Question

What is the magnetic dipole moment of \(Mn^{2+}\;(3d^{5})\)?

• A. \(1.73\,BM\)
• B. \(3.87\,BM\)
• C. \(5.92\,BM\)
• D. \(4.90\,BM\)

Hint:

Always remember the spin-only formula: \[ \boxed{\mu=\sqrt{n(n+2)}\;BM} \] where \(n\) is the number of unpaired electrons. For \(d^{5}\) configuration, \[ \boxed{n=5,\qquad \mu=\sqrt{35}=5.92\,BM.} \]

Answer 1

The Correct Option is C

Concept: The magnetic dipole moment of transition metal ions is generally calculated using the spin-only formula, \[ \boxed{\mu=\sqrt{n(n+2)}\;BM} \] where \[ \mu=\text{Magnetic dipole moment}, \] \[ n=\text{Number of unpaired electrons}. \] The unit of magnetic moment is the Bohr Magneton (BM). For most first-row transition metal ions, the orbital contribution is negligible, so the spin-only formula gives an accurate value.

Step 1: Determine the electronic configuration of \(Mn^{2+}\).
The atomic number of manganese is \[ Z=25. \] The electronic configuration of neutral manganese is \[ Mn=[Ar]\,3d^{5}4s^{2}. \] On losing two electrons, \[ Mn^{2+}=[Ar]\,3d^{5}. \] Thus, \[ \boxed{Mn^{2+}\text{ has five unpaired electrons}.} \] Hence, \[ n=5. \]

Step 2: Apply the spin-only formula.
Using \[ \mu=\sqrt{n(n+2)}, \] we get \[ \mu = \sqrt{5(5+2)}. \] Therefore, \[ \mu = \sqrt{35}. \]

Step 3: Calculate the magnetic moment.
Since, \[ \sqrt{35}\approx5.92, \] therefore, \[ \boxed{\mu=5.92\,BM.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (C)}}. \]