What is the correct electronic configuration | vedclass

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(B) In the complex $K_{4}[Fe(CN)_{6}]$,the oxidation state of $Fe$ is $+2$.The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^{6} 4s^{2}$.For $

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

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(B) In the complex $K_{4}[Fe(CN)_{6}]$,the oxidation state of $Fe$ is $+2$.The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^{6} 4s^{2}$.For $Fe^{2+}$,the configuration is $3d^{6}$.$CN^{-}$ is a strong field ligand $(SFL)$,which causes pairing of electrons in the $d$-orbitals.According to crystal field theory for an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_{g}$ sets.Due to the strong field,all $6$ electrons pair up in the lower energy $t_{2g}$ orbitals,resulting in the configuration $t_{2g}^{6} e_{g}^{0}$.

List-$I$	List-$II$
$P$. $t_{2g}^6 e_g^0$	$1$. $[Fe(H_2O)_6]^{2+}$
$Q$. $t_{2g}^3 e_g^2$	$2$. $[Mn(H_2O)_6]^{2+}$
$R$. $e^2 t_2^3$	$3$. $[Co(NH_3)_6]^{3+}$
$S$. $t_{2g}^4 e_g^2$	$4$. $[FeCl_4]^{-}$
	$5$. $[CoCl_4]^{2-}$

What is the correct electronic configuration of the central atom in $K_{4}[Fe(CN)_{6}]$ based on crystal field theory?

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