The correct order of spin-only magnetic momen | vedclass

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(B) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$1$. $[Fe(CN)_6]^{3-}$: $Fe^{

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$[Fe(CN)_6]^{3-} < [Mn(CN)_6]^{3-} < [CoF_6]^{3-} < [MnBr_4]^{2-}$

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(B) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$1$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$. $CN^-$ is a strong field ligand,so $n = 1$. $\mu = \sqrt{1(3)} = 1.73 \ B.M.$$2$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $d^4$. $CN^-$ is a strong field ligand,so $n = 2$. $\mu = \sqrt{2(4)} = 2.83 \ B.M.$$3$. $[CoF_6]^{3-}$: $Co^{3+}$ is $d^6$. $F^-$ is a weak field ligand,so $n = 4$. $\mu = \sqrt{4(6)} = 4.90 \ B.M.$$4$. $[MnBr_4]^{2-}$: $Mn^{2+}$ is $d^5$. $Br^-$ is a weak field ligand,so $n = 5$. $\mu = \sqrt{5(7)} = 5.92 \ B.M.$The correct order is $[Fe(CN)_6]^{3-} < [Mn(CN)_6]^{3-} < [CoF_6]^{3-} < [MnBr_4]^{2-}$.

List-$I$ (Complex)	List-$II$ (Type of isomerism)
$A$. $[Co(NH_3)_5(NO_2)]Cl_2$	$I$. Solvate isomerism
$B$. $[Co(NH_3)_5(SO_4)]Br$	$II$. Linkage isomerism
$C$. $[Co(NH_3)_6][Cr(CN)_6]$	$III$. Ionization isomerism
$D$. $[Co(H_2O)_6]Cl_3$	$IV$. Coordination isomerism

The correct order of spin-only magnetic moments for the following complex ions is:

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