Magnetic moment of `Mn^(+2)` is 5.9 B.M. Hence the number of unpaired electrons presenting the metal ion is

Options

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

Answer

To find the number of unpaired electrons in the Mn^(+2) ion given that its magnetic moment is 5.9 Bohr Magneton (B.M.), we can use the formula for magnetic moment:

### Step-by-Step Solution:

1. **Understanding the Magnetic Moment Formula**:
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Setting Up the Equation**:
Given that the magnetic moment \( \mu = 5.9 \) B.M., we can set up the equation:
\[ 5.9 = \sqrt{n(n + 2)} \]

3. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ (5.9)^2 = n(n + 2) \]
\[ 34.81 = n(n + 2) \]

4. **Rearranging to Form a Quadratic Equation**:
Rearranging gives us:
\[ n^2 + 2n - 34.81 = 0 \]

5. **Using the Quadratic Formula**:
We can solve for \( n \) using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = -34.81 \).

6. **Calculating the Discriminant**:
First, calculate the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \times 1 \times (-34.81) = 4 + 139.24 = 143.24 \]

7. **Finding the Roots**:
Now substitute back into the quadratic formula:
\[ n = \frac{-2 \pm \sqrt{143.24}}{2 \times 1} \]
\[ n = \frac{-2 \pm 11.99}{2} \]

8. **Calculating the Two Possible Values for \( n \)**:
This gives us two potential solutions:
\[ n_1 = \frac{9.99}{2} \approx 4.995 \quad (\text{approximately } 5) \]
\[ n_2 = \frac{-13.99}{2} \quad (\text{not a valid solution since } n \text{ cannot be negative}) \]

9. **Conclusion**:
The number of unpaired electrons in \( Mn^{+2} \) is approximately 5.

### Final Answer:
The number of unpaired electrons in \( Mn^{+2} \) is **5**.