1. Find oxidation state and $d$-electron count

• For $[MnCl_4]^{2-}$:

  Let oxidation state of Mn be $x$. $x+4(-1)=-2 \Rightarrow x=+2$ So, $Mn^{2+}$ has configuration $3d^5$.

• For $[CoCl_4]^{2-}$:

  Let oxidation state of Co be $x$. $x+4(-1)=-2 \Rightarrow x=+2$ So, $Co^{2+}$ has configuration $3d^7$.

• For $[Fe(CN)_6]^{4-}$:

  Let oxidation state of Fe be $x$. $x+6(-1)=-4 \Rightarrow x=+2$ So, $Fe^{2+}$ has configuration $3d^6$.

2. Determine geometry/field strength and number of unpaired electrons

(i) $[MnCl_4]^{2-}$

• $Cl^-$ is a weak field ligand.
• Four-coordinate chloride complex is tetrahedral.
• Tetrahedral complexes are generally high spin.
• For $d^5$ high spin tetrahedral: number of unpaired electrons, $n=5$.

Thus, $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)}=\sqrt{35}$

(ii) $[CoCl_4]^{2-}$

• $Cl^-$ is weak field.
• This complex is tetrahedral and high spin.
• For $d^7$ tetrahedral high spin: number of unpaired electrons, $n=3$.

Thus, $\mu = \sqrt{3(3+2)}=\sqrt{15}$

(iii) $[Fe(CN)_6]^{4-}$

• $CN^-$ is a strong field ligand.
• Six-coordinate complex is octahedral.
• For $Fe^{2+}$, $d^6$ with strong field gives low spin: $t_{2g}^6 e_g^0$
• Hence, number of unpaired electrons, $n=0$.

Thus, $\mu = 0$

3. Compare magnetic moments

Since $\sqrt{35} > \sqrt{15} > 0$ we get $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$

4. Check options

• A: $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$ ✅
• B: incorrect
• C: incorrect
• D: incorrect

Therefore, the correct option is A.