RI 2024 H2 Chem Y6 TP Suggested Solutions

1 © Raffles Institution 2024 9729/J/24 2024 Y6 H2 Chemistry Term 3 Timed Practice – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C C A D B C A B D D C D B A A MCQ worked solutions 1 Ans: C CO is a neutral oxide which does not react with aq. NaOH. CO2 gas is acidic and reacts with aq. NaOH. Fe xOy + yCO → xFe + yCO2 Amt of FexOy = 0.523 mol (given) Amt of Fe = (146 / 55.8) = 2.616 mol Since volume of CO 2 was measured at r.t.p., amt of CO2 = (75.4 / 24.0) = 3.142 mol FexOy + yCO → xFe + yCO2 Amt/mol 0.523 2.616 3.142 Mole ratio 1 5 6 Hence, x = 5 and y = 6. 2 Ans: C 5 X− + XOn− → halogen-containing product (either X2 or XO− from observing the MCQ options) Step 1: Balance X; 5 X− + XOn− → 3 X2 OR 5 X− + XOn− → 6 XO− Step 2: Check if amt of electrons gained = amt of electrons lost A 5 X− + XO− → 3 X2 amt of electrons gained by 1 mol of XO− to form ½ mol X2 = 1 mol amt of electrons lost by 5 mol of X− to form 5 2 mol of X2 = 5 mol not balanced B 5 X− + XO2− → 6 XO− amt of electrons gained by 1 mol of XO2− to form 1 mol of XO− = 2 mol amt of electrons lost by 5 mol of X− to form 5 mol of XO−= 10 mol not balanced C 5 X− + XO3− → 3 X2 amt of electrons gained by 1 mol of XO3− to form ½ mol X2 = 5 mol amt of electrons lost by 5 mol of X− to form 5 2 mol of X2 = 5 mol balanced D 5 X− + XO4− → 6 XO− amt of electrons gained by 1 mol of XO4− to form 1 mol of XO− = 6 mol amt of electrons lost by 5 mol of X− to form 5 mol of XO− = 10 mol not balanced

2 © Raffles Institution 2024 9729/J/24 3 Ans: A Step 1: Write electronic configuration of Fe atom i.e. [Ar] 3d64s2 Step 2: Remove electrons from atom to form ion, removing 4s before 3d. For Fe3+, two 4s electrons and one 3d electron are removed from Fe i.e. Fe3+ : [Ar] 3d5. 4 Ans: D 1 Not a result of hydrogen bonding. CH3CH2CONH2 is neutral as the lone pair of electrons on the N atom is delocalised into the C=O bond, making it unavailable for dative covalent bond formation to a proton. 2 A result of hydrogen bonding. In vapour state, CH3COOH can exist as a dimer which is held by hydrogen bonds: CH3 C O O H CH3C O OH δ+δ- δ- δ-δ+δ- Mr of dimer = 120 3 Not a result of hydrogen bonding. (CH3)3N has more electron-donating alkyl groups than (CH3)2NH, which increases the electron density on the N atom in (CH3)3N, thus increasing the availability of the lone pair of electrons on N for dative covalent bond formation to a proton. 4 A result of hydrogen bonding. Acid HOOC COOH HOOC COOH Conjugate base HO O O O δ+ δ-O H O O O Remarks Conjugate base is not stabilised by intramolecular hydrogen bond Conjugate base is stabilised by intramolecular hydrogen bond ⇒ conjugate base of cis-acid is more stable ⇒ cis-acid is more acidic than trans-acid ⇒ Ka1 of cis-acid is greater than Ka1 of trans-acid. Hence, Ka1 of trans-acid is smaller than that of Ka1 of cis-acid.

3 © Raffles Institution 2024 9729/J/24 5 Ans: B 1 Option 1 is correct. According to the slow step (rate-determining step), rate = k[G][F]. However, 1 molecule of G (an intermediate) is formed from 2 molecules of E in the preceding fast step. Hence, the overall rate equation should be rate = k[E]2[F]. Note that intermediates should not appear in the overall rate equation. 2 Option 2 is correct. 2E ⇌ G G + F → H H + F → E2F2 By summing up the three equations in the mechanism, the overall equation is: 2E + 2F → E2F2. 3 Option 3 is incorrect. The mechanism involves three transition states. Each step has a transition state. The mechanism involves two intermediates (G and H). Recall that an intermediate is a species that is formed in one step of a reaction mechanism and consumed in a subsequent step. 6 Ans: C The orders of reaction for each reactant are not given. Hence the question is asking for possible shapes of graph for different orders. Option 2: As bromine is a reactant, its concentration will decrease with time. Depending on the orders with respect to the various reactants, the decrease could be linear or as a curve (as shown in the question). Hence, option 2 is possible. Since absorbance ∝ [Br2(aq)], the above explanation for option 2 holds for a graph of absorbance against time. Hence, option 1 is possible. The pressure of the system is contributed by its gaseous components. In this reaction, CO 2 (a product) is the only gaseous species. A graph of pressure against time would be show an increasing trend starting from the origin. Hence, option 3 is not possible. A graph of rate against [Br 2(aq)] would be a horizontal line (if zero order) or show an increasing trend starting from the origin (if 1st or 2nd order). Option 4 is not possible.

4 © Raffles Institution 2024 9729/J/24 7 Ans: A Recall: A Brønsted-Lowry acid is a proton donor; A Brønsted-Lowry base is a proton acceptor. A Lewis acid is an electron-pair acceptor; A Lewis base is an electron-pair donor. A 2NH3(l) + 2Na(s) → 2NaNH2(s) + H2(g) NH3 is a brønsted-Lowry acid as it donates a proton, H+, to form NH2−. [H+ then undergoes redox reaction with Na to form Na+ and H2] B NH3(g) + H2O(l) → NH4+(aq) + OH–(aq) NH3 is a brønsted-Lowry base as it accepts a proton, H+, from H2O to form NH4+. C 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) NH3 is oxidised to NO (oxidation state of N changes from –3 to +2). D 4NH3(aq) + Cu2+(aq) → [Cu(NH3)4]2+(aq) Each NH3 donates a lone pair of electrons to form a dative coordinate bond with Cu2+ in [Cu(NH3)4]2+(aq). Hence, NH3 is a lewis base. 8 Ans: B Let the weak acid be HA. Equivalence point (point B) is reached when 10.0 cm3 of 0.100 mol dm–3 HA is added. Since n(HA) : n(NaOH) = 1 : 1, HA must be a monobasic acid ⇒ option 1 is correct At equivalence point (point B), all the NaOH has reacted with HA to form NaA. The solution contains A−, the conjugate base of HA. A− undergoes hydrolysis to form OH−. A−(aq) + H2O(l) ⇌ HA(aq) + OH−(aq) Hence, the solution is alkaline (pH >7) ⇒ option 2 is correct At point A, only 50% of the NaOH has been reacted with HA to form NaA. The solution contains unreacted NaOH and the salt, NaA formed (n(OH −)unreacted = n(A−)formed). Since the solution does not contain the weak acid, HA, and its salt, NaA, it is not a buffer solution ⇒ option 3 is incorrect [Note: A buffer solution with maximum buffering capacity is formed at point D, where n(HA)excess = n(A−)formed.] 9 Ans: D To prepare a buffer of pH 7.40, choose the acid with pK a closest to 7.40 i.e. 7.20, which is the pKa of H 2PO4–. The species in the buffer would be the weak acid, H 2PO4–, and its conjugate base, HPO42– i.e. option D. The following options are incorrect for the following reasons. A HPO42– reacts with NaOH to form PO43–. If NaOH is limiting, a mixture of HPO42– and PO43– i.e. a mixture of an acid with its conjugate base will form. This resultant mixture has a pH of around 12.38. B This mixture of an acid (H3PO4) and its conjugate base (H2PO4–) results in a buffer solution of pH around 2.15. C H2PO4– reacts with H+ from HCl to form H3PO4. If HCl is limiting, a mixture of an acid (H3PO4) and its conjugate base (H2PO4–) results in a buffer solution of pH around 2.15.

5 © Raffles Institution 2024 9729/J/24 10 Ans: D [IO3 –] in solution Z Solution Z is a saturated solution of Ca(IO3)2. Ca(IO3)2 ⇌ Ca2+(aq) + 2 IO3– Eqm conc/ mol dm–3 – s 2s Ksp = s(2s)2 = 6.47 x 10–6 ⇒ s = 0.01174 mol dm–3 [IO3 –] = 2(0.01174) = 0.0235 mol dm–3 [Pb2+] when Pb(IO3)2 just precipitates Pb(IO3)2 pr