$A$ compound is formed by cation $C$ and anion $A$. The anions form hexagonal close packed $(hcp)$ lattice and the cations occupy $75 \%$ of octahedral voids. The formula of the compound is

In an $fcc$ crystal lattice,atoms $A$ are at the corners of the cube and atoms $B$ are at the face centers. If one atom is removed from one of the face centers,find the formula of the compound.

$A$ crystalline solid is made of $X$,$Y$ and $Z$ elements. Atoms of $X$ form $FCC$ packing; atoms of $Y$ occupy octahedral voids while atoms of $Z$ occupy tetrahedral voids. What will be the simplest formula of the solid if atoms along one body diagonal are removed?

What is the coordination number of a one-dimensional close-packed arrangement?

$A$ solid has $CCP$ arrangement having atoms $A, B, C$. If $A$ atoms are present at face centres,$B$ at corners and $C$ atoms occupy $50\%$ tetrahedral voids,then the molecular formula of the solid will be:

In the face-centered unit cell,the lattice points are present at