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

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine the oxidation state of manganese (Mn) in the ionic state given that its magnetic moment is \(\sqrt{15} \, \text{B.M.}\), we can follow these steps:

### Step 1: Use the Magnetic Moment Formula
The formula for calculating the magnetic moment (\(\mu\)) of an ion is given by:
\[ \mu = \sqrt{n(n + 2)} \, \text{B.M.} \]
where \(n\) is the number of unpaired electrons.

### Step 2: Set Up the Equation
We know from the question that:
\[ \mu = \sqrt{15} \, \text{B.M.} \]
Thus, we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{15} \]

### Step 3: Square Both Sides
Squaring both sides of the equation to eliminate the square root gives:
\[ n(n + 2) = 15 \]

### Step 4: Rearrange the Equation
Rearranging the equation, we get:
\[ n^2 + 2n - 15 = 0 \]

### Step 5: Factor the Quadratic Equation
Now, we can factor the quadratic:
\[ (n + 5)(n - 3) = 0 \]
This gives us two possible solutions for \(n\):
\[ n = -5 \quad \text{or} \quad n = 3 \]
Since \(n\) must be a non-negative integer, we take:
\[ n = 3 \]

### Step 6: Determine the Oxidation State
The value of \(n\) represents the number of unpaired electrons. Manganese (Mn) has an atomic number of 25, and its ground state electronic configuration is:
\[ [Ar] \, 4s^2 \, 3d^5 \]
In the ionic state, Mn loses electrons. In this case, since there are 3 unpaired electrons, it means that 2 electrons from the \(4s\) subshell are lost, and 1 electron from the \(3d\) subshell is also lost.

Thus, the oxidation state of manganese can be calculated as follows:
- The original number of electrons in Mn is 25.
- After losing 4 electrons (2 from \(4s\) and 2 from \(3d\)), the oxidation state is:
\[ \text{Oxidation state} = 25 - 4 = +4 \]

### Final Answer
Therefore, the oxidation state of manganese in this ionic state is \(+4\).

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