
Arrange \[{\text{AgF, AgCl, AgBr and AgI}}\] in the increasing order of solubility in water.
A.\[{\text{AgBr < AgI < AgCl < AgF}}\]
B.\[{\text{AgI < AgBr < AgCl < AgF}}\]
C.\[{\text{AgI < AgCl < AgBr < AgF}}\]
D.None of the above
Answer
179.4k+ views
Hint:For this question we must have the knowledge of the concept of solubility of ion in water. Solubility of ions depends on the hydration energy released when a substance is dissolved in water. Higher the energy released, higher will be solubility.
Complete step by step solution:
The maximum amount of solute that is dissolved in a given amount of solvent is termed as solubility. As we know water is a polar molecule and it has a dipole. Because of more electronegativity of oxygen it has slight negative charge and hydrogen has slight positive charge. Whenever we put any solvent in water it dissociates into its respective cation and anion. A cation is positively charged and oxygen in water has negative charge, so cation will interact with oxygen because opposite charges interact with each other. Similarly, the anion will interact with hydrogen or water. Due to this interaction with water energy is released called hydration energy. The more is hydration energy, the more is interaction of species with hydrogen or say more is solubility.
As cation is same in both the cases that is silver ion, so things will only depend on anion. The size of halogens increases as we go down the group. Bigger is the size of anion, the less strong bond will form obviously. Since there will Be poor interaction between a solvent and solute molecule.
According to this \[{{\text{F}}^ - }\] is smallest and hence has highest solubility and \[{{\text{I}}^ - }\]will have lowest solubility.
The correct option is option B.
Note:The same trend follows for lattice energy of silver halides as well. In spite of the fact that \[{\text{AgF}}\] has higher lattice energy, it is most soluble in water. This is because of the high amount of hydration energy released in water by \[{{\text{F}}^ - }\] ion. Both these factors are responsible for solubility of any compound in water
Complete step by step solution:
The maximum amount of solute that is dissolved in a given amount of solvent is termed as solubility. As we know water is a polar molecule and it has a dipole. Because of more electronegativity of oxygen it has slight negative charge and hydrogen has slight positive charge. Whenever we put any solvent in water it dissociates into its respective cation and anion. A cation is positively charged and oxygen in water has negative charge, so cation will interact with oxygen because opposite charges interact with each other. Similarly, the anion will interact with hydrogen or water. Due to this interaction with water energy is released called hydration energy. The more is hydration energy, the more is interaction of species with hydrogen or say more is solubility.
As cation is same in both the cases that is silver ion, so things will only depend on anion. The size of halogens increases as we go down the group. Bigger is the size of anion, the less strong bond will form obviously. Since there will Be poor interaction between a solvent and solute molecule.
According to this \[{{\text{F}}^ - }\] is smallest and hence has highest solubility and \[{{\text{I}}^ - }\]will have lowest solubility.
The correct option is option B.
Note:The same trend follows for lattice energy of silver halides as well. In spite of the fact that \[{\text{AgF}}\] has higher lattice energy, it is most soluble in water. This is because of the high amount of hydration energy released in water by \[{{\text{F}}^ - }\] ion. Both these factors are responsible for solubility of any compound in water
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