[CopperSmurf]

Hey.

I'm confused, how would one determine an electronic configuration for a complex like IrCl₆^3-? Since it is octahedral, I only know that it is usually represented as t_2g^xe_g^y where x and y are some numbers. I think there's 9 d electrons. What do t_2g and e_g represent anyways?

[cth]

T_2g and E_g are notations from the group theory http://en.wikipedia.org/wiki/Group_theory which deals with symmetry operations, applied to chemistry.

The 5 d-atomic orbitals are gathered into two groups, namely T_2g and E_g, depending on how they behave when symmetry operations (from the octahedral geometry, http://en.wikipedia.org/wiki/Octahedral_symmetry) are applied to them.
T_2g contains the orbitals d_xy, d_xz and d_yz.
E_g contains d_x²-y² and d_z²
The g from T_2g and E_g means that the orbitals considered are symmetric by the inversion centre. It comes from German gerade.
T_2g is lower in energy than E_g for an octahedral geometry.

Iridium metal Ir⁰ has 9 valence electrons. So, Ir³⁺ has 6 electrons.

So, how would you arrange the electrons into T_2g and E_g, to get what you named T_2g^xE_g^y?
 

[CopperSmurf]

The g from T_2g and E_g means that the orbitals considered are symmetric by the inversion centre. It comes from German gerade.
T_2g is lower in energy than E_g for an octahedral geometry.

Iridium metal Ir⁰ has 9 valence electrons. So, Ir³⁺ has 6 electrons.

So, how would you arrange the electrons into T_2g and E_g, to get what you named T_2g^xE_g^y?

if you don't mind me asking, how do you know so quickly that T_2g is lower in energy than E_g for an octahedral? this would be important to know when determining high spin vs low spin right?

Ir³⁺ would have 6 valelnce electrons yes, but would the chlorides (or any other ligands like water or PPh₃) also contribute to these as well? If not, then I would have guessed that the electronic configuration is T_2g⁴E_g², high spin (judging by what you said earlier.

[cth]

if you don't mind me asking, how do you know so quickly that T_2g is lower in energy than E_g for an octahedral?

Simply because it is always like that: T_2g is always lower in energy than E_g for an octahedral geometry.  (Remark, it is not necessarily true for a different geometry.)
Why is it so?
If you look at the orbitals belonging to the group E_g, namely d_z² and d_x²-y², they are pointing directly towards the Cl^- ligands  they are more destabilised.
And for the group T_2g, the 3 orbitals d_xy, d_xz and d_yz that it contains are not pointing directly towards the ligands, but rather towards the space in between them  T_2g is less destabilised.

this would be important to know when determining high spin vs low spin right?

What is important when determining high or low spin, is the energy gap between E_g and T_2g.
If E_g is much higher in energy than T_2g, the electrons will occupy T_2g first. You have a low spin.
If E_g is close in energy from T_2g, the electrons will occupy both T_2g and E_g. You have a high spin.

Ir³⁺ would have 6 valence electrons yes, but would the chlorides (or any other ligands like water or PPh₃) also contribute to these as well?

No, you don't count electrons from ligands. We consider only valence electrons coming from Ir³⁺.

If IrCl₆^3- is low spin, then you have T_2g⁶E_g⁰. The compound would be diamagnetic.
If IrCl₆^3- is high spin, then you have T_2g⁴E_g². The compound would be paramagnetic.
I am not sure which one of the two possibilities is right. I don't know the T_2g-E_g energy gap for that compound.

[CopperSmurf]

If IrCl₆^3- is low spin, then you have T_2g⁶E_g⁰. The compound would be diamagnetic.

If IrCl₆^3- is low spin, then could it also be T_2g⁵E_g¹ ? Two possible low spins?

[cth]

No, there are only two possibilities: one with low spin and one with high spin. T_2g⁵E_g¹ would be an excited state, not the ground state.

Look at this picture taken from http://en.wikipedia.org/wiki/Crystal_field_theory

Up you have E_g, down you have T_2g.

On one hand, putting 2 electrons into the same orbital (on with spin +1/2, the other one with spin -1/2) cost more energy than putting them into two distinct orbitals with the same energy (for example d_xy and d_xz in this case). Indeed, all electrons have a -1 electric charge, so keeping them next to one another is destabilising and costs energy.
On the other hand, putting one electron from T_2g to E_g is also destabilising because E_g is higher in energy.

You need to consider both energies and go for the lower one.
If T_2g-E_g energy gap is smaller than the pairing energy, then electrons will go to E_g rather than having 2 electrons into one orbital.
If T_2g-E_g energy gap is larger than the pairing energy, then you get 2 electrons per T_2g orbitals before putting any into E_g.

For T_2g⁵E_g¹:
- in low spin, E_g is high in energy  the one electron in E_g will go to T_2g because it is more energetically stable. You get T_2g⁶E_g⁰
- in high spin, E_g is not so high in energy  one electron from T_2g will go to E_g. The energy you lose by promoting one electron into a higher energy orbital is more than compensated by the gain in pairing energy. You have T_2g⁴E_g².
So, T_2g⁵E_g¹ is an excited state. There is a way to lower the energy in both situations.