Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: aromatic on July 05, 2006, 06:28:19 PM
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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?
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In complexes, platinum's electrons would be hybridised with the 6p subshell. I frequently see platinum in dsp2 (square planar) hybridised complexes.
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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?
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What would the energy diagram look like with hybridized orbitals.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.unine.ch%2Fchim%2Fchw%2Fhtml%2520files%2Fsplitting%2520sq.jpg&hash=24ddb9de45a057e31c08e7a9c6335147e6ab95dd)
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 (http://courses.cm.utexas.edu/archive/Fall2001/CH301/Labrake/SI/geometryprobs.pdf).
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. :)
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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.
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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?
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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.