The values of the crystal field stabilization energies for a high spin d⁶ metal ion in octahedral and tetrahedral…

🧠 Compute CFSE for HS d⁶ in Each Geometry

Octahedral HS d⁶: $t_{2g}^4 e_g^2$

$\mathrm{CFSE_{oct}} = 4(-0.4\Delta_o) + 2(+0.6\Delta_o) = -1.6\Delta_o + 1.2\Delta_o = -0.4\Delta_o$

Tetrahedral d⁶ (always HS, since Δ_t small): in Td, e is lower (2 orbitals) and $t_2$ is upper (3 orbitals). With splitting energies $-0.6\Delta_t$ for e and $+0.4\Delta_t$ for $t_2$.

Filling 6 electrons in Td: $e^4 t_2^2$? No, wait — let me redo. In Td, fill e first (lower), then $t_2$.

For d⁶: e gets up to 4 electrons (2 orbitals × 2), $t_2$ gets the remaining 2.

But Hund's: $e^3 t_2^3$ (HS distribution) — 6 electrons in HS Td filling minimises pairing.

Actually for Td HS d⁶: $e^3 t_2^3$ → 6 electrons in 5 orbitals → still 1 forced pair, but the HS arrangement minimises it: 2 in e (paired) + 1 in e + 3 in $t_2$ = 6, with 1 pair in e.

Hmm actually let me think again. For Td HS, Hund's says maximise unpaired:

5 orbitals, 6 electrons → 1 forced pair. Where? The pair goes in the lowest energy orbital (one of the e set), and the remaining 4 electrons singly occupy the other e orbital + all three $t_2$ orbitals → $e^3 t_2^3$ (3 in e, 3 in t₂).

$\mathrm{CFSE_{td}} = 3(-0.6\Delta_t) + 3(+0.4\Delta_t) = -1.8\Delta_t + 1.2\Delta_t = -0.6\Delta_t$

🗺️ Double-Check via Hole Formalism

A d⁶ HS Td has 4 unpaired electrons. The configuration is $e^3 t_2^3$ — 3 in e, 3 in $t_2$ (with 1 pair within e).

CFSE: $(3 \times -0.6 + 3 \times 0.4)\Delta_t = -0.6\Delta_t$. ✓

⚡ CFSE in Tetrahedral: e Lower, $t_2$ Upper

In Td, the splitting is inverted vs octahedral:

• $e$ orbitals (lower): $-0.6\Delta_t$ each.
• $t_2$ orbitals (upper): $+0.4\Delta_t$ each.
• $\Delta_t \approx \frac{4}{9}\Delta_o$ (much smaller than oct).

Filling order: e first, then $t_2$.

⚠️ Don't Use Octahedral Formulas in Td

The CFSE coefficients ($-0.4, +0.6$) in oct are inverted in Td ($-0.6, +0.4$). Always check geometry first.

$\boxed{\text{Answer: (1) } -0.4\Delta_o \text{ and } -0.6\Delta_t}$