An ion `(Mn^(a+))` has the magnetic moment equal to `4.9` B.M` What is the value of (a) :

Options

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

Explanation

To solve the problem, we need to determine the value of \( a \) in the ion \( Mn^{a+} \) given that its magnetic moment is \( 4.9 \, B.M \) (Bohr Magneton).

### Step-by-Step Solution:

1. **Understand the Magnetic Moment Formula**:
The magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{N(N + 2)} \, B.M \]
where \( N \) is the number of unpaired electrons.

2. **Set Up the Equation**:
We know that the magnetic moment is \( 4.9 \, B.M \). Therefore, we can set up the equation:
\[ 4.9 = \sqrt{N(N + 2)} \]

3. **Square Both Sides**:
To eliminate the square root, we square both sides of the equation:
\[ (4.9)^2 = N(N + 2) \]
\[ 24.01 = N(N + 2) \]

4. **Rearrange the Equation**:
Rearranging gives us a quadratic equation:
\[ N^2 + 2N - 24.01 = 0 \]

5. **Solve the Quadratic Equation**:
We can use the quadratic formula \( N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -24.01 \):
\[ N = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-24.01)}}{2 \cdot 1} \]
\[ N = \frac{-2 \pm \sqrt{4 + 96.04}}{2} \]
\[ N = \frac{-2 \pm \sqrt{100.04}}{2} \]
\[ N = \frac{-2 \pm 10}{2} \]

This gives us two possible solutions:
\[ N = \frac{8}{2} = 4 \quad \text{(valid solution)} \]
\[ N = \frac{-12}{2} = -6 \quad \text{(not valid)} \]

6. **Determine the Oxidation State**:
Manganese (Mn) has an atomic number of 25 and its electron configuration is:
\[ [Ar] \, 4s^2 \, 3d^5 \]
In the case of \( Mn^{3+} \), we lose 2 electrons from \( 4s \) and 1 from \( 3d \):
\[ 4s^2 \, 3d^5 \rightarrow 4s^0 \, 3d^4 \]
This results in 4 unpaired electrons.

7. **Conclusion**:
Therefore, the value of \( a \) in \( Mn^{a+} \) is:
\[ a = 3 \]