# Elementar cell of Y2O3

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)

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)