When a mananous salt is fused with a mixture of `KNO_(3)` and and solid NaOH, the oxidation number of Mn change from +2 to:

Options

A. `+4`
B. `+3`
C. `+6`
D. `+7`

Answer

To solve the problem of determining the change in the oxidation number of manganese (Mn) when a manganese salt is fused with a mixture of KNO₃ and solid NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Oxidation State of Mn**:
- The manganese salt is given as Mn²⁺, which means the initial oxidation state of manganese is +2.

2. **Understand the Reaction Conditions**:
- The manganese salt is fused with KNO₃ and NaOH. In this reaction, KNO₃ acts as an oxidizing agent in a basic medium provided by NaOH.

3. **Determine the Final Oxidation State of Mn**:
- In the presence of KNO₃ and NaOH, manganese is oxidized to MnO₄⁻ (permanganate ion).
- To find the oxidation state of Mn in MnO₄⁻, we can use the following equation:
\[ X + 4(-2) = -1 \]
where \(X\) is the oxidation state of Mn and there are 4 oxygen atoms each with an oxidation state of -2.

4. **Solve for X**:
- Rearranging the equation gives:
\[ X - 8 = -1 \]
\[ X = -1 + 8 = +7 \]
- Therefore, the oxidation state of Mn in MnO₄⁻ is +7.

5. **Calculate the Change in Oxidation State**:
- The change in oxidation state can be calculated as:
\[ \text{Change} = \text{Final Oxidation State} - \text{Initial Oxidation State} \]
\[ \text{Change} = +7 - (+2) = +5 \]

### Final Answer:
- The oxidation number of Mn changes from +2 to +7.