In the given typical redox reaction : `M^(X+) + MnO_(4)^(-) to MO_(3)^(-) + Mn^(2+) + 1/2 O_(2)` , If one mole of `MnO_(4)^(-)` oxidises 2.5 moles of `M^(x+)` , then the value of x is :

Explanation

To solve the problem, we need to analyze the given redox reaction and the changes in oxidation states of the elements involved.

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- In the reactants, we have:
- \( M^{(X+)} \): The oxidation state of \( M \) is \( +x \).
- \( MnO_4^{-} \): The oxidation state of manganese (Mn) is \( +7 \).
- In the products, we have:
- \( MO_3^{-} \): The oxidation state of \( M \) is \( +5 \).
- \( Mn^{2+} \): The oxidation state of manganese (Mn) is \( +2 \).

2. **Calculate the Change in Oxidation States**:
- For manganese:
- Change in oxidation state from \( +7 \) to \( +2 \):
\[ 7 - 2 = 5 \]
- For \( M \):
- Change in oxidation state from \( +x \) to \( +5 \):
\[ 5 - x \]

3. **Set Up the Stoichiometric Relationship**:
- According to the problem, 1 mole of \( MnO_4^{-} \) oxidizes 2.5 moles of \( M^{(X+)} \).
- The total change in oxidation states for the reaction can be expressed as:
\[ 1 \times 5 = 2.5 \times (5 - x) \]

4. **Solve the Equation**:
- Set up the equation:
\[ 5 = 2.5 \times (5 - x) \]
- Simplify the equation:
\[ 5 = 12.5 - 2.5x \]
- Rearranging gives:
\[ 2.5x = 12.5 - 5 \]
\[ 2.5x = 7.5 \]
- Divide both sides by 2.5:
\[ x = \frac{7.5}{2.5} = 3 \]

5. **Conclusion**:
- The value of \( x \) is \( 3 \).