Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

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

$e_g^4 t_{2g}^2$

$t_{2g}^4 e_g^2$

$t_{2g}^6 e_g^0$

$e_g^3 t_{2g}^3$

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 $Fe^{2+}$ .
Electronic Configuration: The atomic number of Iron is 26 ( $[Ar] 3d^6 4s^2$ ). For $Fe^{2+}$ , the configuration is $[Ar] 3d^6$ .
Ligand Field Strength: According to the spectrochemical series, the cyanide ion ( $CN^-$ ) is a strong field ligand .
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.
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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