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

### Step-by-Step Solution:

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

2. **Substitute the Given Value**: We know that \( \mu_{spin} = 5.92 \) BM. Substitute this value into the formula:
\[ 5.92 = \sqrt{n(n + 2)} \]

3. **Square Both Sides**: To eliminate the square root, square both sides of the equation:
\[ (5.92)^2 = n(n + 2) \]
This simplifies to:
\[ 35.0464 = n(n + 2) \]

4. **Rearrange the Equation**: Rearrange the equation to form a standard quadratic equation:
\[ n^2 + 2n - 35.0464 = 0 \]

5. **Approximate the Constant**: Since dealing with decimals can be complex, we can approximate \( 35.0464 \) to \( 35 \) for easier calculations:
\[ n^2 + 2n - 35 = 0 \]

6. **Factor the Quadratic Equation**: To solve the quadratic equation, we can factor it:
\[ n^2 + 7n - 5n - 35 = 0 \]
This can be factored as:
\[ (n - 5)(n + 7) = 0 \]

7. **Solve for \( n \)**: Set each factor to zero:
- \( n - 5 = 0 \) gives \( n = 5 \)
- \( n + 7 = 0 \) gives \( n = -7 \) (not valid since \( n \) cannot be negative)

8. **Conclusion**: The only valid solution is \( n = 5 \). Therefore, the number of unpaired electrons in the species is:
\[ \boxed{5} \]