In octahedral crystal-field splitting, what is the energy of the $e_g$ orbitals **relative to** the $t_{2g}$ orbitals?

**Geometry of the splitting.** In an octahedral field, six ligands approach the metal along the $\pm x$, $\pm y$, $\pm z$ axes. The $e_g$ pair ($d_{z^2}$ and $d_{x^2-y^2}$) points *toward* the ligands and is destabilised; the $t_{2g}$ trio ($d_{xy}$, $d_{yz}$, $d_{zx}$) points *between* the ligands and is stabilised. **Energy gap.** The energy *difference* between the two sets is called the **crystal-field splitting energy** $\Delta_o$, with $e_g$ **higher** than $t_{2g}$ by exactly $\Delta_o$. - Option C is correct about the $e_g$ rising by $\frac{3}{5}\Delta_o$ above the **unsplit average**, while $t_{2g}$ drops by $\frac{2}{5}\Delta_o$. The total gap is therefore $\frac{3}{5}\Delta_o + \frac{2}{5}\Delta_o = \Delta_o$ — but the question asks for $e_g$ *relative to $t_{2g}$*, not relative to the average. That's $\Delta_o$, not $\frac{3}{5}\Delta_o$. - Option A reverses the direction (in *tetrahedral* splitting the order flips — $t_{2g}$ rises above $e_g$).