1
$\begingroup$

The ionic radius of the $\ce{Ag^+}$ ion is $129$ pm, and that of the $\ce{Na^+}$ ion is $116$ pm.

Since the sodium ion is smaller than the silver ion, it makes sense that it has a stronger polarizing power than the silver ion. Due to the higher charge density, the lattice energy of $\ce{NaX}$ compounds is predicted to be larger than those of $\ce{AgX}$ compounds. However, that is not the case, as shown here:

enter image description here

Furthermore, the enthalpy of hydration of $\ce{Na^+}$ is expected to be more exothermic than that of $\ce{Ag^+}$, because $\ce{Na^+}$ is more polarizing than $\ce{Ag^+}$. However, that is not the case:

enter image description here

Why are the lattice enthalpies of silver compounds larger than those of sodium compounds, and why are hydration enthalpies of silver compounds more exothermic than those of sodium compounds, despite the silver ion being larger than the sodium ion?

Sources:

R. D. Shannon (1976). "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides". Acta Crystallogr A. 32: 751–767. Bibcode:1976AcCrA..32..751S. doi:10.1107/S0567739476001551.

Atkin's Physical Chemistry

$\endgroup$
  • 1
    $\begingroup$ Partly at least because ionic radii are not as uniquely or strictly defined as you would seem to believe. And, partly because lattice enthalpies vs ionic radii is a very tenuous connection. $\endgroup$ – Jon Custer Jul 9 at 18:51
  • 1
    $\begingroup$ The silver ion is larger and can have more interactions at the same time. $\endgroup$ – Karl Jul 9 at 20:00
  • $\begingroup$ ... and because it's relatively larger, it forms a different crystal structure, and then quantitative comparison of molar properties becomes very questionable $\endgroup$ – Karl Jul 9 at 20:52
  • $\begingroup$ Crystal structure shouldn't affect enthalpy of hydration, right? $\endgroup$ – DrPepper Jul 9 at 20:56
  • 1
    $\begingroup$ not "crystal", but (statistically) more water molecules fit around a larger ion $\endgroup$ – Karl Jul 9 at 21:12
1
$\begingroup$

The electron configuration of sodium ion is that of the inert gas neon (1s^2 2s^2 2p^6), a spherical shell with a positive charge at the nucleus. There are 10 electrons inside this shell.

The electron configuration of silver ion is that of krypton plus 4d^10, a spherical shell with a positive charge at the center. There are 46 electrons inside this shell.

While the slightly smaller sodium ion might have slightly more polarizing power than silver ion, silver ion has far more polarizability than sodium ion. Polarizability is the distortion of the electron cloud as a result of a point charge (or dipole). Dispersion forces are a form of polarizability where two polarizable entities polarize each other. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Polarizability

The effects of dispersion forces due to the polarizability of the anion can be seen by comparing the enthalpies. The crystal structures for the fluorides, chlorides and bromides and NaI are FCC, rock salt. AgI is wurtzite. The difference in enthalpies for the fluorides is 43, but as the polarizability of the chloride anion increases, the attraction of dispersion forces increases the difference to 125, and for bromide, 148. Even tho the crystal structure of AgI is different, it still seems to have a polarizablity-polarizability effect (difference = 181).

This is not what you would call a predictable effect, but does a good job of explaining the trend. It is not a first-order attraction, nor a second-order polarizability, but a third-order polarizability-induced polarization effect.

And very perspicacious of you to have observed this trend.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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