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:

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

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The Correct Option is D

Solution and Explanation

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). \vspace{10pt}

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