Question

Calculate the lattice energy of MgO if the charge on the magnesium ion is +2 and the charge on the oxide ion is -2. The ionic radius of Mg2+ is 0.072 nm and the ionic radius of O2- is 0.140 nm.

Answer 1

To calculate the lattice energy of MgO, we can use the Born-Haber cycle or the formula derived from Coulomb's Law and the Born-Lande equation. Here, we will use the formula:Lattice Energy  U  =  N * A * e^2 * |Z1 * Z2|  /  4 *  * 0 * r0 where:N = Avogadro's number  6.022 x 10^23 mol^-1 A = Madelung constant  for MgO, A = 1.748 e = elementary charge  1.602 x 10^-19 C Z1 = charge on Mg2+ ion  +2 Z2 = charge on O2- ion  -2 0 = vacuum permittivity  8.854 x 10^-12 C^2 J^-1 m^-1 r0 = sum of ionic radii  0.072 nm + 0.140 nm First, we need to convert the sum of ionic radii from nm to meters:r0 =  0.072 + 0.140  * 10^-9 m = 0.212 * 10^-9 mNow, we can plug in the values into the formula:U =  6.022 x 10^23 mol^-1 * 1.748 *  1.602 x 10^-19 C ^2 * |-2 * 2|  /  4 *  * 8.854 x 10^-12 C^2 J^-1 m^-1 * 0.212 * 10^-9 m U =  6.022 x 10^23 mol^-1 * 1.748 *  2.566 x 10^-38 C^2  * 4  /  4 *  * 8.854 x 10^-12 C^2 J^-1 m^-1 * 0.212 * 10^-9 m U =  1.748 * 6.022 x 10^23 mol^-1 * 10.264 x 10^-38 C^2  /   * 8.854 x 10^-12 C^2 J^-1 m^-1 * 0.212 * 10^-9 m U =  1.053 x 10^-14 C^2  /  2.356 x 10^-21 J^-1 m U = 4474.5 J/molThe lattice energy of MgO is approximately 4474.5 J/mol.