The spin-only magnetic moments of [Fe(NH_3)_6 | vedclass

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(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.For $[Fe(NH_3)_6]^{3+}$,$F

Vedclass · Accepted Answer

$1.73$ and $5.92$

Vedclass · Answer

(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.For $[Fe(NH_3)_6]^{3+}$,$Fe$ is in the $+3$ oxidation state $(3d^5)$. Since $NH_3$ is a strong field ligand,it causes pairing of electrons,resulting in $n = 1$ unpaired electron.$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \ BM$.For $[FeF_6]^{3-}$,$Fe$ is in the $+3$ oxidation state $(3d^5)$. Since $F^-$ is a weak field ligand,no pairing occurs,resulting in $n = 5$ unpaired electrons.$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \ BM$.Therefore,the values are $1.73 \ BM$ and $5.92 \ BM$.

The spin-only magnetic moments of $[Fe(NH_3)_6]^{3+}$ and $[FeF_6]^{3-}$ in $BM$ are,respectively

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