`MnO_(4)^(-)` is a good oxidising agent in different medium changing to
`MnO_(4)^(-) to Mn^(2+) to MnO_(4)^(2-) to MnO_(2) to Mn_(2)O_(3)`
Changes in oxidation number respectively,are

Options

A. 1,3,4,5
B. 5,4,3,2
C. 5,1,3,4
D. 2,6,4,3

Explanation

To determine the changes in oxidation states of manganese (Mn) during the reduction of permanganate ion (MnO₄⁻) to various manganese species, we will calculate the oxidation number of Mn in each species and find the difference in oxidation states.

### Step-by-Step Solution:

1. **Identify the oxidation state of Mn in MnO₄⁻:**
- The oxidation state of oxygen (O) is -2.
- In MnO₄⁻, there are 4 oxygen atoms:
\[ \text{Total oxidation state of O} = 4 \times (-2) = -8 \]
- Let the oxidation state of Mn be \( x \):
\[ x + (-8) = -1 \quad \Rightarrow \quad x = +7 \]
- **Change:** MnO₄⁻ (Mn is +7)

2. **Identify the oxidation state of Mn in Mn²⁺:**
- The oxidation state of Mn in Mn²⁺ is +2.
- **Change:** From +7 to +2, the change is:
\[ 7 - 2 = 5 \]

3. **Identify the oxidation state of Mn in MnO₄²⁻:**
- Let the oxidation state of Mn be \( x \):
\[ x + 4 \times (-2) = -2 \quad \Rightarrow \quad x - 8 = -2 \quad \Rightarrow \quad x = +6 \]
- **Change:** From +7 to +6, the change is:
\[ 7 - 6 = 1 \]

4. **Identify the oxidation state of Mn in MnO₂:**
- Let the oxidation state of Mn be \( x \):
\[ x + 2 \times (-2) = 0 \quad \Rightarrow \quad x - 4 = 0 \quad \Rightarrow \quad x = +4 \]
- **Change:** From +7 to +4, the change is:
\[ 7 - 4 = 3 \]

5. **Identify the oxidation state of Mn in Mn₂O₃:**
- Let the oxidation state of Mn be \( x \):
\[ 2x + 3 \times (-2) = 0 \quad \Rightarrow \quad 2x - 6 = 0 \quad \Rightarrow \quad 2x = 6 \quad \Rightarrow \quad x = +3 \]
- **Change:** From +7 to +3, the change is:
\[ 7 - 3 = 4 \]

### Summary of Changes:
- From MnO₄⁻ to Mn²⁺: Change = 5
- From MnO₄⁻ to MnO₄²⁻: Change = 1
- From MnO₄⁻ to MnO₂: Change = 3
- From MnO₄⁻ to Mn₂O₃: Change = 4

### Final Result:
The changes in oxidation numbers respectively are: **5, 1, 3, 4**.