2025 RI Prelims H2 Chemistry P1 suggested solutions

1 © Raffles Institution 2025 9729/01/S/25 2025 Y6 H2 Chemistry Preliminary Exams Paper 1 – Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer A D C C B D C B B A D B A A C MCQ worked solutions Q1 (Ans: A) The 37Cl+ ion has 37 nucleons which comprise 17 protons and 20 neutrons. It has one positive charge (having lost one electron) and hence it has a total of 16 electrons. Q2 (Ans: D) angle of deflection ∝ ቚq mቚ For 17O+, ቚq mቚ = ଵ ଵ଻ (= 0.0588) For [16O18O]2–, ቚq mቚ = ଶ (ଵ଺ାଵ଼) = ଵ ଵ଻ (= 0.0588) Hence [16O18O]2– will be deflected to the same extent as 17O+. Option 1 is incorrect. Option 2 is correct since positively charged species will deflect in an opposite direction to negatively charged species. Each beam of charged particles will travel in a curved path, not a straight path. Option 3 is incorrect. Q3 (Ans: C) The cis-isomer (Q) has a higher boiling point as it has a net dipole moment and has permanent dipole-permanent-dipole interactions, whereas the trans-isomer (P) is non-polar. R has instantaneous dipole-induced dipole interactions, as well as the smallest electron cloud and hence has the weakest interactions amongst the three. Thus, it has the lowest boiling point. Correct order of increasing boiling point: R < P < Q. Q4 (Ans: C) BeF2 acts as a Lewis acid (as NH3 donates its lone pair to Be to form H3NBeF2). A is incorrect. It is tetrahedral around the N atom in BeF2 • NH3, hence the molecule is not planar. B is incorrect. In BeF42–, two F– ions form one co-ordinate bond each with BeF2. C is correct. It is possible for the lone pair on F to form a hydrogen bond with the H in H2O. D is incorrect. Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C B C B D B D B C A C D A D A

2 © Raffles Institution 2025 9729/01/S/25 Q5 (Ans: B) pV = nRT p = 1 V (nRT) Since n, R and T are constants, p = 1 V (k) Hence graph is y = kx (straight line that passes through origin) Option A pV = nRT Since n, R and T are constants, pV = constant at a particular T Option C pV = nRT p = mass MV RT = density(RT M) 1 T=density(R Mp) Since R, M and p are constants, 1 T= density(k) Option D pV = nRT pV T= nR Since n and R are constants, pV T= constant at a particular n density 1 T pV T p p pV T pV 0 0 0

3 © Raffles Institution 2025 9729/01/S/25 Q6 (Ans: D) element identity X Si Y Na Z P ✘ A Y (Na) has a smaller atomic radius than Z (P) ✘ B Y (Na) has a lower electrical conductivity than X (Si) ✘ C X (Si) has a higher first ionisation energy than Z (P) ✔ D X (Si) has a higher electronegativity value than Y (Na)

4 © Raffles Institution 2025 9729/01/S/25 Q7 (Ans: C) A E value becomes more negative down the Group. M2+(aq) + 2e ⇌ M(s) E The tendency of backward reaction occurring increases, hence reducing power of metal increases. B The reactivity of Group 2 elements increases down the group as the ease of Group 2 elements losing electrons increases down the group. C Cationic radius increases down the Group, resulting in a lower charge density and weaker polarising power of the cations. Consequently, there is decreasing extent of distortion of the electron cloud of the CO32– anion and hence decreasing extent of weakening of covalent bonds within the CO32– anion. More heat energy is required to break the covalent bonds within the CO32– anion, causing the decomposition temperature to increase. Hence, the thermal stability of the Group 2 carbonates increases. D As cationic radius increases down the Group, |𝐻୦୷ୢ୭ [M+(g)] |  | ୯శ ୰శ | the magnitude of the enthalpy of hydration of the metal ion decreases. Q8 (Ans: B) Let the nitrogen-containing compound be X. Since the mole ratio of NO2 to HNO3 is 3:2, the remaining N atom is in X. Hence, the redox equation involving only the N compounds is: 3 NO 2 → X + 2 HNO3 2 moles of NO2 (where N has an oxidation state of +4) is oxidised to 2 moles of HNO3 (where N has an oxidation state of +5) by the loss of 2e-. Therefore, the remaining 1 mole of NO2 gains 2e- when it is reduced to 1 mole of X. Thus, the oxidation of N in X is +2. Hence, X is NO. Alternatively, the redox equation can be balanced by stoichiometry to deduce the identity of X. Q9 (Ans: B) Ar of Cr = (4.3)(50) + (83.8)(52) + (9.5)(53) + (2.4)(54) (4.3 + 83.8 + 9.5 + 2.4) = 52.06 = 52.1 Note that the sum of the relative abundances may not always add up 100. Hence, it’s necessary to divide the numerator by the sum of the relative abundances given. Q10 (Ans: A) ✔ 1 Sublimation requires energy to overcome the attractive forces holding the particles together in a solid state to change into a gaseous state. ✘ 2 The combustion of all fuels is always exothermic. ✘ 3 The formation of ion-dipole interactions releases energy, making the process exothermic.

5 © Raffles Institution 2025 9729/01/S/25 Q11 (Ans: D) Theoretical lattice energy can be calculated using |lattice energy|  | ୯శ୯ష ୰శ ା ୰ష | which assumes a 100% ionic nature. The discrepancy between the experimental and theoretical lattice energies shows the presence of covalent character in the bonding in the silver halides, arises due to substantial polarisation of the anion by the cation. Since the anionic radius increases down the Group, the polarisability of the anion increases, causing a greater degree of covalency and hence greater deviation of the experimental from the theoretical value. Q12 (Ans: B) A possible method to solving this question without an energy cycle is to use the algebraic method (refer to section 4.7 of your lecture notes for Energetics Part 1). enthalpy change / kJ mol −1 Ba(s) + O2(g)  BaO2(s) s BaO2(s)  Ba2+(g) + O22−(g) −q Ba2+(g) + 2e−  Ba(g) −p Ba(g)  Ba(s) −r O2(g) + 2e−  O22−(g) s – q – p – r Q13 (Ans: A) Statement A is correct. Adding a catalyst to increase the rate of reaction increases the value of the rate constant (by decreasing the activation energy). Statement B is incorrect. In autocatalytic reactions, the concentration of the catalyst increases as the products are formed, thus increasing the rate of reaction initially. However, as the concentration of the reactants decrease, the rate of reaction also decreases. Statement C is incorrect. Heterogeneous catalysts are in a different phase from the reactant molecules. Statement D is incorrect. In enzyme-catalysed reactions, increasing the concentration of the substrate increases the rate of the reaction until all the active sites are saturated, after which, the concentration of the substrate has no effect on the rate of the reaction.

6 © Raffles Institution 2025 9729/01/S/25 Q14 (Ans: A) At constant temperature, pressure is directly proportional to concentration, since pV=nRT => P RT= n V => p ∝ n V . At very low pressures of A, the k2[A] term in the denominator becomes negligible in comparison with the k3 term. rate = k1k3[A]2 k3 = k1[A]2 Hence, the rate equation is second order with respect to A. At very high pressures of A, the k3 term in the denominator becomes negligible in comparison with the k2[A] term. rate = k1k3[A]2 k2[A] = k1k3 k2 [A] Hence, the rate equation is first order with respect to A. Q15 (Ans: C) The end-point volume, V, is independent of temperature as it depends only on the concentrations of HA and NaOH. The pH at 1 2V is equal to the pKa of the weak acid, HA. As pKa is affected by temperature, its value will decrease at a higher temperature (as the dissociation of the weak acid is endothermic). Hence, the pH at 1 2V also decreases. Q16 (Ans: C) When solid CuS