Problem 1

New catalysts for stereoregular polymerisation prompted the study of the system A–B, where A = Te(cryst.) and B =

\ce{TeCl4 + 4AlCl3}

(an analogue of

\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

\ce{TeCl4}

per 2 mol of I. I and II are pinkish-purple, dissociate into three ions in molten

\ce{NaAlCl4}

, and have molar masses

1126 \pm 43

and

867 \pm 48

g mol⁻¹ (cryoscopy in

\ce{NaAlCl4}

). Each shows a single IR band at 133 cm⁻¹ assigned to a Te–Te bond, and a single

^{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

\ce{AlCl3}

is volatile? (1.6) Write the reaction converting one compound into the other on heating.

Step 1 of 4: Minimal Te : Al : Cl ratios

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\mathrm{I}:\ 9\mathrm{Te}:8\mathrm{Al}:32\mathrm{Cl}\ -\ \ce{TeCl4}\ \to\ 2:2:7;\qquad \mathrm{II}:\ 8\mathrm{Te}:4\mathrm{Al}:16\mathrm{Cl}\ \to\ 2:1:4;\qquad \mathrm{III}:\ 12\mathrm{Te}:4\mathrm{Al}:16\mathrm{Cl}\ \to\ 3:1:4

Analysis

77.8 mol % Te means Te:B = 7:2, i.e. $7\ce{Te} + 2\ce{TeCl4} + 8\ce{AlCl3}$ = Te₉Al₈Cl₃₂; subtracting the released $\ce{TeCl4}$ and dividing by 2 gives 2:2:7 for I. Similarly 87.5 % gives 7:1 ( $8\ce{Te} + 4\ce{Al} + 16\ce{Cl}$ , 2:1:4) for II, and 91.7 % gives 11:1 ( $12\ce{Te} + 4\ce{Al} + 16\ce{Cl}$ , 3:1:4) for III.