If a compound (oxide) has an $hcp$ arrangement and the cations $M$ occupy two-thirds of the octahedral voids,the formula of the compound is:

Which of the following metals has an $hcp$ crystal structure?

In a compound $AB_2O_4$,the oxide ions form a cubic close-packed $(CCP)$ lattice. The molar ratio of trivalent cations to bivalent cations is $2:1$. If all the bivalent ions occupy octahedral voids and trivalent ions occupy both tetrahedral and octahedral voids in equal numbers,then the fraction of octahedral voids that remain unoccupied is:

The packing efficiency of the face-centered cubic $(fcc)$,body-centered cubic $(bcc)$,and simple/primitive cubic $(pc)$ lattices follows the order:

In the spinel structure,oxide ions are cubical-closest packed where $1/8$ of tetrahedral voids are occupied by $A^{2+}$ cations and $1/2$ of octahedral voids are occupied by $B^{3+}$ cations. The general formula of the compound having spinel structure is

What is the number of octahedral and tetrahedral voids present respectively in $0.25 \ mol$ of a substance having $hcp$ structure?