RI H2 Chemistry 2020 to 2022 TYS Solutions

© Raffles Institution 2020 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C D B D A B B A D B C C D C B 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 B C D D B C B A D C D A C A D Q1 (C) There are a total of five 3d orbitals. However, only the 3 dx2-y2 , 3d xy, 3d xz and 3d yz orbitals have four lobes. Q2 (D) Since there is a large jump from the 8 th to 9 th ionisation energy, the element has eight valence electrons and is a Group 18 element, i.e. Ar. Q3 (B) When nucleon number = 217, atomic number no. of neutrons no. of electrons Po2+ 84 217 – 84 = 133 84 – 2 = 82 At3+ 85 217 – 85 = 132 85 – 3 = 82 Rn4+ 86 217 – 86 = 131 86 – 4 = 82 Fr5+ 87 217 – 87 = 130 87 – 5 = 82 From the above table, only At3+ has 50 more neutrons than electrons. Q4 (D) The three alkanes are constitutional isomers with the same Mr and number of electrons. All three are simple, non-polar molecules with only instantaneous dipole-induced dipole (id-id) interactions. Since boiling involves overcoming the intermolecular forces of attraction (NOT covalent bonds), the differences in boiling points among the three alkanes are due to differences in the strength of their id-id interactions. Pentane, a s traight–chained hydrocarbon, has greater surface area for intermolecular interactions compared to its branched isomers, 2-methylbutane and 2,2-dimethylpropane. Thus, intermolecular id-id interaction is the strongest in pentane. As branching increases from 2 -methylbutane to 2,2- dimethylpropane, the surface area for intermolecular interaction decreases. Hence the boiling point decreases from pentane to 2–methylbutane to 2,2–dimethylpropane. Q5 (A) Both liquids are initially at 20 oC. Since stronger intermolecular forces between CHC l3 and CH3COCH3 are formed (compared to their original intermolecular forces), energy is released upon mixing and initial temperature of the mixture will be above 20 oC. Since the intermolecular forces between CHCl 3 and CH3COCH3 are stronger (than their original intermolecular f orces) and require more energy to overcome, the boiling point of the mixture will be above 61 oC. Q6 (B) Since all four gases behave as ideal gases and temperature is kept constant, pV = nRT = constant Graph of pV against V is a horizontal straight line. Hence, the gas with a larger n will have a larger pV. Since Mr of CH4 < Ne < N2 < Cl2, for equal masses of the four gases, amount, n, of CH4 > Ne > N2 > Cl2. Hence, graph B corresponds to Ne.

© Raffles Institution Q7 (B) H2(g) + I2(g) ⇌ 2HI(g) initial amt / mol 0 0 0.040 change in amt / mol +x +x –2x eqm amt / mol x x 0.040 – 2x At equilibrium, ntotal = x + x + (0.040 – 2x) = 0.040 mol Ptotal = 1.0 atm Kp = PHI 2 PH2PI2 = ( 0.040 – 2x 0.040 × 1.0) 2 ( x 0.040 × 1.0)2 = 54 x = 0.004279 mol PHI = 0.040 – 2(0.004279) 0.040 × 1.0 = 0.79 atm