I was going through JD Lee and I found certain contradictory statements.


With most compounds, on descending the group, the hydration energy decreases more rapidly than the lattice energy: hence the compounds become less soluble as the metal gets larger. However, with fluorides and hydroxides the lattice energy decreases more rapidly than the hydration energy, and so their solubility increases on descending the group.


Beryllium fluoride is very soluble in water, owing to the high solvation energy of Be in forming $\ce{[Be(H2O)4]^2+}$. The other fluorides MF2 are are all almost insoluble.


The sulphates of calcium, strontium and barium are insoluble, and the carbonates, oxalates, chromates and fluorides of the whole group are insoluble. This is a useful factor in qualitative analysis.

I'm confused about what the order is. Wikipedia says $\ce{BeF2}$ is highly soluble, whereas $\ce{CaF2}$ is insoluble. Are these statements contradictory because of the toxicity of beryllium compounds, which is why they have been studied less and are not spoken about frequently? What is the actual order of solubility of alkaline earth metal fluorides?

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    – Buck Thorn
    Jun 11, 2021 at 3:34
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    – Siddhant
    Jun 11, 2021 at 11:25
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    – Mithoron
    Jun 11, 2021 at 14:19

1 Answer 1


The order of solubility of the alkaline earth fluorides is, after $\ce{BeF2}$ :

$\ce{MgF2 : 0.026 g/L}$

$\ce{CaF2 : ~0.016 g/L}$

$\ce{SrF2 : ~~~0.11 g/L}$

$\ce{BaF2 : ~~~1.7 g/L}$

$\ce{BeF2}$ itself is mentioned as infinitely soluble in water. So, the minimum solubility is in the middle of the series. It is $\ce{CaF2}$

Ref : C. D. Hodgman, R. C. Weast, Handbook of Chemistry and Physics, Chemical Rubber Publishing Co. Cleveland, 1962.


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