To determine the electronic configuration of the central atom in $K_4[Fe(CN)_6]$ based on crystal field theory, we need to follow these steps:$\newline$1. Identify the oxidation state of the central metal atom (Fe):$\newline$• In $K_4[Fe(CN)_6],$ the overall charge of the complex is 0.$\newline$• Potassium (K) has a charge of +1, and there are 4 K atoms, contributing a total charge of +4.$\newline$• Cyanide (CN) is a monodentate ligand with a charge of -1, and there are 6 CN ligands, contributing a total charge of -6.$\newline$• Let the oxidation state of Fe be $x.$ The equation is:$\newline x + 4(+1) + 6(-1) = 0 \newline x + 4 - 6 = 0 \newline x = +2 \newline$Therefore, the oxidation state of Fe is +2.$\newline$2. Determine the electronic configuration of Fe in the +2 oxidation state:$\newline$• The atomic number of Fe is 26, so its electronic configuration is $[Ar] \, 3d^6 \, 4s^2. \newline$• In the +2 oxidation state, Fe loses 2 electrons, resulting in $[Ar] \, 3d^6. \newline$3. Analyze the ligand field:$\newline$• CN is a strong field ligand, which causes pairing of electrons in the d orbitals.$\newline$4. Determine the crystal field splitting and electron configuration:$\newline$• In an octahedral field, the d orbitals split into $t_{2g}$ and $e_g$ levels.$\newline$• Strong field ligands like CN cause a large splitting, leading to pairing of electrons in the $t_{2g}$ orbitals.$\newline$• The 6 electrons will fill the $t_{2g}$ orbitals completely: $t_{2g}^6 \, e_g^0. \newline$Therefore, the correct electronic configuration of the central atom in $K_4[Fe(CN)_6]$ is $t_{2g}^6 \, e_g^0. \newline$This corresponds to Option 2.