Question

Spin only magnetic moment in BM of \( [\text{Fe(CO)}_4(\text{C}_2\text{O}_4)]^+ \) is :

• A. 1
• B. 1.73
• C. 5.92
• D. 0

Hint:

For strong field ligands and octahedral \( d^5 \) systems, always pair electrons in the lower energy levels first. One electron will always remain unpaired in \( d^5 \) low-spin complexes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The magnetic moment depends on the number of unpaired electrons in the central metal ion. This is influenced by the oxidation state of the metal and the strength of the surrounding ligands (Crystal Field Theory).
Step 2: Key Formula or Approach:
Spin-only magnetic moment \( \mu = \sqrt{n(n+2)} \) BM, where \( n \) is the number of unpaired electrons.
Step 3: Detailed Explanation:
1. **Oxidation State of Fe:** Let it be \( x \).
\( x + 4(0) + (-2) = +1 \implies x = +3 \).
2. **Electronic Configuration of \( \text{Fe}^{3+} \):** Ground state is \( [ \text{Ar} ] 3d^5 \).
3. **Ligand Strength:** CO is a very strong field ligand. Oxalate (\( \text{C}_2\text{O}_4^{2-} \)) is also a strong field ligand (especially with \( +3 \) metals).
4. **Pairing:** Coordination number is 6 (4 from monodentate CO and 2 from bidentate oxalate). In an octahedral field with strong field ligands, the 5 electrons in the \( 3d \) orbital will pair up as much as possible in the \( t_{2g} \) orbitals.
Arrangement: \( t_{2g}^{2,2,1} \), \( e_g^{0,0} \).
Number of unpaired electrons \( n = 1 \).
5. **Calculation:**
\( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM} \).
Step 4: Final Answer:
The spin-only magnetic moment is 1.73 BM.