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relative energies of orbitals using CFT


psi * psi

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I have a question that asks me to divide the d orbitals of the metal atom in the trigonal prismatic ML6 complex into their symmetry types. I've answered this part of the question, but I'm not sure I answered the question the problem was asking. Does it just mean look at the character table (D3h) and say:

dz2=A1'

dx2-y2, dxy= E'

 

etc?

 

Then the question asks to predict the relative energies using CFT. I have dz2>dx2-y2>dxz, dyz>dxy, but I'm not sure that's right...

Edited by psi * psi
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you can think about this geometrically. Draw out the trigonal prismatic geometry with the set of d-orbitals overlayed.

 

Think of an easier case. Are you familiar with the crystal field splitting arrangement for the octahedral geometry? Why is the set with t2g symmetry lowest in energy for octahedral complexes?

 

 

Apply that same logic to the trigonal prismatic case remembering that there may be more than two sets of degenerate symmetries in this case.

Edited by mississippichem
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Are bonded states lower in energy than non-bonded states among otherwise degenerate atomic (d) orbitals?

 

If they were, that wouldn't explain why t2g is lower in energy. my notes say that t2g is lower in energy because of the geometrical arrangement of the orbitals- in the t2g arrangements, the orbital lobes are directed in between the bonds of the ligands instead of along them.

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If they were, that wouldn't explain why t2g is lower in energy. my notes say that t2g is lower in energy because of the geometrical arrangement of the orbitals- in the t2g arrangements, the orbital lobes are directed in between the bonds of the ligands instead of along them.

 

I was trying to imply that the t2g orbitals are not non-bonding. They do in fact contribute to the set of molecular orbitals just less so than eg.

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