The magnetic moment of {K_3}[Fe(CN)_6] is fou | vedclass

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Question

(A) In ${K_3}[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$. Electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. Electronic configuration

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(A) In ${K_3}[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$. Electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. The $3d^5$ configuration becomes $t_{2g}^5 e_g^0$,which means there is $1$ unpaired electron. The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. For $n=1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ B.M.$ Thus,there is $1$ unpaired electron present per molecule.

The magnetic moment of ${K_3}[Fe(CN)_6]$ is found to be $1.7 \ B.M.$ How many unpaired electron$(s)$ is/are present per molecule?

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