What is the correct electronic configuration of the central atom in $K_4[Fe(CN)_6]$ based on crystal field the

Question

What is the correct electronic configuration of the central atom in $K_4[Fe(CN)_6]$ based on crystal field theory?

Accepted Answer

1.  **Oxidation State:** The complex is $K_4[Fe(CN)_6]$, which dissociates into $4K^+$ and $[Fe(CN)_6]^{4-}$. The ligand is cyanide ($CN^-$), which has a charge of -1. Let the oxidation state of Iron ($Fe$) be $x$. 
    $x + 6(-1) = -4 \Rightarrow x = +2$. 
    The central metal ion is $Fe^{2+}$.
2.  **Electronic Configuration:** The atomic number of Iron is 26 ($[Ar] 3d^6 4s^2$). For $Fe^{2+}$, the configuration is $[Ar] 3d^6$.
3.  **Ligand Field Strength:** According to the spectrochemical series, the cyanide ion ($CN^-$) is a **strong field ligand** .
4.  **Crystal Field Splitting:** Strong field ligands cause a large crystal field splitting energy ($\Delta_o$), such that $\Delta_o > P$ (pairing energy). Consequently, electrons will pair up in the lower energy orbitals rather than occupying the higher energy orbitals.
5.  **Filling of Orbitals:** In an octahedral field, the $d$-orbitals split into lower energy $t_{2g}$ and higher energy $e_g$ sets. The 6 $d$-electrons of $Fe^{2+}$ will all occupy the lower energy $t_{2g}$ orbitals, filling them completely before any enter the $e_g$ orbitals.
    Configuration: **$t_{2g}^6 e_g^0$** .

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