Question

A certain oxide of metal M crystallizes in such a way that $O^{2-}$ ion occupies hcp arrangement following ABAB pattern. The metal ion, however, occupies $\frac{2}{3}$rd of the octahedral voids. The formula of the compound is :

• A. $M_2{O}_3$
• B. $M_{3}O$
• C. $M_8{O}_9$
• D. $M{O}_2$

Answer 1

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

Let the $O^{2-}$ ions occupies hcp arrangement be having n lattice points. Since metal ion are occupying $\frac{2}{3}rd$ of the octahedral voids. The octahedral voids present in total is also n. So, metal ions are occupying $\frac{2}{3}$n in of octahedral voids.

Hence, The formula of compound will de $M_{\frac{2}{3}}n{O}_n$

Multiplying whole formula with $\frac{3}{n}$ to get general form,

$M_2{O}_3$