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

Step 1: Understanding the magnetic moment. The magnetic moment depends on the number of unpaired electrons. For Cr$^{3+}$ (with atomic number 24), the electron configuration is \( [Ar]3d^3 \), meaning there are 3 unpaired electrons.

Step 2: Calculation. The magnetic moment (\(\mu\)) is calculated using the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] Where \( n \) is the number of unpaired electrons. For Cr$^{3+}$ (3 unpaired electrons), we get: \[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \]

Step 3: Conclusion. Thus, the magnetic moment is 3.87 B.M., corresponding to option (B).