Using the data in the table , calculate the standard enthalpy change, ∆Hrꝋ, for this reaction. ∆Hrꝋ = Σ∆Hfꝋ(product) − Σ∆Hfꝋ(reactant) = −59.4 − 2(−49.8) = +40.2 kJ mol−1 Note: ∆Hfꝋ Cl2(g) = 0 [1] (c) S2Cl2 reacts with water, as shown in the equation below. S2Cl2 + 2H2O ⎯→ SO2 + H2S + 2HCl By considering the oxidation states of sulfur, state and explain the type of reaction taking place. [2] Disproportionation. The oxidation state of S increases from +1 in S2Cl2 to +4 on SO2, and decreases to −2 in H2S, hence it has been oxidized and reduced simultaneously.

4 NJC SH1 Promotional Examination 8873/02/2022 For Examiner’s Use (d) (i) Draw a dot-and-cross diagram to show the bonding present in S2Cl2. given that the atoms are bonded in the order Cl–S–S–Cl. [1] (ii) Predict the shape about the S atom and the Cl–S–S bond angle in S2Cl2. shape: …………………..bent bond angle: …………………………..105⁰ [1] (e) Chloramine, NH2Cl, can be used in the treatment of drinking water to kill bacteria. Draw a labelled diagram to show the interaction formed between NH 2Cl and a H 2O molecule. OR 1) Lone pair electron on N/O interacting with + H of H−O and H−N 2) Indicate + and − 3) Clearly label “hydrogen bonding” [2]

5 NJC SH1 Promotional Examination 8873/02/2022 [Turn over For Examiner’s Use (f) The ionic radii of some ions are shown in the table below. species ionic radius / nm Cl− 0.181 O2− 0.140 S2− 0.184 Using the data given, explain the difference in ionic radius of the following pairs of elements. (i) Cl− and S2− Cl− : 1s2 2s2 2p6 3s2 3p6 S2− : 1s2 2s2 2p6 3s2 3p6 Nuclear charge of Cl− is larger than S2− (due to higher number of protons) Shielding effect remains (relatively) constant (since same number of electrons in the inner principal quantum shells). Nuclear attraction for the valence electrons in Cl− is larger than S2−. The valence electrons are pulled closer to the nucleus, hence ionic radius for Cl− is smaller. [2] (ii) O2− and S2− O2− : 1s2 2s2 2p6 S2− : 1s2 2s2 2p6 3s2 3p6 The valence electron in S2− is further away from the nucleus due to having 1 more filled principal quantum shell. S2− has a larger shielding effect than O2−. These factors outweigh the increase in nuclear charge for S 2−. Nuclear attraction for the valence electrons in S 2− is lesser than O2−, hence valence electrons are less attracted to the nucleus and ionic radius for S2− is larger. [2] [Total: 17]

6 NJC SH1 Promotional Examination 8873/02/2022 For Examiner’s Use 2 (a) Gold can be extracted from low -grade ore around the world by use of the Elsner reaction in which the impure gold ore is reacted with an aqueous solution of sodium cyanide, as shown in the equation below. 4Au + 8NaCN + O2 + 2H2O ⎯→ 4NaAu(CN)2 + 4NaOH The experimental procedure to determine the percentage by mass of gold in the sample of impure gold ore is as follows: • A sample of 35 g of impure gold ore was added to 250 cm 3 aqueous solution of sodium cyanide. • 25.0 cm3 of the resulting solution was neutralised with 15 cm3 of 0.3 mol dm−3 sulfuric acid. • The excess sulfuric acid was then titrated with 0.06 mol dm −3 potassium hydroxide solution. 17.00 cm3 of KOH (aq) was required to reach the end point. Calculate the percentage by mass of gold in the sample of impure gold ore. Reaction 1: 4A