The crystal field stabilization energy (CFSE) of [Fe(H2O)6]Cl2 and K2[NiCl4], respectively, are:

🧠 Compute CFSE for Each Complex

$[\mathrm{Fe(H_2O)_6}]\mathrm{Cl_2}$: Fe(II), d⁶, octahedral with weak-field H₂O → HS.

Configuration: $t_{2g}^4 e_g^2$.

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

$\mathrm{K_2[NiCl_4]}$: Ni(II), d⁸, tetrahedral with Cl⁻.

In Td: e (lower, 2 orbitals) + $t_2$ (upper, 3 orbitals). HS d⁸ filling: $e^4 t_2^4$.

$\mathrm{CFSE} = 4(-0.6\Delta_t) + 4(+0.4\Delta_t) = -2.4\Delta_t + 1.6\Delta_t = -0.8\Delta_t$

🗺️ Td d⁸ Filling Visualised

In Td, 5 d-orbitals split as $e^2 t_2^3$ slots. d⁸ HS:

• e (2 orbitals) → 4 electrons (both paired). Coefficient: $-0.6\Delta_t$ each.
• $t_2$ (3 orbitals) → 4 electrons (1 pair + 2 unpaired). Coefficient: $+0.4\Delta_t$ each.

Total CFSE = $-0.8\Delta_t$. Two unpaired → paramagnetic (matches NiCl₄²⁻ μ ≈ 2.83 BM).

⚡ Td vs Oct CFSE Coefficients

| Geometry | Lower set | Upper set | |---|---|---| | Octahedral | $t_{2g}$ at $-0.4\Delta_o$ | $e_g$ at $+0.6\Delta_o$ | | Tetrahedral | $e$ at $-0.6\Delta_t$ | $t_2$ at $+0.4\Delta_t$ |

Note: the −0.6/−0.4 coefficients are inverted between geometries.

⚠️ HS d⁶ Oct Has Small CFSE

$[\mathrm{Fe(H_2O)_6}]^{2+}$ has CFSE = $-0.4\Delta_o$ — small, partly because the +0.6 from $e_g^2$ cancels much of the $-1.6$ from $t_{2g}^4$. This is one reason Fe²⁺ aquo complexes are kinetically labile.

$\boxed{\text{Answer: (3) } -0.4\Delta_o \text{ and } -0.8\Delta_t}$