Question

Compute the theoretical density of caf2, which has the fluorite structure. ionic radii of ca2+ and f- are 0.100 nm and 0.133 nm respectively.

Answer 1

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 $Ca^{2+}$ and anion is $F^{-1}$. On plugging the given values the edge length of the $CaF_{2}$ 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$N_{A}$), where M is the molar mass of CaF₂= 74.03 g/mol. $N_{A}$ is Avogadro's number= 6.023 X 10²³ number of molecules. density, d= $\frac{ 4 X 74.03}{(5.176 X 10^{-8})^{3}X 6.023 X 10^{23} }$= 3.545 g/cm³.

So, theoretical density of CaF₂ is 3.545 g/cm³.