Final answer:
Solubility is determined by free energy change and dissolution equilibrium, where complex ion formation plays a significant role. Ionic nature is a rudimentary explanation; complex formation constants and equilibria like the common ion effect crucially affect solubility.
Step-by-step explanation:
It is correct to consider that free energy changes are what ultimately determine the solubility of a compound. In the case of silver fluoride (AgF), its higher solubility compared to other silver halides like silver chloride (AgCl) is linked to its ionic nature, but this is a simplistic explanation.
The actual solubility process involves the dissolution equilibrium, where the ionic solid dissolves into its constituent ions. When complex ion formation occurs, as seen with AgCl in presence of ammonia (NH3), the solubility is significantly affected. The complex formation constant, which reflects the stability of the complex ion, influences the free silver ion concentration, thereby affecting the overall solubility according to Le Chatelier's principle.
For instance, AgCl partially dissolves in water, and its solubility can be represented by AgCl(s) ⇒ Ag+(aq) + Cl-(aq). Adding a common ion like Cl-, from KCl for instance, ordinarily would decrease the solubility of AgCl due to the common ion effect. However, it is important to note that silver can also form a two-coordinate complex with chloride ions (AgCl2-), which affects the solubility dynamics in a concentrated solution of KCl.