4
$\begingroup$

Out of curiosity, what would the group orbitals be like in a C4v complex (square planar complexes or octahedral complexes with 5 identical ligands and 1 other) look like? I get a irreducible representation of 2A1 + B1 + E, which raises the question of how the 2 different A1 group orbitals look like.

Can someone explain this to me?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
3
$\begingroup$

I don't have time for a full answer or derivation right now, but here is a rough explanation. We can just consider s-type orbitals on each of the five ligands.

If the group orbital is to transform under the TSIR, then each of the four "equatorial" orbitals must have the same phase as each other. This is because these orbitals can be interconverted by symmetry operations, and if the character under every operation is to be $1$, then every orbital has to have exactly the same contribution to the group orbital. However, the equatorial ones don't need to have the same phase/magnitude as the "axial" orbital, because the equatorial/axial positions are not symmetry-related.

Therefore, you end up with two group orbitals that look somewhat like this:

Very rough illustration of C4v group orbitals

Obviously, the exact relative contribution will depend on the system, but this should be the correct form.

For nearly any question of this nature, an excellent reference is Albright, Burdett, and Whangbo's Orbital Interactions in Chemistry, 2nd ed. Unfortunately, I can't find a diagram for C4v MOs, but in general most things can be found there.

Improve this answer
$\endgroup$
8
  • 2
    $\begingroup$ There is one for $\ce{C5H5+}$ on p. 263. $\endgroup$
    – andselisk ♦
    Commented Feb 14, 2019 at 12:43
  • $\begingroup$ I presume projection operator method breaks down here, since by projection operator all the orbitals won’t have the axial ligand contribution. Could I inquire how we derive this then? $\endgroup$
    – ANZGC FlyingFalcon
    Commented Feb 14, 2019 at 14:15
  • $\begingroup$ Conceptually, you can just start from the Oh ligand groups and remove one ligand. That eliminates one of your t1 orbitals, so the two left are now e. And the loss of symmetry means that your e orbitals from Oh become a1 and b1. The a1 from Oh remains a1. $\endgroup$
    – Andrew
    Commented Feb 14, 2019 at 15:26
  • 1
    $\begingroup$ @ANZGCFlyingFalcon, I am quite rusty with the projection operator, but my gut feeling is if you start with the axial orbital then the projection operator will simply return the axial orbital itself as the group orbital. That is quite valid, too; because in a real system, you would still be able to construct the MOs from linear combinations of those. $\endgroup$
    – orthocresol
    Commented Feb 14, 2019 at 15:47
  • 1
    $\begingroup$ I just ran through the projection method and it worked fairly well. If we number the axial ligand as 1 and the others as 2,3,4 and 5 going around the square, I got 1 alone or 2+3+4+5 for the A1 projections. As orthocresol noted, these can be combined into the two a1 orbitals drawn above. For b1, I get something of the form 5+3-4-2 (ie opposite corners are same phase) and for e I get 2-4 or 3-5 (only two ligands- opposite corners out of phase). The e orbitals match up to the metal px and py, the b1 to d(x2-y2) and the a1's to s and d(z2) or to s and pz $\endgroup$
    – Andrew
    Commented Feb 14, 2019 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.