The magnetic moment of a transition metal ion is found to be 3.87 Bohr Magneton (BM). The number of unpaired electrons present in it is:

Answer

To determine the number of unpaired electrons in a transition metal ion with a magnetic moment of 3.87 Bohr Magneton (BM), we can use the formula for calculating the magnetic moment:

### Step-by-Step Solution:

1. **Understanding the Formula**:
The magnetic moment (μ) in Bohr Magneton is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Substituting the Given Value**:
We know from the problem that:
\[ \mu = 3.87 \, \text{BM} \]
Therefore, we can set up the equation:
\[ \sqrt{n(n + 2)} = 3.87 \]

3. **Squaring Both Sides**:
To eliminate the square root, we square both sides of the equation:
\[ n(n + 2) = (3.87)^2 \]

4. **Calculating the Square**:
Now we calculate \( (3.87)^2 \):
\[ (3.87)^2 = 14.9769 \approx 14.97 \]
So we have:
\[ n(n + 2) = 14.97 \]

5. **Rearranging the Equation**:
This can be rearranged to form a quadratic equation:
\[ n^2 + 2n - 14.97 = 0 \]

6. **Solving the Quadratic Equation**:
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 2, c = -14.97 \):
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-14.97)}}{2 \cdot 1} \]
\[ n = \frac{-2 \pm \sqrt{4 + 59.88}}{2} \]
\[ n = \frac{-2 \pm \sqrt{63.88}}{2} \]
\[ n = \frac{-2 \pm 7.99}{2} \]

7. **Finding the Positive Root**:
We take the positive root:
\[ n = \frac{5.99}{2} \approx 2.995 \approx 3 \]

8. **Conclusion**:
Therefore, the number of unpaired electrons \( n \) is approximately 3.

### Final Answer:
The number of unpaired electrons present in the transition metal ion is **3**.