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.

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 and anion is . On plugging the given values the edge length of the 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), where M is the molar mass of CaF₂= 74.03 g/mol. is Avogadro's number= 6.023 X 10²³ number of molecules. density, d= = 3.545 g/cm³.

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

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

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