To calculate the lattice energy of calcium fluoride CaF2 , we can use the Born-Lande equation:Lattice energy U = N * A * 1 - 1/n * z+ * z- * e^2 / 4 * * 0 * r0 where:N = Avogadro's number 6.022 x 10^23 mol^-1 A = Born-Lande equation constant 2.56 n = Born exponent we can assume n = 9 for most ionic compounds z+ = charge of the cation Ca2+ = +2 z- = charge of the anion F- = -1 e = elementary charge 1.602 x 10^-19 C 0 = vacuum permittivity 8.854 x 10^-12 C/Jm r0 = sum of the ionic radii 100 pm + 133 pm = 233 pm = 2.33 x 10^-10 m Now, we can plug in the values into the equation:U = 6.022 x 10^23 mol^-1 * 2.56 * 1 - 1/9 * 2 * 1 * 1.602 x 10^-19 C ^2 / 4 * * 8.854 x 10^-12 C/Jm * 2.33 x 10^-10 m U = 6.022 x 10^23 mol^-1 * 2.56 * 8/9 * 2 * 2.56 x 10^-38 C / 4 * * 8.854 x 10^-12 C/Jm * 2.33 x 10^-10 m U = 6.022 x 10^23 mol^-1 * 2.56 * 8/9 * 2 * 2.56 x 10^-38 C / 2.93 x 10^-20 Jm U = 2.63 x 10^6 J/molSince we want the lattice energy in kJ/mol, we can convert it:U = 2.63 x 10^6 J/mol * 1 kJ / 1000 J = 2630 kJ/molSo, the lattice energy of calcium fluoride CaF2 is approximately 2630 kJ/mol.