In bcc lattice containing $X$ and $Y$ type of atoms, $X$ type of atoms are present at the corners and $Y$ type of atoms are present at the centers. In its unit cell, if three atoms are missing in the corners, the formula of the compound is

The molecular formula of a crystal is $A B_{2} \mathrm{O}_{4}$. Oxygen atoms form ccp lattice. Atoms of $A$ occupy $x \%$ of tetrahedral voids and atoms of $B$ occupy $y \%$ of octahedral voids. $x$ and $y$ are respectively

An element with molar mass $2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}$ forms a cubic unit cell with edge length of 405 pm . If its density is $2.7 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, the number of atoms present in one unit cell of it is (Given; $N_A=6.023 \times 10^{23} \mathrm{~mol}^{-1}$ )