Correct solubility order of AgF, AgCl, AgBr, AgI in water is:
Show Hint
- AgF<AgCl>AgBr>AgI
- AgI<AgBr<AgCl<AgF
- AgF<AgCl<AgBr<AgI
- AgCl>AgBr>AgF>AgI
The Correct Option is B
Approach Solution - 1
The solubility of silver halides in water generally decreases down the group from fluorine to iodine: \(AgF > AgCl > AgBr > AgI\). This trend can be explained by considering two main factors: the lattice energy and the hydration enthalpy of the ions.
Lattice Energy:
Lattice energy is the energy required to separate one mole of an ionic solid into its gaseous ions. It is directly proportional to the charges of the ions and inversely proportional to the distance between them. The lattice energy of silver halides increases from \(AgF\) to \(AgI\) because the size of the halide ion increases from \(F^-\) to \(I^-\), which decreases the interionic distance. A higher lattice energy means the compound is more stable in the solid state and less likely to dissolve.
Hydration Enthalpy:
Hydration enthalpy is the energy released when one mole of gaseous ions is dissolved in water. It is also influenced by the size and charge of the ions. Smaller ions with higher charge densities have more negative (larger) hydration enthalpies. The hydration enthalpy decreases from \(F^-\) to \(I^-\) because the ionic size increases.
Solubility Trend:
The solubility of an ionic compound depends on the balance between the lattice energy and the hydration enthalpy. If the hydration enthalpy is greater than the lattice energy, the dissolution process is exothermic and the compound is more soluble. If the lattice energy is greater than the hydration enthalpy, the dissolution process is endothermic and the compound is less soluble.
For silver halides:
* \(AgF\) is highly soluble in water because the hydration enthalpy of \(Ag^+\) and \(F^-\) is significantly greater than its lattice energy. Fluoride ion is small and highly hydrating, leading to high solubility. * As we move down the group, the hydration enthalpy of the halide ions decreases, while the lattice energy increases. This causes the solubility to decrease. * \(AgCl\), \(AgBr\), and \(AgI\) are all sparingly soluble or practically insoluble in water because their lattice energies are much greater than their hydration enthalpies. The order of decreasing solubility is \(AgCl > AgBr > AgI\).
Conclusion:
The correct solubility order is: \(AgI < AgBr < AgCl < AgF\)
Approach Solution -2
The solubility of silver halides in water decreases as the size of the halide ion increases. This is because:
- Larger halide ions (e.g., I−) have lower lattice energy compared to smaller halide ions (e.g., F−).
- The lattice energy dominates over the hydration energy, making the solubility lower for larger ions.
Therefore, the solubility order is: AgF>AgCl>AgBr>AgI.
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