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 determining the number of unpaired electrons in a transition metal ion given its 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 calculating the spin-only magnetic moment (μ) in terms of the number of unpaired electrons (n) is given by:
\[ \mu = \sqrt{n(n + 2)} \]

2. **Substitute the Given Value**: We know that the magnetic moment (μ) is 5.92 BM. We can substitute this value into the formula:
\[ 5.92 = \sqrt{n(n + 2)} \]

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:
\[ n(n + 2) = 35.04 \]

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. **Factor the Quadratic Equation**: We need to factor the quadratic equation:
\[ n^2 + 7n - 5n - 35 = 0 \]
This can be factored as:
\[ (n + 7)(n - 5) = 0 \]

6. **Solve for n**: Setting each factor to zero gives us:
\[ n + 7 = 0 \quad \Rightarrow \quad n = -7 \quad (\text{not valid, as number of electrons cannot be negative}) \]
\[ n - 5 = 0 \quad \Rightarrow \quad n = 5 \quad (\text{valid solution}) \]

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