Chapter 24, Problem 66FP

The crystal field stabilization energy (CFSE) can be defined as CFSE = 31 number of electrons the dxy, d_xx, and d_yzorbitals are stabilized (lower energy) with respect to the average energy of thed orbitals and that the d_eand d_r, orbitals are destabilized. As described in Are You Wondering 24-3, the stabilization is -0.4 $\Delta$ - and destabilization is 0.6 $\Delta$ -. The crystal field stabilization energy (CFSE) can be defined as CFSC = [(number Of electrons in the d_xy, d_xx, and d_yxorbitals)×(-0.4 $\Delta$ -)]+[number of electrons in the d_x-_y, and orbitals. The following table contains the enthalpy of hydration for the reaction ${\text{M}}^{\text{2+}}{\text{(g)+6H}}_{\text{2}}\text{O(l)}\to {{\text{[M(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{2+}}\text{(aq)}$

a. Plot the hydration energies as a function of the atomic number of the metals shown.
b. Assuming that all the hexaaqua complexes are high spin, which ions have zero CFSE?
c. If lines are drawn between those ions with CFSE = 0, a line of negative slope is obtained. Can you explain this finding?
d. The ions that do not have CFSE = 0 have heats of hydration that are more negative than the lines drawn in part (c). What is the explanation for this?

e. Estimate the value of L. for the Fe(ll) ion in an octahedral field of water molecules. What wavelength of light would the ion absorb?