2019 TJC H2 Chem Prelim P1 ANS

1 2019 TJC JC2 H2 Chemistry Prelim MCQ Worked Solutions Question 1 Answer: C Particle Neutron Nucleon Proton Electron U 16 33 17 17 V͞ 18 35 17 18 S2͞ 16 32 16 18 T2+ 17 34 17 15 Q3͞ 16 31 15 18 Question 2 Answer: C CH4 + 2O2 CO2 + 2H2O x 0.95x CH4 + 3/2 O2 CO + 2H2O 0.05x If combustion is complete, x dm3 CH4 will require 2x dm3 O2. Combustion of 1 mol CH4 to form CO, requires 3/2 mol O2 ie ½ mol O2 less than complete combustion. To obtain 0.05x mol CO will need 0.05x/2 mol less O2. Amt of O2 needed for incomplete combustion = 2x – 0.05x/2 Or Volume of methane burnt = y dm3 5% of methane is burnt to give CO and the remaining 95% is burnt to give CO2 Vol of methane burnt to give CO = 0.05y Vol of methane burnt to give CO2 = (1-0.05)y CH4 + 2O2  CO2 + 2H2O Volume used (1-0.05)y 2(1-0.05)y CH4 + 3/2O2  CO + 2H2O Volume used 0.05y 3/2(0.05y) Vol of O2 used = 2(1-0.05)y + 3/2(0.05y) = 2y – 2(0.05)y + 3/2(0.05y) = 2y – ½ (0.05y) Question 3 Answer: D Na : [Ne]3s1 (1 unpaired electron) P3͞ : [Ar] (0 unpaired electron) V : [Ar]3d34s2 (3 unpaired electron) Mn2+: [Ar]3d5 (5 unpaired electron) Question 4 Answer: B  1 2 3 4 5 6 7 8 9 10 C C D B A C C B B A 11 12 13 14 15 16 17 18 19 20 C D A D D A A B D C 21 22 23 24 25 26 27 28 29 30 B D B A C B D B A C 1 1

2 Question 5 Answer: A ICl2͞ : 2BP + 3 LP, shape linear, Bond angle 1800, non-polar CO2 : 2 BP, shape linear, Bond angle 1800 , non-polar BCl3 : 3 BP shape trigonal planar, Bond angle 1200 , non-polar ClO2͞ : 2 BP + 2 LP, shape bent, Bond angle <109.50 , polar HCN : 2 BP, shape linear, Bond angle 1800, polar XeF4 : 4 BP + 2 LP, shape square planar, Bond angle 900, non-polar Question 6 Answer: C I Both diamond and silicon have giant molecular structures. The atomic radius of C is smaller than Si. C-C bond length is shorter than Si-Si bond. Diamond has a higher mp than silicon as more energy is needed to break the stronger C-C bonds. II H2O is able to form more hydrogen bonds on average compared to NH3 as it has 2 lone pair of electrons on O. III SiCl4 has a simple molecular structure and Al2O3 has a giant ionic structure. The id-id attractions between SiCl4 molecules is weaker than the strong ionic bonds in Al2O3. IV Both Br2 and ICl have similar Mr. The id-id attractions between Br2 molecules is weaker than the pd-pd attractions between ICl molecules. Question 7 Answer: C 1 Structure of dimer Al2Cl6 Shape of AlCl3 = Trigonal planar hence bond angle is 120o. Shape of Al2Cl6 = Tetrahedral about each Al atom hence bond angle is not 120o. 2 Using ideal gas equation, pV = nRT pV = (m/M) RT where M = molar mass (1.16 x 105) (250 x 10–6) = (x / 214.9) (8.31)(500) X = 1.50 g 3 Size of electron cloud for : Al2Cl6 > AlCl3 Extent of distortion of electron cloud for: Al2Cl6 > AlCl3 Strength of intermolecular id-id : Al2Cl6 > AlCl3 Deviation from ideality for: Al2Cl6 > AlCl3 Question 8 Answer: B 3C(s) + Cr2O3(s)  2Cr(s) + 3CO(g) ∆H = +790 kJ mol−1 ∆Hrxn = ∑∆Hf(products) – ∑∆Hf(reactants) +790 = 3 ∆Hf(CO) – ∆Hf (Cr2O3) +790 = 3 ∆Hf(CO) – (–1120) ∆Hf(CO) = (+790 – 1120) / 3 = –110 kJ mol−1 By definition, C(s) + 1 2 O2(g)  CO(g) ∆Hf(CO) = –110 kJ mol−1 Hence, 2C(s) + O2(g)  2CO(g) 2 ∆Hf(CO) = –220 kJ mol−1

3 Question 9 Answer: B Energy cycle for ∆Hsoln q Hg2SO4 ∆Hsoln o (Hg2SO4) Hg2SO4(s) 2Hg+(aq) + SO42-(aq) -∆Hlatt q 2∆Hhyd q (Hg+) ∆Hhyd q (SO42-) (Hg2SO4) 2Hg+(g) + SO42-(g) By Hess’ law: ∆Hsoln q (Hg2SO4) = -∆Hlatt q (Hg2SO4) + 2∆Hhyd q (Hg+) + ∆Hhyd q (SO42-) = -(-2127) + 2(-625) + (-1160) = - 283 kJ mol-1 Since the calculated value of ∆Hsoln q (Hg2SO4) is a negative value at r.t.p., it means that: 1. Correct. Magnitude is 283 kJ mol-1 2. Correct. ∆Hlatt q ∝ 𝑞+𝑞− 𝑟++ 𝑟− , since Hg+ has a bigger ionic radius as compared to Cd+ (q+, q-, r- same), ∆Hlatt q of Hg2SO4 is less exothermic than that of Cd2SO4. 3. Incorrect. ∆Gsoln o of Hg2SO4 is less negative than ∆Gsoln o of Cd2SO4 since Hg2SO4 is less soluble than Cd2SO4. Question 10 Answer: A E q cell = E q red – E q oxi 0.3 V = EoCu2+/Cu – EoX2+/X 0.5 V = EoCu2+/Cu q – EoY2+/Y 0.4 V = EoZ2+/Z – EoCu2+/Cu q From the data in the terms of increasing positive value: EoZ2+/Z > EoCu2+/Cu > EoX2+/X > EoY2+/Y Tendency to be reduced for : Z2+, Cu2+, X2+, Y2+ Thus, the strongest to the weakest oxidising agents is Z2+, Cu2+, X2+, Y2+ Question 11 Answer: C NH4+ + 3H2O E q = 0.87 V -- (1) NO3– + 10H+ 8e NO2 + H2O E q = 0.81 V --(2) NO3– + 2H+ + e As [H+] increases. By Le Chatelier’s Principle, the position of both equilibria shifts to the right to decrease [H +], favouring reduction Oxidising ability of NO3– increases while reducing abilities of NO2 and NH4+ decreases. Since for eqm (2), a lesser number of moles of H+ is required for one mole of NO3– to be reduced, there is greater tendency for equilibrium (2) to shift to the right hence backward reaction is less favoured. Hence, NO 2 is a weaker reducing agent where [H+] = 10-5 mol dm-3 < standard condition of 1 mol dm-3). Question 12 Answer: D Rate would be slower as concentration of nitric acid is lower. Amount of products would be lesser as nitric acid is the limiting agent and there are less amount of nitric acid used in the 2nd scenario (0.025 mol vs 0.030 mol) Question 13 Answer: A Total pressure increases at constant volume but partial pressure of individual product and reactant remains constant. So position of equilibrium will NOT shift at all.

4 Question 14 Answer: D [H+] = √𝐾𝑎𝐶 When the acid is diluted, the conc of acid drops, so [H+] will drop too  pH increases, following a logarithm function since pH = -lg[H+]. However at infinitely dilute condition, [H+]overall = [H+]acid + [H+]water, where [H+]acid << [H+]water, so [H+]overall ≈ [H+]water = 10-7 mol dm-3 at 25 oC. So the pH at infinitely dilute condition will reach a constant value of 7. Question 15 Answer: D Zn3[Fe(CN)6]2 ⇌ 3Zn2+ + 2[Fe(CN)6]3- At eqm, there are 3y mol of Zn2+ and 2y mol of [Fe(CN)6]3-. Ksp = [Zn2+]3 [Fe(CN)63-]2 = (3y)3(2y)2 = 108y5 = W y = √ W 108 5 [Fe(CN)63-] = 2 x √ W 108 5 = √8W 27 5 Question 16 Answer: A pKa value of –COOH = 1.9, –SH = 8.1 and -NH3+ = 10.3. At isoelectric point, only –COO- and –NH3+ exist, so pI = ½ (1.9+8.1) = 5.0 Hence only Bromocresol Green and Methyl Red can detect the isoelectric point of cysteine as the pI is within the working range of both indicators. Question 17 Answer: A X is Al: - Al2O3 has no reaction with water due to high lattice energy. - AlCl3 undergoes hydrolysis with water to give an acidic solution of pH 3 - Al2O3, an amphoteric oxide, reacts with HCl to form AlCl3 (chloride salt) and water. Y is Na - Na2O reacts with water to forms NaOH (formation of hydroxide). - NaCl undergoes only hydration with water to form a solution with pH = 7 - Na2O, a basic oxide, reacts with HCl to form NaCl and H2O. Z is Si - SiO2 does not react with water due to its giant covalent structure - SiCl4 reacts with water to give an acidic solution pH 1 - SiO2 does not react with HCl. Question 18 Answer: B Thermal stability is increased by