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 3 of 4: Ions in the two salts

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Intuition

The square-planar [Te₄]²⁺ ring is the classic 6 π-electron inorganic aromatic — the Te atoms are equivalent and only Te–Te vibrations appear in the IR.

\mathrm{I}:\ [\ce{Te4}]^{2+}\ 2[\ce{Al2Cl7}]^{-};\qquad \mathrm{II}:\ [\ce{Te4}]^{2+}\ 2[\ce{AlCl4}]^{-}

Analysis

Both compounds dissociate into three ions and share the same colour, so they contain the same cation. The single Te–Te IR band and all-equivalent Te atoms fit a symmetric $\ce{Te4}$ cluster; charge balance forces $[\ce{Te4}]^{2+}$ with two monoanions. For II the only sensible tetrahedral monoanion is $[\ce{AlCl4}]^{-}$ ; for I the extra $\ce{AlCl3}$ units build $[\ce{Al2Cl7}]^{-}$ (two corner-sharing tetrahedra), consistent with a different $^{27}\ce{Al}$ shift.