The oxidation number is changed in which of the following case?

Options

A. `SO_(2)` gas is passed into `Cr_(2)O_(7)^(2-)//H^(+)`
B. Aqueous solution of `CrO_(4)^(2-)` is acidified
C. `CrO_(2)Cl_(2)` is dissolved in `NaOH`
D. `Cr_(2)O_(7)^(2-)` solution is made alkaline.

Answer

To determine where the oxidation number changes among the given options, we will analyze each reaction step by step.

### Step 1: Analyze Option 1
**Reaction:** SO2 gas is passed into Cr2O7²⁻ in presence of H⁺.

1. **Identify oxidation states:**
- In Cr2O7²⁻: Let the oxidation state of Cr be \( x \).
\[ 2x + (-2 \times 7) = -2 \quad \Rightarrow \quad 2x - 14 = -2 \quad \Rightarrow \quad 2x = 12 \quad \Rightarrow \quad x = +6 \]
- In SO2: Let the oxidation state of S be \( y \).
\[ y + (-2 \times 2) = 0 \quad \Rightarrow \quad y - 4 = 0 \quad \Rightarrow \quad y = +4 \]
- In SO4²⁻: Let the oxidation state of S be \( z \).
\[ z + (-2 \times 4) = -2 \quad \Rightarrow \quad z - 8 = -2 \quad \Rightarrow \quad z = +6 \]

2. **Determine changes:**
- Cr changes from +6 (in Cr2O7²⁻) to +3.
- S changes from +4 (in SO2) to +6 (in SO4²⁻).

**Conclusion:** Oxidation numbers change.

### Step 2: Analyze Option 2
**Reaction:** CrO4²⁻ is acidified.

1. **Identify oxidation states:**
- In CrO4²⁻: \( x + (-2 \times 4) = -2 \)
\[ x - 8 = -2 \quad \Rightarrow \quad x = +6 \]
- In Cr2O7²⁻: The oxidation state of Cr is still +6.

2. **Determine changes:**
- No change in oxidation state for Cr.

**Conclusion:** Oxidation numbers do not change.

### Step 3: Analyze Option 3
**Reaction:** CrO2Cl2 is dissolved in NaOH.

1. **Identify oxidation states:**
- In CrO2Cl2: \( x + (-2 \times 2) + (-1 \times 2) = 0 \)
\[ x - 4 - 2 = 0 \quad \Rightarrow \quad x = +6 \]
- In Na2CrO4: \( 2(+1) + x + (-2 \times 4) = 0 \)
\[ 2 + x - 8 = 0 \quad \Rightarrow \quad x = +6 \]

2. **Determine changes:**
- No change in oxidation state for Cr.

**Conclusion:** Oxidation numbers do not change.

### Step 4: Analyze Option 4
**Reaction:** Cr2O7²⁻ solution is made alkaline.

1. **Identify oxidation states:**
- In Cr2O7²⁻: \( x = +6 \).
- In CrO4²⁻: \( x = +6 \).

2. **Determine changes:**
- No change in oxidation state for Cr.

**Conclusion:** Oxidation numbers do not change.

### Final Answer
The oxidation number changes in **Option 1**.

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