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Lattice energy and LFSE

3. A plot of the lattice energies for MCI; salts for M = Ca to Zn and a drawing of the cell are shown to the right. In the electrostatic model for "ionic bonding" U is proportional to (Z+)(Z-)e2A/(r+ + r.) where Z+ and Z- are the ion charges, A is the structure (Madelung) constant and r+ + r are the radii of the ions (see p. 88 of the text for more on this). Since the radius of the cation should decrease systematically as the effective number charge increases on going across the fourth row, one might expect a similar smooth variation in the lattice energy. Clearly, this is not the case. Provide an explanation for the observed variation in lattice energy that accounts for the fact that the points for Ca2+, Mn2+ and Zn2+ define a nearly straight line that would appear to be consistent with the expected gradual decrease in their radii, whereas the other elements have somewhat greater lattice energies than predicted on this basis. The graph is in attachment.

4. Use the experimentally determined values for Delta 0 for V2+ in solid VCl2; (9200 cm") and Ni2+ in solid NiCl2 (6920 cm-1) to support your explanation for the phenomena described in problem 3. This requires calculations of LFSE; example problem 20.3 will provide guidance for those who need help. Estimate the actual deviation in lattice energies from the predicted value based on the line connecting the values for Ca2+, Mn2+ and Zn2+ in the figure. The diagram is in the attachment.
from problem 3.


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3. The lattice energy is the enthalpy change required to convert one mole of solid into its constituent gaseous ions, i.e. MCl2 (s) → M2+ (g) + 2Cl- (g)
As the size of M2+ gets smaller, the charge density increases, so a gradual or smooth variation in lattice energy is expected.

If there was no such thing as crystal field theory, we'd expect the given graph to follow the dashed line, increasing regularly from Ca2+ (which doesn't have any d-electrons) to Zn2+ (which has d10 configuration). But that is not the case! Some M2+ lattice energies lie above the line, which means that these complexes are more stable than we would expect.

What's more, if we look at the ...