The calculated spin-only magnetic moments (BM | vedclass

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(D) For $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons. The configuration i

Vedclass · Accepted Answer

$0$ and $4.9$

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(D) For $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons. The configuration is $t_{2g}^6 e_g^0$. Number of unpaired electrons $(n)$ = $0$. Magnetic moment $\mu = \sqrt{0(0+2)} = 0 \ BM$.For $[Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $H_2O$ is a weak field ligand,no pairing occurs. The configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons $(n)$ = $4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \ BM$.

The calculated spin-only magnetic moments $(BM)$ of the species $[Fe(CN)_6]^{4-}$ and $[Fe(H_2O)_6]^{2+}$,respectively,are:

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