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Justanotherchemist
Justanotherchemist
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I was wondering why the second order Jahn-Teller effect is also said to happen for many d0-systems? It makes sense to me that due to relativistic effects and the stabilization of the s-Orbital and destabilization of the d-orbitals the energetic gap between the let's say eg in an octahedral complex and the above laying unoccupied a1g s-orbital becomes small enough that, for the right amount of electrons filled into the d-orbitals a mixing of both states is of some advantage leading to a second order Jahn-Teller effect like in Hg(II) compounds. But why does it also happen to d0-complexes?

My suggestion would be that we cannot discuss any field splitting in the first place. So first of all, how would any distortion here give us an energetic advantage and then also we would have to take the nephelauxetic effect of the host or ligand more into consideration if we do not consider a real crystal field splitting. I don't really know where a methyl ligand would be considered in the nephelauxetic series (since some heavy transition metal methyl complexes form trigonal prismatic complexes rather than octahedral ones, which is often explained with a second order Jahn-Teller effect) but if I consider some covalency in this metal-carbon bond then it may shift the d-orbitals down enough so they could perhaps mix with the lower occupied p-Orbitals. So that perhaps the mixing is not between occupied d- and unoccupied s-orbitals but between occupied p- and unoccupied d-Orbitals.

EDIT:

I was reading an article on the second order Jahn Teller effect yesterday and found the following line:

enter image description here

I am not sure if I got all those rules correctly but it seems like I have to consider the product of both states and the product has to be of the same symmetry. If I think about the next low lying occupied orbital in an octahedral environment it would be the filled t1g Orbital. So This would mean I have to consider a t1g and t2g mixing here. I have no clue about those irreducible representations and point groups but from the tables you can find on the net I found that a multiplication of T1 and T2 in Oh point-group would end up in an A2 + E + T1 + T2. As this contains both T1 and T2 I guess it will somehow obey those rules.

Perhaps someone with more experience in symmetry and character tables might have suggestion about this.

I was wondering why the second order Jahn-Teller effect is also said to happen for many d0-systems? It makes sense to me that due to relativistic effects and the stabilization of the s-Orbital and destabilization of the d-orbitals the energetic gap between the let's say eg in an octahedral complex and the above laying unoccupied a1g s-orbital becomes small enough that, for the right amount of electrons filled into the d-orbitals a mixing of both states is of some advantage leading to a second order Jahn-Teller effect like in Hg(II) compounds. But why does it also happen to d0-complexes?

My suggestion would be that we cannot discuss any field splitting in the first place. So first of all, how would any distortion here give us an energetic advantage and then also we would have to take the nephelauxetic effect of the host or ligand more into consideration if we do not consider a real crystal field splitting. I don't really know where a methyl ligand would be considered in the nephelauxetic series (since some heavy transition metal methyl complexes form trigonal prismatic complexes rather than octahedral ones, which is often explained with a second order Jahn-Teller effect) but if I consider some covalency in this metal-carbon bond then it may shift the d-orbitals down enough so they could perhaps mix with the lower occupied p-Orbitals. So that perhaps the mixing is not between occupied d- and unoccupied s-orbitals but between occupied p- and unoccupied d-Orbitals.

I was wondering why the second order Jahn-Teller effect is also said to happen for many d0-systems? It makes sense to me that due to relativistic effects and the stabilization of the s-Orbital and destabilization of the d-orbitals the energetic gap between the let's say eg in an octahedral complex and the above laying unoccupied a1g s-orbital becomes small enough that, for the right amount of electrons filled into the d-orbitals a mixing of both states is of some advantage leading to a second order Jahn-Teller effect like in Hg(II) compounds. But why does it also happen to d0-complexes?

My suggestion would be that we cannot discuss any field splitting in the first place. So first of all, how would any distortion here give us an energetic advantage and then also we would have to take the nephelauxetic effect of the host or ligand more into consideration if we do not consider a real crystal field splitting. I don't really know where a methyl ligand would be considered in the nephelauxetic series (since some heavy transition metal methyl complexes form trigonal prismatic complexes rather than octahedral ones, which is often explained with a second order Jahn-Teller effect) but if I consider some covalency in this metal-carbon bond then it may shift the d-orbitals down enough so they could perhaps mix with the lower occupied p-Orbitals. So that perhaps the mixing is not between occupied d- and unoccupied s-orbitals but between occupied p- and unoccupied d-Orbitals.

EDIT:

I was reading an article on the second order Jahn Teller effect yesterday and found the following line:

enter image description here

I am not sure if I got all those rules correctly but it seems like I have to consider the product of both states and the product has to be of the same symmetry. If I think about the next low lying occupied orbital in an octahedral environment it would be the filled t1g Orbital. So This would mean I have to consider a t1g and t2g mixing here. I have no clue about those irreducible representations and point groups but from the tables you can find on the net I found that a multiplication of T1 and T2 in Oh point-group would end up in an A2 + E + T1 + T2. As this contains both T1 and T2 I guess it will somehow obey those rules.

Perhaps someone with more experience in symmetry and character tables might have suggestion about this.

Source Link
Justanotherchemist
Justanotherchemist
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  • 38

Second order Jahn-Teller effect for d0 complexes

I was wondering why the second order Jahn-Teller effect is also said to happen for many d0-systems? It makes sense to me that due to relativistic effects and the stabilization of the s-Orbital and destabilization of the d-orbitals the energetic gap between the let's say eg in an octahedral complex and the above laying unoccupied a1g s-orbital becomes small enough that, for the right amount of electrons filled into the d-orbitals a mixing of both states is of some advantage leading to a second order Jahn-Teller effect like in Hg(II) compounds. But why does it also happen to d0-complexes?

My suggestion would be that we cannot discuss any field splitting in the first place. So first of all, how would any distortion here give us an energetic advantage and then also we would have to take the nephelauxetic effect of the host or ligand more into consideration if we do not consider a real crystal field splitting. I don't really know where a methyl ligand would be considered in the nephelauxetic series (since some heavy transition metal methyl complexes form trigonal prismatic complexes rather than octahedral ones, which is often explained with a second order Jahn-Teller effect) but if I consider some covalency in this metal-carbon bond then it may shift the d-orbitals down enough so they could perhaps mix with the lower occupied p-Orbitals. So that perhaps the mixing is not between occupied d- and unoccupied s-orbitals but between occupied p- and unoccupied d-Orbitals.