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.
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.