The magnetic moment of `._25Mn` in ionic state is `sqrt(15)B.M`, then Mn is in:

Options

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

Explanation

To determine the oxidation state of manganese (Mn) in the ionic state given its magnetic moment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The atomic number of manganese (Mn) is 25.
- The magnetic moment is given as \( \sqrt{15} \) Bohr Magneton (B.M).

2. **Determine the Electronic Configuration:**
- The electronic configuration of Mn is \( [Ar] 3d^5 4s^2 \).

3. **Use the Formula for Magnetic Moment:**
- The magnetic moment (\( \mu \)) is related to the number of unpaired electrons (\( n \)) by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
- Given that \( \mu = \sqrt{15} \), we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{15} \]

4. **Square Both Sides:**
- Squaring both sides gives:
\[ n(n + 2) = 15 \]

5. **Rearrange to Form a Quadratic Equation:**
- Rearranging the equation leads to:
\[ n^2 + 2n - 15 = 0 \]

6. **Solve the Quadratic Equation:**
- We can factor or use the quadratic formula to solve for \( n \):
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-15)}}{2(1)} \]
\[ n = \frac{-2 \pm \sqrt{4 + 60}}{2} = \frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2} \]
- This gives us two potential solutions for \( n \):
\[ n = 3 \quad (\text{positive solution}) \quad \text{and} \quad n = -5 \quad (\text{not valid}) \]

7. **Determine the Oxidation State:**
- Since we have found \( n = 3 \), this means there are 3 unpaired electrons in the ionic state.
- The original configuration of Mn is \( 3d^5 4s^2 \). To have 3 unpaired electrons, we can remove 4 electrons (2 from \( 4s \) and 2 from \( 3d \)), resulting in the configuration \( 3d^3 \).
- Therefore, the oxidation state of Mn is:
\[ +4 \]

### Final Answer:
The oxidation state of manganese (Mn) in the ionic state is **+4**.