Nerro
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Fe(CO)5 and it's orbitals
I have here in front of me a test from last year's course on coordination chemistry, one of the tasks is to describe the electronic environment around
metal cores in complexes.
The ironpentacarbonyl complex is given (Fe(CO)5), the answer given by the teacher is that it's a TBP configuration in low spin with Fe(0) so it's
specified as d8. So far so good but then he says it's got an empty d<sub>z²</sub> orbital.
I just can't seem to figure out why it would have that empty orbital... Since it's low spin it would seem to me that it would have full
a<sub>1g</sub>; three full t<sub>1u</sub> and two full e<sub>g</sub> orbitals and three full (non-bonding)
t<sub>2g</sub> orbitals.
The e<sub>g</sub> orbitals are the d<sub>z²</sub> and the d<sub>x²-y²</sub> orbitals, but as I just said I
think those are full in this complex.
Can someone prove me wrong please?
[Edited on Tue/Oct/2006 by Nerro]
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guy
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I think its because of pi backbonding, it interacts with only the dxy dxz and dyz, and fills it up without having to fill up the dz and dx2-y2. It
donates 6 electrons from that orbital and gains back 6. Fe is a d6 not d8, so therefore the e<sub>g</sub> orbitals sohould be empty.
[Edited on 10/31/2006 by guy]
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Nerro
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But he specifically names the z² and not the x²-y²...
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guy
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Are you doing ligand field theory, or just simple coordiation complexes without MO theory?
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Nerro
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ligand field theory
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guy
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Alright I think I got it figured out.
Fe(CO)5 has 4 pi bonded CO and one sigma bonded CO(following the 18-electron count rule). The sigma bond is created using the s orbital, and the
others are dxy, dxz, dyz, and dx2-y2. Its impossible to form a bond with the dz2 because of its shape (donut). The donut ring cannot possibly
overlap with the CO to form a pi bond.
[Edited on 11/1/2006 by guy]
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Nerro
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Why would all bonds be pi-bonds? I would expect all bonds to be sigma bonds...
The structure of CO suggests that every C would donate one pair of electrons from a full sp hybrid orbital directly opposite from the O in CO.
wouldn't it?
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guy
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No,because you have to follow the 18 electron rule, and plus, CO bonds are known for their pi backbonding abilities.
http://en.wikipedia.org/wiki/18_electron_rule
The only way I can see it is: 4 d orbitals being synergically bonded to CO, and one sigma bonded to the s orbital.
So you have a orbitals created from the s, p, and d orbitals. The d orbitals are gonna be split into 4 (dxy, dxz, dyz, dx2-y2) and 1 (dz2). You have
10 donor electrons from the CO and 10 empty pi* orbitals(only 8 will be filled). As the s and p get filled, the d's get bumped into the antibonding
CO orbitals. So you will have a total of 18 electrons and end up with 4 double bonds and 1 single bond. Most websites don't show the CO bonds
correctly; look it up in Linus Pauling's book "General Chemistry".
[Edited on 11/1/2006 by guy]
[Edited on 11/1/2006 by guy]
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