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I think this picture from wikipedia is wrong because I count:

\begin{align} \left. \begin{array}{ccccr} 1&×&1 &= &\ce{1~Y} \\ 6&×&\frac{1}{8} &= &\ce{\frac{3}{4}O} \end{array} \right\} \text{ gives }\ce{Y4O3} \end{align}

How would be the correct elementary cell of $\ce{Y2O3}$?

(I didn't found any correct elementary cell using google)

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The picture in Wikipedia doesn't give you the primitive cell. In Yttria, $\ce{Y2O3}$, 8 metal ions are in the special positions $\frac{1}{4}$, $\frac{1}{4}$, $\frac{1}{4}$; the remaining 24 metals occupy the sites $u$, $0$, $\frac{1}{4}$. The 48 oxygen ions are in general positions $x$, $y$, $z$, and are arranged in distorted octahedra around the metal ions, the metal-oxygen bonding distances being unequal. So, the stoichiometry is $\ce{Y32O48}$ which gets simplified to $\ce{Y2O3}$. This structure is generally referred to as the C-type metal oxide structure. All the sesquioxides of the rare earths belong to this system, as do $\ce{Ga2O3}$, $\ce{In2O3}$, $\ce{Tl2O3}$ and the mineral bixbyite, $\ce{(Fe, Mn)2O3}$. The primitive cell looks like this (oxygen ions are red, metal ions are gray)

enter image description here

Here another view which might show more easily that indeed the positions of all the 48 oxygen atoms are different in such a way that no smaller primitive cell can be found.

Reference

M. G. Paton, E. N. Maslen, Acta Cryst. 1965, 19, 307-310

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