Filled d orbitals of the metal can have correct symmetry to back donate electron density to the pi* orbital of the CO ligand (this interaction strengthens the bonding with respect to the M-L bond, but weakens its with respect to the CO pi bond - This can actually be seen in the IR spectra of these complexes). The pi* orbital has greater coefficient on the C atom, and so CO often bonds via carbon as it allows for better orbital overlap between the d and pi* orbitals.
(edited 10 years ago)
Original post by Oxazole
Filled d orbitals of the metal can have correct symmetry to back donate electron density to the pi* orbital of the CO ligand (this interaction strengthens the bonding with respect to the M-L bond, but weakens its with respect to the CO pi bond - This can actually be seen in the IR spectra of these complexes). The pi* orbital has greater coefficient on the C atom, and so CO often bonds via carbon as it allows for better orbital overlap between the d and pi* orbitals.
Original post by Big-Daddy
Why is it impossible for O to act as the donor when CO is a ligand in a complex ion? O has a lone pair after all, but for some reason it is always C that coordinates to the metal atom, why?
This is known as synergic bonding. The CO LUMO is of the correct symmetry to overlap with the some of the occupied d orbitals, leading to an overall bonding interaction.
As the largest coefficient of the LUMO is placed on the C, bonding at that site is preferential.
Sorry, I know this is largely repeating your stuff......
Original post by Oxazole
Filled d orbitals of the metal can have correct symmetry to back donate electron density to the pi* orbital of the CO ligand (this interaction strengthens the bonding with respect to the M-L bond, but weakens its with respect to the CO pi bond - This can actually be seen in the IR spectra of these complexes). The pi* orbital has greater coefficient on the C atom, and so CO often bonds via carbon as it allows for better orbital overlap between the d and pi* orbitals.
Thanks, but wouldn't this then suggest that C is generally a better donor than O to metal complexes, which I am feeling a bit sceptical about ... O coordinates all the time, just not from CO, so why in particular is it difficult for O to coordinate here? I would expect some answer which applies only somewhat rarely, since in most situations O does/can provide the dative covalent bond ...
Original post by Big-Daddy
Thanks, but wouldn't this then suggest that C is generally a better donor than O to metal complexes, which I am feeling a bit sceptical about ... O coordinates all the time, just not from CO, so why in particular is it difficult for O to coordinate here? I would expect some answer which applies only somewhat rarely, since in most situations O does/can provide the dative covalent bond ...
The situation in metal complexes is a little more comlex (see what i did there
) than just which atom is the best sigma donor to the metal in terms of forming the dative bond. The ability of a ligand to act as an electon acceptor is also a very important consideration, and ultimatley that is why CO coordinates via C.Carbon is also by no means bad at forming dative bonds to metal centres, there are plently of examples of carbon based ligands coordinating to metals (it's actually a massive field of chemistry called organometallic chemistry).
Original post by Oxazole
The ability of a ligand to act as an electon acceptor is also a very important consideration, and ultimatley that is why CO coordinates via C.
You mean electron donor?
In general, is there a good principle to predict which atom(s) in the ligand will be able to coordinate to the metal atom? (obviously besides the requirement that the coordinator has a lone pair to donate)
Original post by Big-Daddy
You mean electron donor?
In general, is there a good principle to predict which atom(s) in the ligand will be able to coordinate to the metal atom? (obviously besides the requirement that the coordinator has a lone pair to donate)
In general, is there a good principle to predict which atom(s) in the ligand will be able to coordinate to the metal atom? (obviously besides the requirement that the coordinator has a lone pair to donate)
Sadly there aren’t many hard and fast rules that work all the time, but there certainly are concepts which can help you deduce the type of bonding going on in metallic complexes.
The bonding scheme in metal complexes is a little more nuanced compared to other compounds and the answer to your question requires a pretty decent grasp of molecular orbital theory and some knowledge of group theory. This is really 1st/2nd year undergrad stuff so I’m sorry If this answer is a little too in depth/drags on. If you do want to learn more about it I would suggest picking up a copy of Shriver and Atkins’ Inorganic Chemistry (the earlier editions are quite cheap and just as good as the latest edition).
As was mentioned before, this concept of synergistic bonding is very important. There are two main bonding interactions you have to consider. The examples below refer to a 4th row octahedral complex (though the concepts extend to other complexes).
Sigma bonding:
This refers metal-ligand bonds that have sigma symmetry about the bond axis (you'll recognise this as the dative M-L bond). There are 6 ligands each donating an electron pair to the metal, the orbitals on the metal which have correct symmetry to accept these electrons are the 3x4p, 4s and 2x3d orbitals (this result comes from group theory, but for now you’re just gonna have to take my word for it). The result of this interaction is the formation of 12 new molecular orbitals (six bonding MOs and six antibonding MOs). The six bonding MOs are completely filled with the electrons from the ligands and the six antibonding MOs are unfilled. The 3 remaining d orbitals don't have correct symmetry to sigma bond with the ligands and so they remain non-bonding and fully localised on the metal atom.
Something to bear in mind is that for sigma bonding the ligand electrons don’t just have to be lone pairs, they can even be bonds - for example the pi bond of an alkene or the sigma bond of dihydrogen (as long as they have sigma symmetry that is what’s important!)
Pi bonding:
If the ligand also has orbitals of pi symmetry (in the case of CO this would be the pi* orbital) then you get a second bonding interaction between the metal and the ligand. The 3 nonbonding d orbitals have pi symmetry and so they can overlap with the pi orbitals on the ligand. The result of this interaction is the formation of 6 new MOs (three bonding MOs and three antibonding MOs). Now provided the pi orbitals on the ligand were empty (these type of ligands are known as pi acceptor ligands) the electrons from the metal now occupy these new bonding MOs. So overall this pi interaction has lowered the energy of the system by allowing the electrons to occupy lower energy bonding MOs instead of higher energy nonbonding orbitals.
So ultimately if you have a ligand that as two or more potential coordination sites you have to consider both the sigma bonding ability AND pi bonding ability of each of these sites when rationalising how it will actually bond.
(edited 10 years ago)
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