Explain why $[Fe(H_{2}O)_{6}]^{3+}$ has a magnetic moment value of $5.92 \ BM$ whereas $[Fe(CN)_{6}]^{3-}$ has a value of only $1.74 \ BM$.

(N/A) In both complexes,the oxidation state of $Fe$ is $+3$,which corresponds to a $d^5$ electronic configuration.
$1.$ In $[Fe(H_{2}O)_{6}]^{3+}$,$H_{2}O$ is a weak field ligand. It does not cause pairing of electrons in the $d$-orbitals. Thus,the $d^5$ configuration remains as $t_{2g}^3 e_g^2$,resulting in $n = 5$ unpaired electrons.
The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ BM = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$2.$ In $[Fe(CN)_{6}]^{3-}$,$CN^{-}$ is a strong field ligand. It causes pairing of electrons in the $d$-orbitals. Thus,the $d^5$ configuration becomes $t_{2g}^5 e_g^0$,resulting in $n = 1$ unpaired electron.
The magnetic moment is calculated as $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.74 \ BM$.