ACJC 2022 Prelim P1 Answers

© ACJC2022 9729/01/Prelim/2022 ACJC solutions for H2 Chemistry Prelim Paper 1 2022 1 A 11 A 21 C 2 C 12 C 22 A 3 D 13 D 23 A 4 A 14 C 24 C 5 D 15 B 25 B 6 D 16 D 26 C 7 C 17 B 27 B 8 B 18 B 28 A 9 C 19 D 29 C 10 A 20 B 30 B 1 Which species deflects the most in an electric field? A 7Li+ B 24Mg2+ C 32S2− D 27Al3+ Answer: A 7Li+ 24Mg2+ 32S2− 27Al3+ charge mass 1 7 1 12 1 16 1 9 Angle of deflection ∝ charge mass 2 The shape of each p-orbital is represented as two lobes. How many 3d orbitals have four lobes? A 2 B 3 C 4 D 5 Answer: C

© ACJC2022 9729/01/Prelim/2022 3 Which species has the greatest number of unpaired electrons in its ground state? A Cu+ B CH3− C Mg D F Answer: D Cu+: 1s2 2s2 2p6 3s2 3p6 3d10 (no unpaired electrons) CH3−: no unpaired electrons Mg: 1s2 2s2 2p6 3s2 (no unpaired electrons) F: 1s2 2s2 2p5 (1 unpaired electron) 4 Which species contains a dative bond? 1 CO 2 NO3− 3 O3 A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 only Answer: A C O O O O O N O O −

© ACJC2022 9729/01/Prelim/2022 5 Which option correctly describes the shape and polarity of the species? species shape polarity A AlCl3 trigonal planar polar B SiF4 square planar non-polar C BrF3 trigonal pyramidal polar D BeCl2 linear non-polar Answer: D species shape polarity A AlCl3 trigonal planar (3 b.p.) non-polar B SiF4 tetrahedral (4 b.p.) non-polar C BrF3 T shaped (3 b.p. + 2 l.p.) polar D BeCl2 Linear (2 b.p.) non-polar 6 Which graph shows the behaviour of a fixed mass of an ideal gas at a constant temperature? [p = pressure, V = volume, Mr = molar mass, ρ = density] A B C D p ρ pV p pMr V p

© ACJC2022 9729/01/Prelim/2022 Answer: D A pV = nRT pV = r m M RT pMr = k m V B pV = nRT = k C pV = nRT p = 1nRT V   p = k 1 V   D pV = r m M RT p = m V   r RT M    = ρk 7 Which statements explain the difference in ionic radius between Na+ and F−? 1 Outermost electrons of F− experience weaker nuclear charge than those of Na+. 2 Outermost electrons of Na+ experience greater shielding effect than those of F−. 3 Outermost electrons of Na+ are nearer to the nucleus than those of F−. A 3 only B 1 and 2 only C 1 and 3 only D 1, 2 and 3 Answer: C Statement 1: Correct. F− contains fewer protons compared to Na+. Statement 2: Incorrect. Valence electrons of Na+ experience the same shielding effect as those of F− as both species contain the same number of inner shell electrons. pMr V pV p p

© ACJC2022 9729/01/Prelim/2022 Statement 3: Correct. Valence electrons of Na+ are nearer the nucleus than those of F−, resulting in a smaller ionic radius. 8 0.1 mol of compound X dissolves in 1 dm3 of water to give a solution with a pH of 1. What is the identity of X? A AlCl3 B CH3COCl C NH4NO3 D CH3COOH Answer: B AlCl3 : AlCl3 (g) → Al3+ (aq) + 3Cl− (aq) [Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+ (pH > 1) CH3COCl + H2O → CH3COOH + HCl HCl dissociates completely to give a [H+] of 0.1 moldm−3. CH3COOH contributes negligible H+ to the solution as it is a weak acid and its dissociation is suppressed by the strong acid. pH = 1 NH4NO3(s) → NH4+(aq) + NO3−(aq) NH4NO3 dissolves in water to give NH4+ and NO3−. NH4+ hydrolyses in water NH4+ + H2O ⇌ NH3 + H3O+ (pH > 1) CH3COOH + H2O ⇌ CH3COO− + H3O+ (pH > 1) 9 Wüstite, containing both Fe 2+ and Fe3+ ions, has the formula Fe 20Ox. Fe2+ constitutes 90% of the iron ions present in the compound. What is the value of x? A 18 B 19 C 21 D 22 Answer: C Total charge on iron ions: (18 × 2) + (2 × 3) = +42 Total charge on iron ions = Total charge on oxide ions 42 = 2x x = 21 10 Which option involves a positive entropy change? A The homolytic fission of gaseous chlorine. B The lattice energy of sodium chloride. C The contraction of an ideal gas at a constant temperature. D Cooling a copper strip from 373 K to 273 K. Answer: A

© ACJC2022 9729/01/Prelim/2022 A: There is an increase in the number of gaseous particles as the reaction proceeds. Cl2(g) → 2Cl(g) B: Na+(g) + Cl−(g) → NaCl(s) There is a decrease in the number of gaseous particles. C: Gas particles have fewer ways of arrangement with a smaller volume. D: Cooling a metal results in less disorder as the particles vibrate less. In addition, there are fewer energy quanta available for distribution so there are fewer ways to distribute them in the metal. 11 The reaction between NO and Br2 is proposed to proceed via the following mechanism: Step 1: NO + Br2 NOBr2 (fast) Step 2: NOBr2 + NO 2NOBr (slow) Which statements are correct? 1 NOBr2 is a radical. 2 The rate equation for this reaction is rate = k[Br2][NO]2. 3 NOBr2 is a transition state. A 1 and 2 only B 1, 2 and 3 C 1 and 3 only D 2 and 3 only Answer: A Statement 1: Correct. There is one unpaired electron on the N atom in the molecule. Statement 2: Correct. From the slow step, Rate = k[NOBr2][NO] --- (1) From step 1, Kc = 2 2 [NOBr ] [NO][Br ] [NOBr] = Kc[NO][Br2] --- (2) Sub. (2) into (1), Rate = k(Kc[NO][Br2])[NO] = k’[NO]2[Br2] Statement 3: Incorrect. NOBr2 is an intermediate as it appears in the reaction mechanism but does not appear in the overall equation. 12 Which statement regarding catalysts is correct? A Catalysts change the ∆H value of a reaction. B Catalysts increase the yield of product in a reaction. C Catalysts provide a different mechanism for a reaction. D Catalysts change the Kc value of a reaction. Answer: C

© ACJC2022 9729/01/Prelim/2022 A: Catalysts do not change the reactants and products of a reaction. Hence the ∆H value remains unchanged. B: Catalysts do not change in the position of equilibrium in a reaction. Hence the yield of the product does not change. C: Catalysts reduce the ac tivation energy of a reaction by providing an alternative mechanism for the reaction to proceed by. D: The rate constant values of both the forward and backward reaction are changed to the same extent by a catalyst. Hence the Kc value of a reaction remains unchanged. 13 In aqueous solution, an equilibrium is established between chromate , CrO42− (yellow) and dichromate ions, Cr2O72− (orange). 2CrO42−(aq) + 2H+(aq) Cr2O72−(aq) + H2O(l) Which statement regarding the ions and the equilibrium is correct? A The oxidation number of chromium in both chromium-containing ions is different. B The difference in colour between CrO 42− and Cr 2O72− is due to a difference in energy gap between the 3d orbitals. C The Kc expression for the equilibrium is Kc = 2 2 7 2 2 2 + 2 4 [Cr O ][H O] [CrO ] [H ]   . D Increasing the pH turns the solution yellow. Answer: D A: Cr is in the same oxidation number of +6 for both species. Let x be the oxidation number of Cr. x + 4(−2) = −2 x = +6 (CrO42−) 2x + 7(−2) = −2 x = +6 (Cr2O72−) B: There are no electrons in the 3d subshell for both chromium-containing species. Hence the difference in colour is not due to the difference in energy gap between the 3d orbitals. C: The Kc expression for the equilibrium is Kc = 2 27 2 2 + 2 4 [Cr O ] [CrO ] [H ]   . [H2O] is not included in the expression as it is a solvent and its concentration remains constant. D: Increasing the pH results in a decrease in [H+]. Position of equilibrium shifts to the left and the major species in solution is CrO42−.

© ACJC2022 9729/01/Prelim/2022 14 Nitrogen dioxide dimerises in a