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Topic: metal properties and d-d transitions  (Read 5666 times)

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Offline vogelfänger

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metal properties and d-d transitions
« on: February 05, 2011, 06:38:32 PM »
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?

Offline jdy07

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Re: metal properties and d-d transitions
« Reply #1 on: February 06, 2011, 03:38:28 AM »
Actually, the larger polarization of the Pt<sup>2+</sup> 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.

Offline vogelfänger

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Re: metal properties and d-d transitions
« Reply #2 on: February 06, 2011, 06:45:01 PM »
Thank you for your reply. Can you explain what you mean by polarisation in this context?

Offline Schrödinger

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Re: metal properties and d-d transitions
« Reply #3 on: February 17, 2011, 11:20:53 PM »
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. :)
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Offline jdy07

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Re: metal properties and d-d transitions
« Reply #4 on: February 19, 2011, 09:22:46 PM »
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.

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