Choose the correct statement from the following:

First we recall that the ease with which an ionic solid dissolves in water is governed by the thermodynamic relation

$$\Delta H_{\text{solution}} \;=\; \Delta H_{\text{lattice}} \;+\; \Delta H_{\text{hydration}},$$

where

$$\Delta H_{\text{lattice}} > 0 \quad (\text{energy required to break the crystal}),$$

and

$$\Delta H_{\text{hydration}} < 0 \quad (\text{energy released when ions are hydrated}).$$

A salt dissolves readily when the magnitude of the exothermic hydration enthalpy outweighs the endothermic lattice enthalpy, giving a small or negative $$\Delta H_{\text{solution}}$$. Conversely, very large positive lattice enthalpy or very small (less negative) hydration enthalpy will make the solid sparingly soluble.

For an alkali metal halide $$\text{M}^+\text{X}^-$$ both ions may vary in size. The lattice enthalpy is inversely proportional to the distance between the ion centres, stated qualitatively by

$$\Delta H_{\text{lattice}} \;\propto\; \dfrac{z_+\,z_-}{r_+ + r_-},$$

where $$z_+$$ and $$z_-$$ are the ionic charges (both unity in the present case) and $$r_+$$, $$r_-$$ are the ionic radii. Hence, smaller ions give larger (more positive) lattice enthalpies.

Within the alkali metal ions the order of size is

$$\text{Li}^+ \lt \text{Na}^+ \lt \text{K}^+ \lt \text{Rb}^+ \lt \text{Cs}^+,$$

while within the halide ions

$$\text{F}^- \lt \text{Cl}^- \lt \text{Br}^- \lt \text{I}^-.$$

Therefore for the fluoride series $$\text{LiF},\,\text{NaF},\,\text{KF},\,\dots$$ the compound $$\text{LiF}$$ contains the smallest cation and the smallest anion, so $$r_+ + r_-$$ is minimum and

$$\Delta H_{\text{lattice}}(\text{LiF})$$

is the largest (most positive) among all alkali metal fluorides. A very large positive lattice term requires a large negative hydration term before the salt can dissolve. Although $$\text{Li}^+$$ is indeed strongly hydrated, the extreme lattice enthalpy still dominates, giving a very small (and actually positive) value of $$\Delta H_{\text{solution}}$$. Hence $$\text{LiF}$$ is sparingly soluble in water.

Now we examine each assertion:

Option A claims that $$\text{LiF}$$ has the least negative standard enthalpy of formation among the fluorides. Because of its very high lattice enthalpy, the enthalpy of formation of $$\text{LiF}$$ is, in fact, the most negative; therefore the statement is false.

Option B says that the enthalpy of formation of alkali metal bromides becomes less negative down the group. In reality the trend is irregular; lattice enthalpy decreases, but other terms in the Born-Haber cycle change as well, so the statement cannot be generalised as written. Thus the option is incorrect.

Option C attributes the low solubility of $$\text{CsI}$$ to a high lattice enthalpy. Here both $$\text{Cs}^+$$ and $$\text{I}^-$$ are large; consequently $$r_+ + r_-$$ is large and $$\Delta H_{\text{lattice}}$$ is comparatively small. The real reason for poor solubility is the very low (less negative) hydration enthalpy of these large ions. Hence this statement is also false.

Option D states that, among the alkali metal halides, $$\text{LiF}$$ is least soluble in water. We have already reasoned that its extraordinarily high lattice enthalpy overrides the hydration energy advantage, giving the smallest solubility. This statement is true.

Therefore the only correct choice is the fourth one.

Hence, the correct answer is Option 4.