The reagent which can't be used to detect the presence of both `CO_(3)^(2-)` and `HCO_(3)^(-)` in a mixture is -

Options

A. `CaCl_(2)`
B. `SrCl_(2)`
C. `AgNO_(3)`
D. `MgCl_(2)`

Answer

To determine the reagent that cannot be used to detect the presence of both carbonate ions (CO₃²⁻) and bicarbonate ions (HCO₃⁻) in a mixture, we will analyze the provided reagents one by one.

### Step-by-Step Solution:

1. **Identify the Reagents:**
- The reagents given are:
- A) CaCl₂ (Calcium chloride)
- B) SrCl₂ (Strontium chloride)
- C) AgNO₃ (Silver nitrate)
- D) MgCl₂ (Magnesium chloride)

2. **Reagent A: CaCl₂**
- When CaCl₂ is added to a solution containing CO₃²⁻ or HCO₃⁻, it reacts to form calcium carbonate (CaCO₃), which is a precipitate (PPT).
- Therefore, CaCl₂ can detect both carbonate and bicarbonate ions.

3. **Reagent B: SrCl₂**
- Strontium chloride (SrCl₂) also reacts with both CO₃²⁻ and HCO₃⁻ to form strontium carbonate (SrCO₃), which is a precipitate.
- Thus, SrCl₂ can detect both ions as well.

4. **Reagent C: AgNO₃**
- Silver nitrate (AgNO₃) reacts with both carbonate and bicarbonate ions to form silver carbonate (Ag₂CO₃), which is a precipitate.
- Hence, AgNO₃ can also detect both carbonate and bicarbonate ions.

5. **Reagent D: MgCl₂**
- Magnesium chloride (MgCl₂) reacts with CO₃²⁻ to form magnesium carbonate (MgCO₃), which is a white precipitate.
- However, when MgCl₂ is added to a solution containing HCO₃⁻, it does not form a precipitate under normal conditions. It requires heating to produce magnesium carbonate, which means it cannot detect bicarbonate ions at room temperature.

6. **Conclusion:**
- Among the given reagents, MgCl₂ is the only reagent that cannot be used to detect the presence of both CO₃²⁻ and HCO₃⁻ in a mixture under normal conditions. Therefore, the answer is **D) MgCl₂**.