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

The correct option is A $M_2{O}_3$
A certain oxide of metal M crystallises in such a way that $O^{2-}$ ions occupy hcp arrangement following AB AB... pattern. The metal ions, however, occupy $\frac{2}{3}$rd of the octahedral voids. The formula of the compound is $M_2{O}_3$
$O^{2-}$ ions occupy hcp arrangement. There are six $O^{2-}$ ions per unit cell. There are 6 octahedral voids per unit cell out of which $\frac{2}{3}$rd or $\frac{2}{3}\times 6=4$ octahedral voids are occupied with metal ions.
The ratio of number of atoms $M:O=4:6=2:3$