4
$\begingroup$

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. enter image description here

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?)

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Well, you seem to don't understand that in oxalate all 4 atoms are identical. Write mesomeric structures if you need. $\endgroup$
    – Mithoron
    Sep 2, 2017 at 22:12
  • 1
    $\begingroup$ @Mithoron, so in an oxalate ion there is no two "double-bonded" oxygen atoms as such, rather all of the oxygen atoms have partial negative charge and also partial double-bond character? Then did the answer in the provided link just ignore this in order to give a simple explanation and in fact the two pairs of electrons can get donated from Any two adjacent O atoms? $\endgroup$
    – HeWhoMustBeNamed
    Sep 3, 2017 at 3:13
  • $\begingroup$ Yeah, it's pretty much as you say. $\endgroup$
    – Mithoron
    Sep 3, 2017 at 15:50
  • 1
    $\begingroup$ I am very tempted to find a cool crystal structure with tetradentate oxalate (but it’s past midnight and I need to sleep). However, for coordination to a single metal cation, the four-membered ring that would form if you were to use both oxygens of a carboxy group is too small. Hence bidentate ligands always form at least a five-membered ring with the metal. $\endgroup$
    – Jan
    Sep 5, 2017 at 15:18

1 Answer 1

Reset to default
-1
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I don't understand this answer. Where do the tables come from? $\endgroup$
    – wizzwizz4
    Nov 25, 2018 at 11:32
  • $\begingroup$ It's a metaphor, the four oxygen atoms can be likened to the four corners of a table $\endgroup$
    – electronpusher
    Mar 2, 2019 at 3:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.