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(D) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$I.$ $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$.

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$III < I < II < IV$

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(D) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$I.$ $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand,causing pairing. $n = 2$. Magnetic moment $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.$II.$ $[MnCl_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $Cl^-$ is a weak field ligand,no pairing. $n = 4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.$III.$ $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. $n = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$IV.$ $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,no pairing. $n = 5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Comparing the values: $1.73 (III) Thus,the increasing order is $III < I < II < IV$.

Arrange the following in the increasing order of their magnetic moments:
$I.$ $[Mn(CN)_6]^{3-}$
$II.$ $[MnCl_6]^{3-}$
$III.$ $[Fe(CN)_6]^{3-}$
$IV.$ $[FeF_6]^{3-}$

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Arrange the following in the increasing order of their magnetic moments:$I.$ $[Mn(CN)_6]^{3-}$$II.$ $[MnCl_6]^{3-}$$III.$ $[Fe(CN)_6]^{3-}$$IV.$ $[FeF_6]^{3-}$