`X("satl")+AgNO_(3)(aq.) to underset(("yellow ppt."))(Y darr)` (soluble in excess of `NH_(3)` solution)
Salt X, does not contain:

Options

A. `PO_(4)^(3-)`
B. `Br^(-)`
C. `I^(-)`
D. `AsO_(3)^(3-)`

Answer

To solve the question, we need to determine which anion cannot be present in the salt X based on the reaction with AgNO3, which produces a yellow precipitate that is soluble in excess NH3.

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is:
\[ X + \text{AgNO}_3 \rightarrow Y \quad (\text{yellow ppt.}) \]
where Y is a yellow precipitate that is soluble in excess NH3.

2. **Analyze Possible Anions**: We need to analyze various anions to see if they can be part of salt X.

3. **Examine PO4^3- (Phosphate Ion)**:
- When reacting PO4^3- with AgNO3, we get Ag3PO4.
- Ag3PO4 is a yellow precipitate and is soluble in excess NH3.
- Therefore, PO4^3- **can be** part of X.

4. **Examine Br^- (Bromide Ion)**:
- When reacting Br^- with AgNO3, we get AgBr.
- AgBr is a yellow precipitate and is soluble in excess NH3.
- Therefore, Br^- **can be** part of X.

5. **Examine I^- (Iodide Ion)**:
- When reacting I^- with AgNO3, we get AgI.
- AgI is a yellow precipitate but is **insoluble** in excess NH3.
- Therefore, I^- **cannot be** part of X.

6. **Examine AsO3^3- (Arsenite Ion)**:
- When reacting AsO3^3- with AgNO3, we get Ag3AsO3.
- Ag3AsO3 is a yellow precipitate and is soluble in excess NH3.
- Therefore, AsO3^3- **can be** part of X.

7. **Conclusion**: Since I^- is the only anion that does not fit the criteria of being soluble in excess NH3, we conclude that salt X does not contain I^-.

### Final Answer:
Salt X does not contain: **I^- (iodide ion)**.

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