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


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. )
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
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.
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.