Why can't oxalate ion donate two pairs of electrons from the two double-bonded oxygen atoms So, I read that an oxalate ion is a bidentate ligand but since I saw that it has 4 oxygen atoms I was confused as to why it isn't a tetradentate ligand instead.


I searched it on Google and found this:
Oxalate ion dentiticity - why is it bidentate and not tetradentate?

Now, in the link it's implied that the oxalate ion can donate the 2 pairs of electrons either from the 2 oxygen atoms having negative formal charge or from 1 "double-bonded" O atom and 1 negatively charged O atom. 

But why can't the two pairs of electrons be donated from Both the "double-bonded" O atoms? 
 Why isn't this "third" bonding mode possible?

EDIT:  From the comment below, I infer that because of resonance, there is no double-bonded or negatively charged O atom in the Oxalate ion; rather all of the O atoms have a partial negative formal charge and double-bond character, is that correct? 

So, can electrons be donated from any O atom in the ion? 

(And so did the answer in the provided link just ignore this for the sake of simplicity?)
Oxalate is a planar molecule. The four oxygen atoms share the pi electron density through resonance, as noted in comments, and thus are all identical. Imagine those four oxygen as four corners of a table. Now you are a Lewis acid, perhaps a transition metal cation, and you walk up to the table. How many of the four corners ligate you? Perhaps you imagine the planar ligand dropping on top of you, so that the four corners are around you. For effective overlap with all four corners of the table, you'd need to be a pretty big cation. In almost all cases you'd achieve better orbital overlap, and a more energetically favorable coordination complex, if you just cozy up to two of the donor oxygens.