In an ionic solid,anions $(B)$ are arranged in an $hcp$ array and cations $(A)$ occupy $\frac{2}{3}$ of the octahedral voids. What is the formula of the ionic compound?

What is the total number of tetrahedral voids in $0.6 \ mole$ of a compound that forms a $hcp$ structure?

In a crystalline compound,$A$ atoms occupy all corners,$B$ atoms occupy all face-centered positions,$C$ atoms occupy all edge centers,and $D$ atoms occupy all tetrahedral voids formed by $A$ and $B$ atoms. Find the formula of the compound.

In a close-packed lattice,the number of tetrahedral voids compared to the number of octahedral voids is:

In an $fcc$ arrangement of $A$ and $B$ atoms, $A$ atoms are at the corners of the unit cell and $B$ atoms are at the face centers. If one of the $A$ atoms is missing from one corner, what is the simplest formula of the compound?

In a crystal lattice,$C$ anions form a cubic close-packed $(CCP)$ structure. Cations $(A)$ occupy $50\%$ of the tetrahedral voids and cations $(B)$ occupy $50\%$ of the octahedral voids. Determine the molecular formula of the compound.