The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

Options

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

Answer

To determine the metal ion from the given magnetic moment of 3.87 Bohr magneton (B.M.), we will follow these steps:

### Step 1: Understand the Magnetic Moment Formula
The magnetic moment (μ) is calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons in the d-orbitals of the transition metal ion.

### Step 2: Set Up the Equation
Given that the magnetic moment is 3.87 B.M., we can set up the equation:
\[ 3.87 = \sqrt{n(n + 2)} \]

### Step 3: Square Both Sides
To eliminate the square root, we square both sides:
\[ (3.87)^2 = n(n + 2) \]
Calculating \( (3.87)^2 \):
\[ 14.9769 \approx 15 \quad \text{(for simplicity)} \]
So, we have:
\[ 15 = n(n + 2) \]

### Step 4: Solve the Quadratic Equation
Rearranging gives us:
\[ n^2 + 2n - 15 = 0 \]
Now, we can factor this quadratic:
\[ (n + 5)(n - 3) = 0 \]
This gives us two possible solutions for \( n \):
\[ n = -5 \quad \text{(not possible)} \quad \text{or} \quad n = 3 \]

### Step 5: Identify the Metal Ion
Now that we have \( n = 3 \), we need to identify the transition metal ion that has 3 unpaired electrons.

- Looking at the transition metals, we find that Vanadium (V) in the +2 oxidation state has the following electron configuration:
\[ \text{V: } [Ar] 3d^3 4s^2 \quad \text{(in +2 state: } 3d^3\text{)} \]
This configuration shows that Vanadium in the +2 oxidation state has 3 unpaired electrons.

### Conclusion
The metal ion with a magnetic moment of 3.87 B.M. is Vanadium in the +2 oxidation state.