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 under what condition the equivalent weight of \( \text{MnCl}_2 \) is half of its molecular weight.

### Step-by-Step Solution:

1. **Determine the Molecular Weight of \( \text{MnCl}_2 \)**:
- The molecular weight of \( \text{MnCl}_2 \) can be calculated as follows:
- Manganese (Mn) has an atomic weight of approximately 55 g/mol.
- Chlorine (Cl) has an atomic weight of approximately 35.5 g/mol.
- Therefore, the molecular weight of \( \text{MnCl}_2 \) is:
\[ \text{Molecular Weight} = 55 + 2 \times 35.5 = 55 + 71 = 126 \text{ g/mol} \]

2. **Understanding Equivalent Weight**:
- The equivalent weight of a substance is defined as the molecular weight divided by the number of equivalents (n-factor).
- The problem states that the equivalent weight of \( \text{MnCl}_2 \) is half of its molecular weight:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]
\[ \frac{\text{Molecular Weight}}{n} = \frac{1}{2} \times \text{Molecular Weight} \]
- This implies:
\[ n = 2 \]

3. **Identify the Oxidation States**:
- The n-factor is determined by the change in oxidation state of manganese during the reaction.
- We will analyze the different options provided in the question to find out which one gives an n-factor of 2.

4. **Calculate the Change in Oxidation State**:
- **Option 1: \( \text{MnCl}_2 \) to \( \text{MnO}_4^- \)**:
- Change in oxidation state: \( +2 \) to \( +7 \) (change of 5).
- n-factor = 5 (not suitable).

- **Option 2: \( \text{MnCl}_2 \) to \( \text{Mn}_2\text{O}_3 \)**:
- Change in oxidation state: \( +2 \) to \( +3 \) (change of 1).
- n-factor = 1 (not suitable).

- **Option 3: \( \text{MnCl}_2 \) to \( \text{MnO}_2 \)**:
- Change in oxidation state: \( +2 \) to \( +4 \) (change of 2).
- n-factor = 2 (suitable).

5. **Conclusion**:
- The equivalent weight of \( \text{MnCl}_2 \) is half of its molecular weight when it is converted to \( \text{MnO}_2 \).

### Final Answer:
The equivalent weight of \( \text{MnCl}_2 \) is half of its molecular weight when it is converted to \( \text{MnO}_2 \).