Question

Calculated magnetic moment value for \(\mathrm{Fe}^{2+}\) ion in BM is:

• A. \(3.87\)
• B. \(4.90\)
• C. \(2.84\)
• D. \(1.73\)

Hint:

For spin-only magnetic moment, use: \[ \mu=\sqrt{n(n+2)} \] where \(n\) is the number of unpaired electrons.

Answer 1

The Correct Option is B

Step 1: Write the electronic configuration of Fe and Fe²⁺.
Fe (Z = 26): [Ar] 3d⁶ 4s². Fe²⁺ loses 2 electrons from the 4s orbital first: Fe²⁺: [Ar] 3d⁶.
Step 2: Determine the number of unpaired electrons in 3d⁶.
Fill the 3d orbitals using Hund's rule: \[ 3d: \uparrow\downarrow | \uparrow | \uparrow | \uparrow | \uparrow \] The 3d⁶ configuration in a high-spin, weak-field environment (free ion): first 5 electrons each go into separate d orbitals (all parallel spins), the 6th pairs up in the first orbital. Unpaired electrons = 4.
Step 3: Apply the spin-only magnetic moment formula.
\[ \mu = \sqrt{n(n+2)} \text{ BM} \] where $n$ = number of unpaired electrons = 4 for Fe²⁺.
Step 4: Calculate $\mu$.
\[ \mu = \sqrt{4 \times (4+2)} = \sqrt{4 \times 6} = \sqrt{24} = 2\sqrt{6} \approx 4.899 \text{ BM} \]
Step 5: Round to appropriate significant figures.
$\mu \approx 4.90$ BM. This is the spin-only magnetic moment for Fe²⁺, which has 4 unpaired 3d electrons.
Step 6: State the answer.
Fe²⁺ has 4 unpaired electrons in its 3d⁶ configuration, giving a spin-only magnetic moment of $\sqrt{24} \approx 4.90$ BM.
\[ \boxed{\mu = 4.90 \text{ BM (option 2)}} \]