Consider the following sequence of reaction:
`M^(2+)(aq.) overset(NH_(4)Cl(s)+(NH_(4))_(2)CO_(3)sol.)to Q darr underset("Followed by addition of "(NH_(4))_(2)C_(2)O_(4))overset(CH_(3)COOH)toR darr`
Which of the following cation can form ppt. Q but does not form ppt. 'R' ?

Options

A. `Mg^(2+)(aq.)`
B. `Ca^(2+)(aq.)`
C. `Sr^(2+)(aq.)`
D. `Ba^(2+)(aq.)`

Explanation

To solve the problem, we need to analyze the sequence of reactions involving the cation \( M^{2+} \) and determine which cation can form precipitate \( Q \) but does not form precipitate \( R \).

### Step-by-Step Solution:

1. **Identify Possible Cations**:
The cations \( M^{2+} \) can be \( Mg^{2+}, Ca^{2+}, Sr^{2+}, \) or \( Ba^{2+} \).

2. **First Reaction**:
The first reaction involves \( M^{2+} \) reacting with \( NH_4Cl \) and \( (NH_4)_2CO_3 \):
\[ M^{2+} + NH_4Cl + (NH_4)_2CO_3 \rightarrow Q \]
This reaction typically forms a carbonate precipitate of the cation involved.

3. **Formation of Precipitate \( Q \)**:
- For \( Mg^{2+} \): \( Mg^{2+} + (NH_4)_2CO_3 \) forms \( MgCO_3 \), which is soluble.
- For \( Ca^{2+} \): \( Ca^{2+} + (NH_4)_2CO_3 \) forms \( CaCO_3 \), which is a precipitate.
- For \( Sr^{2+} \): \( Sr^{2+} + (NH_4)_2CO_3 \) forms \( SrCO_3 \), which is a precipitate.
- For \( Ba^{2+} \): \( Ba^{2+} + (NH_4)_2CO_3 \) forms \( BaCO_3 \), which is a precipitate.

4. **Identify Precipitate \( R \)**:
The next step involves adding \( CH_3COOH \) followed by \( (NH_4)_2C_2O_4 \) to \( Q \):
\[ Q + CH_3COOH + (NH_4)_2C_2O_4 \rightarrow R \]
This reaction typically forms an oxalate precipitate.

5. **Formation of Precipitate \( R \)**:
- For \( Mg^{2+} \): Since \( Q \) is soluble, no \( R \) is formed.
- For \( Ca^{2+} \): \( CaCO_3 \) reacts to form \( CaC_2O_4 \), which is a precipitate.
- For \( Sr^{2+} \): \( SrCO_3 \) reacts to form \( SrC_2O_4 \), which is a precipitate.
- For \( Ba^{2+} \): \( BaCO_3 \) reacts to form \( BaC_2O_4 \), which is soluble.

6. **Conclusion**:
- \( Mg^{2+} \) does not form \( Q \) (as it is soluble).
- \( Ca^{2+} \), \( Sr^{2+} \), and \( Ba^{2+} \) all form precipitates \( Q \) and \( R \).
- Only \( Ba^{2+} \) forms precipitate \( Q \) but does not form precipitate \( R \) (as \( BaC_2O_4 \) is soluble).

### Final Answer:
The cation that can form precipitate \( Q \) but does not form precipitate \( R \) is \( Ba^{2+} \).