The correct order of magnetic moments (spin v | vedclass

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Question

(D) $Cl^-$ is a weak field ligand,while $CN^-$ is a strong field ligand.$1$. In $[MnCl_4]^{2-}$,$Mn$ is in $+2$ oxidation state ($d^5$ configuration).

Vedclass · Accepted Answer

$[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$

Vedclass · Answer

(D) $Cl^-$ is a weak field ligand,while $CN^-$ is a strong field ligand.$1$. In $[MnCl_4]^{2-}$,$Mn$ is in $+2$ oxidation state ($d^5$ configuration). Since $Cl^-$ is a weak ligand,it has $5$ unpaired electrons.$2$. In $[CoCl_4]^{2-}$,$Co$ is in $+2$ oxidation state ($d^7$ configuration). Since $Cl^-$ is a weak ligand,it has $3$ unpaired electrons.$3$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state ($d^6$ configuration). Since $CN^-$ is a strong ligand,it forms a low spin complex with $0$ unpaired electrons.The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Thus,the order of magnetic moments is $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$.

The correct order of magnetic moments (spin values in $BM$) among the following is:

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

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