[vogelfänger]

In my last homework there was a question asking to compare the d-d transition energy between metals in the same group - e.g., which has smaller d-d split, [Pd(NH3)4]2+ or [Pt(NH3)4]2+.

Since the nuclear charge and ligand field strength are the same in both cases, I rationalised that the larger ion (Pt) would have a smaller split because its d-orbitals are too large to overlap effectively with the small Ligand orbitals. But the answer keys shows Pd as a weaker splitter. Could anyone explain why that is? Is it still due to size or some other property?

[jdy07]

Actually, the larger polarization of the Pt²⁺ allows for better overlap with the ligand orbitals. This stabilizes (lowers the energy) the M-L bonding orbital, thereby, increasing the d-d transition energy.

[vogelfänger]

Thank you for your reply. Can you explain what you mean by polarisation in this context?

[Schrödinger]

According to wiki, the Pt is smaller than Pd (and I believe the same holds with Pt²⁺ and Pd²⁺ as well). Hence, the charge density on Pt²⁺ is greater than that on Pd²⁺. This means that the Pt²⁺cation is more polarized than Pd²⁺. That is, more charge sitting on a smaller sphere. Hence, it has greater ability to overlap with the ligand orbitals. (Fajan's rule that connects covalency/ability to overlap and polarization of the ion)

I hope my reason is justified. Please correct me if I am wrong.

[jdy07]

Sorry about the late reply, I haven't been on the site for a couple of weeks. Polarizability is basically how diffuse the valence electrons are. A way to visualize this is that the d-orbitals are "larger" and therefore can overlap better with the ligand orbitals. The valence electrons are more diffuse in Pt(II) vs. Pd(II) because there is a whole entire extra shell of electrons between Pd and Pt (row 5 vs. row 6). This means the valence electrons in Pt(II) are more shielded from the positively charged nucleus, and therefore, the d-orbitals are more diffuse. Let me know if you need further clarification.