Question

In $BeO$ (Zinc Blende structure), $M{g}^{2+}$ is introduced in available tetrahedral voids. Then ions are removed from a single body diagonal of the unit cell. What will be the molecular formula of the unit cell?

Answer 1

In $ZnO$, $O^{-}$ ions are present on the corners and face centers of the cubic unit cell and $Z{n}^{2+}$ ions are present in the alternate tetrahedral voids.

The tetrahedral voids are two in number on each body diagonal. So, in total there are $4$ tetrahedral voids. Now, $4$ tetrahedral voids are already filled by $Z{n}^{2+}$ and the $4$ which are left are filled by $M{g}^{2+}$.

So, the new composition is :-

$M{g}^{2+}:Z{N}^{2+}:O^{2-}$

$4$ $:4:4$

Now, ions of a single body diagonal are removed. So, one $Z{n}^{2+}$ ion and one $M{g}^{2+}$ ion is removed. So, now the composition of ions is:

$M{g}^{2+}:Z{N}^{2+}:O^{2-}$

$3$ $:3:4-\frac{2}{8}$ (Since, $O^{2-}$ are at corners of body diagonal)

$\Rightarrow M{g}^{2+}:Z{n}^{2+}:O^{2-}$ $\Rightarrow$ $M{g}^{2+}:Z{N}^{2+}:O^{2-}$

$3$ $:3:\frac{15}{4}$ $4$ $:$ $4$ $:$ $5$

So, new formula of unit cell is $M{g}_{4}Z{n}_4{O}_5$