$\ce{CaF2}$ is insoluble in water, but $\ce{CaCl2}$, $\ce{CaBr2}$, $\ce{CaI2}$ are soluble. Why is this so?
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2 Answers
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Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.
Fluoride is very electronegative, and thus the bonding forces between $\ce{Ca^2+}$ and each fluoride anions are very strong, they are tightly held together in the crystal form. A few water molecules do not have enough strength to get in between and dissolve it.
Chloride is less electronegative, thus $\ce{CaCl2}$ has weaker bonds, then comes $\ce{CaBr2}$ and at last $\ce{CaI2}$, the weakest of them all. You may also expect increase in bond length depending on the crystal structure, which may also contribute to the solubility properties of the given compound. So weaker the bond strength, greater the solubility – at least that’s the expected.
In fact, $\ce{CaCl2}$ is more soluble than $\ce{CaF2}$, and $\ce{CaBr2}$ is even more – as expected. However, I did some research and the Wikipedia article on $\ce{CaI2}$ claims it’s less soluble than $\ce{CaBr2}$, which is not the expected, but it does not provide a source for that information. It might be true, though, since there are a variety of factors that may or may not contribute to it’s solubility in water, such as it’s crystal structure and the particular characteristics of iodide.
You can do the experiment yourself to find out which one is more soluble, since all solubility facts and numbers we know come from experiments. I couldn’t find anything on the solubility of $\ce{CaI2}$ in my CRC Handbook of Chemistry and Physics either, so I can’t provide you with a precise answer on why it’s less soluble than $\ce{CaBr2}$ (if in fact it is), but I hope you can now understand some of the factors that make $\ce{CaF2}$ the least soluble compound on that list.
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I have a thought of the crystal structures. When we look at $\ce{BeF2}$, $\ce{Be^2+}$ being a tiny cation couldn't simply accommodate 2 strong $\ce{F-}$ in its fluorite fitting. It would lead to negative repulsions and even the hydration enthalpy of $\ce{Be^2+}$ being high would consequently increase the chances of solubility. On the other hand, when the cation size increases, let it be $\ce{Mg^2+}$ or $\ce{Ca^2+}$, they could be large enough to accommodate $\ce{F-}$ ions in their tetrahedral voids making the lattice stable. This would reduce the chances of ionic dissociation. Now when we replace $\ce{F-}$ with larger $\ce{Cl-}$ and $\ce{Br-}$there would be again a change in the limiting radius ratio of the crystal which would enhance dissociation and hence solubility.
