Why does the nature and oxidation state of a central metal affect field split? I know the following factors affect the value of field splitting in a coordination compound:


No. of ligands surrounding the central atom.
Strength of the ligand (as per spectrochemical series)
Nature of the metal ion
Oxidation State/charge on central atom 


I have understood why the first two are contributing factors, but the 3rd and 4th points do not ring a bell.

For the 4th point, i.e., ox. state of central atom, my guess was that if there is a lower oxidation state present, then less number of electrons will be present in the orbitals of the central atom, and hence due to lesser repulsion between ligands, value of field splitting will decrease.

Am I correct in this?

As for the 3rd point, i.e., the nature of the metal ion, I am completely at a loss. What does nature specify here, and how does it affect the value of field splitting?
Let me start this answer — as usually for an answer I give about coordination complexes — by showing the molecular orbital scheme of a typical octahedral complex ignoring $\ce{L\bond{-&gt;}M}$ &#x3c0; interactions.


Figure 1: Octahedral $\ce{[ML6]}$ complex with no &#x3c0; interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

It is easy to see where field splitting comes from: two of the ligand group orbitals transform as $\mathrm{e_g}$ as do two of the metal’s $\mathrm{d}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). These interact in a bonding/antibonding manner to give $\mathrm{e_g}$ (considered a ligand-centred orbital due to the electronegativity difference) and $\mathrm{e_g^*}$ (a metal-centred orbital).

It is important to note that this scheme depends strongly on the relative positions of the orbitals. The closer the energies of the ligand and metal group orbitals, the larget the split will be.

If you consider different aquacomplexes, you can always be sure that the ligand orbitals start off from the same energy (they are the same water, after all). However, the metal orbitals will be differently positioned. The higher the first ionisation enthalpy (typically corresponding to the removal of an electron from $\mathrm{4s}$ — here $\mathrm{a_{1g}}$) the lower these orbitals will be in this scheme. Loosely, the ionisation enthalpy depends on the number of protons in the nucleus, so we might expect nickel to have lower-lying orbitals than titanium. Whatever the actual difference is (there are many more effects), no two different metals will ever have the same orbital energies and therefore any two different metals will always have different field splits corresponding to a different colour. You know this: hexaaquachromium(III) is green while hexaaquairon(III) is yellow-orange.

The oxidation state influences the orbital scheme in a similar way. Each oxidation corresponds to the removal of an electron meaning less electronic repulsion meaning lower-lying orbitals (in slightly larger contexts, this corresponds to the contraction of cations). Thus, if you change the oxidation state of a metal, e.g. from $\mathrm{+II}$ to $\mathrm{+III}$, the metal orbitals will be lowered in energy and the field split will most likely increase since there is better overlap with the ligand orbitals. Therefore, iron(II) and iron(III) have different colours in solution.