In an ionic solid, anions $(B)$ are arranged in an $hcp$ array and cations $(A)$ occupy $1/2$ of the tetrahedral voids. What is the formula of the ionic compound?

$A$ compound made up of elements $A$ and $B$ (with a general formula $A_x B_y$), where $B$ forms an $hcp$ lattice and $A$ occupies $2/3$ of the tetrahedral voids. The formula of the compound is

In a close-packed arrangement,$A$ type of atoms are at corners,$B$ type of atoms are at face centers,$C$ type of atoms are at tetrahedral voids,and $D$ type of atoms are at octahedral voids. If all the atoms along any one body diagonal are removed,what will be the formula of the compound?

If the anions $(A)$ form hexagonal closest packing and cations $(C)$ occupy only $2/3$ of the octahedral voids in it,then the general formula of the compound would be

Calculate the total volume occupied by all particles in an $fcc$ unit cell if the volume of the unit cell is $6.4 \times 10^{-23} \ cm^3$.

$A$ two-dimensional solid is made by alternating circles with radius $a$ and $b$ such that the sides of the circles touch. The packing fraction is defined as the ratio of the area under the circles to the area under the rectangle with sides of length $x$ and $y$. The ratio $r=b/a$ for which the packing fraction is minimised is closest to $.....$