1. Find oxidation state and electronic configuration of Fe

In both complexes, $\left[\mathrm{Fe}(\mathrm{H_2O})_6\right]^{3+} \quad \text{and} \quad \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ iron is in the oxidation state $+3$.

Neutral Fe atom: $\mathrm{Fe}: Z=26 = [\mathrm{Ar}]\,3d^6 4s^2$

So, $\mathrm{Fe}^{3+} = [\mathrm{Ar}]\,3d^5$

Thus in both complexes, the metal ion is $d^5$.

2. Nature of ligands and spin state

(i) $\left[\mathrm{Fe}(\mathrm{H_2O})_6\right]^{3+}$

• $\mathrm{H_2O}$ is a weak field ligand.
• Therefore, the complex is high spin.
• For octahedral high-spin $d^5$: $t_{2g}^3 e_g^2$ Number of unpaired electrons, $n=5$

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{B.M.}$

(ii) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$

• $\mathrm{CN^-}$ is a strong field ligand.
• Therefore, the complex is low spin.
• For octahedral low-spin $d^5$: $t_{2g}^5 e_g^0$ Number of unpaired electrons, $n=1$

Spin-only magnetic moment: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732\ \text{B.M.}$

3. Match with options

Thus the magnetic moments are: $5.92\ \text{B.M. and } 1.732\ \text{B.M.}$

This matches Option B.

4. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

So, the stored answer is correct.