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 value for the Cr³⁺ ion, we can follow these steps:

### Step 1: Determine the electronic configuration of Cr³⁺
Chromium (Cr) has an atomic number of 24. Its ground state electronic configuration is:
\[ \text{Cr: [Ar] 3d}^5 4s^1 \]
When chromium loses three electrons to form Cr³⁺, the electrons are removed first from the 4s orbital and then from the 3d orbital. Thus, the electronic configuration for Cr³⁺ is:
\[ \text{Cr}^{3+}: \text{[Ar] 3d}^3 \]

### Step 2: Identify the number of unpaired electrons
In the 3d subshell, Cr³⁺ has three electrons (3d³). The arrangement of these electrons in the d orbitals will be as follows:
- 3d orbital can hold a maximum of 10 electrons, and the three electrons will occupy three separate orbitals due to Hund's rule, which states that electrons will fill degenerate orbitals singly before pairing up.

Thus, all three electrons in Cr³⁺ are unpaired.

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

### Step 4: Substitute the value of unpaired electrons
For Cr³⁺, we have:
- \( n = 3 \) (since there are three unpaired electrons)

Now, substituting \( n \) into the formula:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Step 5: Calculate the magnetic moment
Calculating \( \sqrt{15} \):
\[ \mu \approx 3.87 \, \text{BM} \]
(BM stands for Bohr Magneton, the unit of magnetic moment)

### Conclusion
The spin-only magnetic moment value for the Cr³⁺ ion is approximately **3.87 BM**.

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