RI 2016 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Changes to 2016 H2 Chemistry A Level Question Paper Dear students, The TYS you have purchased is based on the 9647 (old) syllabus. You will be sitting for the 9729 (new) syllabus papers. This document will instruct you on the changes you need to make to the TYS questions. The Planning question for the 9729 syllabus will be in the Paper 4 (Practical). Some concepts are no longer tested in the 9729 syllabus and the values used for some calculations are now different (e.g. molar volume at s.t.p.) which will affect your choice of the answers. You are advised to make the changes on your question papers before you attempt it . Do inform your tutors if you notice any differences which were not highlighted in this document. ---------------------------------------------------------------------------------------------------------------------------------------------- Paper 1 17 Amend question The equation for the thermal decomposition of Mg(NO3)2 is no longer in syllabus. The following equation will help you solve this question: Mg(NO3)2 → MgO + 2NO2 + ½O2 18 Amend question – Options C & D not in syllabus C is not true. Mg(OH)2 is not very soluble in water (QA knowledge) D is not true. Mg reacts very slowly with cold water. 20 Amend options A Y is in Group 2. B Y is in Group 13. C Y is in Group 15. Paper 2 4b(i) Amend question “… show the feasibility spontaneity…” 5(b)(ii) Not in syllabus Paper 3 No amendments

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Suggested Solutions to 2016 ‘A’ Level H2 Chemistry Paper 1 ( 9647/01) 1 Answer: B Amount of Q = 1.0 g ÷ x g mol–1 = 1/x mol Number of molecules = L mol–1 x (1/x) mol = L/x Number of atoms = 2(L/x) = 2L/x 2 Answer: D 2H 2S + 3O2 → 2SO2 + 2H2O CS2 + 3O2 → CO2 + 2SO2 Combining both equations: 2H2S + CS2 + 6O2 → 4SO2 + CO2 + 2H2O Mole ratio of SO2 : CO2 = 4 : 1 3 Answer: C Ar for Cu = [65(63) + 29(65)] ÷ (65 + 29) = 63.6 4 Answer: D 36S2–: 16 protons, 20 neutrons, 18 electrons 37Cl–: 17 protons, 20 neutrons, 18 electrons Option A: The nucleon numbers for S and Cl are 36 and 37 respectively. Option B: Both ions have an outer electronic configuration of 3s23p6. Option C: Both ions have fewer electrons than neutrons. Option D: Both ions have 20 neutrons in their nuclei. 5 Answer: A 6 Answer: C There are three possible structures.  C1–C2≡N (C-1 has 1 lone pair and 1 unpaired electron; N has a lone pair)  C1=C2=N (C-1 has 1 lone pair; N has 1 lone pair and 1 unpaired electron)  C1≡C2–N (C-1 has 1 unpaired electron; N has 2 lone pairs) All three possible structures have 2 lone pairs of electrons and 1 unpaired electron. 7 Answer: C By conservation of mass, mass of liquid = mass of vapour pV = nRT (101 x 10 3)[(78 – 2) x 10–6] = (0.293/M)(8.31)(97 + 273) Molar mass, M = 117.4 g mol–1 Mr ≈ 117 8 Answer: D Lattice energy is the enthalpy change when one mole of ionic compound is formed from its constituent gaseous ions under standard conditions. Note: Lithium fluoride is a solid under standard conditions. 9 Answer: B ΔH f of KCl = 90 + ½(242) + 418 + (–355) + (–710) = –436 kJ mol–1

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 10 Answer: D ΔG = ΔH – TΔS When ΔG is more negative at a higher temperature, ΔS must be positive. OR 1 mol of gas + 1 mol of solid → 2 mol of gases ⇒ ΔS > 0 because a gas has greater entropy than a solid At a lower temperature, when “–TΔS” is a small negative value, ΔG is positive. This implies that ΔH is positive. OR +78000 = ΔH – 378ΔS Since ΔS > 0, – 378ΔS < 0. Thus, ΔH must be positive. 11 Answer: C Ecell = E(Ag+/Ag) – E(Fe3+/Fe2+) = (+0.80) – (+0.77) = +0.03 V To obtain a cell potential of 0.00 V, E(Ag+/Ag) needs to become less positive or E(Fe3+/Fe2+) needs to become more positive, or both electrode potentials have to be adjusted to the same value. Option A: Increase in [Ag+] will make E(Ag+/Ag) > +0.80 V Option B: Increase in [Fe2+] will make E(Fe3+/Fe2+) < +0.77 V Option C: Increase in [Fe3+] will make E(Fe3+/Fe2+) > +0.77 V (can be adjusted to +0.80 V) Option D: Increase in the surface of the electrode does not change the electrode potential. 12 Answer: A Anode reaction: 2O2– → O2 + 4e– Amount of electricity passed = 8 C s–1 x (100 x 60) s = 48000 C When 4 x 96500 C are passed, 1 mol of O2 is liberated. When 48000 C are passed, 48000 / (4 x 96500) = 0.1244 mol of O2 is liberated. Volume of O2 liberated at s.t.p. = 0.1244 x 22.7 = 2.8 dm3 (1 d.p.) 13 Answer: C X is a saturated solution of ZnF 2. ZnF2(s)  Zn2+(aq) + 2F–(aq) Ksp = [Zn2+][F–]2 = 3.2 x 10–2 mol3 dm–9 Let [Zn2+] be y mol dm–3 and [F–] be 2y mol dm–3. y(2y) 2 = 3.2 x 10–2 [F–] = 2y = 4 x 10–1 mol dm–3 When BaF2 just precipitates, [Ba2+][F–]2 = 1.6 x 10–7 mol3 dm–9 Since [F–] = 4 x 10–1 mol dm–3, [Ba2+] = 1.6 x 10–7 ÷ (4 x 10–1)2 = 1 x 10–6 mol dm–3 14 Answer: B A l(H2O)63+(aq) + H2O(l)  Al(OH)(H2O)52+(aq) + H3O+(aq) [H+] = √(1.0 x 10–5 x 0.1) = 1.0 x 10–3 mol dm–3 pH = –log (1.0 x 10–3) = 3.0 15 Answer: D y = k (a) (a)2 (a)2 Hence, k = y/a5 rate = (y/a5) (a/2) (2a)2 (3a)2 = 18y 16 Answer: A Magnesium oxide, though having a giant ionic lattice, has less covalent character than aluminium oxide. Phosphorus pentoxide and silicon dioxide are predominantly covalent, so they do not exist as giant ionic lattices.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 17 Answer: A Mg(NO3)2(s) → MgO(s) + 2NO2(g) + ½O2(g) Amount of Mg(NO3)2 = 10.4 / 148.3 = 0.07013 mol Amount of O2 (neutral gas) = ½ (0.07013) = 0.03507 mol Mass of O2 (neutral gas) = 0.03507 x 32.0 = 1.12 g 18 Answer: A O ption B: Magnesium (m.p. 650 °C) has a higher melting point than sulfur (m.p. 115.2 °C) because the energy needed to overcome the metallic bonds in the giant metallic lattice of magnesium is greater than that needed to overcome the id- id interactions between S 8 molecules in the simple molecular lattice of sulfur. Option C: Magnesium hydroxide is only sparingly s oluble in water. Option D: Magnesium reacts slowly with cold water. 19 Answer: D White silver chloride formed dissolves in concentrated aqueous ammonia to give a colourless solution. Cream silver bromide formed dissolves in concentrated aqueous ammonia to give a colourless solution. Yellow silver iodide formed is insoluble in concentrated aqueous ammonia. 20 Answer: D Element Y is vanadium. 21 Answer: B 2 VO 2+ + SO2 → 2VO2+ + SO42– Ecell = 1.00 – 0.17 = +0.83 V (feasible) 2VO2+ + SO2 → 2V3+ + SO42– Ecell = 0.34 – 0.17 = +0.17 V (feasible) 2V3+ + SO2 + 2H2O → 2V2+ + SO42– + 4H+ Ecell = –0.26 – 0.17 = –0.43 V (not feasible) 22 Answer: B NH3 ligand has no charge, while Cl– ligand has a charge of 1–. Let the number of NH3 ligands be 6 – n and the number of Cl– ligands be n. Charge on cation in platinum(IV) compound = (+4) + (6 – n)(0) + n(–1) = 2 Solving, n = 2, i.e. there are 2 Cl– ligands. Hence, there are 4 NH3 ligands. PtCl4 + 4NH3 → [Pt(NH3)4Cl2]2+ + 2Cl– Option A: The cation, [Pt(NH3)3Cl3]+, has a 1+ charge. Option C: The oxidation state of platinum in [Pt(NH3)6]2+ is +2. Option D: The cation, [Pt(NH3)6]4+, has a 4+ charge. 23 Answer: B There are 2 π bonds found in the C≡C bond. Both C atoms in the C≡C bond are sp hybridised. The remaining C in CH 3 is sp3 hybridised. 24 Answer: A A termination step involves the collision of 2 free radicals. Hence, option B is incorrect. An H● free radical is not formed, so options C and D are incorrect. 25 Answer: D The rate-determining step of this S N1 m