1. Find oxidation state and d-electron count

For all complexes, let the oxidation state of the metal be $x$.

• A. $[FeF_6]^{3-}$ $x+6(-1)=-3 \Rightarrow x=+3$ So metal is $Fe^{3+}$.

  Iron: $Z=26$, so $Fe^{3+} = 3d^5$.

• B. $[Fe(CN)_6]^{3-}$ $x+6(-1)=-3 \Rightarrow x=+3$ So again $Fe^{3+}=3d^5$.

• C. $[MnCl_6]^{3-}$ $x+6(-1)=-3 \Rightarrow x=+3$ So $Mn^{3+}$.

  Manganese: $Z=25$, so $Mn^{3+}=3d^4$.

• D. $[Mn(CN)_6]^{3-}$ $x+6(-1)=-3 \Rightarrow x=+3$ So again $Mn^{3+}=3d^4$.

2. Determine strong-field / weak-field nature of ligands

• $F^-$ and $Cl^-$ are weak field ligands $rightarrow$ high spin complexes.
• $CN^-$ is a strong field ligand $rightarrow$ low spin complexes.

Also, the question explicitly says C is high spin.

3. Find number of unpaired electrons

A. $[FeF_6]^{3-}$

$Fe^{3+}$ is $d^5$, with weak field ligand $F^-$, so high spin octahedral: $t_{2g}^3 e_g^2$ Number of unpaired electrons: $n=5$

B. $[Fe(CN)_6]^{3-}$

$Fe^{3+}$ is $d^5$, with strong field ligand $CN^-$, so low spin octahedral: $t_{2g}^5 e_g^0$ Number of unpaired electrons: $n=1$

C. $[MnCl_6]^{3-}$

$Mn^{3+}$ is $d^4$, high spin: $t_{2g}^3 e_g^1$ Number of unpaired electrons: $n=4$

D. $[Mn(CN)_6]^{3-}$

$Mn^{3+}$ is $d^4$, strong field, so low spin: $t_{2g}^4 e_g^0$ Number of unpaired electrons: $n=2$

4. Compare magnetic moments

Using spin-only formula: $\mu = \sqrt{n(n+2)}\ \text{B.M.}$

So:

• For $n=1$: $\mu=\sqrt{3}$
• For $n=2$: $\mu=\sqrt{8}$
• For $n=4$: $\mu=\sqrt{24}$
• For $n=5$: $\mu=\sqrt{35}$

Hence increasing order of magnetic moment is: $B < D < C < A$

5. Match with options

The correct option is: $\boxed{B < D < C < A}$ This corresponds to Option B.