A metal ion from the first transition series has a magnetic moment (calculated) as `3.87 B.M.` How many unpaired electrons are expected to be present in the ion?

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To find the number of unpaired electrons in a metal ion from the first transition series with a magnetic moment of 3.87 Bohr magnetons (B.M.), we can use the formula for the magnetic moment:

\[ \mu = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons.

### Step-by-Step Solution:

1. **Set up the equation**:
Given that the magnetic moment \( \mu = 3.87 \, \text{B.M.} \), we can write:
\[ 3.87 = \sqrt{n(n + 2)} \]

2. **Square both sides**:
To eliminate the square root, we square both sides of the equation:
\[ (3.87)^2 = n(n + 2) \]
Calculating \( (3.87)^2 \):
\[ 3.87^2 \approx 14.9769 \quad \text{(approximately 15)} \]
So we can simplify our equation to:
\[ 15 = n(n + 2) \]

3. **Rearrange the equation**:
Rearranging gives us:
\[ n^2 + 2n - 15 = 0 \]

4. **Solve the quadratic equation**:
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = -15 \):
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} \]
\[ n = \frac{-2 \pm \sqrt{4 + 60}}{2} \]
\[ n = \frac{-2 \pm \sqrt{64}}{2} \]
\[ n = \frac{-2 \pm 8}{2} \]

This gives us two possible solutions:
\[ n = \frac{6}{2} = 3 \quad \text{(positive solution)} \]
\[ n = \frac{-10}{2} = -5 \quad \text{(not valid)} \]

5. **Conclusion**:
The number of unpaired electrons \( n \) is 3.

### Final Answer:
The metal ion has **3 unpaired electrons**.