`K_(sp)(AgCl) gt K_(sp)(AgBr) gt K_(sp)(AgI)`. This means that :

Options

A. AgCl is less ionised than AgBr and AgI
B. Both AgBr and AgI are less soluble than AgCl
C. AgI is most soluble
D. AgBr is more soluble than AgCl but less soluble than AgI

Explanation

To solve the question regarding the solubility products of AgCl, AgBr, and AgI, we can follow these steps:

### Step 1: Understand the Concept of Solubility Product (Ksp)
The solubility product (Ksp) is an equilibrium constant that applies to the solubility of sparingly soluble ionic compounds. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation.

### Step 2: Analyze the Given Information
The question states that:
\[ K_{sp}(AgCl) > K_{sp}(AgBr) > K_{sp}(AgI) \]
This indicates that AgCl has the highest solubility product, followed by AgBr, and then AgI.

### Step 3: Relate Ksp to Solubility
Since the solubility product increases with increasing solubility, we can conclude that:
- AgCl is more soluble than AgBr.
- AgBr is more soluble than AgI.

### Step 4: Draw Conclusions
From the order of the solubility products, we can deduce:
- AgCl is the most soluble among the three.
- AgBr is less soluble than AgCl but more soluble than AgI.
- AgI is the least soluble among the three.

### Step 5: Evaluate the Options
Based on the conclusions drawn:
1. **AgCl is less ionized than AgBr and AgI** - Incorrect.
2. **Both AgBr and AgI are less soluble than AgCl** - Correct.
3. **AgI is more soluble than AgCl** - Incorrect.
4. **AgBr is more soluble than AgCl** - Incorrect.

### Final Answer
The correct interpretation of the given information is that **both AgBr and AgI are less soluble than AgCl**.

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