What will be n-factor for `Ba(MnO_(4))_(2)` in acidic medium? (Where it behaves as oxidant)

Options

A. 5
B. 10
C. 6
D. 3

Answer

To find the n-factor for \( Ba(MnO_4)_2 \) in acidic medium where it behaves as an oxidant, we can follow these steps:

### Step 1: Identify the oxidation states
In \( Ba(MnO_4)_2 \), the oxidation state of manganese (Mn) in \( MnO_4^- \) is +7. Since there are two manganese atoms in the formula, we will consider both.

### Step 2: Determine the change in oxidation state
When \( Ba(MnO_4)_2 \) acts as an oxidant in acidic medium, manganese is reduced from +7 to +2. The change in oxidation state for one manganese atom is:
\[ \text{Change in oxidation state} = +7 - +2 = +5 \]

### Step 3: Calculate the total change for all manganese atoms
Since there are 2 manganese atoms in \( Ba(MnO_4)_2 \), the total change in oxidation state for both manganese atoms is:
\[ \text{Total change} = 5 \times 2 = 10 \]

### Step 4: Define the n-factor
The n-factor is defined as the total change in oxidation state per formula unit of the compound. Therefore, the n-factor for \( Ba(MnO_4)_2 \) is:
\[ \text{n-factor} = 10 \]

### Conclusion
Thus, the n-factor for \( Ba(MnO_4)_2 \) in acidic medium, where it behaves as an oxidant, is **10**.

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