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What is the lattice energy of CaF2 if the distance between calcium and fluoride ions in the crystal lattice is 2.40 Å and the charges on calcium and fluoride ions are +2 and -1, respectively?

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To calculate the lattice energy of CaF, we can use the Born-Lande equation:Lattice Energy =  N * A *  1 - 1/n  * q * q  /  4 * r Where:- N is the Avogadro's number  6.022 x 10 mol - A is the Madelung constant  for CaF, A = 2.977 - n is the Born exponent  for CaF, n  10 - q and q are the charges on the ions  for Ca, q = +2; for F, q = -1 -  is the vacuum permittivity  8.854 x 10 C /  N m  - r is the distance between the ions  2.40 , which is equal to 2.40 x 10 m Now, let's plug in the values and calculate the lattice energy:Lattice Energy =  6.022 x 10 * 2.977 *  1 - 1/10  *  +2  *  -1   /  4 * 8.854 x 10 * 2.40 x 10 Lattice Energy =  6.022 x 10 * 2.977 * 0.9 *  -2   /  4 * 8.854 x 10 * 2.40 x 10 Lattice Energy  -2631 kJ/molThe negative sign indicates that the lattice energy is released when the crystal lattice is formed. Therefore, the lattice energy of CaF is approximately 2631 kJ/mol.

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