The Crystal Field Stabilization Energy (CFSE) of [CoF3(H2O)3] (0 < P) is:

🧠 Find Co's Oxidation State and d-Count

$[\mathrm{CoF_3(H_2O)_3}]$:

• 3 F⁻ contribute −3.
• 3 H₂O contribute 0.
• Complex is neutral overall: $\mathrm{Co} - 3 + 0 = 0 \Rightarrow \mathrm{Co} = +3$.

So Co(III), which is d⁶.

🗺️ Spin State and Configuration

Given $\Delta_o < P$ → high-spin.

HS d⁶ in octahedral: $t_{2g}^4 e_g^2$.

🗺️ CFSE Calculation

For $t_{2g}^4 e_g^2$:

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

⚡ Pairing-Energy Note

In HS d⁶, the $t_{2g}^4$ already requires pairing 1 electron (4 in 3 orbitals → 1 forced pair). The free-ion d⁶ also has 1 forced pair (5 orbitals, 6 electrons → 1 forced pair). So no extra pairing is induced by the field — the P-correction term is 0.

For LS d⁶ ($t_{2g}^6$): pairs = 3 in field, 1 in free ion → extra 2 pairs → CFSE = −2.4Δ₀ + 2P.

⚠️ Mixed-Ligand Field with $\Delta_o < P$

In $[\mathrm{CoF_3(H_2O)_3}]$, the field is mixed (3 F⁻ + 3 H₂O), and the parenthetical "$\Delta_o < P$" forces HS treatment regardless of how you'd estimate the average ligand strength. The CFSE formula then applies as above.

$\boxed{\text{Answer: } -0.4\Delta_o}$