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 :

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the number of unpaired electrons in a transition metal ion with a spin-only magnetic moment of 5.92 Bohr magnetons (BM), we can use the formula for calculating the spin-only magnetic moment:

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

where \( \mu \) is the magnetic moment in Bohr magnetons and \( n \) is the number of unpaired electrons.

### Step-by-step Solution:

1. **Set up the equation**: We know that the spin-only magnetic moment \( \mu \) is given as 5.92 BM. Therefore, we can set up the equation:
\[ 5.92 = \sqrt{n(n + 2)} \]

2. **Square both sides**: To eliminate the square root, we square both sides of the equation:
\[ (5.92)^2 = n(n + 2) \]
\[ 35.0464 = n(n + 2) \]

3. **Rearrange the equation**: This can be rearranged into a standard quadratic equation:
\[ n^2 + 2n - 35.0464 = 0 \]

4. **Use the quadratic formula**: The quadratic formula is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = -35.0464 \).

5. **Calculate the discriminant**:
\[ b^2 - 4ac = 2^2 - 4 \times 1 \times (-35.0464) = 4 + 140.1856 = 144.1856 \]

6. **Calculate \( n \)**:
\[ n = \frac{-2 \pm \sqrt{144.1856}}{2 \times 1} \]
\[ n = \frac{-2 \pm 12.01}{2} \]

Calculating the two possible values for \( n \):
- \( n = \frac{10.01}{2} = 5.005 \) (approximately 5)
- \( n = \frac{-14.01}{2} \) (not a valid solution since \( n \) cannot be negative)

7. **Conclusion**: Since \( n \) must be a whole number, we round 5.005 to 5. Therefore, the number of unpaired electrons present in the species is **5**.

### Final Answer:
The number of unpaired electrons present in the species is **5**.