To calculate the lattice energy of calcium fluoride CaF2 , we can use the Born-Lande equation:Lattice Energy = N * A * Z+ * Z- * e^2 / 4 * * 0 * r0 Where:N = Avogadro's number 6.022 x 10^23 mol^-1 A = Madelung constant for CaF2, A = 1.74756 Z+ = charge of the cation for Ca2+, Z+ = 2 Z- = charge of the anion for F-, Z- = 1 e = elementary charge 1.602 x 10^-19 C 0 = vacuum permittivity 8.854 x 10^-12 C^2 J^-1 m^-1 r0 = distance between the ions 231 pm = 231 x 10^-12 m Now, we can plug in the values and calculate the lattice energy:Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 2 * 1 * 1.602 x 10^-19 C ^2 / 4 * * 8.854 x 10^-12 C^2 J^-1 m^-1 * 231 x 10^-12 m Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 1.602 x 10^-19 C ^2 / 4 * * 8.854 x 10^-12 C^2 J^-1 m^-1 * 231 x 10^-12 m Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 2.566 x 10^-38 C^2 / 4 * * 8.854 x 10^-12 C^2 J^-1 m^-1 * 231 x 10^-12 m Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 2.566 x 10^-38 C^2 / 4 * * 2.044 x 10^-39 C^2 J^-1 m^-1 Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 2.566 x 10^-38 C^2 / 4 * * 2.044 x 10^-39 C^2 J^-1 m^-1 Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 2.566 x 10^-38 C^2 / 2.044 x 10^-39 C^2 J^-1 m^-1 Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 2.566 x 10^-38 C^2 / 2.044 x 10^-39 C^2 J^-1 m^-1 Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 2.566 x 10^-38 C^2 / 2.044 x 10^-39 C^2 J^-1 m^-1 Lattice Energy = 6.022 x 10^23 mol^-1 * 1.74756 * 4 * 2.566 x 10^-38 C^2 / 2.044 x 10^-39 C^2 J^-1 m^-1 Lattice Energy = 1.255 x 10^10 J/molSince we are given the lattice enthalpy of formation of Ca2+ and F-, we can also calculate the lattice energy using the following equation:Lattice Energy = Lattice enthalpy of formation of Ca2+ + 2 * Lattice enthalpy of formation of F-Lattice Energy = +178 kJ/mol + 2 * -328 kJ/mol Lattice Energy = +178 kJ/mol - 656 kJ/molLattice Energy = -478 kJ/molThe lattice energy of calcium fluoride CaF2 is -478 kJ/mol.