inorganic chemistry - What is the molecular structure of polymetatelluric acid?

Telluric Acid, $\ce{Te(OH)6}$, dehydrates to form polymetatelluric acid: $\ce{(H2TeO4)10}$. However, I have been unable to locate the exact structure of polymetatelluric acid, and it does not seem entirely obvious.

Are the tellurium centers connected directly to each other or are they separated by oxygen? Are these polymeric bonds single bonds or double bonds or some mixture of the two? My best guess is that the $\ce{Te}$ centers are separated by oxygens with three other oxygens surreounding each $\ce{Te}$ center, two of which are $\ce{-OH}$ groups and the third being a $\ce{=O}$ group: this would make the oxygens between the $\ce{Te}$ centers alternating single and then double bonds — possibly conjugated on some level. Is there any chance polymetatelluic acid has direct $\ce{Te-Te}$ bonds?

Answer

It would be ideal to check out x-ray analysis data, but it seems like polymetatelluric acid is a rather amorphous substance and does not form suitable crystals. As of now, I haven't discovered anything relevant neither in ICSD, nor in CCDC.

Another method of structural analysis would be vibrational spectroscopy. There is an article "Ultrarotspektren von Tellursäuren, Telluraten und Antimonaten" [1, p. 165] (in German) containing an extensive research of various $\ce{Te}$-containing substances performed with infrared and Raman-spectroscopy (I reduced the table to the frequencies for polymetatelluric acid exclusively):

$$\begin{array}{ll} \hline \ce{(H2TeO4)_x} &\text{Zuordnung} \\ \hline 450~\mathrm{(m, b)} & \delta~(\ce{TeO}) \\ \hline 600~\mathrm{(st, b, Sch)} & \nu~(\ce{TeO}) \\ 720~\mathrm{(sst, b)} & \\ 600~\mathrm{(sst, b, Sch)} & \\ \hline 1085~\mathrm{(mst, b)} & \delta~(\ce{TeOH}) \\ \hline 1618~\mathrm{(s, b)} & \delta~(\ce{H2O}) \\ \hline 2360~\mathrm{(m)} & 2\delta~(\ce{TeOH}) \\ \hline 3200~\mathrm{(sst, b)} & \nu~(\ce{TeOH}) \\ 3360~\mathrm{(sst, b, Sch)} & \nu~(\ce{H2O}) \\ \hline \end{array}$$

Polymetatellursäure $\ce{(H2TeO4)_x}$ ist eine feste amorphe Substanz. In ihrem UR-Spektrum sind alle Banden sehr breit, wie man es bei amorphen Substanzen meist antrifft. Das Tellur ist in dieser Verbindung wahrscheinlich wie im $\ce{Te(OH)6}$ sechsfach koordiniert, wie die annähernd gleiche Lage der $\ce{TeO}$-Valenz- und Deformationsschwingungen in beiden Verbindungen zeigt. Hieraus folgt, daß $\ce{TeOTe}$-Bindungen vorhanden sein müssen. Bei Viererkoordination würden die Valenzschwingungen höher liegen, wie es in den vergleichbaren Molekeln $\ce{IO}$, und $\ce{H5IO6}$ der Fall ist.[...]

Im Gegensatz zum $\ce{Te(OH)6}$ tritt im Spektrum des $\ce{(H2TeO4)_x}$ eine Frequenz bei $\pu{1618 cm-1}$ auf, die nur von freiem Wasser herrühren kann. Ebenso beobachtet man neben der $\ce{OH}$-Valenzschwingung der $\ce{TeOH}$-Gruppen ($\pu{3200 cm-1}$) die Bande $\pu{3360 cm-1}$ des freien Wassers. Aus den Intensitäten dieser Banden ist zu entnehmen, daß $\ce{H2O}$ und $\ce{TeOH}$-Gruppen in größenordnungsmäßig gleicher Menge vorhanden sind. Eine stöchiometrische Konstitutionsformel wird man dieser Substanz wohl nicht zuordnen können.

Briefly, the IR-spectra bands of $\ce{[H2TeO4}]_x$ bands are very broad, suggesting amorphous non-stoichiometric compound (as Mithoron suggested) with polymeric structure where $\ce{Te}$ atom is 6-coordinated. Bands were only assigned for $\ce{Te-O}$, $\ce{Te-OH}$ bonds and $\ce{H2O}$. IR-spectra do not show the presence of $\ce{Te-Te}$ bonds. Most likely there are $\ce{[TeO]6}$ units with edge- or corner-sharing, linked together in a chain-alike fashion.

Reference

1. Siebert, H. Z. anorg. allg. Chem. 1959, 301 (3–4), 161–170. DOI: 10.1002/zaac.19593010305 (in German).