Question

A certain oxide of metal M crystallizes in such a way that $O^{2-}$ ions adopt an HCP arrangement. The metal ions, occupy $\frac{1}{3}rd$ of the tetrahedral voids. The formula of the compound is:

• A. $M_3{O}_4$
• B. $M{O}_3$
• C. $M_2{O}_3$
• D. $M{O}_2$

Answer 1

The correct option is C $M_2{O}_3$

The number of oxide ions in hexagonal unit cell $=6$
$\therefore$ tetrahedral voids $=2\times 6=12$
Given, metal ions occupyt one third of the total tetrahedral voids.
So, $M=12\times \frac{1}{3}=4$
So the formula is $M_4{O}_6$ or $M_2{O}_3$