If(X) turns acidified `K_(2)Cr_(2)O_(7)` solution green , then X may be

Options

A. `SO_(2)`
B. `CO_(2)`
C. `NO_(2)^(Θ)`
D. `Fe^(2+)`

Answer

To determine which species (X) can turn acidified potassium dichromate (K₂Cr₂O₇) solution green, we will analyze the possible candidates: SO₂, CO₂, NO₂⁻, and Fe²⁺.

### Step-by-Step Solution:

1. **Identify the Reaction with SO₂:**
- When sulfur dioxide (SO₂) is treated with acidified K₂Cr₂O₇, it reduces the dichromate ion (Cr₂O₇²⁻) to chromium(III) ions (Cr³⁺), which are green in color.
- The reaction can be represented as:
\[ SO_2 + K_2Cr_2O_7 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O \]
- The product, chromium(III) sulfate (Cr₂(SO₄)₃), imparts a green color to the solution.

2. **Evaluate CO₂:**
- Carbon dioxide (CO₂) does not reduce K₂Cr₂O₇ and therefore does not change the color of the solution.
- Thus, CO₂ cannot be X.

3. **Identify the Reaction with NO₂⁻:**
- When nitrite ion (NO₂⁻) reacts with acidified K₂Cr₂O₇, it also reduces the dichromate ion to chromium(III) ions, resulting in a green solution.
- The reaction can be represented as:
\[ 3KNO_2 + K_2Cr_2O_7 + 4H_2SO_4 \rightarrow 3KNO_3 + Cr_2(SO_4)_3 + 4H_2O \]
- Again, the chromium(III) sulfate formed is green.

4. **Evaluate Fe²⁺:**
- Iron(II) ions (Fe²⁺) do not produce a green solution with K₂Cr₂O₇. Instead, they react to form iron(III) ions (Fe³⁺) and chromium(III) ions, but the solution does not appear green.
- The reaction can be represented as:
\[ 6Fe^{2+} + Cr_2O_7^{2-} + 14H^{+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O \]
- This indicates that Fe²⁺ cannot be X.

### Conclusion:
Based on the analysis, the species X that can turn acidified K₂Cr₂O₇ solution green are:
- **SO₂**
- **NO₂⁻**

### Final Answer:
X may be SO₂ or NO₂⁻.

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