The complex having the highest spin-only magn | vedclass

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

Vedclass · Accepted Answer

$[MnF_6]^{4-}$

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(C) The spin-only magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$(A)$ $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand,causing pairing of electrons. $n = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM$.$(B)$ $[Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $H_2O$ is a weak field ligand,no pairing occurs. $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM$.$(C)$ $[MnF_6]^{4-}$: $Mn$ is in $+2$ oxidation state $(3d^5)$. $F^-$ is a weak field ligand,no pairing occurs. $n = 5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ BM$.$(D)$ $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,no pairing occurs. $n = 2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.Comparing the values,$[MnF_6]^{4-}$ has the highest number of unpaired electrons $(n=5)$,hence the highest magnetic moment.

The complex having the highest spin-only magnetic moment is

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