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I got that multidentate ligands form more stable coordination complexes than monodentate ligands, but why?


marked as duplicate by Mithoron, M.A.R. ಠ_ಠ, airhuff, Todd Minehardt, Community Jul 16 '17 at 19:49

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The chelate effect is generally agreed to be a thermodynamic effect caused by the change in entropy upon binding of a bidentate ligand.

General description of the chelate effect:

Consider a coordination complex, [M]L2, where L = a standard monodentate ligand. If we now add a bidentate ligand, two monodentate ligands are released. enter image description here

The ligand substitution reaction in the forward direction is therefore generally favourable: ∆S, the entropy term is positive due to two molecules reacting to become three, with this favourable entropy term having the effect of tending to make ∆G negative (negative ∆G is favourable, i.e. the reaction likely to proceed).[*]

Once a bidentate ligand is bound, it then becomes less favourable to go in the other direction (the entropy term would be negative due to entropy decreasing in the system, which in turn would tend to make ∆G positive, and hence the reaction unfavourable).

Example and thermodynamic data:

To give a more concrete example demonstrating the significance of the entropy term, consider the following reaction and associated thermodynamic data (taken from Physical Inorganic Chemistry, S. F. A. Kettle, 1996):

enter image description here

enter image description here

Clearly, the entropy term dominates, causing ∆G to be negative in both cases (hence causing the forward reaction to be favoured in both cases).

[*]: The caveat here is that we're assuming that enthalpy really isn't contributing a whole lot, this is clearly a vast over-simplification, and there are cases where enthalpy becomes important (i.e. when the complex is significantly more stable than the starting materials). We're also ignoring the fact that the act of binding a bidentate ligand itself has some unfavourable entropy, since we're containing the ligand and removing some conformational flexibility

  • $\begingroup$ This answer seems convincing thermodynamically, but is there a more to-the-point explanation? An explain that involves the binding of the chelate ligand? $\endgroup$ – Pritt Balagopal Jul 16 '17 at 15:39
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    $\begingroup$ Feel free to answer Prit - th thermodynamic argument is the one I've most commonly came across. The binding enthlpy is, to my knowledge, negligible in relative terms $\endgroup$ – NotEvans. Jul 16 '17 at 15:41
  • $\begingroup$ I'd love to, too bad I'm not aware of such an answer anyway. I wouldn't ask you to answer if I could myself would I? :) $\endgroup$ – Pritt Balagopal Jul 16 '17 at 15:47

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