The magnetic moment ($\mu$) is calculated using the 'spin-only' formula: $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

(a) $[Fe(CN)_6]^{3-}$: $Fe$ is in +3 oxidation state ($3d^5$). Since $CN^-$ is a strong field ligand, electrons pair up, giving a low spin $t_{2g}^5 e_g^0$ configuration. Number of unpaired electrons ($n$) = 1. $\mu = \sqrt{1(1+2)} = 1.73$ BM. (b) $[Fe(H_2O)_6]^{3+}$: $Fe$ is in +3 oxidation state ($3d^5$). Since $H_2O$ is a weak field ligand, electrons do not pair up, giving a high spin $t_{2g}^3 e_g^2$ configuration. Number of unpaired electrons ($n$) = 5. $\mu = \sqrt{5(5+2)} = 5.92$ BM. (c) $[Fe(CN)_6]^{4-}$: $Fe$ is in +2 oxidation state ($3d^6$). Since $CN^-$ is a strong field ligand, electrons pair up completely, giving a low spin $t_{2g}^6 e_g^0$ configuration. Number of unpaired electrons ($n$) = 0. $\mu = 0$ BM. (d) $[Fe(H_2O)_6]^{2+}$: $Fe$ is in +2 oxidation state ($3d^6$). Since $H_2O$ is a weak field ligand, electrons do not pair up, giving a high spin $t_{2g}^4 e_g^2$ configuration. Number of unpaired electrons ($n$) = 4. $\mu = \sqrt{4(4+2)} = 4.90$ BM.

Therefore, the correct match is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).