Chemistry Labs

Problem 1

New catalysts for stereoregular polymerisation prompted the study of the system A–B, where A = Te(cryst.) and B = TeClX4+4 AlClX3\ce{TeCl4 + 4AlCl3} (an analogue of Te[AlClX4]X4\ce{Te[AlCl4]4}, which cannot be isolated). Interaction of A and B gives three compounds I, II, III for initial mixtures containing 77.8, 87.5 and 91.7 mol % of A respectively. For II and III no side products form; formation of I releases 1 mol of volatile TeClX4\ce{TeCl4} per 2 mol of I. I and II are pinkish-purple, dissociate into three ions in molten NaAlClX4\ce{NaAlCl4}, and have molar masses 1126±431126 \pm 43 and 867±48867 \pm 48 g mol⁻¹ (cryoscopy in NaAlClX4\ce{NaAlCl4}). Each shows a single IR band at 133 cm⁻¹ assigned to a Te–Te bond, and a single 27Al^{27}\ce{Al} NMR signal of tetrahedral Al, at different shifts for I and II. (1.1) Determine the minimal Te : Al : Cl ratio for I, II, III. (1.2) Write the molecular formulae of I and II. (1.3) Give the cation and anion formulae in I and II. (1.4) Describe the stereochemistry of the ions, assuming the cations are inorganic aromatic systems. (1.5) Which compound is thermally more stable, given AlClX3\ce{AlCl3} is volatile? (1.6) Write the reaction converting one compound into the other on heating.
Step 2 of 4: Molecular formulae from molar mass
I=TeX4[AlX2ClX7]X2 (Mcalc=1115 g mol−1);II=TeX4[AlClX4]X2 (Mcalc=848 g mol−1)\mathrm{I} = \ce{Te4[Al2Cl7]2}\ (M_{calc} = 1115\ \mathrm{g\,mol^{-1}});\qquad \mathrm{II} = \ce{Te4[AlCl4]2}\ (M_{calc} = 848\ \mathrm{g\,mol^{-1}})
Analysis

The simplest formulae Te₂Al₂Cl₇ (M=557M = 557) and Te₂AlCl₄ (M=424M = 424) are far below the measured values, but their doubles fit: TeX4AlX4ClX14\ce{Te4Al4Cl14} (1115 vs 1126 ± 43) and TeX4AlX2ClX8\ce{Te4Al2Cl8} (848 vs 867 ± 48). Writing them as salts: I = TeX4[AlX2ClX7]X2\ce{Te4[Al2Cl7]2}, II = TeX4[AlClX4]X2\ce{Te4[AlCl4]2}.