How to assign symmetry to water molecule using D4h symmetry?

  • Thread starter fangrz
  • Start date
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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

Science Advisor
6,010
748
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.
 

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