2022 TJC Prelims H2 Paper 1 MCQ Ans Only

2 9729 / TJC Prelims / 2022 TJC 2022 H2 Chemistry Paper 1 Answer Key 1 2 3 4 5 6 7 8 9 10 D C A B D B B C A A 11 12 13 14 15 16 17 18 19 20 D C B D B A C A B D 21 22 23 24 25 26 27 28 29 30 C B D C A D C B C A 1 Answer: D Worked Solution: Number of neutrons in 31P3– = 31 – 15 = 16 Number of neutrons in 32S2– = 32 – 16 = 16 Thus, 31P3– are 32S2– are isotonic. Angle of deflection of species in electric field  𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 31P3– has a higher charge and a smaller mass than 32S2–. The angle of deflection of 31P3– will be greater than that for 32S2–. 2 Answer: C Worked Solution: 3rd electron of B comes from an inner shell, B has 3 valence electron and is from Group 13. C is from Group 15. 3 Answer: A Worked Solution: Ga Ge As Se Br Kr Rb Sr 200 400 600 800 1000 1200 1400 30 31 32 33 34 35 36 37 38 39 First Ionisation Energy Atomic Number

3 9729 / TJC Prelims / 2022 1: correct, from diagram, IE of Ga < Ge < As 2: wrong, Se < As < Br 3: correct, Se < Br < Kr 4: wrong, Sb < As < P 4 Answer: B Worked Solution:  A O is the most electronegative element in Group 16. Thus, the C=O bond is the most polar amongst the three bonds. ✓ B O is the most electronegative element in Group 16. Since C= Z bond is less polar than C=Y bond, this suggests that Y is more electronegative than Z. (See diagrams above.)  C While this statement is true, it does not explain the observed trend.  D Bond length has little bearing on net molecular dipole. 5 Answer: D Worked Solution: SO2 has bent shape, bond angle < 120o. SO3 has trigonal planar shape, bond angle is 120o. 6 Answer: B Worked Solution: PV= nRT A. Incorrect. The number of moles of L is greater than number of moles of M B. Correct. At constant T, PV= n x constant C. Incorrect. At constant T, V = nRT/P, straight line pass through origin, different gradient for L & M D. Incorrect. At constant P, V = (nR)/p x T = constant x T net molecular dipole = 0 equal bond polarity C OO C YO net molecular dipole = 0.71 C=O bond is more polar than C= Y bond, which results in net molecular polarity. C ZO net molecular dipole = 0.73 C=O bond is more polar than C= Z bond, which results in net molecular polarity. However, C= Z bond is less polar than C= Y bond, which results in a larger net molecular dipole.

4 9729 / TJC Prelims / 2022 7 Answer: B Worked Solution: element pH of solution when oxide of element is added to water pH of solution when chloride of element is added to water X Na Na2O + H2O → 2NaOH pH = 13 NaCl(aq), pH = 7 Y P P4O10 + 6H2O → 4H3PO4 pH = 2 PCl5 + 4H2O → H3PO4 + 5HCl pH = 2 Z Al Al2O3 is insoluble in water pH = 7 AlCl3 undergoes hydration and hydrolysis. pH of solution is about 3. 8 Answer: C Worked Solution: Ionic radii of group 2 cations increases down the group. Charge density and polarising power decrease down the group, thermal stability increases down the group. 9 Answer: A A. Incorrect statement - Chlorine has a greater reducing strength than bromine. Bromine will not be able to displace chloride. B. AgCl is insoluble in nitric acid. C. Chlorine oxidises thiosulfate to sulfate, which forms white ppt of barium sulfate. D. Chlorine displaces iodide, brown iodine solution is formed. 10 Answer: A Worked Solution: Let 𝑉𝐶𝐻4 cm3 be the volume of CH4 in mixture.  Volume of CO in mixture = (15 – 𝑉𝐶𝐻4) cm3 CH4 : O2 : CO2 mole ratio 1 : 2 : 1 volume ratio 𝑉𝐶𝐻4 : 2 𝑉𝐶𝐻4 𝑉𝐶𝐻4 CO : O2 : CO2 mole ratio 1 : 1 2 : 1 volume ratio 15 – 𝑉𝐶𝐻4 : 1 2 (15 – 𝑉𝐶𝐻4) : 15 – 𝑉𝐶𝐻4 Since total volume of CH 4 and CO combusted = 15 cm 3, and equimolar ratios of CH 4 to CO2, and of CO to CO2 are formed, total volume of CO2 formed = 15 cm3 Total volume of O2 added into reaction vessel = (𝑉𝑂2,𝑟𝑒𝑎𝑐𝑡𝑒𝑑 + 𝑉𝑂2,𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑) cm3

5 9729 / TJC Prelims / 2022 15 + 𝑉𝑂2,𝑟𝑒𝑎𝑐𝑡𝑒𝑑 + 𝑉𝑂2,𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = 15 + (15 + 𝑉𝑂2,𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑)  𝑉𝑂2,𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = 15 cm3  2 𝑉𝐶𝐻4 + 1 2 (15 – 𝑉𝐶𝐻4) = 15  𝑉𝐶𝐻4 = 5 cm3, which is 5 15  100 = 33.3 % of the mixture 11 Answer: D Worked Solution: Hr = ∑ mHf(pdts) − ∑ nHf(rxts) Hf(MgCl2) = Hr + Hf(HCl) The standard enthalpy change of formation of elements in their standard states at 298 K and 1 bar is, by definition, zero 12 Answer: C Worked Solution: The rate of the reaction depends on the rate of the slow step, i.e. rate = k’ [B] [C] However, the rate equation for the overall reaction should be in terms of reactant concentrations only. (Note that C and D are reaction intermediates.) Since step 1 of the mechanism is an equilibrium, 2 A ⇌ C, we can write an expression for its equilibrium constant, Kstep 1 = [C] [A]2 Rearranging, [C] = Kstep 1 [A]2 The rate equation then becomes Rate = k’ [B] (Kstep 1 [A]2) = k [A]2 [B] where k = k’  Kstep 1 CH4 + CO + O2 2CO2 + 2H2O 15 cm3 15 + ( + ) + = 15 + Total gas volume before reaction Total gas volume after reaction

6 9729 / TJC Prelims / 2022 13 Answer: B Worked Solution: Graph of [W] against time is a downward-sloping straight line with constant gradient.  Rate of reaction is independent of [W].  Reaction is zero order wrt [W]. [W] / mol dm−3 [V] / mol dm−3 Rate of reaction / mol dm−3 min−1 0.10 1.0 0.10 20 = 0.0050 0.10 2.0 0.10 10 = 0.010 When [V]  2, while [W] is unchanged, rate of reaction  2  Reaction is first order wrt [V]. Order of reaction wrt [X] cannot be determined from the graph.  Only Statements 1 & 2 are correct. 14 Answer: D Worked Solution: BiCl3(aq) + H2O(l) ⇌ BiOCl(s) + 2HCl(aq) Initial amt / mol 0.1 − − 0 Change in amt / mol −0.02 − +0.02 +2(0.02) Eqm amt / mol 0.08 − 0.02 2(0.02) 𝐾𝑐 = [𝐻𝐶𝑙]2 [𝐵𝑖𝐶𝑙3] = (2 × 0.02 2 )2 0.08 2 15 Answer: B Worked Solution: When temp increases, by LCP, position of eqm shifts to the right to favour the endothermic reaction to absorb heat. So K a increases. Since K a = α2c, degree of dissociation also increases. 16 Answer: A Worked Solution: Solubility product is only affected by changes in temperature.

7 9729 / TJC Prelims / 2022 17 Answer: C Worked Solution: Products formed: 18 Answer: A Worked Solution: Due to the delocalisation of  electrons, the  electron cloud in benzene is less susceptible to an electrophilic attack than a localised  electron cloud in cyclohexene. Thus, cyclohexene reacts readily with a weak electrophile like a polarised Br 2, while benzene requires a Lewis acid catalyst to generate a stronger electrophile, Br +, before a reaction with bromine can occur. 19 Answer: B Worked Solution: Isomer A and B are the same. No. of constitutional isomers: 3 A B C D Primary chloroalkane secondary chloroalkane secondary chloroalkane

8 9729 / TJC Prelims / 2022 20 Answer: D Worked Solution: Each mole of dichloroethane provides 2x the amount of AgC l as chloroethane, for the same amount of time of the experiment. For bromoethane and iodoethane, the rate of reaction is faster due to weaker C −X bond, but the amount of AgX produced is the same as chloroethane, so the graph should be: 21 Answer: C Worked Solution: −OH ≡ Na ≡ ½H2 −CO2H ≡ NaHCO3 ≡ CO2 alcohol √ X phenol √ X acid √ √ Option H2 from Na CO2 from NaHCO3 A 1 0 B 1 0 C 1 1 D 1 2 22 Answer: B Worked Solution: A × 2,4-DNPH: positive test for carbonyl group in vanillin and cinnamaldehyde B √ Fehling's solution: positive test for aliphatic aldehyde group in cinnamaldehyde and negative for benzaldehyde in vanillin C × Hot acidified K 2Cr2O7: positive test for aldehyde group in vanillin and cinnamaldehyde. D × Tollens’ reagent: positive test for both

9 9729 / TJC Prelims / 2022 23 Answer: D Worked Solution: Order of decreasing acidity: carboxylic acids > phenols > alcohols [Refer to page 5, Section 3.1 (Comparison of Acid strength between RCOOH, phenol and alcohols) in the Carboxylic Acids lecture no