5
$\begingroup$

You would expect solubility of Group $2$ fluorides to increase down the group, as lattice energy plummets much more sharply than hydration energy does. For the most part, this is true: $\ce{BaF2}$ is more soluble than $\ce{SrF2}$, which is more soluble than $\ce{CaF2}$, and so on. However, $\ce{BeF2}$ is several times more soluble than even $\ce{BaF2}$. Why is that?

Beryllium forms the complex $\ce{[Be(H2O)4]^{2+}}$ in water. According to Wikipedia, $\ce{M-O}$ distance expectedly increases down the group and enthalpy of solvation becomes less and less negative, but beryllium aquo complex is the black sheep:

\begin{array}{c|c|c} & \ce{[Be(H2O)4]^{2+}} & \ce{[Mg(H2O)6]^{2+}} & \ce{Ca^{2+}(aq)} & \ce{Sr^{2+} (aq)} & \ce{Ba^{2+} (aq)}\\ \hline \ce{M-O distance (pm)} & 167 & 209 & 242 & 263 & 281\\ \ce{\Delta H- solvation (kJ mol^{-1})} & 2494 & 1921 & 1577 & 1443 & 1305 \end{array}

I found these solubility figures from JD Lee's Concise Inorganic Chemistry, Fifth Edition:

\begin{array}{c|c|c} & \ce{BeF2} & \ce{MgF2} & \ce{CaF2} & \ce{SrF2} & \ce{BaF2}\\ \hline \ce{solubility (grams solute per 100 g water} & \ce{very soluble} & 0.008 & 0.0016 & 0.012 & 0.12 \end{array}

$\endgroup$
5
  • $\begingroup$ So are you asking why Be has a high hydration energy or are you asking why BeF2 is more soluble than BaF2? I did a bit of researching through a few books and, it appears that there are conflicting beliefs over the relative solubility. Can you confirm the solubility data? $\endgroup$
    – C_Lycoris
    Mar 23, 2021 at 6:40
  • $\begingroup$ Huh. Although my teacher said that the solubility follows a clear trend: $\ce{BeF2 > BaF2 > SrF2 > CaF2 > MgF2}$, the trend of solubility according to JD Lee is more here-and-there $\endgroup$ Mar 23, 2021 at 7:01
  • 1
    $\begingroup$ @C_Lycoris I am asking why BeF2 is more soluble than BaF2. I surmised it was because of beryllium's high hydration enthalpy. $\endgroup$ Mar 23, 2021 at 7:50
  • 1
    $\begingroup$ I believe that the high hydration enthalpy is due to the points correctly stated by you. Furthermore, I find that the general trend is for the solubility to increase down the group when the size of the cation is near about that of the anion. [The lattice enthalpy usually dictates the solubility in this case] for eg: $$\ce{NaF<KF<CsF}$$. While it is true that the Hydration energy dominates the solubility in this case, I have yet to find a satisfactory reason for this. $\endgroup$
    – C_Lycoris
    Mar 23, 2021 at 8:53
  • $\begingroup$ The problem with solubility of ionic compounds is that it hinges on two different factors [Hydration enthalpy and lattice enthalpy], which further hinge on 2 other factors:[size of cation and anion]. Another factor that comes into play is Fajan's polarization rule. All this culminates into varying, often confusing trends. Ultimately, what decides solubility is ΔG= ΔH-TΔS. That's the surefire way of comparing solubility, but given that this data is not always available during exams, it isn't very useful. $\endgroup$
    – C_Lycoris
    Mar 23, 2021 at 9:06

0

Your Answer

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