NJC Prelim P1 Solutions

1 NJC H2 Chemistry Prelim Paper 1 Suggested Answers 1 C 6 B 11 B 16 C 21 D 26 D 2 D 7 D 12 D 17 B 22 B 27 B 3 A 8 D 13 A 18 A 23 D 28 A 4 C 9 C 14 C 19 D 24 C 29 C 5 C 10 A 15 C 20 B 25 C 30 C 1 Ans: C SO2 reacts with acidified KMnO4 Total vol of SO2 = 10 cm3 Left over CO2 reacts with NaOH(aq) Vol of CO2 = 2 cm3 Vol of SO2 produced from CS2 = 4 cm3 Vol of SO2 produced from H2S = 10  4 = 6 cm3 Vol of CS2 : Vol of H2S = 2 : 6 mole fraction of H2S in the mixture = 6 8 = 0.750 2 Ans: D Step I: Oxidation state of Cr increases from +3 in Cr2O42− to +6 in CrO42−. Step II: Oxidation state of Cr remains as +6 in both CrO42− and Cr2O72−. 2CrO42− + 2H+  Cr2O72− + H2O Step III: Oxidation state of Cr decreases from +6 in Cr2O72− to +3 in Cr2O3.

2 3 Ans: A 5Fe2+ + MnO4− + 8H+  5Fe3+ + Mn2+ + 4H2O Amount of MnO4− required = 20 1000 × 0.40 = 0.008 mol Amount of Fe2+ = 5 × 0.008 = 0.040 mol At point X, the reaction is only half completed, [Fe2+] = [Fe3+]. Ecell when connected to standard hydrogen electrode = +0.77 V The colour change at end point is from yellow to first permanent pink. Vol KMnO4 added Species present in conical flask Colour of solution 0 Fe2+(aq) Pale green Before eqv pt Fe2+(aq), Fe3+(aq), Mn2+(aq) Yellow-green At eqv pt Fe3+(aq), Mn2+(aq) Yellow Just after eqv pt Fe3+(aq), Mn2+(aq), small amt of MnO4−(aq) Pink In excess Fe3+(aq), Mn2+(aq), large amt of MnO4−(aq) Purple 4 Ans: C Making reference to the I .E. values from Data Booklet, we can conclude that I is potassium. Element C is Al, it is in Group 13. Element F is S and it exists as S8 molecules. The lowest boiling point is Ar gas (element H). Ion of E (P3−, 0.212 nm) is larger than that of J (Ca2+, 0.099nm) Element D and G are Si and Cl respectively. The compound formed is SiCl4. 5 Ans: C The ionic compound can form strong ion-dipole interaction with polar solvent, making it soluble in polar solvents. The long chain hydrocarbon can also form strong temporary dipole-induced dipole interaction with non-polar organic solvents, making it soluble in organic solvents. Around S atom, there are 4 bond pair regions and 0 lone pair, the shape is tetrahedral. All the C atoms are sp3 hybridised with tetrahedral shape and bond angle of 109.5°

3 6 Ans: B pV = nRT, with n and p kept constant. As T increases, Option A: V increases. [ V = nR p ×T] Option B: pV T remains constant. [ pV T = nR] Option C: Density of the gas deceases [ m V = pMr RT ] Option D: pV increases. [pV = nRT] 7 Ans: D Hrxn =  Hfo (products)   Hfo (reactants) −27 = 3×(−394) − [Hfo Fe2O3 + 3×(−110)] Hfo Fe2O3 = −825 kJ mol−1 8 Ans: D The following processes are involved in the Born-Haber cycle for ionic compounds: 1) Enthalpy change of formation of BaF3(s) 2) Enthalpy change of atomisation of Ba(s) 3) Bond energy of F−F 4) 1st + 2nd + 3rd ionisation energies of Ba(g) 5) First electron affinity of F(g) Ba3+ and F− forms ionic compounds and would not have Ba−F bond energy. 9 Ans: C Note since pressure is kept constant, ICE table should NOT be about change in pressure. X(g) Y(g) + 2Z(g) Initial / mol x 0 0 Change / mol y +y +2y Eqm / mol xy y 2y PY + PZ = p  0.25p = 0.75p Since Y and Z are in a mol ratio of 1:2, PY = 0.25p, PZ = 0.5p Kp = (0.25p)(0.5p)2 0.25p = 0.25p2

4 10 Ans: A Titration of a strong acid (H 2SO4) with a weak base (NH 3). Equivalent point pH is less than 7 as NH4+ is a weakly acidic cation. Methyl orange will be a suitable indicator as the working pH range of methyl orange coincides with the region of sharp pH change at equivalent point of this titration. 11 Ans: B Since Ca(OH)2 exist as aq and Mg(OH)2 exist as a solid, Ksp of Mg(OH)2 must be lower than Ca(OH)2. |ΔHhyd|  charge density of ion, Mg2+ has a more exothermic ΔHhyd. |L.E.|      rr qq , magnitude of L.E. for Mg(OH)2 is greater. 12 Ans: D Since V total is kept constant, volume of reactant used is proportional to its concentration in the final reaction mixture. Comparing expt 3 and 4, when [P] × 2, initial rate also × 2. It is first order w.r.t P. Comparing expt 4 and 2, when [R] × 2, initial rate × 4. It is second order w.r.t R. Comparing expt 1 and 3, let rate = k[P][Q]y[R]2 rate 1 rate 3 = k(10)(10)y(10)2 k(5)(5)y(10)2 4 = (2)(2)y (2)y=2 It is first order w.r.t Q 13 Ans: A Since E o(MnO4,H+/Mn2+) = + 1.52 V is greater than E o(Cu2+/Cu) = + 0.34 V, the reduction half-cell is MnO4,H+/Mn2+ and oxidation half-cell is Cu2+/Cu. Eocell = Eo(MnO4,H+/Mn2+)  Eo(Cu2+/Cu) A When excess NH 3 is added to Cu 2+/Cu half-cell, the half -cell become [Cu(NH3)4]2+/Cu, Eo = 0.05 V and Eocell increases. B When additional H + is added into MnO 4,H+/Mn2+ half-cell, the position of equilibrium for Mn O4 + 8H + + 5e  Mn2+ + 4H 2O shifts to the right, Eo(MnO4,H+/Mn2+) increases and Eocell increases. C Size of electrode does not affect Eo(Cu2+/Cu) as solids does not affect position of equilibrium. D When alloy of copper and zinc is present, Zn reacts with Cu 2+ to give Zn 2+ and Cu(s). The concentration of Cu2+ decreases and position of equilibrium for Cu2+ + 2e Cu shifts to the left, Eo(Cu2+/Cu) decreases and Eocell increases.

5 14 Ans: C 15 Ans: C This is the simplified electrolytic cell. Species present: Na+ (aq), NO3 (aq), H2O, Zn cathode, Cu anode At cathode: 2H2O + 2e  H2 + 2OH At anode: Cu  Cu2+ + 2e The filter paper will turn blue due to the formation of Cu2+(aq) Note: this is not a litmus paper. 16 Ans: C A Melting point: SiO2 > Na2O> P4O10 SiO2 (covalent lattice with strong covalent bond) Na2O (ionic lattice with strong ionic bond) P4O10 (simple molecular with weak IMF) B Covalent character: P4O10 = SiO2 > Na2O C pH when mixed with water: Na2O (pH = 14) > SiO2 (pH = 7) > P4O10 (pH = 3) D Na2O is soluble in aq alkali due to reaction of Na2O with H2O to give NaOH. SiO2 is insoluble in aq alkali due to strong covalent bond in the covalent lattice. P4O10 is soluble in aq alkali due to reaction of P4O10 with NaOH to give Na3PO4. e < e <

6 17 Ans: B M(NO3)2(s)  MO(s) + 2NO2(g) + ½ O2(g) (option 2 is correct) Ionic radius of Mg2+ is smaller than Ba2+. Mg2+ has a higher charge density and stronger polarising power. Mg 2+ is able to distort electron cloud of NO 3 to a larger extent and NO covalent bond in Mg(NO3)2 is weakened significantly. Hence Mg(NO3)2 decompose at a lower temperature. (option 1 is correct). |L.E.|      rr qq , magnitude of L.E. for Mg(NO 3)2 is greater. However, N O covalent bond is broken during thermal decomposition and hence we should not even be comparing the lattice energy. 18 Ans: A Before heating, YCln reacts with AgNO3 to produce 8.368 × 103 mol of AgCl. After heating, the product reacts with AgNO3 to produce 5.021 × 103 mol of AgCl. This shows that Y can form two chlorides with different oxidation state. Y must be from Group 15 (e.g. PCl5 and PCl3) 1 mol of YCl5 gives 5 mol of Cl Amount of YCl5 = (8.368 × 103)  5 = 1.674 × 103 mol Mr of YCl5 = 0.50  1.674 × 103 = 298.8 Ar of Y = 298.8  5×35.5 = 121.3 Y is Sb 19 Ans: D Down Group 17, - Thermal stability of HX decreases due to weaker bond energy - Enthalpy change of formation becomes less exothermic (calculate using BE values) - The ease of oxidation increases as Eo(X2/X) decreases down the group.

7 20 Ans: B A The tartrate ion acts as a Bronsted acid, it donates 2 H+ to react with 2 OH and form the new ligand. B There are 2 chiral carbons for tartrate ion with 4 possible stereoisomers. However due to internal line of symmetry, 2 of the isomers are identical (they are known as the meso compound). Hence there are only 3 stereoisomers. C C H H HO OH COO COO C C H H HO COO