For the reaction
`M^(x+)+MnO_(4)^(ө)rarrMO_(3)^(ө)+Mn^(2+)+(1//2)O_(2)`
if `1 "mol of" MnO_(4)^(ө)` oxidises `1.67 "mol of" M^(x+) "to" MO_(3)^(ө)`, then the value of `x` in the reaction is

Answer

To solve the problem, we need to analyze the redox reaction given and determine the oxidation state of the element M in the compound M^(x+).

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- In the reaction, M^(x+) is oxidized to MO3^(ө). The oxidation state of M in MO3 is +5. Therefore, we have:
- M^(x+) → MO3^(5+)
- This means that M goes from an oxidation state of x to +5.

2. **Determine the Change in Oxidation State**:
- The change in oxidation state for M can be calculated as:
\[ \text{Change in oxidation state} = 5 - x \]

3. **Reduction of MnO4^-**:
- The permanganate ion (MnO4^-) is reduced to Mn^(2+). The oxidation state of manganese changes from +7 in MnO4^- to +2 in Mn^(2+). This reduction involves the gain of 5 electrons:
\[ \text{MnO4}^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \]

4. **Relate Electrons Transferred**:
- From the stoichiometry of the reaction, we know that 1 mole of MnO4^- can accept 5 moles of electrons.
- For every mole of M^(x+) oxidized to MO3^(5+), it loses (5 - x) moles of electrons.

5. **Set Up the Equation**:
- According to the problem, 1 mole of MnO4^- oxidizes 1.67 moles of M^(x+). Therefore, we can set up the relationship:
\[ \frac{5 \text{ moles of electrons}}{5 - x} = 1.67 \]

6. **Cross-Multiply and Solve for x**:
- Cross-multiplying gives:
\[ 5 = 1.67(5 - x) \]
- Expanding this:
\[ 5 = 8.35 - 1.67x \]
- Rearranging to isolate x:
\[ 1.67x = 8.35 - 5 \]
\[ 1.67x = 3.35 \]
- Dividing both sides by 1.67:
\[ x = \frac{3.35}{1.67} \approx 2 \]

7. **Conclusion**:
- The value of x is 2.

### Final Answer:
The value of x in the reaction is **2**.