`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 problem, we need to analyze the reaction of the salt \( X \) with \( AgNO_3 \) and determine which anions can be present in \( X \) based on the formation of a yellow precipitate and its solubility in excess \( NH_3 \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The salt \( X \) reacts with \( AgNO_3 \) to form a yellow precipitate \( Y \). The yellow precipitate indicates the formation of a silver salt.

2. **Determine Possible Anions**:
We need to consider common anions that can form yellow precipitates with silver nitrate:
- Phosphate ion (\( PO_4^{3-} \))
- Bromide ion (\( Br^{-} \))
- Iodide ion (\( I^{-} \))
- Chromate ion (\( CrO_4^{2-} \))

3. **Analyze Each Anion**:
- **Phosphate (\( PO_4^{3-} \))**:
- Forms \( Ag_3PO_4 \) which is a yellow precipitate and is soluble in excess \( NH_3 \).
- **Bromide (\( Br^{-} \))**:
- Forms \( AgBr \) which has a pale yellow color and is also soluble in excess \( NH_3 \).
- **Iodide (\( I^{-} \))**:
- Forms \( AgI \) which is a yellow precipitate but is **insoluble** in excess \( NH_3 \).
- **Chromate (\( CrO_4^{2-} \))**:
- Forms \( Ag_2CrO_4 \) which is a yellow precipitate and is soluble in excess \( NH_3 \).

4. **Conclusion**:
Since the salt \( X \) must be soluble in excess \( NH_3 \) and the only anion that does not meet this criterion is \( I^{-} \), we conclude that \( X \) does not contain the iodide ion.

### Final Answer:
The salt \( X \) does not contain: **Iodide ion (\( I^{-} \))**.