I know that Be has a tendency to create bonds with an increase covalent character so I can tell that BeF2 is more covalent than let's say MgF2. So why is BeF2 soluble in water while MgF2 is barely soluble ? Shouldn't the ionic compound be more soluble ? My guess is that the lattice enthalpy of BeF2 is not as great as the enthalpies of the other fluorides so it can be break down with much more ease and the energy gained from hydration outweighs the energy needed to separate the ions.

  • $\begingroup$ Because sucrose (sugar) is also soluble in water, maybe? Water can dissolve plenty of covalent compounds, especially those with polar bonds or having hydrogen-bonding capability with the water. $\endgroup$ Commented Feb 11, 2020 at 1:10

1 Answer 1


It seems that for BeF2 the hydration enthalpy of Be2+ is really high (due to Be2+ being a small cation and having a high charge density) so the energy required to break the lattice is smaller than that gained from hydration of Be2+ while for something like MgF2 the enthalpy of the lattice is greater than the energy you would get from hydration of Mg2+. In general when dealing with compounds of the group II with F- because of the small radius of F- the lattice enthalpy is great and only the hydration energy released by the hydration of Be2+ outweigh it thus making BeF2 soluble while all the other fluorides formed with the group II elements are quite insoluble in water


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.