In which one of the following changes there are transfer of five electrons?

Options

A. `MnO_(4)^(-) to Mn^(2+)`
B. `CrO_(4)^(2-) to Cr^(3+)`
C. `MnO_(4)^(-) to MnO_(2)`
D. `Cr_(2)O_(7)^(2-) to 2Cr^(3+)`

Answer

To determine the change in oxidation states and the number of electrons transferred in the given options, we will analyze each option step by step.

### Step-by-Step Solution:

**Step 1: Analyze the first option - MnO4⁻ to Mn²⁺**

1. **Determine the oxidation state of Mn in MnO4⁻:**
- Let the oxidation state of Mn be \( x \).
- The oxidation state of oxygen (O) is -2.
- The equation for the oxidation state is:
\[ x + 4(-2) = -1 \]
- Simplifying this gives:
\[ x - 8 = -1 \implies x = +7 \]
- Therefore, Mn in MnO4⁻ has an oxidation state of +7.

2. **Determine the oxidation state of Mn in Mn²⁺:**
- The oxidation state of Mn in Mn²⁺ is +2.

3. **Calculate the change in oxidation state:**
- Change = +7 to +2, which involves the transfer of:
\[ 7 - 2 = 5 \text{ electrons} \]

**Step 2: Analyze the second option - CrO4²⁻ to Cr³⁺**

1. **Determine the oxidation state of Cr in CrO4²⁻:**
- Let the oxidation state of Cr be \( x \).
- The equation is:
\[ x + 4(-2) = -2 \]
- Simplifying gives:
\[ x - 8 = -2 \implies x = +6 \]
- Therefore, Cr in CrO4²⁻ has an oxidation state of +6.

2. **Determine the oxidation state of Cr in Cr³⁺:**
- The oxidation state of Cr in Cr³⁺ is +3.

3. **Calculate the change in oxidation state:**
- Change = +6 to +3, which involves the transfer of:
\[ 6 - 3 = 3 \text{ electrons} \]

**Step 3: Analyze the third option - MnO4⁻ to MnO2**

1. **Determine the oxidation state of Mn in MnO4⁻:**
- As calculated before, Mn in MnO4⁻ is +7.

2. **Determine the oxidation state of Mn in MnO2:**
- Let the oxidation state of Mn in MnO2 be \( x \).
- The equation is:
\[ x + 2(-2) = 0 \]
- Simplifying gives:
\[ x - 4 = 0 \implies x = +4 \]
- Therefore, Mn in MnO2 has an oxidation state of +4.

3. **Calculate the change in oxidation state:**
- Change = +7 to +4, which involves the transfer of:
\[ 7 - 4 = 3 \text{ electrons} \]

**Step 4: Analyze the fourth option - Cr2O7²⁻ to Cr³⁺**

1. **Determine the oxidation state of Cr in Cr2O7²⁻:**
- Let the oxidation state of Cr be \( x \).
- The equation is:
\[ 2x + 7(-2) = -2 \]
- Simplifying gives:
\[ 2x - 14 = -2 \implies 2x = 12 \implies x = +6 \]
- Therefore, Cr in Cr2O7²⁻ has an oxidation state of +6.

2. **Determine the oxidation state of Cr in Cr³⁺:**
- The oxidation state of Cr in Cr³⁺ is +3.

3. **Calculate the change in oxidation state:**
- Change = +6 to +3, which involves the transfer of:
\[ 6 - 3 = 3 \text{ electrons} \]

### Conclusion:
From the analysis, the only option that involves the transfer of 5 electrons is the first option: **MnO4⁻ to Mn²⁺**.