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The Re2Cl8 2- ion is composed of two square planar ReCl4 units joined by four Re- Re bonds (quadruple bonding) and has D4h symmetry.

(a) Determine the symmetries (irreducible components) of the MOs responsible for the d-d metal-metal quadruple bonding in this molecule.

This is the entire question. However, I only need help genereating the irreducible representations. There are four sets of irreps that need to be generated that are based on the dz^2 orbital, dxy orbital, dyz orbital, and dxz orbital. I have only been able to determine the irrep for the dz^2 orbital. Here is what I did:

  1. I transformed the dz^2 orbital under every symmetry operation from the D4h point group. (E, 2C4, C2, 2C2', 2C2'', i, 2S4, sigma h, 2 sigma v, and 2 sigma d. )

  2. I got a reducible representation of E=2, 2C4=2, C2=2, 2C2'=0, 2C2''=0, i=0, 2S4=0, sigma h=0, 2 sigma v=2, 2 sigma d=2

  3. Lastly, I used tabular reduction to get that dz^2 = A1g +A2u

For the other three d orbitals I can not seem to get the correct irreps.

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  • $\begingroup$ I'm confused how you are getting your reducible representation. Since dz^2 is a single orbital, it should only correspond to a irrep. I'm not sure how you are getting 2 for the character of the identity operator. The symmetry of the Cartesian functions can be found in the last two columns of most character tables if you want to check your work. $\endgroup$
    – Tyberius
    May 1 '20 at 21:22
  • $\begingroup$ I can add a picture of the work for the dz^2 irreps. The dz^2 irreps were solved by my instructor. I know how to look at the character table to get each irreps corresponding to the d-orbital. But I am asked to find the irreps by hand. This is also very confusing to me as far as how to do it. I know that you treat the d orbital with each operation and see how the orbital changes. If it stays the same then the value=1, if it moves the value is =0, if it inverts the value is =-1 $\endgroup$
    – Lu san
    May 1 '20 at 21:36
  • $\begingroup$ The procedure you describe sounds correct. So when you apply E to the dz^2 orbital, what happens to it? As to the character table, you can just use that to check that your by hand result is correct. $\endgroup$
    – Tyberius
    May 1 '20 at 21:39
  • $\begingroup$ Under the E operation nothing happens so I would have a value of +1, or in this case +2 because I am looking at both d z^2 oprbitals on the Re2 $\endgroup$
    – Lu san
    May 1 '20 at 22:39
  • $\begingroup$ Okay sorry I see what you are doing now. Your method will work fine. Could you should what you are getting for the reducible representation of one of the other orbitals. $\endgroup$
    – Tyberius
    May 1 '20 at 23:00
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I'll try to walk through how it works for $d_{xy}$. You will need to consider $d_{yz}$ and $d_{xz}$ as a pair later when you attempt those.

Obviously $\chi(E)=2$. Now for the $C_4$ rotation, just think about rotating one of the orbitals in the $xy$ plane (the same thing will happen to the other orbital, so we will then double the character). When we do this, the orbital remains in place, but it's positive and negative lobes invert. So it will contribute -1 and for both orbitals $\chi(C_4)=-2$. For $C_2$, the rotation again leaves the orbital in place and leaves the lobes in each location with the same sign so $\chi(C_2)=2$.

$\chi(C_2')=\chi(C_2'')=\chi(i)=\chi(S_4)=\chi(\sigma_h)=0$ since all of these operations involve a rotation/reflection around a parallel line/plane so they will move the orbitals.

$\sigma_v$ runs along the $\ce{Re-Cl}$ bonds. Along and on either side of this plane, the sign of the lobes is the same and so $\chi(\sigma_v)=2$. $\sigma_d$ runs between the $\ce{Re-Cl}$ bonds and so this plane inverts the sign and $\chi(\sigma_d)=-2$.

Plugging these into a calculator I get a $b_{1g}$ and a $b_{2u}$ irrep. I will leave it to you to figure out which of these is the bonding combination and which is antibonding.

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  • $\begingroup$ Thank you, this makes a lot more sense. One question, what do you mean by consider dyz and dxz as a pair? I know on the character table they are a degenerate set, but how will that affect how I go about finding their irreps? $\endgroup$
    – Lu san
    May 2 '20 at 0:22
  • $\begingroup$ So when I apply the C4 operation to dxz I get dyz, so what does that mean as far as the character? (is it +2, +4?) Also when I do the C2 operation for the dxz orbital, I get the inverted dxz orbital so is the character -2? $\endgroup$
    – Lu san
    May 2 '20 at 0:30
  • $\begingroup$ Basically, I just mean that the irreps you eventually get will be degenerate and if you just try to break the reducible representation of one orbital or the other into irreps it won't work. $\endgroup$
    – Tyberius
    May 2 '20 at 1:15

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