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(C) To determine the number of unpaired electrons,we analyze the oxidation state of the metal and the nature of the ligand:$I.$ $[MnCl_6]^{3-}$: $Mn^{

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$IV < III < I < II$

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(C) To determine the number of unpaired electrons,we analyze the oxidation state of the metal and the nature of the ligand:$I.$ $[MnCl_6]^{3-}$: $Mn^{3+}$ $(3d^4)$,$Cl^-$ is a weak field ligand. Unpaired electrons = $4$.$II.$ $[FeF_6]^{3-}$: $Fe^{3+}$ $(3d^5)$,$F^-$ is a weak field ligand. Unpaired electrons = $5$.$III.$ $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ $(3d^4)$,$CN^-$ is a strong field ligand. Unpaired electrons = $2$.$IV.$ $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ $(3d^5)$,$CN^-$ is a strong field ligand. Unpaired electrons = $1$.Comparing the number of unpaired electrons: $IV (1) Thus,the increasing order is $IV < III < I < II$.

Arrange the following in the increasing order of the number of unpaired electrons present in the central metal ion:
$I.$ $[MnCl_6]^{3-}$
$II.$ $[FeF_6]^{3-}$
$III.$ $[Mn(CN)_6]^{3-}$
$IV.$ $[Fe(CN)_6]^{3-}$

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Arrange the following in the increasing order of the number of unpaired electrons present in the central metal ion:$I.$ $[MnCl_6]^{3-}$$II.$ $[FeF_6]^{3-}$$III.$ $[Mn(CN)_6]^{3-}$$IV.$ $[Fe(CN)_6]^{3-}$