Question

The spin only magnetic moment of Mn\(^{2+}\) ion is _______.

• A. \( 4.901 \, \text{BM} \)
• B. \( 5.916 \, \text{BM} \)
• C. \( 3.873 \, \text{BM} \)
• D. \( 2.846 \, \text{BM} \)

Answer 1

The Correct Option is B

The magnetic moment of a transition metal ion is due to the presence of **unpaired electrons** in its \( d \)-orbitals.

The **spin-only magnetic moment** is given by the formula: \[ \mu = \sqrt{n(n+2)} \, \text{BM} \] where \( n \) = number of unpaired electrons.

For \( \text{Mn}^{2+} \): Atomic number of Mn = 25 Electronic configuration of Mn = [Ar] 3d\(^5\) 4s\(^2\)

In \( \text{Mn}^{2+} \), two electrons are removed from the 4s orbital: \[ \text{Mn}^{2+} = [\text{Ar}] \, 3d^5 \]

Hence, the number of unpaired electrons \( n = 5 \).

Substituting in the formula: \[ \mu = \sqrt{5(5 + 2)} = \sqrt{35} = 5.916 \, \text{BM} \]

Therefore, the spin-only magnetic moment of \( \text{Mn}^{2+} \) ion is: \[ \boxed{5.916 \, \text{BM}} \]