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 > y > z
B. y < z < X
C. y < z < X
D. X= y = Z

Answer

To solve the problem, we need to analyze the solubility products (Ksp) of the silver halides: AgCl, AgBr, and AgI. The question states that when AgNO3 is added to an equimolar solution of Cl^-, Br^-, and I^- ions, AgI precipitates first, followed by AgBr, and then AgCl.

### Step-by-Step Solution:

1. **Understanding Precipitation**:
- When a salt is added to a solution containing its ions, a precipitate forms if the ionic product (IP) exceeds the solubility product (Ksp) of that salt.
- The order of precipitation indicates the relative Ksp values of the salts involved.

2. **Identifying the Order of Precipitation**:
- According to the problem, AgI precipitates first, followed by AgBr, and finally AgCl. This means that AgI has the lowest Ksp, while AgCl has the highest Ksp.

3. **Establishing Relationships**:
- Since AgI precipitates first, it has the smallest Ksp value, which we denote as Z.
- AgBr precipitates next, indicating that its Ksp value (Y) is higher than that of AgI but lower than that of AgCl.
- AgCl precipitates last, meaning it has the highest Ksp value (X).

4. **Conclusion**:
- From the order of precipitation, we can conclude:
- Ksp(AgI) < Ksp(AgBr) < Ksp(AgCl)
- In terms of our variables, this translates to: Z < Y < X.

5. **Final Relation**:
- Rearranging gives us the relation: X > Y > Z.

### Final Answer:
The relationship between the Ksp values is: **X > Y > Z**.