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(D) To find the magnetic moment $(\mu = \sqrt{n(n+2)} \ BM)$,we determine the number of unpaired electrons $(n)$ for each complex:$I$. $[Fe(CN)_6]^{4-

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$I < II < IV < III$

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(D) To find the magnetic moment $(\mu = \sqrt{n(n+2)} \ BM)$,we determine the number of unpaired electrons $(n)$ for each complex:$I$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ $(d^6)$. $CN^-$ is a strong field ligand,so electrons pair up. $n = 0$. $\mu = 0 \ BM$.$II$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ $(d^5)$. $CN^-$ is a strong field ligand,so electrons pair up. $n = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$III$. $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ $(d^3)$. $n = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$IV$. $[Ni(H_2O)_4]^{2+}$: $Ni^{2+}$ $(d^8)$. $H_2O$ is a weak field ligand,tetrahedral geometry. $n = 2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.Comparing the values: $0 (I) Thus,the increasing order is $I < II < IV < III$.

The increasing order of magnetic moments for the following complexes is:
$I$. $[Fe(CN)_6]^{4-}$
$II$. $[Fe(CN)_6]^{3-}$
$III$. $[Cr(NH_3)_6]^{3+}$
$IV$. $[Ni(H_2O)_4]^{2+}$

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The increasing order of magnetic moments for the following complexes is:$I$. $[Fe(CN)_6]^{4-}$$II$. $[Fe(CN)_6]^{3-}$$III$. $[Cr(NH_3)_6]^{3+}$$IV$. $[Ni(H_2O)_4]^{2+}$