Atomic number of a transition metal is 25 then its magnetic moment in its +3 oxidation state is

Options

A. `1.73 BM`
B. `2.84 BM`
C. `4.9 BM`
D. `5.9 BM`

Explanation

To find the magnetic moment of a transition metal with atomic number 25 in its +3 oxidation state, we can follow these steps:

### Step 1: Identify the transition metal
The element with atomic number 25 is manganese (Mn).

### Step 2: Determine the electron configuration of manganese
The electron configuration of manganese (Mn) is:
\[ \text{Mn: } [Ar] 4s^2 3d^5 \]

### Step 3: Determine the electron configuration in the +3 oxidation state
In the +3 oxidation state, manganese loses three electrons. The electrons are removed first from the 4s orbital and then from the 3d orbital. Therefore, the electron configuration in the +3 oxidation state is:
\[ \text{Mn}^{3+}: [Ar] 3d^4 \]

### Step 4: Determine the number of unpaired electrons
In the 3d^4 configuration, we can distribute the electrons in the 3d orbitals. The distribution will be as follows:
- 3d orbital can hold a maximum of 10 electrons and has five orbitals (dxy, dyz, dzx, dx2-y2, dz2).
- The 4 electrons will occupy the orbitals singly before pairing occurs, leading to the following arrangement:
- dxy: ↑
- dyz: ↑
- dzx: ↑
- dx2-y2: ↑

Thus, there are 4 unpaired electrons in the 3d^4 configuration.

### Step 5: Calculate the magnetic moment
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

Substituting \( n = 4 \):
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \]

### Step 6: Simplify the expression
\[ \mu = \sqrt{24} = 2\sqrt{6} \]

### Final Answer
The magnetic moment of manganese in its +3 oxidation state is \( 2\sqrt{6} \) Bohr magnetons.

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