To solve this problem, we need to determine the magnetic moments of the given complexes.$\newline$The magnetic moment is given by the formula:$\newline \mu = \sqrt{n(n+2)} \,$BM, where $n$ is the number of unpaired electrons.$\newline$Let's analyze each complex:$\newline$(a) $[Fe(CN)_6]^{3-}: \newline$• Iron is in the +3 oxidation state, so $Fe^{3+}$ has 5 d-electrons.$\newline$• CN$^-$ is a strong field ligand, causing pairing of electrons.$\newline$• All 5 electrons are paired, so $n = 0. \newline$• Magnetic moment $\mu = \sqrt{0(0+2)} = 0 \,$BM.$\newline$(b) $[Fe(H_2O)_6]^{3+}: \newline$• Iron is in the +3 oxidation state, so $Fe^{3+}$ has 5 d-electrons.$\newline$• H$_2O$ is a weak field ligand, so electrons remain unpaired.$\newline$• There are 5 unpaired electrons, so $n = 5. \newline$• Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \,$BM.$\newline$(c) $[Fe(CN)_6]^{4-}: \newline$• Iron is in the +2 oxidation state, so $Fe^{2+}$ has 6 d-electrons.$\newline$• CN$^-$ is a strong field ligand, causing pairing of electrons.$\newline$• There are no unpaired electrons, so $n = 0. \newline$• Magnetic moment $\mu = \sqrt{0(0+2)} = 0 \,$BM.$\newline$(d) $[Fe(H_2O)_6]^{2+}: \newline$• Iron is in the +2 oxidation state, so $Fe^{2+}$ has 6 d-electrons.$\newline$• H$_2O$ is a weak field ligand, so electrons remain unpaired.$\newline$• There are 4 unpaired electrons, so $n = 4. \newline$• Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \,$BM.$\newline$Matching the calculated magnetic moments with the given options:$\newline$• (a) $[Fe(CN)_6]^{3-} \rightarrow$(ii) 0 BM$\newline$• (b) $[Fe(H_2O)_6]^{3+} \rightarrow$(i) 5.92 BM$\newline$• (c) $[Fe(CN)_6]^{4-} \rightarrow$(ii) 0 BM$\newline$• (d) $[Fe(H_2O)_6]^{2+} \rightarrow$(iii) 4.90 BM$\newline$Therefore, the correct option is Option 4:$\newline$(a) -> (ii), (b) -> (i), (c) -> (ii), (d) -> (iii).