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

Options

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

Explanation

To solve the problem, we need to determine the change in the oxidation number of manganese (Mn) when a manganous salt (MnO) is fused with a mixture of KNO₃ and solid NaOH.

### Step-by-Step Solution:

1. **Identify the Initial Oxidation State of Mn:**
- In the manganous salt, which is MnO, we need to find the oxidation state of manganese (Mn).
- The oxidation state of oxygen (O) is -2.
- Let the oxidation state of manganese be \( x \).
- The overall charge of the compound is 0, so we can set up the equation:
\[ x + (-2) = 0 \]
- Solving for \( x \):
\[ x = +2 \]
- Therefore, the oxidation state of Mn in MnO is +2.

2. **Write the Reaction:**
- The reaction when MnO is fused with KNO₃ and NaOH can be represented as:
\[ \text{MnO} + \text{NaOH} + \text{KNO}_3 \rightarrow \text{Na}_2\text{MnO}_4 + \text{KNO}_2 + \text{H}_2\text{O} \]

3. **Identify the Final Oxidation State of Mn:**
- In the product Na₂MnO₄, we need to determine the oxidation state of manganese.
- Let the oxidation state of manganese in Na₂MnO₄ be \( y \).
- The oxidation state of sodium (Na) is +1 and that of oxygen (O) is -2.
- The equation for the overall charge is:
\[ 2(+1) + y + 4(-2) = 0 \]
- Simplifying this gives:
\[ 2 + y - 8 = 0 \]
\[ y - 6 = 0 \]
\[ y = +6 \]
- Therefore, the oxidation state of Mn in Na₂MnO₄ is +6.

4. **Determine the Change in Oxidation State:**
- The oxidation state of Mn changes from +2 in MnO to +6 in Na₂MnO₄.
- This indicates an increase in oxidation state, which means Mn is oxidized.

5. **Conclusion:**
- The final oxidation number of manganese after the reaction is +6.

### Final Answer:
The oxidation number of Mn changes from +2 to +6.