How to assign symmetry to water molecule using D4h symmetry?

[fangrz]

Could you explain why an H2O that attacks Ni(CN)4 ^(2-) (in square planar) (which has D4h symmetry) at one of the axial sites would have A2u symmetry? I know that the "HOMO" for the p orbital of the H2O would be a p orbital. I also tried to link that electron pair to an irreducible representation in the D4h character table.

Going across the row:

E 2C4 (z) C2 2C'2 2C''2 i 2S4 σh 2σv 2σd

I got the following numbers: +1, +1, +1, 0, 0, 0, 0, 0, +1, +1...which doesn't correspond with any of the irreducible representations. (When the H2O flips because of a C2' or C2'' rotation (and others...), it gives 0s. Was there something wrong in my approach? I used the metal center as the center for all operations.

 

[DrDu]

The water molecule would lower the symmetry of the complex to at least C2v. Then the p orbital has simply the symmetry it has in water alone.