NJC 2020 A Level Chem Solution

H2 Chemistry TYS 2020 suggested answers Paper 1 Answer Key 1 C 11 C 21 C 2 D 12 C 22 B 3 B 13 D 23 A 4 D 14 C 24 D 5 A 15 B 25 C 6 B 16 B 26 D 7 B 17 C 27 A 8 A 18 D 28 C 9 D 19 D 29 A 10 B 20 B 30 D 2020 A level Paper 1 suggested answers 1) dx2-y2 orbital dz2 orbital dxy orbital dxz orbital dyz orbital Out of the d orbitals, 4 of them have four lobes. dx2-y2, dxy, dxz, dyz. Ans: C 2) The largest increase in successive I.E. is from the 8th to 9th I.E. This shows that the 9th electron is removed from an inner principle quantum shell which experiences a stronger nuclear attraction. Hence the valence shell contains 8 valence electron and is most likely to be from Group 18. Ans: D 3) Nucleon no. Atomic no. No. of neutrons No. of electrons Po2+ 217 84 133 82 At3+ 217 85 132 82 Rn4+ 217 86 131 82 Fr5+ 217 87 130 82 Ans: B

2020 A level Paper 1 suggested answers 4) The three alkanes are constituent isomers with the same molecular formula of C5H12. Alkanes are non-polar molecules with instantaneous dipole-induced dipole (id−id) attraction between molecules. During melting and boiling, energy is required to overcome the id−id attraction between the molecules. The strength of id−id is stronger for straight chain isomer due to larger surface area for interaction. For the branched-chain isomers, the id−id is weaker and hence lower boiling point. Ans: D 5) Question stated that upon mixing, stronger intermolecular forces are formed between CH3Cl and CH3COCH3 molecules. The mixing process is exothermic and hence the temperature of the mixture increases to above 20 C. More energy would be required to overcome the intermolecular forces during boiling, hence the boiling point of the mixture will be higher than the boiling point of individual compound. (Qn states that mixture has stronger IMF than their original IMF) Ans: A 6) Amount of gas = mass molar mass Molar mass of the gases : CH4 (16.0) < Ne (20.2) < N2 (28.0) < Cl2 (71.0) For ideal gas, pV = nRT, pV vs V graph gives a constant y-axis value (Y = nRT). For equal masses of the four gases, amount of Ne is the second highest. Hence is correspond to line B. Ans: B 7) H2 (g) + I2 (g) 2 HI (g) Initial / mol 0 0 0.04 Change / mol +x +x −2x Equilibrium / mol x x 0.04−2x Total amount of gas at eqm = 0.04 – 2x + x + x = 0.04 mol According to Dalton’s law of partial pressure, PA = nA ntotal mol of gas × Ptotal Kp = (PHI)2 PH2 × PI2 = (0.04 - 2x 0.04 × 1.0)2 (x 0.04 × 1.0)2 54 = (0.04 - 2x)2 x2 7.348 = 0.04 - 2x x 9.348x = 0.04 x = 0.004279 PHI = 0.04 - 2(0.004279) 0.04 × 1.0 = 0.786 atm Ans: B

2020 A level Paper 1 suggested answers 8) Down the group, cations of Group 2 metals have the same charge but their size increases (make reference to their ionic radii). The charge size ratio of the cation decreases. This leads to a decrease in polarising power