In a solid,oxide ions are arranged in $HCP$. Cations $A$ occupy one-sixth of $THV$ and cations $B$ occupy one-third of the $OHV$. The empirical formula of the compound is

In an $fcc$ arrangement of $A$ and $B$ atoms, $A$ atoms are at the corners of the unit cell and $B$ atoms are at the face centers. If one of the $A$ atoms is missing from one corner, what is the simplest formula of the compound?

Fill in the blanks:
$(a)$ In $ccp$ structure,$a = \dots \dots r$.
$(b)$ In $hcp$ structure,packing efficiency $= \dots \dots \%$.
$(c)$ In $bcc$ structure,percentage of packing efficiency $= \dots \dots \%$.

In a compound made of atoms $A$ and $B$,atoms $A$ form an $fcc$ lattice,while atoms $B$ occupy $25\%$ of the tetrahedral voids and $50\%$ of the octahedral voids. If the crystal is cut such that all face-centered atoms along one axis are removed,the molecular formula of the resulting solid will be ...........

Which of the following crystal lattice will have a void fraction of $0.476$?

How many unit cells share a corner of a unit cell in a crystal lattice?