The Crystal Field Stabilisation Energy (CFSE) for $[\mathrm{CoCl}{}_6]{}^{4-}$ is $18000{\text{cm}}^{-1}$. The CFSE for $[\mathrm{CoCl}{}_4]{}^{2-}$ will be:

Solution & Explanation

\textbf{Answer:} (D) $8000{\text{cm}}^{-1}$ \textbf{Solution:} For the same metal ion and ligand, the tetrahedral splitting is related to the octahedral splitting by ${\Delta }_t=\frac{4}{9}{\Delta }_o$. Taking the octahedral value as ${\Delta }_o=18000{\text{cm}}^{-1}$ for $[\mathrm{CoCl}{}_6]{}^{4-}$, the tetrahedral $[\mathrm{CoCl}{}_4]{}^{2-}$ gives ${\Delta }_t=\frac{4}{9}\times 18000=8000{\text{cm}}^{-1}$. Hence the answer is option D.