Question

Which of the following ions shows spin only magnetic moment of 4.9 B.M.?

• A. Mn\(^{2+}\)
• B. Cr\(^{2+}\)
• C. Fe\(^{3+}\)
• D. Co\(^{2+}\)

Hint:

The magnetic moment of transition metal ions depends on the number of unpaired electrons, and can be calculated using the above formula.

Answer 1

The Correct Option is A

Magnetic moment \( \mu \) for a transition metal ion is given by: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] Where \( n \) is the number of unpaired electrons. For Mn\(^{2+}\), the electronic configuration is \( [Ar] 3d^5 \). Therefore, it has 5 unpaired electrons. Substituting \( n = 5 \) into the formula: \[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.} \] Therefore, Mn\(^{2+}\) shows a magnetic moment of approximately 4.9 B.M.