Q.$[\text{CoF}_6]^{3-}$ is a high-spin octahedral complex with configuration $t_{2g}^4e_g^2$. Calculate its Crystal Field Stabilization Energy (CFSE) in terms of $\Delta_o$.

[!TLDR] $\text{CFSE} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\Delta_o$ with $t_{2g}^4 e_g^2$. [!ANSWER] $\text{CFSE} = -0.4(4)\Delta_o + 0.6 …

$[\text{CoF}_6]^{3-}$ is high-spin $d^6$ (Example 5, Section 5.6), with the configuration $t_{2g}^4 e_g^2$: four electrons in the three lower $t_{2g}$ orbitals, two electrons in the two higher $e_g$ orbitals. Substituting $n_{t_{2g}}=4$ and $n_{e_g}=2$ into the CFSE formula: $\text{CFSE} = (-0.4\times4 + 0.6\times2)\Delta_o = (-1.6 + 1.2)\Delta_o = -0.4\,\Delta_o$. The negative sign indicates a net stabilization (lower energy) relative to the hypothetical unsplit, spherical-field arrangement — but the stabilization is modest, only $0.4\Delta_o$, because most of the extra stability gained by the four $t_{2g}$ electrons ( …

Identify the electron distribution across $t_{2g}$ and $e_g$ (from the known high-spin/low-spin configuration) and substitute directly int …

Using the wrong sign convention (writing $+0.4$ for $t_{2g}$ and $-0.6$ for $e_g$, the reverse of the correct stabilizing/destabilizing signs) — $t_{2g}$ occupation is always stabilizing (negative cont …