Atoms of an element $Z$ form a hexagonal closed packed $(hcp)$ lattice and atoms of element $X$ occupy all the tetrahedral voids. The formula of the compound is

An ionic solid is $HCP$ of $B^{2-}$ ions and $A^{x+}$ ions are in one-fourth of tetrahedral voids. The value of $x$ should be

In a solid compound,the oxide ions are arranged in $CCP$. Cations $A$ occupy $1/6^{th}$ of the tetrahedral voids and cations $B$ occupy one third of the octahedral voids. The probable formula of the compound is

$A$ compound is formed by two elements $X$ and $Y$. The element $Y$ forms a cubic close-packed $(CCP)$ arrangement and the atoms of element $X$ occupy one-third of the tetrahedral voids. What is the formula of the compound?

$A$ metal $M$ forms a hexagonal close-packed structure. The total number of voids in $0.02 \ mol$ of it is $......... \times 10^{21}$ (Nearest integer) (Given $N_{A} = 6.02 \times 10^{23}$)

The molecular formula of a compound is $AB_2O_4$. Atoms of $O$ form a $ccp$ lattice. Atoms of $A$ (cation) occupy $\frac{1}{8}$ of the tetrahedral voids. Atoms of $B$ (cation) occupy a fraction of the octahedral voids. What is the fraction of vacant octahedral voids?