Question

The magnetic moment of \( \text{Mn}^{3+} \) is:

• A. \(1.73 \, \text{BM} \)
• B. \(2.83 \, \text{BM} \)
• C. \(4.90 \, \text{BM} \)
• D. \(5.92 \, \text{BM} \)

Hint:

Use the formula \( \mu = \sqrt{n(n+2)} \, \text{BM} \) for calculating magnetic moment, where \(n\) is the number of unpaired electrons.

Answer 1

The Correct Option is C

Manganese (Mn) has an atomic number of 25.
The electronic configuration of Mn is: \([Ar] \, 3d^5 \, 4s^2\).
Mn\(^{3+}\) loses 3 electrons, resulting in the configuration \([Ar] \, 3d^4\).

The number of unpaired electrons in \(3d^4\) is 4.

The magnetic moment \( \mu \) is calculated using the formula \( \mu = \sqrt{n(n+2)} \, \text{BM} \), where \(n\) represents the number of unpaired electrons.

\[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM} \]