For example, silver chloride, silver sulfide, and silver oxide are three exceedingly stable silver salts in water solution.

Why is it that silver forms such stable salts?

Does this have to do with charge density of the component ions? This doesn't seem to hold water because we know that silver oxide is much more soluble than silver sulfide, and the oxide anion is obviously a lot smaller than the sulfide ion, so the oxide ion should have a much higher negative charge density.

So simple charge density rationalizations go out the window. I've Googled, I've looked in books, etc., but I can't find an explanation.

On the other hand, can a Lewis acid/base model be used to rationalize the stability of the silver compounds? For example, the sulfur, by nature of being less electronegative than oxygen, is a better Lewis base/electron pair donor/nucleophile; the sulfide ion is more willing to "donate" its electrons. Also the sulfide ion should be bigger and has empty 3d valence orbitals which may or may not be accessible to a great extent. Nonetheless, metals, by nature of possessing a large number of valence d-electrons, may be able to increase the stability of the sulfide ion by populating its d-valence.

This Lewis acid/base model seems to get a lot of mileage. I also know that silver is a good Lewis acid, as are many metal cations, because of the poor shielding afforded by the d electrons. Also, the silver cation has an empty 5s orbital. Thus, we can form a sulfide-to-cation sigma coordinate covalent bond.

We may also form a pi-type coordinate covalent bond as the far-away d-electron density "drifts" to empty 3d valence orbitals on the sulfide anion, making the sulfide anion now more nucleophilic than ever.

Let me know if my rationalizations are incorrect or if there are better explanations of silver salt stability.


1 Answer 1


This is just an unproven guess on my part. Whether compounds are soluble or not depends on 1) how strong the bonding (including lattice) forces are that hold the compound together (e.g. how much energy do we have to put in to liberate the ions, and 2) how much energy is recovered through solvation of the resulting ions. I don't see anything odd about the ionic radius of the silver ion so I'm guessing that the energy of solvation of the resultant ions would be fairly normal; so it is probably not factor "2" that makes most silver salts insoluble. Here is a link to an article that suggests that bonding in silver compounds may have a covalent component, it says, for silver salts "the structure has some contribution from covalent bonding. This appears to be bonding the structure together more tightly, as in a giant covalent structure." The words "more tightly" catch my eye. This seems to suggest that factor "1" may be in play in that silver salts may have increased bond\lattice energies. If true, this would help explain the insolubility of many silver salts. Finally I'll add that some of the soluble silver salts ($\ce{AgF, AgNO3, Ag(CH3CO2)}$) seem to involve anions that would be expected to be strongly hydrated. The extra energy generated through solvation would further assist in solubilizing these silver salts.

  • $\begingroup$ Interesting comment about the covalent nature of the bonding; would this have anything to do the d-electron density "shifting" from the silver to the sulfur? Regarding solvation effects, I know that the fluoride ion would be solvated by water quite nicely, but what about the oxide or even sulfide ions? Both are much, much stronger bases than the fluoride ion in water solution. $\endgroup$
    – Dissenter
    Commented Jun 5, 2014 at 23:41
  • $\begingroup$ I think atomic radii may be more important than basicity. A higher charge density (as would result on a smaller ion) is probably an important factor in terms of solvation energy. This argument is used to explain decreasing solubility down the halide salt series. Dunno what role d-electrons play in the sulfur case. $\endgroup$
    – ron
    Commented Jun 5, 2014 at 23:59
  • $\begingroup$ Perhaps we can go back to Bronsted acidity/basicity; consider the H+ ion, O(2-), and S(2-). The H+ ion can only form a sigma bond. S(2-) is a stronger base than O(2-). This implies that S(2-) donating an electron pair is more thermodynamically favorable than O(2-) donating an electron pair. Perhaps this can account for why Ag2S is much more stable than Ag2O. As you mention, there is some covalent character. Perhaps sigma bond strength is one factor. $\endgroup$
    – Dissenter
    Commented Jun 6, 2014 at 0:07
  • 2
    $\begingroup$ @Dissenter Partial covalent bonding exists even in CsI, it's part usually is several tens of percents even in strongly ionic compounds. As for silver, the atom has high elecronegativity, making formation of $Ag^+$ very unfavorable. $\endgroup$
    – permeakra
    Commented Jun 6, 2014 at 6:03
  • 2
    $\begingroup$ @Dissenter, $\ce{CsI}$ = caesium iodide. I think the point is even highly electropositive elements exhibit some covalent bonding. As for the EN of $\ce{Ag}$, 2.0 is high; only a handful of trans. metals surpass it and, ceteris paribus, that may favor covalent interactions. Finally, I'm fairly sure sulfide is not a stronger base than oxide; the greater polarizability of sulfur allows it to stabilize a negative charge more effectively. Hence, $\ce{H2S}$ is a stronger acid than $\ce{H2O}$. As for $\ce{Ag2O}$ vs. $\ce{Ag2S}$, I would personally look to HSAB theory, even though it may be handwavy. $\endgroup$
    – Greg E.
    Commented Jun 6, 2014 at 11:00

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