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 b