In a reaction, `K_2 MnO_4` is converted into `KMnO_4` . The change in the oxidation number of Mn is :

Options

A. 0
B. `+1`
C. `-1`
D. `+7`

Explanation

To determine the change in the oxidation number of manganese (Mn) when \( K_2MnO_4 \) is converted into \( KMnO_4 \), we need to follow these steps:

### Step 1: Determine the oxidation state of Mn in \( K_2MnO_4 \)
In \( K_2MnO_4 \):
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

Let the oxidation state of Mn be \( x \). The formula can be represented as:
\[ 2(+1) + x + 4(-2) = 0 \]
This simplifies to:
\[ 2 + x - 8 = 0 \implies x - 6 = 0 \implies x = +6 \]
So, the oxidation state of Mn in \( K_2MnO_4 \) is +6.

### Step 2: Determine the oxidation state of Mn in \( KMnO_4 \)
In \( KMnO_4 \):
- Again, K has an oxidation state of +1.
- Oxygen has an oxidation state of -2.

Let the oxidation state of Mn be \( y \). The formula can be represented as:
\[ 1(+1) + y + 4(-2) = 0 \]
This simplifies to:
\[ 1 + y - 8 = 0 \implies y - 7 = 0 \implies y = +7 \]
So, the oxidation state of Mn in \( KMnO_4 \) is +7.

### Step 3: Calculate the change in oxidation state
Now, we can find the change in oxidation state of Mn:
\[ \text{Change in oxidation state} = \text{Final oxidation state} - \text{Initial oxidation state} = +7 - (+6) = +1 \]

### Conclusion
The change in the oxidation number of Mn when \( K_2MnO_4 \) is converted into \( KMnO_4 \) is +1.

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