RI 2025 Y6 T3 TP Suggested Solutions for Students

1 © Raffles Institution 2025 9729/J/25 2025 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 A C C B C A A B C D B D C D A MCQ worked solutions 1 Ans: A A Correct. 63Cu+ : [Ar]3d10. With a fully filled d-subshell, it has no unpaired electrons. 65Cu3+ : [Ar]3d8 i.e. has 2 unpaired electrons. B Incorrect. 63Cu has 29 electrons and has the electronic configuration [Ar]3d104s1. Hence, 63Cu+ has the electronic configuration [Ar]3d10. C Incorrect. No. of neutrons in 63Cu+ = 63 – 29 = 34 No. of neutrtons in 65Cu3+ = 65 – 29 = 36 D Incorrect. The number of protons determine the identity of the element. Since both 65Cu2+ and 65Cu+ are ions of copper, they both have 29 protons. 2 Ans: C Volume strength of 10 ⇒ 10 dm3 of O2 evolved from 1 dm3 of H2O2(aq) ⇒ 10 22.7 = 0.4405 mol of O2 evolved from 1 dm3 of H2O2(aq) Since 2H2O2 → 2H2O + O2, amt of H2O2 in 1 dm3 of H2O2 solution = 2(0.4405) = 0.8810 mol 10 cm3 of H2O2 contains 3 22 10 (0.8810) 8.810 10 mol of H O1000 −= × Amt of MnO4– reacted = 32(8.810 10 )5 −× = 3.524 x 10–3 mol Volume of MnO4– = 3 33.524 10 0.176 dm0.0200 −× =

2 © Raffles Institution 2025 9729/J/25 3 Ans: C butan-1-ol (C4H10O) 2-methylpropan-2-ol (C4H10O) Butan-1-ol is a straight-chain isomer of C4H10O and has a greater surface area for instantaneous dipole–induced dipole (id- id) interactions to occur compared to 2- methylpropan-2-ol which is a branched-chain isomer of C4H10O. Thus, id- id interactions are stronger in butan -1-ol than in 2- methylpropan-2-ol. ( Option 1 is correct) The strength of the hydrogen bonds in both molecules are the same as the hydrogen atom involved in hydrogen bonding is bonded to the same element, oxygen. The extent of hydrogen bonding is also the same as they have same number of lone pairs and number of H atoms attached to O. (Options 2 and 3 are incorrect) 4 Ans: B G H Remarks A O C S net dipole: S C S no net dipole G has a greater net dipole than H. B Cl P Cl Cl net dipole: F P F F net dipole: P−F bond is more polar than P−Cl bond. G has a smaller net dipole than H. C I F Br Cl I−F bond is more polar than Br−Cl bond. G has a greater net dipole than H. D F Xe F F F no net dipole Cl P Cl Cl Cl Cl no net dipole Both G and H do not have a net dipole. OH OH

3 © Raffles Institution 2025 9729/J/25 5 Ans: C In the second reaction, the same number of moles of reactants were used, therefore the same amount of heat, q, will be evolved. However the total volume of solution was halved, therefore there is half the mass of solution to heat up, and the temperature increase will hence be doubled according to the equation q = mcΔT 6 Ans: A For the process C6H12(s) → C6H12(l), • ∆H is positive because energy is required to overcome the id- id interactions between molecules of C6H12 to melt the compound. • ∆S is positive as there is a greater disorder in the liquid state than the solid state i.e. there is an increase in disorder during melting. 7 Ans: A By considering the slow step, the rate equation for step 2 is rate = k2[O•][O3] but O• is an intermediate, thus [O•] cannot be in the overall rate equation. Using the equilibrium constant of step 1, 𝐾𝐾1 = [O •][O2] O3 [O •] = 𝐾𝐾1[O3] [O2] Substituting this into the rate equation from step 2, rate = 𝑘𝑘2 𝐾𝐾1[O3] [O2] [O3] rate = 𝑘𝑘2𝐾𝐾1 [O3]2 [O2] Thus, statements 1 and 2 are correct. Statement 3 is correct because [O 2] is the denominator , thus when [O 2] increases, rate decreases. 8 Ans: B All the 4 C in the ring are sp2 hybridised. The C in the –C≡N group is sp hybridised. Each C=C contains one π bond and the –C≡N group contains 2 π bonds.

4 © Raffles Institution 2025 9729/J/25 9 Ans: C A The conjugate base may be a weak base. Example: CH 3COOH ( Ka = 10 –4.75, a weak acid) and its conjugate base CH 3COO– (Kb = 10–9.25, a weak base). B A buffer solution contains a weak acid (or weak base) and its conjugate base (or conjugate acid). Addition of a limited amount of strong base to a weak acid will produce a buffer containing the (excess) weak acid and the conjugate base. However, if the strong base is added in excess, the weak acid will be completely reacted and the solution will not be a buffer. C HA ⇌ H+ + A– Ka = [H+][A–] / [HA] As water is added, the concentrations of HA, H + and A – decrease and the reaction quotient decreases. The system will try to increase the reaction quotient by favouring the dissociation reaction until the new equilibrium is established (i.e. when reaction quotient is equal to Ka). Hence, the extent of dissociation increases with increasing dilution. D The dissociation of a weak acid in water is usually endothermic. As temperature increases, the system will try to counteract the increase in temperature by favouring the forward endothermic reaction in order to absorb heat. Hence, the extent of dissociatio n increases and K a increases with increasing temperature. 10 Ans: D A Incorrect. There are 0.015 mol of CH 3COO– and 0.030 mol of H + from HCl i.e. H+ is in excess. Since CH3COO– + H+ → CH3COOH, the resultant mixture contains 0.015 mol of H+ and 0.015 mol of CH3COOH. This is not a buffer solution. B There are 0.020 mol of CH3COO– and 0.010 mol of H+ from HCl i.e. CH3COO– is in excess. Since CH3COO– + H+ → CH3COOH, the resultant mixture contains 0.010 mol of CH3COO– and 0.010 mol of CH3COOH. This is a buffer solution. C Incorrect. There are 0.010 mol of CH3COOH and 0.020 mol of OH– from NaOH i.e. OH– is in excess. Since CH3COOH + OH– → CH3COO– + H2O, the resultant mixture contains 0.010 mol of OH– and 0.010 mol of CH3COO–. This is not a buffer solution. D There are 0.030 mol of CH3COOH and 0.015 mol of OH– from NaOH i.e. CH3COOH is in excess. Since CH3COOH + OH– → CH3COO– + H2O, the resultant mixture contains 0.015 mol of CH3COOH and 0.015 mol of CH3COO–. This is a buffer solution. Comparing the buffer solutions in B and D, the solution in D contain greater amounts of the buffering species (i.e. CH3COOH and CH3COO–) to react with any small amounts of acid or base added. Hence, for the same amount of acid or base added, the pH change in D will be smaller, making it more effective in resisting pH change.

5 © Raffles Institution 2025 9729/J/25 11 Ans: B A Incorrect. CH3CH2CH2CH3 + Cl2 → CH3CH2CH2CH2Cl + HCl ∆Hr = bonds broken – bonds formed = BE(C–H) + BE(Cl–Cl) – [BE(C-Cl) + BE(H-Cl)] = 410 + 244 – (340 + 431) = –117 kJ mol–1 < 0 i.e. the reaction is exothermic. B Correct. In the propagation step, 2 different alkyl radicals can be formed. + Cl + HCl + HCl In the termination step, the 2 different radicals can combine to form 3-methylheptane. + C Incorrect. The reaction takes place only in the absence of water i.e. reaction will not take place in the presence of aqueous chlorine. D Incorrect. A total of three isomers are formed (including stereoisomers) as 2-chlorobutane is chiral and exists as a pair of enantiomers. 12 Ans: D This reaction involves the nitration of chlorobenzene. A Incorrect. The catalyst is concentrated H2SO4. B Incorrect. The intermediate is Cl H NO2 (or the 2-isomer). C Incorrect. Cl is a deactivating group (refer to the Data Booklet) which makes the ring less electron rich and less reactive towards electrophilic attack. Thus a higher temperature is required for reaction compared to benzene. D Correct. The NO2+ electrophile is produced from the reaction between conc HNO3 and conc H2SO4. 13 Ans: C All three substances are conjugate bases of weak acids. Since the same amount of each substance is dissolved separately in the same volume of water, the higher the