WBJEE 2024 Physics & Chemistry Paper

The correct answer is Option B:$\mathrm{AgI} < \mathrm{AgBr} < \mathrm{AgCl} < \mathrm{AgF}$Here's why: Solubility is the ability of a substance (solute) to dissolve in a solvent. In this case, we're looking at the solubility of silver halides (AgF, AgCl, $\mathrm{AgBr}, \mathrm{Agl})$ in water.Lattice Energy: The strength of the ionic bond in the silver halide crystal lattice plays a major role in solubility. As the size of the halide ion increases $(\mathrm{F}^{-} < \mathrm{Cl}^{-} < \mathrm{Br}^{-} < \mathrm{I}^{-})$, the lattice energy decreases. This is because the larger ions have weaker electrostatic attractions.Hydration Energy: The energy released when ions are surrounded by water molecules is called hydration energy. Smaller ions with higher charge density have stronger hydration energies.In the case of silver halides:AgF: Fluoride ion is the smallest and has the highest charge density. It has a strong hydration energy, overcoming the relatively weak lattice energy. This makes AgF highly soluble. AgCl: Chloride ion is larger than fluoride, so the lattice energy is weaker. The hydration energy is also somewhat reduced. AgCl is less soluble than AgF .AgBr: Bromide ion is even larger, resulting in even weaker lattice energy and further reduced hydration energy. AgBr is less soluble than AgCl .Agl: lodide ion is the largest, leading to the weakest lattice energy and the lowest hydration energy. Agl is the least soluble among the silver halides.Therefore, the solubility order is:$\mathrm{AgI} < \mathrm{AgBr} < \mathrm{AgCl} < \mathrm{AgF}$