The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of `Cr^(3+)` ion is

Options

A. 2.87 B.M.
B. 3.87 B.M.
C. 3.47 B.M.
D. 3.57 B.M.

Explanation

To find the spin-only magnetic moment of the Cr³⁺ ion, we can follow these steps:

### Step 1: Determine the Electronic Configuration of Chromium
- The atomic number of chromium (Cr) is 24. Its electronic configuration is:
\[ \text{Cr: } [\text{Ar}] 3d^5 4s^1 \]

### Step 2: Determine the Electronic Configuration of Cr³⁺
- When chromium loses three electrons to form Cr³⁺, it loses one electron from the 4s orbital and two from the 3d orbital:
\[ \text{Cr}^{3+}: [\text{Ar}] 3d^3 \]

### Step 3: Identify the Number of Unpaired Electrons
- In the 3d subshell of Cr³⁺, we have three electrons. The electron configuration can be represented as:
\[ \uparrow \quad \uparrow \quad \uparrow \]
- All three electrons are unpaired. Therefore, the number of unpaired electrons \( n \) is 3.

### Step 4: Use the Formula for Spin-Only Magnetic Moment
- The formula for calculating the spin-only magnetic moment (\( \mu \)) is:
\[ \mu = \sqrt{n(n + 2)} \]
- Substituting the value of \( n \):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Step 5: Calculate the Value of the Magnetic Moment
- Now we calculate \( \sqrt{15} \):
\[ \mu \approx 3.87 \text{ Bohr magnetons} \]

### Final Answer
- The spin-only magnetic moment value of the Cr³⁺ ion is approximately **3.87 Bohr magnetons**.

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