The magnetic moment of `M^(2+)` is 3.87 BM. The species is :

Options

A. `Mn^(2+)`
B. `V^(2+)`
C. `Ti^(2+)`
D. `Cr^(2+)`

Answer

To determine the species corresponding to the magnetic moment of \( M^{2+} \) being 3.87 Bohr magneton (BM), we can follow these steps:

### Step 1: Identify the Element
The magnetic moment is given as 3.87 BM. We need to consider common transition metals that can exist in a +2 oxidation state.

### Step 2: Determine the Ground State Electronic Configuration
For manganese (Mn), the ground state electronic configuration is:
- \( \text{Mn: } [Ar] 3d^5 4s^2 \)

### Step 3: Find the Configuration for \( M^{2+} \)
When manganese loses two electrons to form \( Mn^{2+} \), the configuration becomes:
- \( \text{Mn}^{2+}: 3d^5 \)

### Step 4: Calculate the Number of Unpaired Electrons
In the \( 3d^5 \) configuration, all five electrons are unpaired (due to Hund's rule). Thus:
- Number of unpaired electrons, \( n = 5 \)

### Step 5: Use the Formula for Magnetic Moment
The formula for calculating the magnetic moment (\( \mu \)) is:
\[ \mu = \sqrt{n(n + 2)} \]
Substituting \( n = 5 \):
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \text{ BM} \]
This value is higher than the given magnetic moment, indicating that manganese is not the correct species.

### Step 6: Check Other Transition Metals
Next, we check vanadium (V):
- Ground state configuration: \( \text{V: } [Ar] 3d^3 4s^2 \)
- For \( V^{2+} \): \( 3d^3 \)
- Number of unpaired electrons in \( 3d^3 \): \( n = 3 \)

Calculating the magnetic moment for \( V^{2+} \):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \text{ BM} \]
This matches the given magnetic moment.

### Conclusion
The species corresponding to the magnetic moment of \( M^{2+} = 3.87 \) BM is:
- **Vanadium (V)** in the +2 oxidation state.