The equivalent weight of`MnCl_2`, is half of its molecular weight when it is converted to

Options

A. `MnO_4^-`
B. `Mn_2O_3`
C. `MnO_2`
D. `MnO_4^(2-)`

Answer

To solve the problem, we need to determine the compound to which `MnCl2` converts such that its equivalent weight is half of its molecular weight. The equivalent weight is defined as the molecular weight divided by the n-factor.

### Step-by-Step Solution:

1. **Determine the Molecular Weight of MnCl2**:
- The molecular weight of `MnCl2` can be calculated as follows:
- Manganese (Mn) = 54.94 g/mol
- Chlorine (Cl) = 35.45 g/mol
- Therefore, the molecular weight of `MnCl2` = 54.94 + (2 × 35.45) = 54.94 + 70.90 = 125.84 g/mol.

2. **Understanding Equivalent Weight**:
- The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]
- According to the problem, the equivalent weight of `MnCl2` is half of its molecular weight:
\[ \text{Equivalent Weight} = \frac{M}{2} \]
- This implies:
\[ \frac{M}{n} = \frac{M}{2} \implies n = 2 \]

3. **Identifying the Oxidation States**:
- The oxidation state of manganese in `MnCl2` is +2.
- We need to find a compound where the change in oxidation state of manganese results in an n-factor of 2.

4. **Evaluating the Options**:
- **Option 1: MnO4-**
- Oxidation state of Mn = +7 (calculated as X - 8 = -1, thus X = +7).
- Change in oxidation state = 7 - 2 = 5 (n-factor = 5).
- **Option 2: Mn2O3**
- Oxidation state of Mn = +3 (calculated as 2X - 6 = 0, thus X = +3).
- Change in oxidation state = 3 - 2 = 1 (n-factor = 1).
- **Option 3: MnO2**
- Oxidation state of Mn = +4 (calculated as X - 4 = 0, thus X = +4).
- Change in oxidation state = 4 - 2 = 2 (n-factor = 2).
- **Option 4: MnO4 2-**
- Oxidation state of Mn = +6 (calculated as X - 8 = -2, thus X = +6).
- Change in oxidation state = 6 - 2 = 4 (n-factor = 4).

5. **Conclusion**:
- The only compound where the change in oxidation state results in an n-factor of 2 is `MnO2`.
- Therefore, the correct answer is **MnO2**.

### Final Answer:
The equivalent weight of `MnCl2` is half of its molecular weight when it is converted to **MnO2**.

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