NJC 2022 A Level Chem Solution

H2 Chemistry TYS 2022 suggested answers Paper 1 Answer Key 1 C 11 D 21 A 2 C 12 C 22 B 3 D 13 B 23 B 4 C 14 B 24 B 5 A 15 C 25 D 6 B 16 C 26 D 7 A 17 A 27 A 8 B 18 C 28 A 9 B 19 B 29 D 10 D 20 30 D 2022 A level Paper 1 suggested answers 1 Ans: C NaF is an ionic compound with strong ionic bonds between Na+ and F− ions. Largest amount of energy is required to overcome strong ionic bond and it has the highest boiling point. CH3CH2CH3, CH3CH2NH2 and CH3CH2OH are covalent compounds with simple molecular structure. CH3CH2CH3 is non-polar with instantaneous dipole-induced dipole (id-id) interaction between its molecules while CH3CH2NH2 and CH3CH2OH are polar with id-id and hydrogen bonding (H-bond) between molecules. Least amount of energy is required to overcome the id-id in CH3CH2CH3 and it has the lowest boiling point. As O is more electronegative than N, the dipole of O−H bond is greater than N−H bond. More energy is required to overcome the stronger H -bond between CH3CH2OH molecules as compared to CH3CH2NH2. Hence boiling point of CH3CH2OH is greater than CH3CH2NH2. 2 Ans: C Electronegativity decreases down the Group. Cl is more electronegative than Br. Statement A is incorrect. Statement B is correct but it does not help to explain the polarity of the bond. Outer shell electrons in Br experience greater shielding effect due to greater number of INNER shell electrons than Cl. Down the Group, the increase in shielding effect and increase in distance of outer shell electron from the nucleus outweighs the increase in nuclear charge. Nuclear attraction towards outer shell electrons decreases down the Group and hence Br is less electronegative than Cl, leading to a polar covalent bond between Br and Cl. Statement C is correct and explain the polarity. Statement D is incorrect, the repulsion between electrons in outer shell is negligible. E.g. repulsion between 4s and 4p electrons. 3 Ans: D pKa = −lgKa, hence Ka = 10−pKa. Data suggests that thioacetic acid has a larger Ka value and dissociates into H+ more readily than ethanoic acid. Statements 1&2 are incorrect. If the solutions are of the same concentration, thioacetic acid (larger Ka) will give a higher [H+]. We can use this information to deduce that the S−H bond is broken more readily than O−H bond. Bond energy values in Data Booklet also support this (O−H 460 kJ mol−1 vs S−H 347 kJ mol−1) Statement 3 is correct.

4 Ans: C pV = nRT (Note: p in Pa, V in m3, T in K) total mol of gas, n = pV RT = 3.55 ´ 107 ´ 5 ´ 10-3 8.31 ´ (20+273) = 72.9 mol Since the gas contains equal amount of O2 and N2O, amount of N2O = 36.45 mol Mass of N2O used = 36.45  (14.0  2 + 16.0) = 1604 g  1.60 kg 5 Ans: A Solution of AlCl3 is pH 3 due to significant cation hydrolysis, dissociated H+ can react with Na2CO3 to give effervescence of CO2(g). Solution of MgCl2 is pH 6.5 due to slight cation hydrolysis, it is weakly acidic. It might or might not react with a weak base Na2CO3. Solution of NaCl is pH 7, no reaction with Na2CO3. The best available option is A, only AlCl3 6 Ans: B Electronegativity decreases down the Group. Since X is more electronegative than As, X must be P (Phosphorus). Electronegativity increases across the Period. Since Y is more electronegative than P, Y must be S (Sulfur), proton number 16. 7 Ans: A In 100 g of solder glass, mass of B2O3 = 16 g, mass of PbO = 84 g. Amount of B2O3 = 16 10.8 ´ 2 + 16.0 ´ 3 = 0.2299 mol Amount of B = 2  0.2299 = 0.4598 mol Amount of PbO = 84 207.2 + 16.0 = 0.3763 mol Amount of Pb = 0.3763 mol Mol ratio of Pb : B = 0.3763 0.4598 = 0.818 8 Ans: B Sixth I.E. involves the removal of the 6th most loosely held electron. X5+(g) ⎯→ X6+(g) + e− For Group 15 elements, the sixth I.E. would be significantly larger than that of Group 16 elements. The sixth electron for a Group 15 element would come from an inner principal quantum shell and experiences stronger nuclear attraction, thus a large amount of energy is required to remove the 6th electron as compared to Group 16 elements. We can conclude that J and M are from Group 15 while G and H are from Group 16. I.E. decreases down the Group as the electrons experiences weaker nuclear attraction. Hence G is from Period 4 and H is from Period 3. 9 Ans: B Cs+, I− and Xe are isoelectronic species (same number of electrons). The nuclear attraction towards the most loosely held electron is affected by the number of protons in the species. Since proton number increases from I− to Xe to Cs+, most energy is required to remove the most loosely held electron in Cs+. ∆H1 > ∆H3 > ∆H2

10 Ans: D |L.E.| ∝ |q+ ´ q- r+ + r- | The product of charges (q+  q−) is the same for the 4 compounds. Hence, we should compare the inter-ionic distance of the 4 ionic compounds, Zn2+ and Cl− has the smallest inter-ionic distance, and this would mean that the lattice energy for ZnCl2 is expected to be the most exothermic. 11 Ans: D ∆S for the reaction is a positive value as the number of gas particles increases after the reaction. ∆G = ∆H – T∆S With ∆H = −ve and ∆S = +ve, the reaction is spontaneous at all temperature. 12 Ans: C Information shows that PbO2 function as a catalyst for the decomposition of H2O2. - Rate of reaction (gradient of vol vs time graph) is faster in presence of PbO2. - PbO2 remains chemically unchanged at the end of the reaction. We can conclude that activation energy is lowered for experiment 2. And this caused the rate constant, k, to be higher for experiment 2. FYI: k = A𝑒−𝐸𝑎 𝑅𝑇 13 Ans: B 2SO2(g) + O2(g) 2SO3(g) Initial /mol 2.00 2.00 0 Change /mol −1.80 −0.90 +1.80 Eqm /mol 0.20 1.10 1.80 Kc = [SO3]2 [SO2]2 ´ [O2]⬚ = [1.80 0.5] 2 [0.20 0.5] 2 ´ [1.10 0.5] ⬚ = 36.8 mol−1 dm3 14 Ans: B For a buffer solution, pH = pKa + lg[HCO3 -] [H2CO3] 7.4 = −lg(2.5  10−4) + lg [HCO3 -] [H2CO3] 3.8 = lg [HCO3 -] [H2CO3] [HCO3 -] [H2CO3] = 6280 [H2CO3] [HCO3 -] = 1.59  10−4

15 Ans: C AgCl(s) + aq Ag+(aq) + Cl−(aq) ---- eqm 1 When NH3 is added, Ag+(aq) reacts with NH3(aq) to form the soluble complex ion [Ag(NH3)2]+. This causes [Ag+(aq)] to decrease, by LCP, position of eqm 1 would shifts to the right to partially offset the decrease in [Ag+(aq)], and solubility of AgCl(s) increases. When NaCl(aq) is added, this causes [Cl−(aq)] to increase, by LCP, position of eqm 1 would shifts to the left to partially offset the increase in [Cl− (aq)], and solubility of AgCl(s) decreases. 16 Ans: C When Ag2SO4(s) is dissolved in water, Let the solubility of Ag2SO4(s) be s mol dm−3 Ag2SO4(s) + aq 2Ag+(aq) + SO42−(aq) Initial /mol dm−3 - 0 0 Change /mol dm−3 −s - +2s s Eqm /mol dm−3 - 2s s Given that [Ag+(aq)] = 0.032 mol dm−3, s = 0.016 Ksp of Ag2SO4(s) = [Ag+]2  [SO42−] = (0.032)2  (0.016) = 1.638  10−5 mol3 dm−9 In the second eqm when [SO42−(aq)] = 0.50 mol dm−3, the resulting solution is saturated with Ag2SO4(aq). Ionic Product = Ksp [Ag+(aq)]2  [SO42−] = 1.638  10−5 [Ag+(aq)]2  0.50 = 1.638  10−5 [Ag+(aq)] = 5.72  10−3 mol dm−3 17 Ans: A Option 1 would lead to formation of a C−Cl sigma bond, the product is (CH3)3C−Cl. Option 2 shows a heterolytic cleavage of polar C−Cl bond, the product is (CH3)3C+, a carbocation. Option 3 shows a heterolytic cleavage of polar C−O bond, the product is (CH3)3C+, a carbocation. Option 4 shows a heterolytic cleavage of polar C−Cl bond, AND formation of C=O pi bond. The product is (CH3)2C=O. 18 Ans: C Option 1 is correct. Enantiomers are stereoisomers that are non-superimposable mirror images of each other. They have different interaction with plane of plane-polarised light and different interaction with stereo specific reactant (requires specific 3D arrangement). Biological receptors are similar to enzymes and only inter