Question

Calculate the spin-only magnetic moment of a \( Mn^{2+} \) ion (Atomic number = 25).

• A. \( 3.87 \, BM \)
• B. \( 4.90 \, BM \)
• C. \( 5.92 \, BM \)
• D. \( 6.93 \, BM \)

Hint:

Always remove electrons from the \(4s\) orbital before the \(3d\) orbital when forming transition metal cations.

Answer 1

The Correct Option is C

Concept: The spin-only magnetic moment of a transition metal ion is calculated using the formula: \[ \mu = \sqrt{n(n+2)} \, BM \] where:

• \( n \) = number of unpaired electrons
• \( BM \) = Bohr Magneton

Step 1: {Write the electronic configuration of Mn.} Atomic number of Mn = 25 \[ Mn: [Ar]\,3d^5\,4s^2 \]
Step 2: {Determine the configuration of \( Mn^{2+} \).} Two electrons are removed first from the \(4s\) orbital: \[ Mn^{2+} : [Ar]\,3d^5 \]
Step 3: {Count the number of unpaired electrons.} In \(3d^5\), each orbital contains one electron according to Hund's rule. \[ n = 5 \]
Step 4: {Substitute in the magnetic moment formula.} \[ \mu = \sqrt{n(n+2)} \] \[ \mu = \sqrt{5(5+2)} \] \[ \mu = \sqrt{35} \] \[ \mu \approx 5.92 \, BM \]