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 BM
B. 3.87 BM
C. 3.47 BM
D. 3.57 BM

Explanation

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

### Step 1: Determine the electron configuration of Cr³⁺
Chromium (Cr) has an atomic number of 24. Its electron configuration in the ground state is:
\[ \text{Cr: [Ar] 4s}^1 \text{3d}^5 \]

When chromium loses three electrons to form Cr³⁺, it loses one electron from the 4s subshell and two electrons from the 3d subshell. Therefore, the electron configuration for Cr³⁺ is:
\[ \text{Cr}^{3+}: [\text{Ar}] \text{3d}^3 \]

### Step 2: Identify the number of unpaired electrons
In the 3d subshell, the three electrons will occupy the three available d orbitals singly before pairing occurs (Hund's rule). Thus, the configuration can be represented as:
\[ \text{3d: } \uparrow \quad \uparrow \quad \uparrow \]

This indicates that there are **3 unpaired electrons** in Cr³⁺.

### Step 3: Use the formula for spin-only magnetic moment
The formula for calculating the spin-only magnetic moment (\( \mu_s \)) is given by:
\[ \mu_s = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

Substituting \( n = 3 \):
\[ \mu_s = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Step 4: Calculate the numerical value
To find the numerical value in Bohr magnetons:
\[ \mu_s = \sqrt{15} \approx 3.87 \text{ Bohr magnetons} \]

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

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