Question

The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

• A. $T_1^{2+}$
• B. $Mn ^{2+}$
• C. $V ^{2+}$
• D. $Cr ^{2+}$

Answer 1

The Correct Option is C

The magnetic moment (\( \mu \)) of a transition metal ion can be calculated using the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] where \( n \) is the number of unpaired electrons. Let us calculate the magnetic moment for each option: (1) \( \text{Cr}^{2+} \): Electronic configuration: \( [\text{Ar}], 3d^4, 4s^0 \) Unpaired electrons (\( n \)) = 4 \[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \, \text{B.M.} \] (2) \( \text{Mn}^{2+} \): Electronic configuration: \( [\text{Ar}], 3d^5, 4s^0 \) Unpaired electrons (\( n \)) = 5 \[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \, \text{B.M.} \] (3) \( \text{V}^{2+} \): Electronic configuration: \( [\text{Ar}], 3d^3, 4s^0 \) Unpaired electrons (\( n \)) = 3 \[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \] This matches the given magnetic moment. (4) \( \text{Ti}^{2+} \): Electronic configuration: \( [\text{Ar}], 3d^2, 4s^0 \) Unpaired electrons (\( n \)) = 2 \[ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \, \text{B.M.} \] Conclusion: The only ion with a calculated magnetic moment of approximately 3.87 B.M. is \( \text{V}^{2+} \).