On reduction of `KMnO_(4)` by oxalic acid in acidic medium, the oxidation number of Mn. What is the magnitude of is change?

Options

A. 7 to 2
B. 6 to 2
C. 5 to 2
D. 7 to 4

Answer

To determine the change in the oxidation number of manganese (Mn) during the reduction of potassium permanganate (KMnO₄) by oxalic acid in acidic medium, we can follow these steps:

### Step 1: Identify the oxidation state of Mn in KMnO₄
- In KMnO₄, potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2. Since there are 4 oxygen atoms, their total contribution is 4 × (-2) = -8.
- Let the oxidation state of manganese (Mn) be represented as x.

Using the formula for the sum of oxidation states in a neutral compound:
\[ +1 + x + (-8) = 0 \]

### Step 2: Solve for x
- Rearranging the equation gives:
\[ x - 7 = 0 \]
\[ x = +7 \]

Thus, the oxidation state of Mn in KMnO₄ is +7.

### Step 3: Determine the oxidation state of Mn after reduction
- During the reaction with oxalic acid in acidic medium, KMnO₄ is reduced to Mn²⁺.
- The oxidation state of Mn in Mn²⁺ is +2.

### Step 4: Calculate the change in oxidation state
- The change in oxidation state can be calculated as follows:
\[ \text{Change} = \text{Final oxidation state} - \text{Initial oxidation state} \]
\[ \text{Change} = +2 - (+7) \]
\[ \text{Change} = +2 - 7 = -5 \]

### Step 5: Determine the magnitude of the change
- The magnitude of the change in oxidation state is the absolute value:
\[ \text{Magnitude} = |-5| = 5 \]

### Conclusion
The magnitude of the change in the oxidation number of Mn during the reduction of KMnO₄ by oxalic acid in acidic medium is **5**.

---