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I read that B3+ ions do not exist in aqueous solution, because hydration energy cannot compensate for the sum of first three ionisation energies.

This leads me to the following questions:

  1. If boron (III) ion really doesn't exist, how was the hydration energy data calculated? and ionisation potentials?

  2. Boron being small in size and carrying a high charge of +3 (if it exists) should be greatly hydrated. This high hydration energy should be able to compensate for the ionisation energies, but it is not so.

Could someone please explain? Thanks!

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    $\begingroup$ The formation of an ordered hydration shell decreases the entropy of the system. The hydration of the ion may result in a huge decrease in entropy, which makes the overall process unfavourable. $\endgroup$ – Tan Yong Boon Sep 17 '17 at 14:15
  • $\begingroup$ Why should all hydration shells be ordered? $\endgroup$ – arya_stark Sep 17 '17 at 14:16
  • $\begingroup$ Also, if ∆S < 0, the Gibbs free energy would be negative for highly negative values of ∆H. How do we rule out or explain that possibility? $\endgroup$ – arya_stark Sep 17 '17 at 14:17
  • $\begingroup$ According to most chemical literature I have read, the hydration of any ion has its own associated ordered hydration shell necessarily. This is because of the attraction between the partial negative charge on the water molecule and the positive charge of the cation, in the case of positive ions. This attraction sort of fixes the position of the water molecules. $\endgroup$ – Tan Yong Boon Sep 17 '17 at 14:18
  • $\begingroup$ We can rule it out based on our observation that it does not exist. For these kind of thermodynamics things, we can only use data and empirical observations to substantiate. $\endgroup$ – Tan Yong Boon Sep 17 '17 at 14:21