Identify the correct order of the spin-only m | vedclass

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Question

(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.$1$. In $[

Vedclass · Accepted Answer

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

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(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.$1$. In $[MnCl_4]^{2-}$,$Mn$ is in $+2$ oxidation state: $[Ar] 3d^5$. Since $Cl^-$ is a weak field ligand,$n = 5$. Thus,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, BM$.$2$. In $[CoCl_4]^{2-}$,$Co$ is in $+2$ oxidation state: $[Ar] 3d^7$. Since $Cl^-$ is a weak field ligand,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, BM$.$3$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state: $[Ar] 3d^6$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in $n = 0$. Thus,$\mu = 0 \, BM$.Therefore,the correct order is $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$.

Compound	Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$	$P$. $2$
$B$. $[Co(en)_3]Cl_3$	$Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$	$R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$	$S$. $5$

Identify the correct order of the spin-only magnetic moment.

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Compound Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$ $P$. $2$
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$C$. $[Pt(NH_3)_2Cl_2]$ $R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$ $S$. $5$

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Which of the following will have a spin-only magnetic moment of $2.86 \ BM$?

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