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 D

Step 1: Magnetic Moment Determination. The magnetic moment is contingent upon the count of unpaired electrons. For \( \text{Cr}^{3+} \) (atomic number 24), the electronic configuration is \( [Ar]3d^3 \), indicating the presence of 3 unpaired electrons.

Step 2: Calculation Procedure. The magnetic moment (\(\mu\)) is computed via the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] Here, \( n \) represents the number of unpaired electrons. For \( \text{Cr}^{3+} \) (3 unpaired electrons), the calculation yields: \[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \]

Step 3: Final Result. Consequently, the magnetic moment is determined to be 3.87 B.M., aligning with option (B).