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For many molecules, the gas-phase molecular polarizability, $\alpha_{gas}$ is approximately the same as that in the liquid, $\alpha_{liq.}$. For instance, $\ce{H2O}$, has roughly the same average $\alpha_{liq.}$ as $\alpha_{gas}$. Many ions, especially anions, however, have much smaller $\alpha_{liq}$ than $\alpha_{gas}$, often by greater than 50%.

My question, is whether or not there is any simple theoretical framework, for understanding the decrease. I am well aware of the connection between polarizability and a measure of the atomic volume. It makes sense the atomic volume of an anion will be smaller in the condensed phase, but it is not clear what effects overcome the energetic penalty of shrinking the atomic volume.

There are many thoughts that come to mind:

  1. By decreasing the volume, one increases the charge density allowing for stronger coulomb interactions
  2. The polarizability is decreased due to charge transfer out of the anion (that is, the anion charge is smaller than in the gas-phase)
  3. The decreased anion volume is a result of balancing out the cavitation energy (that is, by shrinking, the solvent is better able to interact with itself)
  4. Some linear combination of the above
  5. Additional effects I am missing

Any references about the leading effect (if there is one) explaining why polarizabilities decrease in the condensed phase would be appreciated.

Also, any discussion of the relationship between partial charges and atomic polarizability would be appreciated. My pet theory, which I have no evidence to support, is that charge transfer from anion to solvent, is a large portion of the decrease. On that note, if I had an anion with a partial charge of -0.9 instead of -1.0, how much smaller should I expect the polarizability to be? Does it decrease linearly, quadratically, cubically? Put another way, what is the relationship between partial charge of an atom and its atomic radius?

Finally, any references to anion polarizability damping in non-water solvents would be appreciated.

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    $\begingroup$ It seems like an interesting question, but there is a lot going on and its difficult for me to determine what the core issue you are trying to solve is. Would it be possible to simplify/narrow down your question? $\endgroup$
    – Tyberius
    Commented Dec 12, 2022 at 15:12
  • $\begingroup$ I agree with Tyberius, are you able to simplify the question? Also let us know if you've figured out the answer! Or are you no longer actively in need of an answer? $\endgroup$ Commented May 28, 2023 at 2:58

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