Oxidation state exhibited by Mn in `KMnO_(4)` is

Answer

To find the oxidation state of manganese (Mn) in potassium permanganate (KMnO₄), we can follow these steps:

### Step 1: Identify the oxidation states of known elements
In KMnO₄, we have potassium (K), manganese (Mn), and oxygen (O).

- The oxidation state of potassium (K) is +1.
- The oxidation state of oxygen (O) is typically -2.

### Step 2: Write the equation for the total charge
Since KMnO₄ is a neutral compound, the sum of the oxidation states of all the atoms must equal zero.

Let the oxidation state of manganese be represented as x. The equation can be set up as follows:

\[ \text{Total charge} = \text{(oxidation state of K)} + \text{(oxidation state of Mn)} + \text{(oxidation state of O)} \]

Substituting the known values:

\[ +1 + x + 4(-2) = 0 \]

### Step 3: Simplify the equation
Now, simplify the equation:

\[ +1 + x - 8 = 0 \]

### Step 4: Solve for x
Rearranging the equation gives:

\[ x - 7 = 0 \]

Thus,

\[ x = +7 \]

### Conclusion
The oxidation state of manganese (Mn) in potassium permanganate (KMnO₄) is +7.

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