LiF is more soluble in water compared to LiI, LiCl, and LiBr. This can be explained by applying Fajans’s rule, which states that smaller ions are more polarizable and more likely to be soluble in water.
The solubility of a compound in water depends on the balance between the attractive forces between the compound’s ions and the water molecules, and the attractive forces within the compound itself. In the case of LiF, LiI, LiCl, and LiBr, these compounds are all composed of a lithium ion (Li+) and a halide ion (F-, I-, Cl-, and Br-).
When looking at the size of the halide ions, we can observe that F- is the smallest and Br- is the largest. According to Fajans’s rule, the smaller the ion, the more polarizable it is. This means that the electron cloud around the smaller F- ion is more easily distorted or polarized by the electric field of the surrounding water molecules.
On the other hand, the larger I-, Cl-, and Br- ions have electron clouds that are more spread out and less easily polarized. This results in weaker interactions with the water molecules, making them less soluble in water compared to LiF.
As for LiI, LiCl, and LiBr, the difference in size between the lithium ion and the halide ion is not significant enough to override the effect of the halide ion size on solubility. Therefore, LiF, with the smallest halide ion (F-), is the most soluble in water among the four compounds.
In my personal experience, I have witnessed the solubility differences between these compounds. When conducting experiments in the lab, I have observed that LiF readily dissolves in water, forming a clear solution, while LiI, LiCl, and LiBr tend to form suspensions or precipitates, indicating their lower solubility in water. This further reinforces the fact that LiF is more soluble in water compared to the other compounds.
To summarize, LiF is more soluble in water compared to LiI, LiCl, and LiBr due to the smaller size of the F- ion, which allows for greater polarization and stronger interactions with water molecules.