Arrange the complex ions in the increasing order of their magnetic moments A. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ B. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ C. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ D. $\left[\mathrm{FeF}_6\right]^{3-}$

The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons.

A. In $\left[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}$, Fe is in $+2$ oxidation state ($3d^6$). $\mathrm{H}_2\mathrm{O}$ is a weak field ligand, so electrons remain unpaired: $t_{2g}^4 e_g^2$. Number of unpaired electrons $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ B.M.

B. In $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$, Fe is in $+2$ oxidation state ($3d^6$). $\mathrm{CN}^{-}$ is a strong field ligand, causing pairing: $t_{2g}^6 e_g^0$. Number of unpaired electrons $n = 0$. $\mu = 0$ B.M.

C. In $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$, Fe is in $+3$ oxidation state ($3d^5$). $\mathrm{CN}^{-}$ is a strong field ligand, causing pairing: $t_{2g}^5 e_g^0$. Number of unpaired electrons $n = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ B.M.

D. In $\left[\mathrm{FeF}_6\right]^{3-}$, Fe is in $+3$ oxidation state ($3d^5$). $\mathrm{F}^{-}$ is a weak field ligand, so electrons remain unpaired: $t_{2g}^3 e_g^2$. Number of unpaired electrons $n = 5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ B.M.

Comparing the magnetic moments: $0 < 1.73 < 4.90 < 5.92$, which corresponds to $\mathrm{B} < \mathrm{C} < \mathrm{A} < \mathrm{D}$.

Answer: $\mathrm{B} <\mathrm{C} <\mathrm{A} <\mathrm{D}$