Spin only magnetic moment of dipositive ion of Mn is:

Options

A. 0
B. `sqrt(8) BM `
C. `sqrt(24) BM`
D. `sqrt(35) BM`

Answer

To find the spin-only magnetic moment of the dipositive ion of manganese (Mn²⁺), we can follow these steps:

### Step 1: Determine the electronic configuration of manganese (Mn)
Manganese has an atomic number of 25. Its ground state electronic configuration is:
\[ \text{Mn: } [Ar] 3d^5 4s^2 \]

### Step 2: Determine the electronic configuration of Mn²⁺
When manganese loses two electrons to form Mn²⁺, the electrons are removed first from the 4s orbital, followed by the 3d orbital. Thus, the electronic configuration of Mn²⁺ is:
\[ \text{Mn}^{2+}: [Ar] 3d^5 4s^0 \]

### Step 3: Identify the number of unpaired electrons
In the 3d subshell, there are 5 electrons. According to Hund's rule, these electrons will occupy separate orbitals before pairing up. Therefore, all 5 electrons in the 3d subshell are unpaired:
- 3d: ↑ ↑ ↑ ↑ ↑ (5 unpaired electrons)

So, the number of unpaired electrons (n) in Mn²⁺ is:
\[ n = 5 \]

### Step 4: Use the formula for spin-only magnetic moment
The formula for calculating the spin-only magnetic moment (μ) is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where n is the number of unpaired electrons.

### Step 5: Substitute the value of n into the formula
Now substituting the value of n:
\[ \mu = \sqrt{5(5 + 2)} \]
\[ \mu = \sqrt{5 \times 7} \]
\[ \mu = \sqrt{35} \]

### Step 6: Express the result in terms of Bohr magnetons
The spin-only magnetic moment is expressed in units of Bohr magnetons (μB):
\[ \mu = \sqrt{35} \, \text{Bohr magnetons} \]

### Final Answer
The spin-only magnetic moment of the dipositive ion of manganese (Mn²⁺) is:
\[ \sqrt{35} \, \text{Bohr magnetons} \]

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