`S^(2-) and SO_(3)^(2-)` can be distinguished by using

Options

A. `(CH_(2)COO)_(2)Pb`
B. `Na_(2)[Fe(CN)_(5)NO]`
C. `Cr_(2)O_(7)^(2-)` solution
D. `CaCI_(2)`

Answer

To distinguish between the sulfide ion (S²⁻) and the sulfite ion (SO₃²⁻), we can use specific reagents that react differently with these ions. Here is a step-by-step solution to the problem:

### Step 1: Understanding the Ions
- **Sulfide Ion (S²⁻)**: This ion is a simple anion that can react with certain reagents to produce distinct products.
- **Sulfite Ion (SO₃²⁻)**: This ion contains sulfur in a different oxidation state and will react differently compared to sulfide.

### Step 2: Choosing the Reagents
We need to choose reagents that can react with either S²⁻ or SO₃²⁻ to produce observable differences.

1. **Lead Acetate (CH₃COO)₂Pb**: When lead acetate is added to a solution containing sulfide ions, it reacts to form lead sulfide (PbS), which is a black precipitate. This indicates the presence of sulfide ions.

**Reaction**:
\[ \text{S}^{2-} + \text{Pb(CH}_3\text{COO)}_2 \rightarrow \text{PbS (black precipitate)} + \text{other products} \]

2. **Sodium Ferricyanide (Na₂[Fe(CN)₆])**: This reagent can react with sulfide ions to form a purple complex, which is another indication of the presence of sulfide ions.

**Reaction**:
\[ \text{S}^{2-} + \text{Na}_2[\text{Fe(CN)}_6] \rightarrow \text{Purple complex} \]

3. **Calcium Chloride (CaCl₂)**: When calcium chloride is added to a solution containing sulfite ions, it reacts with sulfur dioxide (SO₂) produced from the sulfite ion to form calcium sulfate (CaSO₄), which appears as a milky white precipitate.

**Reaction**:
\[ \text{SO}_3^{2-} + \text{CaCl}_2 \rightarrow \text{CaSO}_4 \text{ (milky white precipitate)} + \text{other products} \]

### Step 3: Analyzing the Results
- If you observe a **black precipitate** upon adding lead acetate, you can conclude the presence of sulfide ions (S²⁻).
- If you observe a **purple color** upon adding sodium ferricyanide, this also indicates the presence of sulfide ions (S²⁻).
- If you observe a **milky white precipitate** after adding calcium chloride, this indicates the presence of sulfite ions (SO₃²⁻).

### Conclusion
Thus, the reagents that can be used to distinguish between S²⁻ and SO₃²⁻ are:
- Lead Acetate (produces black precipitate for S²⁻)
- Sodium Ferricyanide (produces purple complex for S²⁻)
- Calcium Chloride (produces milky white precipitate for SO₃²⁻)

### Final Answer
The reagents that can be used to distinguish between S²⁻ and SO₃²⁻ are Lead Acetate, Sodium Ferricyanide, and Calcium Chloride.

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