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What is the correct electronic configuration of the central atom in $K_4[Fe(CN)_6]$ based on crystal field theory?
A
$t_{2g}^4e_g^2$
B
$t_{2g}^6e_g^0$
C
$e^3t_2^3$
D
$e^4t_2^2$
Detailed Solution
In $K_4[Fe(CN)_6]$, Fe is in +2 state.
Fe ground state: $[Ar]3d^64s^2$; $Fe^{2+}$: $3d^64s^0$
$CN^-$ is a strong field ligand, so in the octahedral field pairing occurs and all 6 electrons go into $t_{2g}$.
Configuration: $t_{2g}^6e_g^0$