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 the change in oxidation number in the given cases, we will analyze each option step by step.

### Step-by-Step Solution:

**Option 1: SO2 gas is passed into Cr2O7^2- in acidic medium.**

1. **Write the reaction:**
\[ \text{Cr}_2\text{O}_7^{2-} + \text{SO}_2 + 2\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-} + \text{H}_2\text{O} \]

2. **Determine the oxidation states:**
- For \(\text{Cr}_2\text{O}_7^{2-}\):
- Let the oxidation state of Cr be \(x\).
- \(2x + 7(-2) = -2\)
- \(2x - 14 = -2 \Rightarrow 2x = 12 \Rightarrow x = +6\)
- For \(\text{SO}_2\):
- Each O is -2, so \(S + 2(-2) = 0\)
- \(S - 4 = 0 \Rightarrow S = +4\)
- For \(\text{Cr}^{3+}\):
- Oxidation state is +3.
- For \(\text{SO}_4^{2-}\):
- Each O is -2, so \(S + 4(-2) = -2\)
- \(S - 8 = -2 \Rightarrow S = +6\)

3. **Change in oxidation states:**
- Cr changes from +6 in \(\text{Cr}_2\text{O}_7^{2-}\) to +3 in \(\text{Cr}^{3+}\).
- Thus, the oxidation number changes.

**Conclusion:** Oxidation number changes in this case.

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**Option 2: Aqueous solution of CrO4^2- is acidified.**

1. **Write the reaction:**
\[ \text{CrO}_4^{2-} + 2\text{H}^+ \rightarrow \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \]

2. **Determine the oxidation states:**
- For \(\text{CrO}_4^{2-}\):
- Let the oxidation state of Cr be \(x\).
- \(x + 4(-2) = -2\)
- \(x - 8 = -2 \Rightarrow x = +6\)
- For \(\text{Cr}_2\text{O}_7^{2-}\):
- As calculated before, Cr is +6.

3. **Change in oxidation states:**
- There is no change in the oxidation state of Cr (remains +6).

**Conclusion:** Oxidation number does not change in this case.

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**Option 3: CrO2Cl2 is dissolved in NaOH.**

1. **Write the reaction:**
\[ \text{CrO}_2\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{CrO}_4 + 2\text{HCl} \]

2. **Determine the oxidation states:**
- For \(\text{CrO}_2\text{Cl}_2\):
- Let the oxidation state of Cr be \(x\).
- \(x + 2(-2) + 2(-1) = 0\)
- \(x - 4 = 0 \Rightarrow x = +4\)
- For \(\text{Na}_2\text{CrO}_4\):
- \(x + 4(-2) = 0\)
- \(x - 8 = 0 \Rightarrow x = +6\)

3. **Change in oxidation states:**
- Cr changes from +4 in \(\text{CrO}_2\text{Cl}_2\) to +6 in \(\text{Na}_2\text{CrO}_4\).

**Conclusion:** Oxidation number changes in this case.

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**Option 4: Cr2O7^2- solution is made alkaline.**

1. **Write the reaction:**
\[ \text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^- \rightarrow \text{CrO}_4^{2-} + \text{H}_2\text{O} \]

2. **Determine the oxidation states:**
- For \(\text{Cr}_2\text{O}_7^{2-}\):
- Cr is +6.
- For \(\text{CrO}_4^{2-}\):
- Cr is also +6.

3. **Change in oxidation states:**
- There is no change in the oxidation state of Cr (remains +6).

**Conclusion:** Oxidation number does not change in this case.

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### Final Answer:
The oxidation number changes in **Option 1** and **Option 3**.