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:
Spin only magnetic moment of Fe in [Fe(H2O)6] 3+ and [Fe(CN)6] 3– complexes respectively is:
- 5.92 B.M. and 1.732 B.M
- 6.92 B.M. in both
- 3.87 B.M. and 1.732 B.M.
- 4.89 B.M. and 6.92 B.M.
The Correct Option is A
Solution and Explanation
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.
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