1. Lattice Energy (continued)

2. 1. Born-Mayer Equation

3. The equations that we will use to predict lattice energies for crystalline solids is the Born-Mayer equation, This equation is simple model that calculate the attraction and repulsion for a given arrangement of ions.

4. Born-Mayer Equation: Where: e is the charge of the electron = 1.6022 × 10 −19 C  0 is the permittivity of a vacuum z + is the charge on ion “A” z - is the charge on ion “B” 4 π ε0 = 1.112 × 10 −10 C²/(J m) r is the distance between the cations and anions (in Å) = r+ + r- E: is the electrostatic attraction energy between two ions of charge z+ and z -, its value is –ve and decreases as r value decreases.

5. E: energy change for the formation of 1 mol of an ionic solid from its gaseous ions –measure of the electrostatic attractions and repulsions of the ions in the crystal lattice Na + (g) + Cl - (g)  Na + Cl - (s)

6. 2. Madelung Constant, A

7. The Madelung constant is a numerical constant used in determining the electrostatic energy of each ion in a crystal. Summation of all the attraction and repulsion terms in a crystal lattice Madelung constant

8. For NaCl, Each cation is surrounded by 12 cation at a distance of, (2) 1/2 r. This is a repulsive force. Each cation is surrounded by 8 more anions at a distance of, (3) 1/2 r. This is a attractive force.

9. Each cation is surrounded by six anions at a distance of, r. This is an attractive force. Each cation is surrounded by 6 more cations at a distance of, 2r. This is a repulsive force.

12. Sodium chloride (NaCl) and magnesium oxide (MgO) both crystallize in the cubic rock salt structure. How does the value of the Madelung constant (A) compare for the two compounds? Explain. identical the Madelung constant is a geometric quantity and unrelated to the specific charges on the ions

13. 3. Born-Landé Equation

14. 2. proposed that the lattice energy could be derived from the electrostatic potential of the ionic lattice and a repulsive potential energy term. 1. is a means of calculating the lattice energy of a crystalline ionic compoundlattice energyionic compound

15. determines a theoretical value of the lattice energy, U (or E lat. ) Where N = Avogadro’s number (6.023 x 10 23 mol -1 ) A = Madelung constant z + = cation charge z - = anion charge e = electron charge (1.602 x 10 -19 C)  0 = permittivity of free space (8.854 x 10 -23 C 2 /(J·m)) d 0 (or r 0 ) = sum of the ionic radii n = average Born exponent Joules / mol

16. Born Exponent (n) accounts for the repulsions arising as the ions two nuclei approach each other Electronic Configuration of Ion Born Exponent, nExamples He 5Li + Ne 7Na +, Mg 2+, O 2-, F - Ar 9K +, Ca 2+, S 2-, Cl -, Cu + Kr 10Rb +, Br -, Ag + Xe 12Cs +, I -, Au +

17. Example NaCl –N = 6.023 x 10 23 mol -1 –A = 1.748 –z + = +1 and z - = -1 –e = 1.602 x 10 -19 C –  = 3.142 –  0 = 8.854 x 10 -12 C 2 /(J·m) –r 0 = 116 pm + 167 pm = 283 pm –n = (7+9)/2 = 8

18. NaCl experimental value: -770 kJ/mol if there is significant difference in the theoretical and experimental values, it indicates a significant covalent character to the complex

19. 4. Kapustinskii Equation

20. used if the crystal structure of the compound is not known –n is the number of ions in the empirical formula CaF 2, n = 3 Furthermore, one is able to determine the atomic radii using the Kapustinskii equation when the lattice energy is known. atomic radii K = units conversion = 1.21 MJ·Å/mol d* = parameter to account for nuclear-nuclear repulsion = 0.345 Å d = inter-ion distance = r+ + r–

21. Estimate the lattice energy of BaN 2 using the Kapustinskii equation. The Kapustinskii equation is: n = number of ions in the formula unit = 3 Z+ = cation charge = +2 Z– = anion charge = –1

22. d = inter-ion distance = r+ + r– r+ = 1.49 Å r– ~ 1.35 Å Then, d ~ 1.49 + 1.35 = 2.84 Å d* = parameter to account for nuclear-nuclear repulsion = 0.345 Å K = units conversion = 1.21 MJ·Å/mol Inserting all of the values into the equation:

23. The lattice energy increases as oxidation state increases more than its increment with the ionization energy. The most stable existence of Al in ionic crystals as Al 3+ although the removal of 3 electrons needs 5140 kJ / mol. CaF 2 most common than CaF why? 2CaF CaF 2 + Ca So CaF on its preparation convert to CaF 2 + Ca in exothermic reaction