Compounds with cations having pseudo-inert gas electron configurations have more covalent character than those that don't, therefore they should be less soluble.

Here are some solubility products:

$$ \begin{array}{lcc} \hline \text{Anions} & \ce{Cu^{2+}} & \ce{Zn^{2+}} \\ \hline \text{Arsenate} & \color{red}{ \pu{7.95e-36} } & \color{red}{ \pu{2.8e-28} } \\ \text{Hydroxide} & \color{red}{ \pu{4.8e-20} } & \color{red}{ \pu{3e-17} } \\ \text{Sulfide} & \color{red}{ \pu{8e-37} } & \color{red}{ \pu{2e-25} } \\ \text{Phosphate} & \color{red}{ \pu{1.4e-37} } & \color{red}{ \pu{9e-33} } \\ \hline \end{array} $$

Zinc ($\ce{Zn^{2+}}$) has a pseudo-inert gas configuration ($\ce{[Ar] 3d^{10}}$), and its salts are more soluble than those of copper ($\ce{Cu^{2+}}$), which does not ($\ce{[Ar] 3d^{9}}$). Why?

  • $\begingroup$ I don't know. I've never seen a comparison between Cu and Zn covalency simply based on electronic configuration. Now if we were talking Ca vs Zn, then yeah, I can agree that $\ce{Zn^2+}$ is more polarising than $\ce{Ca^2+}$, hence more covalency. Incidentally, the example given on Wikipedia is $\ce{Ca^2+}$ vs $\ce{Hg^2+}$. If there's a reason for the difference in solubilities listed in the question, I doubt it has much to do with pseudo inert gas configurations - but I may be proven wrong. $\endgroup$
    – orthocresol
    Jan 29, 2018 at 22:50
  • $\begingroup$ Why would complexes of cations with pseudo-inert gas electron configuration be more covalent? I am lost here. If anything, I would expect the opposite. $\endgroup$
    – Greg
    Sep 13, 2018 at 11:00
  • $\begingroup$ C.F.S.E., maybe? $\endgroup$ Sep 18, 2020 at 1:41

1 Answer 1


It is because the difference in electronegativity. The higher the difference in electronegativity between the cation and the anion, the stronger the ionic bond. Since we are comparing species with the same anions and different cations, we only have to look at the values of the latter ones.

cation    electronegativity   configuration
  Cu             1.9         [Ar] 3d^10 4s^1
  Zn             1.65        [Ar] 3d^10 4s^2

$\ce{Cu}$ is more electronegative so the bonds have less ionic character which makes for stronger bonds.

The same happens in the same period between $\ce{K}$ and $\ce{Ca}$, but here the situation with regard to the electronic configuration is reverse. Potassium salts are much more soluble than calcium salts, and again it is explained by the electronegativity:

cation    electronegativity   configuration
  K              0.82           [Ar] 4s^1
  Ca             1.0            [Ar] 4s^2
  • 3
    $\begingroup$ Never thought of interpreting that $\ce{K+}$ salts are more soluble than $\ce{Ca^2+}$ from this perspective. Interesring point, just wondering whether there are any counterexamples (fluorides don't count). $\endgroup$ Jan 31, 2018 at 17:22
  • $\begingroup$ I wonder if it makes any sense to compare cations with different charge this way. They form different kind of lattices, and with eg organic anions there is a clear difference based on the charge instead of electronegativity. $\endgroup$
    – Greg
    Sep 13, 2018 at 11:05

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.