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. The electron clouds in the carbonate anion is being distorted to a smaller extent down the group and the extent of weakening of C–O bond decreases down the group. More energy is required to break the C–O bonds in the carbonate anion. Therefore, the decomposition temperature increases down the group Ans: A 9) Amount of B2O3 = 2.50 10.8 × 2 + 16.0 × 3 = 0.03592 mol Amount of B = 20.03592 = 0.07184 mol Amount of CO2 = 0.80 12.0 + 16.0 × 2 = 0.01818 mol = Amount of C Mol ratio of B : C = 0.07184 : 0.01818 = 4 : 1 Ans: D 10) For zero order reaction, as reaction proceeds, [I2] decreases, rate of reaction (gradient of conc vs time graph) remains constant. Ans: B 11) ∆G = ∆H – T∆S Statement 1 is incorrect. ∆H = +ve and ∆S = −ve, reaction is NOT spontaneous at all temperature. Statement 2 is correct. ∆H = −ve and ∆S = +ve, reaction is spontaneous at all temperature. Statement 3 is incorrect. ∆H = −ve and ∆S = −ve, reaction is spontaneous at low temperature. Ans: C 12) Amount of MnO4− = 25.0 1000 × 0.01 = 0.00025 mol Amount of C2O42− = 25.0 1000 × 0.01 = 0.00025 mol C2O42− is the limiting reagent. Amount of CO2 produced = 2 × 0.00025 = 0.00050 mol Maximum volume of CO2 = 0.00050 × 22.7 = 0.01135 dm3 = 11.4 cm3 Rate = dVCO2 dt = gradient of vol vs time graph. Rate at t2 is faster than rate at t1. Ans: C

2020 A level Paper 1 suggested answers 13) For first order reaction, half-life is constant Time [Reactant] [Product] 0 100% 0% 1st t1/2 50% 50% 2nd t1/2 25% 75% 3rd t1/2 12.5% 87.5% After 3 half−lifes (40 mins × 3), 6.00 dm3 of O2 = 87.5% of O2 produced. Maximum volume of O2 = 6.00 × 100 87.5 = 6.857 dm3 Maximum amount of O2 at r.t.p. = 6.857 24.0 = 0.2857 mol Initial amount of H2O2 = 0.2857 × 2 = 0.5714 mol Initial conc of H2O2 = 0.5714 2001000⁄ = 2.857 mol dm−3 Ans: D 14) Kc = [R] ´ [S]2 [P] ´ [Q] Experiment 1 Experiment 2 Experiment 3 Value of Kc 0.0375 0.0510 0.0375 Comparing experiments 1 and 2, when temperature of the system increases, the value of Kc increases. This shows that the position of equilibrium shifter to the forward reaction at higher temperature. By Le Chatelier’s Principle, endothermic reaction is favoured at higher temperature to absorb the excess heat. Hence, we can conclude that the forward reaction is endothermic. Ans: C 15) A suitable indicator for acid base titration should have a working range pH that coincide with the region of sharp change of pH at the equivalence point. For first equivalence point : sharp change in pH is 2.5-7, the best indicator is naphthyl red For second equivalence point : sharp change in pH is 8-11, the best indicator is thymol blue Ans: B 16) Option A is wrong. Point X is the reaction intermediate (product of step 1) Option B is correct. The reaction pathway diagram shows that step 1 has a higher activation energy than step 2. Option C is wrong. Step 1 is endothermic but that involves the breaking of C=O  bond and formation of C−C bond. Option D is inconclusive. Based on the reaction pathway diagram alone, we cannot deduce whether the reaction is reversible or not. Ans: B

2020 A level Paper 1 suggested answers 17) Option A is wrong. C4H10 has two constitutional isomers: CH3CH2CH2CH3 and CH3CH(CH3)CH3. Option B is wrong. But-1-ene and pent-1-ene are not isomers, they have different molecular formula. Option C is correct. Constitutional isomers that are functional group isomers have different chemical properties. E.g. CH3CH2COOH (carboxylic acid) and CH3COOCH3 (ester) Option D is wrong. Isomers have the same molecular formula. Ans: C 18) Androstenolone undergoes mild oxidation with cold KMnO4, addition of two −OH groups across C=C. OCH3 CH3 OH OH OH * ** * * * * * Ans: D 19) Statement 1 is wrong. The C−Cl bond is weaker than C−F bond, the major free radical formed during initiation should be Cl• and •CClF2. Statement 2 is correct. The halogen radical takes part in the reaction and is regenerated at the end of the reaction. Statement 3 is wrong. Halogen radical is formed during the initiation and propagation step. Termination step produces molecules, not radical. Ans: D 20) Statement 1 is wrong. CO2 is produced in the reactions in the catalytic converter. Statement 2 is correct. NO + CO ⎯→ CO2 + ½ N2. Statement 3 is correct. Unburnt hydrocarbon undergoes oxidation to give CO2 + H2O. Ans: B 21) Chlorobenzene is least reactive towards nucleophile due to the two reasons. 1) C atom of C–Cl is electron rich due to delocalized π electrons of benzene. This repels nucleophile. 2) Partial double bond character of C–Cl bond due to the overlapping of the p orbital of Cl with the  electron cloud of the benzene ring. Ans: C

2020 A level Paper 1 suggested answers 22) V contains an aldehyde gro