Question

The Al2O3 crystal structure (corundum) consists of an HCP arrangement of O2- ions; the Al3+ ions occupy octahedral positions. What fraction of the available octahedral positions are filled with Al3+ ions? In corundum, the Al3+ ions do not fill the tetrahedral positions.

Answer 1

Answer:

The fraction of the available octahedral positions are filled with $Al^{3+}$ ions is $\frac{2}{3}$

Explanation:

If there are n numbers of spheres in HCP in packing:

Number of octahedral void = n

Number of tetrahedral void = 2n

Total number of voids = n + 2n = 3n

The $Al_2O_3$ crystal structure consists of an HCP arrangement of $O^{2-}$ions.

Number of spheres in HCP = 6

Number of $O^{2-}$ = 6

Number of octahedral voids = 6

Number of tetrahedral void = 12

Number of aluminum ions when there are 6 $O^{2-}$ ion =$\frac{2}{3}\times 6= 4$

Number of voids in which $Al^3+$ ions are present = 4

The fraction of the available octahedral positions are filled with $Al^{3+}$ ions = $\frac{4}{6}=\frac{2}{3}$