The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted to

Options

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

Explanation

To determine the equivalent weight of \( \text{MnSO}_4 \) when it is converted to a different oxidation state, we can follow these steps:

### Step 1: Determine the Molecular Weight of \( \text{MnSO}_4 \)

The molecular weight of \( \text{MnSO}_4 \) can be calculated by adding the atomic weights of its constituent elements:
- Manganese (Mn): approximately 54.94 g/mol
- Sulfur (S): approximately 32.07 g/mol
- Oxygen (O): approximately 16.00 g/mol (4 oxygen atoms)

Calculating the total:
\[ \text{Molecular weight of } \text{MnSO}_4 = 54.94 + 32.07 + (4 \times 16.00) = 54.94 + 32.07 + 64.00 = 150.01 \text{ g/mol} \]

### Step 2: Understand the Concept of Equivalent Weight

The equivalent weight of a substance is defined as the mass of the substance that will combine with or displace 1 mole of hydrogen atoms (or 1 mole of electrons in redox reactions).

### Step 3: Determine the Change in Oxidation State

When \( \text{MnSO}_4 \) is converted to a different compound, the manganese ion (Mn) can change its oxidation state. The equivalent weight is half the molecular weight when the change in oxidation state involves the transfer of 2 electrons.

### Step 4: Calculate the Equivalent Weight

Since the equivalent weight of \( \text{MnSO}_4 \) is given to be half its molecular weight during the conversion, we can calculate it as follows:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
Where \( n \) is the number of electrons transferred. In this case, \( n = 2 \).

Thus,
\[ \text{Equivalent weight of } \text{MnSO}_4 = \frac{150.01}{2} = 75.005 \text{ g/equiv} \]

### Conclusion

The equivalent weight of \( \text{MnSO}_4 \) is half its molecular weight when it is converted to a species where manganese undergoes a change in oxidation state that involves the transfer of 2 electrons.

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