The correct order regarding the number of unp | vedclass

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(A) $I. [Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons. Number of unpaired el

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$II > III > I$

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(A) $I. [Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons. Number of unpaired electrons = $0$.$II. [Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state $(d^6)$. $H_2O$ is a weak field ligand,no pairing occurs. Configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons = $4$.$III. [Co(H_2O)_6]^{2+}$: $Co$ is in $+2$ oxidation state $(d^7)$. $H_2O$ is a weak field ligand. Configuration is $t_{2g}^5 e_g^2$. Number of unpaired electrons = $3$.Thus,the correct order is $II (4) > III (3) > I (0)$.

The correct order regarding the number of unpaired electrons present in the following complexes is
$I. [Fe(CN)_6]^{4-} \quad II. [Fe(H_2O)_6]^{2+} \quad III. [Co(H_2O)_6]^{2+}$

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The correct order regarding the number of unpaired electrons present in the following complexes is$I. [Fe(CN)_6]^{4-} \quad II. [Fe(H_2O)_6]^{2+} \quad III. [Co(H_2O)_6]^{2+}$