The solubility order for alkali metal fluoride in water is :

The solubility of alkali metal fluorides in water depends on the balance between lattice energy and hydration energy. Lattice energy is the energy released when gaseous ions form a solid crystal lattice, and hydration energy is the energy released when ions are surrounded by water molecules.

As we move down the group from lithium (Li) to rubidium (Rb), the size of the cation increases. The lattice energy decreases because the distance between the ions increases, weakening the electrostatic attraction. The hydration energy also decreases because larger cations are less effectively hydrated by water molecules.

For fluorides, the small size of the fluoride ion (F⁻) leads to high lattice energies, especially for smaller cations. Lithium fluoride (LiF) has a very high lattice energy due to the small size of the Li⁺ ion, making it less soluble. As the cation size increases from Na⁺ to K⁺ to Rb⁺, the lattice energy decreases more rapidly than the hydration energy, resulting in increased solubility.

Therefore, the solubility order is: LiF < NaF < KF < RbF.

Now, comparing with the options:

• Option A: LiF < RbF < KF < NaF → Incorrect, because RbF should be more soluble than KF and NaF.
• Option B: RbF < KF < NaF < LiF → Incorrect, as it suggests decreasing solubility down the group.
• Option C: LiF > NaF > KF > RbF → Incorrect, as it also suggests decreasing solubility.
• Option D: LiF < NaF < KF < RbF → Correct, matches the expected increasing solubility order.

Hence, the correct answer is Option D.