Question

The magnetic moment is measured in Bohr Magneton (BM).
Spin only magnetic moment of Fe in [Fe(H₂O)₆] ³⁺ and [Fe(CN)₆] ^3– complexes respectively is:

• A. 5.92 B.M. and 1.732 B.M
• B. 6.92 B.M. in both
• C. 3.87 B.M. and 1.732 B.M.
• D. 4.89 B.M. and 6.92 B.M.

Answer 1

The Correct Option is A

Step 1: Electron Configuration of \( \text{Fe}^{3+} \) in \([ \text{Fe}(\text{H}_2\text{O})_6 ]^{3+}\)

• In \([ \text{Fe}(\text{H}_2\text{O})_6 ]^{3+}\), Fe is in the +3 oxidation state.
• The electronic configuration of Fe is \( [ \text{Ar} ] 3d^6 4s^2 \).
• For \( \text{Fe}^{3+} \): \( [ \text{Ar} ] 3d^5 \).
• Water (\( \text{H}_2\text{O} \)) is a weak field ligand, so it does not cause pairing of electrons. The five \( 3d \) electrons are distributed as follows:
• Number of unpaired electrons, \( n = 5 \).

Step 2: Spin-Only Magnetic Moment for \([ \text{Fe}(\text{H}_2\text{O})_6 ]^{3+}\)

The spin-only magnetic moment \( \mu \) is given by:

\[ \mu = \sqrt{n(n+2)} \, \text{BM} \]

Substituting \( n = 5 \):

\[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{BM}. \]

Step 3: Electron Configuration of \( \text{Fe}^{3+} \) in \([ \text{Fe}(\text{CN})_6 ]^{3-}\)

• In \([ \text{Fe}(\text{CN})_6 ]^{3-}\), Fe is also in the +3 oxidation state, so its electronic configuration is \( [ \text{Ar} ] 3d^5 \).
• Cyanide (\( \text{CN}^- \)) is a strong field ligand, causing pairing of electrons. The five \( 3d \) electrons are arranged as follows:
• Number of unpaired electrons, \( n = 1 \).

Step 4: Spin-Only Magnetic Moment for \([ \text{Fe}(\text{CN})_6 ]^{3-}\)

Substituting \( n = 1 \) into the formula for magnetic moment:

\[ \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \, \text{BM}. \]

Final Answer:

The spin-only magnetic moments are 5.92 B.M. and 1.732 B.M., respectively.