Calcium fluoride is a face centered cubic (FCC) lattice.The edge length (a) of the FCC lattice is equivalent to the 2×(radius of cation) + 4×(radius of anion)/1.414.For calcium fluoride, the cation is [tex]Ca^{2+}[/tex] and anion is [tex]F^{-1}[/tex]. On plugging the given values the edge length of the [tex]CaF_{2}[/tex] crystal lattice is {(2×1.00)+(1.33×4)/1.414} = 5.176 A°= 5.176X10⁻⁸cm (as 1nm=10A°).For FCC lattice number of atoms present per unit cell (Z)= 4.
Density of crystal, d= (Z X M)/(a³[tex]N_{A}[/tex]), where M is the molar mass of CaF₂= 74.03 g/mol. [tex]N_{A}[/tex] is Avogadro's number= 6.023 X 10²³ number of molecules. density, d= [tex]\frac{ 4 X 74.03}{(5.176 X 10^{-8})^{3}X 6.023 X 10^{23} }[/tex]= 3.545 g/cm³.
So, theoretical density of CaF₂ is 3.545 g/cm³.
