What is the correct electronic configuration of the central atom in $\mathrm{K}{}_4[\mathrm{Fe}(\mathrm{CN}){}_6]$ based on crystal field theory?

Solution & Explanation

\textbf{Answer:} (B) $t_{2g}^6{e}_g^0$ \textbf{Solution:} In $\mathrm{K}{}_4[\mathrm{Fe}(\mathrm{CN}){}_6]$ the complex ion is $[\mathrm{Fe}(\mathrm{CN}){}_6]{}^{4-}$. With $\mathrm{CN}{}^{-}$ as $-1$ each, iron carries a $+2$ charge, so it is $\mathrm{Fe}{}^{2+}$, a $d^6$ ion. $\mathrm{CN}{}^{-}$ is a strong-field ligand, giving a large octahedral splitting ${\Delta }_o$ that exceeds the pairing energy. The six $d$ electrons therefore pair up in the lower $t_{2g}$ set (low-spin): $t_{2g}^6{e}_g^0$. Options C and D use the $e/t_2$ labels of a \emph{tetrahedral} field and are distractors. Hence the answer is B.