3
$\begingroup$

Why doesn't the field splitting parameter increase steadily along the period for 3rd-row transition metals, as one would expect due to a better energy match between ligands and the metal as $Z_\mathrm{eff}$ increases?

The series is (in the order of increasing $\Delta$: $\ce{Mn^2+}, \ce{Ni^2+}, \ce{Co^2+}, \ce{Fe^2+}, \ce{V^2+}$ etc. Source: Spectrochemical series – Wikipedia, confirmed by Shriver and Atkins.

I would guess that this is to do with $\mathrm{3d}$ orbitals getting too contracted on the right of the period and, and hence, having a poorer size match, but I couldn't find this explanation mentioned anywhere.

$\endgroup$

1 Answer 1

3
$\begingroup$

A greater octahedral field split would be associated with a shorter $\ce{L\bond{->}M}$ bond length. Bond length is a function of:

  • size of both partners
  • strength of the bond

In particular, as we are examining metals in $\mathrm{+II}$ oxidation state, the metal ion’s size should increase somewhat steadily along the period while the ligand size should stay the same. This means that we need to be looking at bond lengths to understand the discrepancy.

For reference, here is a typical octahedral $\ce{[ML6]}$ complex ignoring π effects.

octahedral
Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

We should immediately be able to see how field split is generated: it is the difference between the $\mathrm{t_{2g}}$ orbitals and the $\mathrm{e_g^*}$ orbitals; the latter of which are antibonding. Thus, having any electrons in $\mathrm{e_g^*}$ will weaken the $\ce{L\bond{->}M}$ bonds. This corresponds nicely with vanadium(II), a $\mathrm{d^3}$ ion, having the largest field split, and manganese(II) ($\mathrm{d^5}$) having the lowest.

However, the final value is dependent on very subtle differences and not easily predicted a priori.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.