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(D) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$I$. $[Fe

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$I < III < IV < II$

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(D) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$I$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so electrons pair up. $n = 0$,$\mu = 0 \ B.M.$$II$. $[MnCl_4]^{2-}$: $Mn^{2+}$ is $3d^5$. $Cl^-$ is a weak field ligand,so electrons do not pair. $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ B.M.$$III$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,so electrons pair up. $n = 1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 \ B.M.$$IV$. $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ B.M.$Comparing the values: $0 (I) Thus,the increasing order is $I < III < IV < II$.

Arrange the following complexes in the increasing order of their spin only magnetic moment (in $B.M$):
$I$. $[Fe(CN)_6]^{4-}$
$II$. $[MnCl_4]^{2-}$
$III$. $[Mn(CN)_6]^{4-}$
$IV$. $[Cr(NH_3)_6]^{3+}$

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Arrange the following complexes in the increasing order of their spin only magnetic moment (in $B.M$):$I$. $[Fe(CN)_6]^{4-}$$II$. $[MnCl_4]^{2-}$$III$. $[Mn(CN)_6]^{4-}$$IV$. $[Cr(NH_3)_6]^{3+}$