Arrange the following coordination compounds | vedclass

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(B) The magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$(A)$ $[FeF_{6}]^{3-}$: $Fe^{3+}$ is $3d^

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$B < D < C < A$

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(B) The magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$(A)$ $[FeF_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $F^-$ is a weak field ligand,so electrons remain unpaired. $n = 5$.$(B)$ $[Fe(CN)_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $CN^-$ is a strong field ligand,causing pairing. $n = 1$.$(C)$ $[MnCl_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. High spin complex means $n = 4$.$(D)$ $[Mn(CN)_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. $CN^-$ is a strong field ligand,causing pairing. $n = 2$.Comparing $n$ values: $B(n=1) Therefore,the increasing order of magnetic moments is $B < D < C < A$.

Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: $Mn = 25$; $Fe = 26$)
$(A)$ $[FeF_{6}]^{3-}$
$(B)$ $[Fe(CN)_{6}]^{3-}$
$(C)$ $[MnCl_{6}]^{3-}$ (high spin)
$(D)$ $[Mn(CN)_{6}]^{3-}$

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Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: $Mn = 25$; $Fe = 26$)$(A)$ $[FeF_{6}]^{3-}$$(B)$ $[Fe(CN)_{6}]^{3-}$$(C)$ $[MnCl_{6}]^{3-}$ (high spin)$(D)$ $[Mn(CN)_{6}]^{3-}$