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 solve the problem, we need to calculate the spin-only magnetic moment of the Cr$^{3+}$ ion. The formula for the spin-only magnetic moment is given by:

1. Determining the Electron Configuration of Cr$^{3+}$:
The atomic number of Cr is 24, so the ground-state electron configuration of Cr is [Ar] 3d⁵ 4s¹. For Cr$^{3+}$, we remove three electrons. The configuration becomes [Ar] 3d³, meaning Cr$^{3+}$ has 3 unpaired electrons in its 3d orbital.

2. Formula for Spin-Only Magnetic Moment:
The spin-only magnetic moment is given by:

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

3. Substituting the Value of $n$:
Substituting $n = 3$ into the formula:

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

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