The enthalpies of hydration of d-metal complexes
The trend in observed hydration enthalpies of the hexa-aquo transition metal complexes of the first row transition metals can be explained in terms of the ligand field stabilization energy.
| The enthalpies of hydration for [M(H2O)6]2+The arrow denotes the LFSE for the hexa-aquo VII complex. | |
In the plot, the dashed line is the enthalpy of hydration expected on the basis of its being proportional to z2/r, where z is the effective nuclear charge and r is the ionic radius. The number of d electrons changes as the period is crossed, and the enthalpy of hydration changes also.
The extra hydration enthalpy of some of the complexes compared to that predicted on a purely electrostatic basis corresponds to the ligand field stabilization energy for that configuration.
On going from the gas phase to the condensed phase, the symmetry of the ion changes from spherical in the free ion to octahedral in the complex, and so the energies of the d-orbitals split, and the number of electrons determines the total LFSE that arises. Hence certain configurations have extra stability in the octahedral hexa-aquo complex.
In [Mn(H2O)6]2+, there are 5 d-electrons, and so zero LFSE, and we can see that the hydration energy is that predicted on a purely electrostatic basis.
The structure of Spinels
Spinels are metallic oxides with the formula AB2O4, such as Fe3O4, Co3O4, and Mn3O4. They have the structure of spinel itself, MgAl2O4 (Mg = A, Al = B), which is based on a face-centered cubic array of oxide anions with A cations occupying one eighth of the tetrahedral holes and B cations occupying half the octahedral holes.
However, some spinels crystallize with the inverse spinel structure B(AB)O4. The reason for this is the variation in the ligand field stabilization energy between the two arrangements.
If we consider AB2O4, we see that the stoichiometry of the compound requires that one of the cations (A) have a charge of +2, and the other two cations (B) have a charge of +3. The normal spinel has the A cations in tetrahedral holes and the B cations in the octahedral holes, whereas the inverse spinel has the A cations in octahedral holes, and the B cations with half in octahedral holes and half in tetrahedral holes. We investigate why this is so by considering the structure of Fe3O4.
| The Structure of Fe3O4 | |
| Normal Spinel | Inverse Spinel |
| Fe2+[Fe3+]2O4 | Fe3+[Fe2+Fe3+]O4 |
| Fe (1): d6 in Td holeLFSE = 0 | Fe (1): d5 in Td holeLFSE = +0.4Δt |
| Fe (2): d5 in Oh holeLFSE = -2Δo | Fe (2): d6 in Oh holeLFSE = -2.4Δo |
| Fe (3): d5 in Oh holeLFSE = -2Δo | Fe (3): d5 in Oh holeLFSE = -2Δo |
| Total LFSE= (0) + (-2Δo) + (-2Δo)
= -4Δo |
Total LFSE= (+0.4Δt) + (-2.4Δo) + (-2Δo)
= -4Δo + 0.4(Δt – Δo) |
|
Δt < Δo, and the ligands in the tetrahedral orientation are always in the weak-field limit, and so the inverse spinel structure has a greater ligand field stabilization energy than the normal spinel structure. |
|
| So, Fe3O4 is predicted to have the inverse spinel structure, and this is as observed. | |
In general if the sum of the ligand field stabilization energies for (A2+)tet(B3+)oct is greater than those for (B3+)tet(A2+)oct, then the normal spinel structure will be adopted, and vice versa.
The crystal structure adopted by a given compound is that which has the largest lattice enthalpy, and so we can see that the ligand field stabilization energy has an effect on the value of the lattice enthalpies, with the LFSE being large enough to change the lattice enthalpy so as to drive the conversion of a spinel to an inverse spinel, as in the case of Fe3O4, for example.