Compounds with spin-only magnetic moment equi | vedclass

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(B, C, D) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons. For $n=5$,$\mu = \

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$K_{4}[MnF_{6}]$

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(B, C, D) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons. For $n=5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 	ext{ BM}$.$(A)$ In $K_{4}[Mn(CN)_{6}]$,$Mn$ is in $+2$ oxidation state $(3d^{5})$. $CN^{-}$ is a strong field ligand,causing pairing of electrons. Thus,$n=1$.$(B)$ In $[Fe(H_{2}O)_{6}]Cl_{3}$,$Fe$ is in $+3$ oxidation state $(3d^{5})$. $H_{2}O$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.$(C)$ In $K_{3}[FeF_{6}]$,$Fe$ is in $+3$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.$(D)$ In $K_{4}[MnF_{6}]$,$Mn$ is in $+2$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.Therefore,the compounds with five unpaired electrons are $(B)$,$(C)$,and $(D)$.

Compounds with spin-only magnetic moment equivalent to five unpaired electrons are

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