When `KMnO_(4)` reacts with KBr in acidic medium then oxidation state of Mn changes from +7 to :-

Options

A. `+2`
B. `+4`
C. `+3`
D. `+6`

Answer

To determine the change in the oxidation state of manganese (Mn) when potassium permanganate (KMnO₄) reacts with potassium bromide (KBr) in an acidic medium, we can follow these steps:

### Step 1: Identify the initial oxidation state of Mn in KMnO₄
In KMnO₄, the oxidation state of manganese (Mn) can be calculated as follows:
- The oxidation state of oxygen (O) is -2.
- There are 4 oxygen atoms, contributing a total of -8.
- The overall charge of the permanganate ion (MnO₄⁻) is -1.

Using the formula:
\[ \text{Oxidation state of Mn} + 4(-2) = -1 \]
Let the oxidation state of Mn be \( x \):
\[ x - 8 = -1 \]
\[ x = +7 \]

### Step 2: Write the balanced chemical equation for the reaction
The balanced equation for the reaction of KMnO₄ with KBr in acidic medium is:
\[ 2 \text{KMnO}_4 + 10 \text{KBr} + 8 \text{H}_2\text{SO}_4 \rightarrow 2 \text{MnSO}_4 + 5 \text{Br}_2 + 8 \text{K}_2\text{SO}_4 + 8 \text{H}_2\text{O} \]

### Step 3: Determine the oxidation state of Mn in the products
In the product, manganese is present as MnSO₄. To find the oxidation state of Mn in MnSO₄:
- The oxidation state of sulfur (S) in sulfate (SO₄²⁻) is +6.
- The oxidation state of oxygen (O) is -2, with 4 oxygen atoms contributing -8.
- The overall charge of the sulfate ion (SO₄²⁻) is -2.

Using the formula:
\[ \text{Oxidation state of Mn} + 6 + 4(-2) = -2 \]
Let the oxidation state of Mn be \( y \):
\[ y + 6 - 8 = -2 \]
\[ y - 2 = -2 \]
\[ y = +4 \]

### Step 4: Conclusion
The oxidation state of Mn changes from +7 in KMnO₄ to +4 in MnSO₄.

Thus, the answer is:
**The oxidation state of Mn changes from +7 to +4.**