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 reactions step by step and identify which cation can form precipitate Q but does not form precipitate R.

### Step 1: Identify the Reaction with M²⁺
We start with a cation M²⁺ in an aqueous solution. The first reaction involves adding NH₄Cl and (NH₄)₂CO₃.

**Hint:** Look for cations that can form insoluble carbonates when reacted with ammonium carbonate.

### Step 2: Determine the Precipitate Q
When M²⁺ reacts with NH₄Cl and (NH₄)₂CO₃, we are looking for a cation that forms a precipitate.

- **Mg²⁺**: Does not form a precipitate with ammonium carbonate.
- **Ca²⁺**: Forms CaCO₃, which is a precipitate.
- **Sr²⁺**: Also forms a precipitate similar to Ca²⁺.
- **Ba²⁺**: Forms BaCO₃, which is a precipitate.

**Hint:** Identify which of these cations forms a precipitate with ammonium carbonate.

### Step 3: Analyze the Reaction with CH₃COOH
Next, we need to consider what happens when we take the precipitate Q and react it with acetic acid (CH₃COOH).

- **CaCO₃** (from Ca²⁺) will react with CH₃COOH to form soluble products, thus not forming precipitate R.
- **SrCO₃** (from Sr²⁺) will also react similarly to form soluble products.
- **BaCO₃** (from Ba²⁺) will dissolve in acetic acid, leading to no precipitate R.

**Hint:** Determine which of the precipitates formed from Q dissolves in acetic acid.

### Step 4: Conclusion
From the analysis:
- **Mg²⁺** does not form precipitate Q.
- **Ca²⁺** forms both Q and R.
- **Sr²⁺** forms both Q and R.
- **Ba²⁺** forms precipitate Q (BaCO₃) but does not form precipitate R when reacted with acetic acid.

Thus, the cation that can form precipitate Q but does not form precipitate R is **Ba²⁺**.

**Final Answer:** Ba²⁺