The K of AgCl, AgBr and Agl are X, y and z respectively. On addition of `AgNO_3`, in the equimolar solution of `CI^-` `Br^-` and `I^-` ions, Agl Ist appears as precipitate followed by AgBr, and then AgCl. The relation between x y and z is

Options

A. `x lt y lt z`
B. `y gt zgtX`
C. `y gt zgtX`
D. `X= y = Z`

Answer

To solve the problem, we need to establish the relationship between the solubility product constants (Ksp) of silver halides: AgCl, AgBr, and AgI. We denote their Ksp values as follows:

- Ksp of AgCl = X
- Ksp of AgBr = Y
- Ksp of AgI = Z

### Step-by-Step Solution:

1. **Understanding Precipitation Order**:
- When AgNO3 is added to a solution containing Cl^-, Br^-, and I^- ions, the silver halides will precipitate in the order of their solubility products. The halide with the lowest solubility product will precipitate last.

2. **Identifying the Precipitation Sequence**:
- According to the problem, AgI precipitates first, followed by AgBr, and finally AgCl. This indicates that AgI is the least soluble, while AgCl is the most soluble among the three.

3. **Establishing Ksp Relationships**:
- Since AgI precipitates first, it must have the highest Ksp value, meaning:
\[ Z > Y \quad (\text{AgI is less soluble than AgBr}) \]
- Next, AgBr precipitates, which means it is less soluble than AgCl, leading to:
\[ Y > X \quad (\text{AgBr is less soluble than AgCl}) \]
- Combining these inequalities gives us:
\[ Z > Y > X \]

4. **Conclusion**:
- The relationship between the Ksp values of AgCl, AgBr, and AgI is:
\[ Z > Y > X \]

### Final Answer:
The relation between X, Y, and Z is:
\[ Z > Y > X \]