`F_(2)` can be prepared by reacting hexfluoro magnante (IV) with antimony pentafluoride as:
`K_(2)KnF_(6)+SbF_(5)overset(150^(@)C)toKSbF_(6)+MnF_(3)+F_(2)` ltBrgt The number of equivalent of `K_(2)MnF_(6)` requried to react completely with one " mol of "`SbF_(5)` in the given reaction is

Options

A. `1.52`
B. `5.0`
C. `0.5`
D. `4.0`

Explanation

To solve the problem, we need to determine the number of equivalents of \( K_2MnF_6 \) required to react completely with one mole of \( SbF_5 \) in the given reaction. Let's break down the steps:

### Step 1: Write the balanced chemical equation
The balanced reaction provided is:
\[ K_2MnF_6 + 2SbF_5 \rightarrow 2KSbF_6 + MnF_3 + \frac{1}{2}F_2 \]

### Step 2: Identify the oxidation states
In this reaction, we need to identify the changes in oxidation states to determine the n-factor for \( K_2MnF_6 \).

- **For Manganese (Mn)**:
- In \( K_2MnF_6 \), Mn is in the +4 oxidation state.
- In \( MnF_3 \), Mn is in the +3 oxidation state.
- Therefore, Mn goes from +4 to +3, which means it gains 1 electron.

- **For Fluorine (F)**:
- In \( SbF_5 \), F is in the -1 oxidation state.
- In \( F_2 \), F is in the 0 oxidation state.
- Thus, each F atom goes from -1 to 0, which means it loses 1 electron. Since there are 2 F atoms produced, a total of 2 electrons are lost.

### Step 3: Calculate the n-factor for \( K_2MnF_6 \)
The n-factor is defined as the total number of moles of electrons gained or lost per mole of the substance.

- For \( K_2MnF_6 \):
- Mn contributes 1 electron (from +4 to +3).

- For \( SbF_5 \):
- 2 moles of \( SbF_5 \) contribute a total of 2 electrons (2 F atoms).

Thus, the total change in electrons is:
- \( K_2MnF_6 \) provides 1 electron.
- \( SbF_5 \) consumes 2 electrons.

### Step 4: Determine the relationship between \( K_2MnF_6 \) and \( SbF_5 \)
From the balanced equation, we see that:
- 1 mole of \( K_2MnF_6 \) reacts with 2 moles of \( SbF_5 \).
- Therefore, 1 mole of \( SbF_5 \) will react with \( \frac{1}{2} \) mole of \( K_2MnF_6 \).

### Step 5: Calculate the equivalents
Since we have established that:
- 1 mole of \( K_2MnF_6 \) is equivalent to 1 equivalent (because it provides 1 electron),
- Thus, \( \frac{1}{2} \) mole of \( K_2MnF_6 \) will be equivalent to \( \frac{1}{2} \) equivalent.

### Final Answer
The number of equivalents of \( K_2MnF_6 \) required to react completely with one mole of \( SbF_5 \) is:
\[ \text{0.5 equivalents} \]