In a $ccp$ lattice of $X$ and $Y$,$X$ atoms are present at the corners while $Y$ atoms are at face centers. What would be the formula of the compound if one of the $X$ atoms from a corner is replaced by a $Z$ atom?

$A$ crystal is made up of metal ions $M_1$ and $M_2$ and oxide ions. Oxide ions form a $ccp$ lattice structure. The cation $M_1$ occupies $50 \%$ of octahedral voids and the cation $M_2$ occupies $12.5 \%$ of tetrahedral voids of the oxide lattice. The oxidation numbers of $M_1$ and $M_2$ are,respectively:

$A$ cubic structure is formed where atoms of element $X$ are occupied at the corners of the cube and also at the face centers. Atoms of element $Y$ are present at the body center and at the edge centers. If all the atoms are removed along a plane passing through the middle of the cube (bisecting the four edges), the formula will become

The number of octahedral voids in a unit cell of a cubical closest packed structure is

The number of atoms contained in one face-centred cubic $(FCC)$ unit cell of a monoatomic substance is:

For an ionic crystal of the type $AB$,the value of the limiting radius ratio is $0.40$. This value suggests that the crystal structure should be: