The equivalent mass of `MnSO_4` is half of its molecular mass when it is converted to

Options

A. A)`Mn_(2)O_(3)`
B. `B)MnO_(2)`
C. C)`MnO_(4)^(-)`
D. D)`MnO_(4)^(2-)`

Explanation

To solve the problem of determining when the equivalent mass of `MnSO4` is half of its molecular mass, we will follow these steps:

### Step 1: Understanding Equivalent Mass
The equivalent mass of a substance can be calculated using the formula:
\[ \text{Equivalent Mass} = \frac{\text{Molecular Mass}}{\text{Valency Factor}} \]
In this case, we need to find when the equivalent mass of `MnSO4` is half of its molecular mass. This means:
\[ \text{Equivalent Mass} = \frac{1}{2} \times \text{Molecular Mass} \]

### Step 2: Finding the Molecular Mass of `MnSO4`
The molecular mass of `MnSO4` can be calculated as follows:
- Manganese (Mn) = 54.94 g/mol
- Sulfur (S) = 32.07 g/mol
- Oxygen (O) = 16.00 g/mol (4 O atoms)

Calculating the total:
\[ \text{Molecular Mass of } MnSO4 = 54.94 + 32.07 + (4 \times 16.00) = 54.94 + 32.07 + 64.00 = 150.01 \, \text{g/mol} \]

### Step 3: Setting Up the Equation
Given that the equivalent mass is half of the molecular mass, we can set up the equation:
\[ \frac{150.01}{\text{Valency Factor}} = \frac{1}{2} \times 150.01 \]
This simplifies to:
\[ \text{Valency Factor} = 2 \]

### Step 4: Analyzing the Oxidation States
Now, we need to determine the oxidation states of manganese in the products to find the correct conversion:
1. **MnO2**:
- Oxidation state of Mn = +4
2. **Mn2O3**:
- Oxidation state of Mn = +3
3. **MnO4^-**:
- Oxidation state of Mn = +7
4. **MnO4^{2-}**:
- Oxidation state of Mn = +6

### Step 5: Identifying the Correct Product
The equivalent mass of `MnSO4` being half of its molecular mass indicates that the valency factor must be 2. This occurs when Mn is reduced from +2 to +4 (as in MnO2).

Thus, the correct conversion is:
\[ \text{MnSO4} \rightarrow \text{MnO2} \]

### Conclusion
The correct answer is **MnO2**, which corresponds to **Option B**.

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