Calculate the CFSE for $K_4\left[Fe{\left(CN\right)}_6\right]$

In ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-},$ Fe has +2 oxidation state and its outer electronic configuration becomes 3d⁵. As CN^- is a strong ligand, so all electrons of 3d gets paired in t_2g orbitals and e_g orbitals remains empty, $\left({\mathrm{t}}_2{\mathrm{g}}^6{\mathrm{eg}}^0\right)$.

$$\begin{gathered} \mathrm{CFSE}=-0.4(\mathrm{number} \mathrm{of} \mathrm{electrons} \mathrm{in} {\mathrm{t}}_{2\mathrm{g}})+0.6(\mathrm{number} \mathrm{of} \mathrm{electrons} \mathrm{in} {\mathrm{e}}_{\mathrm{g}}) \\ =-0.4\left(6\right)+0.6\left(0\right) \\ =-2.4{\mathrm{\Delta }}_0 \end{gathered}$$