Question

Arrange the complex ions in the increasing order of their magnetic moments
\[ \text{A. } [Fe(H_2O)_6]^{2+} \quad \text{B. } [Fe(CN)_6]^{4-} \quad \text{C. } [Fe(CN)_6]^{3-} \quad \text{D. } [FeF_6]^{3-} \]

• A. \( D < A < B < C \)
• B. \( B < C < A < D \)
• C. \( D < C < B < A \)
• D. \( A < D < C < B \)

Hint:

Magnetic moment depends on number of unpaired electronsStrong field ligands cause pairing (low spin), weak field ligands give high spin.

Answer 1

The Correct Option is B

Step 1: Determine oxidation states and electronic configuration.
For all complexes, Fe oxidation states:
\[ [Fe(H_2O)_6]^{2+} \Rightarrow Fe^{2+} : d^6 \]
\[ [Fe(CN)_6]^{4-} \Rightarrow Fe^{2+} : d^6 \]
\[ [Fe(CN)_6]^{3-} \Rightarrow Fe^{3+} : d^5 \]
\[ [FeF_6]^{3-} \Rightarrow Fe^{3+} : d^5 \]

Step 2: Identify ligand strength.
\[ CN^- \rightarrow \text{strong field (low spin)} \]
\[ H_2O \rightarrow \text{weak field (high spin)} \]
\[ F^- \rightarrow \text{weak field (high spin)} \]

Step 3: Find number of unpaired electrons.
\[ [Fe(CN)_6]^{4-} : d^6 \text{ low spin} \Rightarrow 0 \text{ unpaired} \]
\[ [Fe(CN)_6]^{3-} : d^5 \text{ low spin} \Rightarrow 1 \text{ unpaired} \]
\[ [Fe(H_2O)_6]^{2+} : d^6 \text{ high spin} \Rightarrow 4 \text{ unpaired} \]
\[ [FeF_6]^{3-} : d^5 \text{ high spin} \Rightarrow 5 \text{ unpaired} \]

Step 4: Use magnetic moment relation.
\[ \mu \propto \sqrt{n(n+2)} \]
where \( n \) = number of unpaired electrons.

Step 5: Arrange in increasing order.
\[ 0 < 1 < 4 < 5 \]
\[ B < C < A < D \]

Step 6: Match with options.
Correct option is (B).

Step 7: Final conclusion.
\[ \boxed{B < C < A < D} \]