The spin only magnetic moment of transition metal ion found to be 5.92 BM. The number of unpaired electrons present in the species is :

Answer

To solve the problem of finding the number of unpaired electrons in a transition metal ion with a spin-only magnetic moment of 5.92 Bohr Magnetons (BM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**: The formula for the spin-only magnetic moment (μ_spin) is given by:
\[ \mu_{\text{spin}} = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Substitute the Given Value**: We know that:
\[ \mu_{\text{spin}} = 5.92 \, \text{BM} \]
Therefore, we can set up the equation:
\[ 5.92 = \sqrt{n(n + 2)} \]

3. **Square Both Sides**: To eliminate the square root, we square both sides of the equation:
\[ (5.92)^2 = n(n + 2) \]
Calculating \( (5.92)^2 \):
\[ 5.92^2 = 35.0464 \approx 35.04 \]
So we have:
\[ 35.04 = n(n + 2) \]

4. **Rearrange the Equation**: Rearranging gives us a quadratic equation:
\[ n^2 + 2n - 35.04 = 0 \]
For simplicity, we can approximate this to:
\[ n^2 + 2n - 35 = 0 \]

5. **Solve the Quadratic Equation**: We can factor or use the quadratic formula to solve for \( n \):
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 2, c = -35 \):
\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-35)}}{2 \cdot 1} \]
\[ n = \frac{-2 \pm \sqrt{4 + 140}}{2} \]
\[ n = \frac{-2 \pm \sqrt{144}}{2} \]
\[ n = \frac{-2 \pm 12}{2} \]

6. **Calculate Possible Values for n**: This gives us two potential solutions:
\[ n = \frac{10}{2} = 5 \quad \text{and} \quad n = \frac{-14}{2} = -7 \]
Since the number of unpaired electrons cannot be negative, we take:
\[ n = 5 \]

7. **Conclusion**: Therefore, the number of unpaired electrons present in the species is:
\[ \boxed{5} \]