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Topic: Transition Metal Hybridization  (Read 11437 times)

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Offline aromatic

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Transition Metal Hybridization
« on: July 05, 2006, 06:28:19 PM »
When orbitals for transition metal complexes are discussed, often only the d-orbitals are considered.  I think that realistically, these metal orbitals ought to be hybridized with p orbitals, but which ones?  If I have a platinum complex will the 5d orbitals hybridize with the filled 5p or with the empty 6p?

Offline Will

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Re: Transition Metal Hybridization
« Reply #1 on: July 05, 2006, 06:57:56 PM »
In complexes, platinum's electrons would be hybridised with the 6p subshell. I frequently see platinum in dsp2 (square planar) hybridised complexes.

Offline aromatic

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Re: Transition Metal Hybridization
« Reply #2 on: July 05, 2006, 08:00:02 PM »
When you say square planar I think of the energy diagram of the d orbitals from ligand field theory.  i.e. d(x^2-y^2) > dz^2 > dxy, dyz, dxz.  What would the energy diagram look like with hybridized orbitals.  How do you know what the hybridization is?  Is it reasonable to assume my square pyramidal iridium complex is d^2Sp^3 just because of the shape?

Offline Will

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Re: Transition Metal Hybridization
« Reply #3 on: July 05, 2006, 08:46:32 PM »
What would the energy diagram look like with hybridized orbitals.


Hope this is what you're looking for!

How do you know what the hybridization is?

I just learnt the hybridisation of the atom with the corresponding geometry. There is a basic table here.

Is it reasonable to assume my square pyramidal iridium complex is d^2Sp^3 just because of the shape?

Not really; square pyramidal complexes are d2sp2 hybridised and octahedral complexes are d2sp3 hybridised.

I hope that answers your questions. :)

Offline aromatic

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Re: Transition Metal Hybridization
« Reply #4 on: July 06, 2006, 02:09:06 PM »
I think that using the geometry to find the hybridization assumes that you have a homoleptic complex.  I'm specifically interested in the hybridization of the orbitals in the metal fragment of a metallabenzene.  I have a rather strong disagreement with my major professor on this issue.  He continues to claim that p orbitals aren't involved in the pi system.  When I ask him where this comes from he can't answer except that nobody has ever mentioned the p orbitals.

Offline Will

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Re: Transition Metal Hybridization
« Reply #5 on: July 06, 2006, 02:29:52 PM »
I think that using the geometry to find the hybridization assumes that you have a homoleptic complex.

You probably know more about this than me but I would be interested to know how different ligands would change the hybridisation of the central metal atom/ion.

I'm specifically interested in the hybridization of the orbitals in the metal fragment of a metallabenzene. I have a rather strong disagreement with my major professor on this issue. He continues to claim that p orbitals aren't involved in the pi system. When I ask him where this comes from he can't answer except that nobody has ever mentioned the p orbitals.

Hehe, you have a pretty cr*ppy professor! I would get another professor's opinion or wait until someone smart enough here can answer you.

Can you say what the metallabenzene is you are referring to?

Offline aromatic

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Re: Transition Metal Hybridization
« Reply #6 on: July 06, 2006, 04:56:12 PM »
I think that without a homoleptic complex, one ligand may have better overlap with one type of orbital so that all of the hybridizations would distribute differently.

I work with both platinabenzenes and iridabenzenes.  The platinabenzene has a Cp* ligand and the Iridium has two phosphines and a carbonyl.  Technically the iridabenzenes are a cross between trigonal bipyramidal and square pyramidal but they crystalize as square pyramidal.

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