2019 JPJC Prelim H2 Chem P2 ANS

2019 JPJC JC2 H2 Chemistry (9729) 1 2019 Prelim Paper 2 Answers JURONG PIONEER JUNIOR COLLEGE 2019 JC2 H2 CHEMISTRY (9729) Preliminary Examination Paper 2 (Suggested Answers) 1 (a) (i) Particles: Cu2+ cation Interactions: metallic bonding (ii) Particles: I2 molecule Interactions: instantaneous dipoleinduced dipole interactions (b) The layers of Cu2+ cations can slide past each other without breaking the metallic bond (or with the mobile valence electrons holding them together). (c) Electrical conductivity: Mobile valence electrons available to conduct electricity in copper but no mobile charge carriers available in iodine to conduct electricity. or Solubility in organic solvent: Iodine is able to form favourable instantaneous dipoleinduced dipole interactions with the organic solvent, but copper is unable to form favourable interactions with the organic solvent. (d) (i) Atomic mass is the weighted average of the mass of isotopes. (ii) Percentage abundance of the isotopes in this sample differs from what is normally obtained. or There are more than 2 types of isotopes of copper present in the sample. (e) Since angle of deflection  charge mass , angle of deflection of I− = 1127 263    (7.0 ) = 1.7  (1dp) (f) X Cl Br I E(X2/X) / V +1.36 +1.07 +0.54 Since E(X2/X–) becomes less positive down the group , it implies that the tendency of X2 to be reduced to X  decreases and hence, the oxidising p ower of X 2 decreases down the group. 127I 63Cu2+

2019 JPJC JC2 H2 Chemistry (9729) 2 2019 Prelim Paper 2 Answers (g) (i) [O]: 2IO3 + 12H+ + 10e  I2 + 6H2O x V [R]: H2O2 + 2H+ + 2e  2H2O +1.77 V E cell = E red  E ox +0.57 = (+1.77)  E(IO3/ I2) E(IO3/ I2) = +1.20 V (ii) Note the E cell value on the voltmeter at first instance of cell being connected. 2 (a) soln hyd ve ve ve LEHH         Down the group, both LE of MCO 3 and Hhyd of M2+ becomes less exothermic (or more endothermic or magnitude/value of LE of MCO 3 and Hhyd of M2+ decreases) since the radius of M2+ increases down the group. However, the decrease in LE is less than the decrease in Hhyd since M 2+ is smaller than CO32. Hence, Hsoln becomes less exothermic and the solubility of MCO3 decreases down the group. (b) (i) Ksp of SrCO3 = [Sr2+][CO32] = (1.05  105)2 = 1.10  1010 mol2 dm6 (ii) For precipitation to occur, ionic product  Ksp. ionic product of SrCO3 = [Sr2+]min[CO32]  Ksp. [Sr2+]min(0.02) = 1.1 × 1010 minimum [Sr2+] to precipitate SrCO3 = 5.50 × 109 mol dm3 ionic of SrF2 = [Sr2+]min[F]2  Ksp. [Sr2+] min(0.1)2 = 2.5 × 109 minimum [Sr2+] to precipitate SrF2 = 2.50 × 107 mol dm3 SrCO3 will precipitate first because a lower [Sr 2+] is requir ed to form the precipitate. 1 mol dm3 H+(aq), I2(aq), IO3(aq), 298 K salt bridge V e