Question

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

• A. $M_2{O}_3$
• B. $M_{3}O$
• C. $M_3{O}_3$
• D. $M{O}_2$

Answer 1

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

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