Calculate the magnetic moment of the Fe^{3+} | vedclass

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(A) In the complex $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a

Vedclass · Accepted Answer

$1.73 \ BM$

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(A) In the complex $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons. The $3d$ orbitals have $1$ unpaired electron. The magnetic moment $\mu$ is calculated as $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

Calculate the magnetic moment of the $Fe^{3+}$ ion in the $[Fe(CN)_6]^{3-}$ complex.

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