Question

Calculated magnetic moment value for \( \text{Fe}^{2+} \) ion in Bohr Magnetons (BM) is:

• A. 3.87
• B. 4.90
• C. 2.84
• D. 1.73

Hint:

Use \( \mu = \sqrt{n(n+2)} \) BM, where \( n \) = number of unpaired electrons.

Answer 1

The Correct Option is B

Electronic configuration of \( \text{Fe}^{2+} \) is: \[ [\text{Ar}] \, 3d^6 \] Number of unpaired electrons \( n = 4 \) Magnetic moment \( \mu = \sqrt{n(n+2)} \) \[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM} \]