Question

The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr$^{3+}$ ion (Atomic no. : Cr = 24) is:

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

Hint:

The magnetic moment can be calculated using the formula \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons.

Answer 1

The Correct Option is B

To determine the spin-only magnetic moment of the Cr$^{3+}$ ion, we perform the following calculations:

1. Electron Configuration of Cr$^{3+}$:
The atomic number of Chromium (Cr) is 24. Its ground-state electron configuration is [Ar] 3d⁵ 4s¹. For Cr$^{3+}$, three electrons are removed, resulting in the configuration [Ar] 3d³. This indicates that Cr$^{3+}$ possesses 3 unpaired electrons in its 3d orbital.

2. Spin-Only Magnetic Moment Formula:
The spin-only magnetic moment ($\mu_{\text{spin}}$) is calculated using the formula:

$ \mu_{\text{spin}} = \sqrt{n(n+2)} $
where $n$ represents the number of unpaired electrons. In this case, $n = 3$.

3. Calculation:
Substituting $n = 3$ into the formula yields:

$ \mu_{\text{spin}} = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \, \mu_{\text{B}} $

Conclusion:
The spin-only magnetic moment of the Cr$^{3+}$ ion is approximately $3.87 \, \mu_{\text{B}}$.