$A$ 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?

In chrysoberyl,a compound containing beryllium,aluminium and oxygen,oxide ions form a cubic close-packed structure. Aluminium ions occupy $\frac{1}{4}$ of the octahedral voids and beryllium ions occupy $\frac{1}{8}$ of the tetrahedral voids. The formula of the compound is

$A$ body-centred cubic $(BCC)$ lattice is made up of two different types of atoms $X$ and $Y$. Atom $X$ occupies the body centre and atoms $Y$ occupy the corner positions. One of the corners is left unoccupied per unit cell. The empirical formula of it is

Calculate the void volume of a $bcc$ unit cell if the volume of the unit cell is $8.0 \times 10^{-23} \text{ cm}^3$.

The $CORRECT$ statement$(s)$ for cubic close packed $(ccp)$ three dimensional structure is(are)
$(A)$ The number of the nearest neighbours of an atom present in the topmost layer is $12$
$(B)$ The efficiency of atom packing is $74 \%$
$(C)$ The number of octahedral and tetrahedral voids per atom are $1$ and $2$,respectively
$(D)$ The unit cell edge length is $2\sqrt{2}$ times the radius of the atom

The molecular formula of a crystalline solid is $X_3Y_2$. Atoms of $Y$ form $ccp$ lattice,and atoms of $X$ occupy $50\%$ octahedral voids and $x\%$ of tetrahedral voids. What is the percentage of unoccupied tetrahedral voids?