Acidic `K_(2)Cr_(2)O_(7)` reacts with `H_(2)S` to produce:

Options

A. `Cr^(6+)` ions
B. `Cr^(3+)` ions
C. `SO_(2)`
D. S

Answer

To determine the products of the reaction between acidic potassium dichromate (K₂Cr₂O₇) and hydrogen sulfide (H₂S), we can follow these steps:

### Step 1: Write the Reactants
The reactants of the reaction are:
- Potassium dichromate: K₂Cr₂O₇
- Hydrogen sulfide: H₂S

### Step 2: Identify the Conditions
The reaction occurs in acidic conditions, which means we will need to include H⁺ ions in our reaction.

### Step 3: Write the Reaction Equation
The reaction can be represented as:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{S} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{S} + \text{H}_2\text{O} \]

### Step 4: Balance the Reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

1. **Chromium (Cr)**: There are 2 Cr atoms in K₂Cr₂O₇, so we will produce 2 Cr³⁺ ions.
2. **Oxygen (O)**: There are 7 O atoms in K₂Cr₂O₇, which will combine with H⁺ ions to form water.
3. **Sulfur (S)**: The H₂S will provide sulfur, and we need to balance the number of sulfur atoms produced.

The balanced equation is:
\[ \text{Cr}_2\text{O}_7^{2-} + 3\text{H}_2\text{S} + 8\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{S} + 7\text{H}_2\text{O} \]

### Step 5: Identify the Products
From the balanced equation, we can identify the products:
- Chromium ions: \( \text{Cr}^{3+} \)
- Sulfur: \( \text{S} \)
- Water: \( \text{H}_2\text{O} \)

### Step 6: Determine the Oxidation States
- Chromium changes from +6 in K₂Cr₂O₇ to +3 in \( \text{Cr}^{3+} \) (reduction).
- Sulfur changes from -2 in H₂S to 0 in elemental sulfur (oxidation).

### Conclusion
The products of the reaction between acidic potassium dichromate and hydrogen sulfide are:
- \( \text{Cr}^{3+} \) ions (Chromium ions)
- Elemental sulfur (S)

### Final Answer
The correct products are \( \text{Cr}^{3+} \) ions and elemental sulfur (S).