MCQ Detailed View

Silver halides are compounds formed between silver (Ag⁺) ions and halide ions (F⁻, Cl⁻, Br⁻, I⁻). Their solubility in water depends on the strength of the ionic bonds and lattice energies compared to the hydration energy of the ions. Among silver halides, AgF (silver fluoride) is the only one that is freely soluble in water. The other halides—AgCl, AgBr, and AgI—are sparingly soluble or practically insoluble.

The high solubility of AgF can be explained by the very strong hydration energy of the fluoride ion. Fluoride is the smallest halide ion with the highest charge density, which allows water molecules to stabilize it effectively through hydrogen bonding and ion–dipole interactions. The hydration energy of F⁻ is greater than the lattice energy of AgF, making dissolution energetically favorable.

On the other hand, AgCl, AgBr, and AgI have much lower solubility because their lattice energies are very high and dominate over hydration energies. For example, AgCl has a very low solubility product (Ksp ≈ 1.8 × 10⁻¹⁰), AgBr is even less soluble, and AgI is practically insoluble in water (Ksp ≈ 8.3 × 10⁻¹⁷). This makes them appear as precipitates in aqueous reactions.

This difference in solubility is used in analytical chemistry, especially in qualitative analysis of halide ions. When silver nitrate (AgNO₃) is added to a solution containing halides, it precipitates AgCl (white), AgBr (pale yellow), and AgI (yellow), but no precipitate is formed with fluoride because AgF remains dissolved.

Therefore, the only silver halide that is soluble in water is AgF, while AgCl, AgBr, and AgI are insoluble and form characteristic precipitates used in halide detection tests.