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The following is written in my book Concise Inorganic Chemistry by J. D. Lee.

Hydration energy, hydrated radius and hydration number of a particular ion depends upon charge per unit area. Hence their order is the same for the following cations in the given examples:

$\ce{Li+(aq) > Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)}$

And other similar examples are also given. My question is, why does charge to surface area ratio matter? The force felt by the water molecules due to any of these cations should be the same due to shell theorem. Also, a cation of bigger size can accommodate more water molecules in its vicinity due to its size. Then still why does an increase in charge to surface area ratio corresponds to an increase in hydration tendency?

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  • $\begingroup$ "My question is, why does charge to surface area ratio matter?" (+1) I've always wondered that myself, but never quite figured it out :( $\endgroup$ – Gaurang Tandon Feb 13 '18 at 9:47
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    $\begingroup$ Related: chemistry.stackexchange.com/questions/2883/… $\endgroup$ – Gaurang Tandon Feb 13 '18 at 9:48
  • $\begingroup$ Could be that hydrogen bonds need polarity to work and charge density simply means that the polarity is high. $\endgroup$ – Avnish Kabaj Feb 13 '18 at 10:23
  • $\begingroup$ @Avatar I am unable to understand what do you mean. $\endgroup$ – Apoorv Potnis Feb 13 '18 at 10:27
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Many people have done experimental, theoretical, statistical, molecular dynamics, etc. studies of solvation free energies of ions. I encourage you to just search solvation free energies of ions on google scholar and you'll find a bunch of very interesting and quite readable papers.

One of these papers ref. [1] (here for pdf) provides a lot of data and a very simple model which can basically be used to answer your question. In this paper, they derive a simple model which has only one fitting parameter. From this, they can determine the effective cavity size of an ion in solution. This cavity size is given by $r+\Delta r$, where $r$ is the radius of the ion and $\Delta r$ is the thickness of the shell.

Essentially, the value of $\Delta r$ is determined by the fitting parameter, and also determines the number of water molecules within the cavity of the ion. A portion of the data table is shown below with actual experimental values on the far right and parameters determined from the model in the other columns.

Solvation free energies

Just focus on the alkali metals which you list in your question. Notice as we go further down, the value of $\Delta r$ approaches zero and the actual radius of the ion is also the size of the cavity. This, however, is very much not true for the smaller ions. The Hydrogen ion being an extreme example where a solvation number of 12.0 is obviously unphysical. This, however, does provide a clue for a reasonable interpretation of $\Delta r$. $\Delta r$ is a measure of how far away a molecule must be for it to not "notice" the effect of an ion in solution. Thus, somewhat counter-intuitively, smaller ions interact effectively with a larger number of water molecules. The reason for this is that as the ions get larger, solvating them requires a greater reorganization of the hydrogen-bonding network in liquid water. This means that one must lose hydrogen bonds in order to solvate the ion.

Thus, the reason the charge to area ratio dictates the solvation energy and the solvation number is because a smaller ion will bring the same charge into solution, but due to its smaller size, will only require a small reorganization of the hydrogen bonding between waters, which minimizes losses in already stable interactions. This lack of reorganization is what leads to large values of $\Delta r$ and hence large values of $n$ in the model described above.

Let me be clear, $n$ is not simply the number of water molecules pointing towards an ion. This would not be measurable experimentally, or well-defined theoretically. Rather, $n$ will always be some kind of average value which tells you the effective number of water monomers which energetically benefit from the presence of the ion in solution. So, now the picture is clear. The smaller cations do not cause too large of a first hydration shell, so that the second hydration shell is closer to the ion than for a larger ion. Hence, the free energy and the solvation number are larger for smaller ions of the same charge.

As a bonus, ref. [2] is an interesting paper which approaches this problem from molecular dynamics. They give plots showing the relative effects which van der Waal's forces and electrostatics have. Note that the smaller ions are the only ones which are stabilized effectively by van der Waal's interactions as these scale closer to $r^{-5}$ rather than $r^{-1}$


References:

[1] Marcus, Y. (1991). Thermodynamics of solvation of ions. Part 5.—Gibbs free energy of hydration at 298.15 K. Journal of the Chemical Society, Faraday Transactions, 87(18), 2995-2999.

[2] : Grossfield, A., Ren, P., & Ponder, J. W. (2003). Ion solvation thermodynamics from simulation with a polarizable force field. Journal of the American Chemical Society, 125(50), 15671-15682.

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