2022 RI Prelim P1 (Ans)

1 © Raffles Institution 2022 9729/01/S/22 2022 RI H2 Chemistry Prelim Paper 1 – Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C C A A A B B B D B A A D D C MCQ worked solutions Q1 (Ans: C) Option A is incorrect. (NH4)2Fe(SO4)2 has 5 ions per formula unit  total number of ions = 5  2  6.02  1023 K2Cr2O7 has 3 ions per formula unit  total number of ions = 3  4  6.02  1023 Option B is incorrect. Number of NO2 molecules = 46 14 + (16  2)  6.02  1023 Number of N2 molecules = 14 14  2  6.02  1023 Option C is correct. Number of electrons in 1 mol of N2 = (7 + 7)  6.02  1023 Number of electrons in 1 mol of CO = (6 + 8)  6.02  1023 Option D is incorrect. Let Vm be the molar volume of gas. At the same temperature and pressure, number of atoms in 5 dm3 of O2 = 5 Vm  2  6.02  1023 number of atoms in 10 dm3 of Ar = 10 Vm  6.02  1023 Q2 (Ans: C) Let a be the fraction of 63Cu and (1 – a) be the fraction of 65Cu in naturally occurring copper. From the Data Booklet, the relative atomic mass of Cu is 63.5. 63.5 = 63a + 65(1 – a)  a = 0.75 Hence, in the sample of bronze, % composition of Cu63 % composition of Cu65 = d 88 – d = 0.75 0.25 d = 66 Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C D B A B D C C D C D B B A D

2 © Raffles Institution 2022 9729/01/S/22 Q3 (Ans: A) 30Si4+ 31P3− 32S2− Number of electrons 10 18 18 Number of neutrons 16 16 16 Number of protons 14 15 16 Statement 1 is correct. The ions have the same number of neutrons. Statement 2 is incorrect. As the ions have different number of electrons, their electronic configurations are different. Statement 3 is incorrect. The ionic radii increase in the order 30Si4+ < 32S2− < 31P3−. The valence electrons of 30Si4+ are found in the n=2 shell while the valence electrons of 31P3– and 32S2− are found in the n=3 shell. Hence the ionic radius of 30Si4+ is smaller than that of 31P3– and 32S2−. Since 31P3− and 32S2− are isoelectronic with the same electronic configuration, their valence electrons experience the same shielding effect. However, nuclear charge of 31P3− is smaller than that of 32S2−. Hence, the effective nuclear charge of 31P3− is smaller than that of 32S2−, and therefore 31P3− has a larger ionic radius than 32S2−. Q4 (Ans: A) Angle of deflection  charge mass ions 7Li2− 15O2+ 15N4− 11C3+ charge mass 0.286 0.133 0.267 0.273 Q (11C3+) should be a cation as it is attracted to the negatively charged plate, and its charge/mass ratio should be between that of the two anions P (7Li2−) and R (15N4−). Q5 (Ans: A) Indium is a Group 13 element with 3 valence electrons. Hence, there should be a large jump between the third and fourth IE as the fourth electron is removed from an inner electronic shell. Hence, option A is correct.

3 © Raffles Institution 2022 9729/01/S/22 Q6 (Ans: B) G H Remarks A G has a greater net dipole than H. B P-F bond is more polar than P-Cl bond. G has a smaller net dipole than H. C I-F bond is more polar than Br-Cl bond. G has a greater net dipole than H. D Both G and H do not have a net dipole. Q7 (Ans: B) Statement 1 is correct. Since compounds M and N have the same electron cloud size, the strength of their intermolecular instantaneous dipole-induced dipole interactions is similar. However, the hydrogen bonds present between molecules of M are stronger than the p ermanent dipole-permanent dipole interactions between molecules of N. Hence, M would have a higher boiling point and lower volatility than N. Statement 2 is correct. Compounds M and N are constitutional isomers as both have the same molecular formula C5H10O but different structural formula. Statement 3 is incorrect. M has 16  bonds whereas N has 15  bonds. (Note: Remember to include the C-H and O-H  bonds.) Q8 (Ans: B) For ionic compounds, the extent of polarisation/distortion of the electron cloud of the anion and hence the degree of covalent character increases with: • higher polarising power of the cation • higher polarisability of the anion Compared to Ca2+, Mg2+ has a smaller ionic radius and hence a higher charge density and polarising power. Mg2+ is able to distort the electron cloud of the anion to a greater extent. Compared to O2−, S2− has a larger and more polarisable electron cloud. Hence, MgS has the greatest covalent character.

4 © Raffles Institution 2022 9729/01/S/22 Q9 (Ans: D) After mixing at constant temperature Applying p1V1 = p2V2 (since n and T are constant), (20 kPa)(1 m3) = (pHe)(3 m3) pHe upon mixing = 6.67 kPa (10 kPa)(2 m3) = (pNe)(3 m3) pNe upon mixing = 6.67 kPa Total pressure upon mixing = 6.67 + 6.67 = 13.3 kPa After decreasing temperature • Total pressure will decrease below 13.3 kPa since p  T (at constant V and n)  options A and B are incorrect. • pHe will still be equal to pNe as the number of moles of each gas remains the same  option C is incorrect. Q10 (Ans: B) Element D is silicon as it has the highest melting point. Since the elements are consecutively arranged, B is sodium. Q11 (Ans: A) Option A is correct. The solubility of silver halides in aqueous ammonia decreases down the group (due to decreasing Ksp values). Since silver iodide is insoluble in aqueous ammonia, silver astatide is expected to be also insoluble in aqueous ammonia. Option B is incorrect. The melting and boiling points of halogens increase down the group. Since iodine is a solid, astatine is expected to be also a solid at room temperature and pressure. Option C is incorrect. H-X bond energy decreases down the group due to decreasing effectiveness of the valence orbital overlap. Hence the bond energy of HAt is expected to be smaller than that of HI. Option D is incorrect. Oxidising power of halogens decreases down the group. Hence, a halogen can oxidise a halide below (but not above) it. Bromine is expected to be able to oxidise sodium astatide to give astatine. Q12 (Ans: A) Hr = nHc (reactants) − mHc (products) = −3268 + 3(−286) – (−3754) = −372 kJ mol−1 From the calculations, pHe = pNe upon mixing

5 © Raffles Institution 2022 9729/01/S/22 Q13 (Ans: D) Option A is incorrect. The standard enthalpy change of atomisation of chlorine is the energy absorbed when 1 mole of gaseous C l atoms is formed from C l2(g) under standard conditions. The correct equation should be 1 2Cl2(g) ⎯→ Cl(g) Option B is incorrect. The standard enthalpy change of combustion of a substance is the energy released when 1 mole of the substance is completely burnt in excess oxygen under standard conditions. The correct equation should be H2S(g) + 3 2O2(g) ⎯→ H2O(g) + SO2(g) Note: H2 cannot be the end product of a combustion reaction as it can be combusted to give H2O. Option C is incorrect. The standard enthalpy change of formation of a substance is the energy change when 1 mole of the pure substance in a specified state is formed from its constituent elements in their standard states under standard conditions. The correct equation should be H2(g) + 2O2(g) + S(s) ⎯→ H2SO4(l) Note: The elements, h ydrogen and oxygen , do not exist in the monoatomic form under standard conditions (i.e. 1 bar and 298 K). Option D is correct. The standard enthalpy change of solution of a substance is the energy change when 1 mole of the substance (ionic compound) is completely dissolved in a solvent to form an infinitely dilute solution (containing aqueous ions) under standard conditions. Q14 (Ans: D) Comparing the first and third experiments, When [R]  3 while keeping [S] constant, initial rate  3  rate  [R]. Hence, order of reaction with respect to R is 1. Comparing the