Question

Determine the spin-only magnetic moment value for a central divalent manganese gaseous ion (\( \text{Mn}^{2+} \)) based on its d-orbital electron distribution. (Atomic Number of \( \text{Mn} = 25 \))

• A. \( 4.90 \text{ BM} \)
• B. \( 1.73 \text{ BM} \)
• C. \( 3.87 \text{ BM} \)
• D. \( 5.92 \text{ BM} \)

Hint:

Use this fast estimation shortcut for your exam: the first decimal digit of a spin-only magnetic moment value always matches the number of unpaired electrons (\( n \)). For example, if \( n=3 \), \( \mu \approx 3.8 \text{ BM} \); if \( n=5 \), \( \mu \approx 5.9 \text{ BM} \).

Answer 1

The Correct Option is D

Concept: The spin-only magnetic moment (\( \mu \)) of a transition metal ion depends on the total number of unpaired d-orbital electrons (\( n \)) present in its valence shell, calculated using the formula: \[ \mu = \sqrt{n(n + 2)} \text{ Bohr Magnetons (BM)} \]

Step 1: Determine the electronic configuration of the divalent ion.
The neutral ground state configuration of Manganese (\( Z = 25 \)) is: \[ \text{Mn} = [\text{Ar}] \, 3\text{d}^5 \, 4\text{s}^2 \] Forming the divalent ion (\( \text{Mn}^{2+} \)) requires removing the two outer valence electrons from the \( 4\text{s} \) orbital first: \[ \text{Mn}^{2+} = [\text{Ar}] \, 3\text{d}^5 \, 4\text{s}^0 \]

Step 2: Count the number of unpaired electrons using Hund's Rule.
The \( \text{d} \) subshell contains 5 separate degenerate orbitals. Distributing 5 electrons across these orbitals following Hund's rule gives 1 electron per orbital: \[ \text{Number of unpaired electrons } (n) = 5 \]

Step 3: Substitute \( n \) into the magnetic moment equation.
Plug the value \( n = 5 \) into our formula: \[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \] Since \( \sqrt{36} = 6 \), the value of \( \sqrt{35} \) evaluates to just below 6: \[ \mu \approx 5.92 \text{ BM} \]