Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value: A:…

🧠 CFSE Magnitude Depends on (a) d-Filling and (b) Δ Itself

| Complex | Metal/OS | d-count | Spin | CFSE (in own Δ_o) | |---|---|---|---|---| | (A) $[\mathrm{CoF_6}]^{3-}$ | Co(III), d⁶ | HS (F⁻ weak) | $-0.4\Delta_{o,A}$ | | (B) $[\mathrm{Co(H_2O)_6}]^{2+}$ | Co(II), d⁷ | HS | $-0.8\Delta_{o,B}$ | | (C) $[\mathrm{Co(NH_3)_6}]^{3+}$ | Co(III), d⁶ | LS | $-2.4\Delta_{o,C}$ | | (D) $[\mathrm{Co(en)_3}]^{3+}$ | Co(III), d⁶ | LS | $-2.4\Delta_{o,D}$ |

In absolute energy terms, multiply each Δ-coefficient by the metal's actual $\Delta_o$:

• A: HS d⁶, weak field → small Δ × small coefficient → smallest |CFSE|.
• B: HS d⁷, weak field → small Δ × moderate coefficient.
• C: LS d⁶, NH₃ field → large Δ × large coefficient.
• D: LS d⁶, en field (en > NH₃) → largest Δ × large coefficient → biggest |CFSE|.

🗺️ Why D > C

Both C and D are Co(III) d⁶ LS, both give the same coefficient $-2.4\Delta_o$ in their own Δ. But the spectrochemical series puts en above NH₃ (en is bidentate with chelate boost). So $\Delta_{o,D} > \Delta_{o,C}$ → |CFSE_D| > |CFSE_C|.

⚡ CFSE Magnitude in Increasing Order

$|\mathrm{CFSE_A}| < |\mathrm{CFSE_B}| < |\mathrm{CFSE_C}| < |\mathrm{CFSE_D}|$

In real numbers (rough estimates):

• A ≈ −0.4 × 13,000 cm⁻¹ ≈ −5 kJ/mol
• B ≈ −0.8 × 9,000 cm⁻¹ ≈ −9 kJ/mol
• C ≈ −2.4 × 23,000 cm⁻¹ ≈ −66 kJ/mol
• D ≈ −2.4 × 24,000 cm⁻¹ ≈ −69 kJ/mol

⚠️ CFSE Coefficients Alone Don't Settle the Order

If you stop at "$-2.4\Delta_o$" for both C and D and conclude they're equal, you miss the point. The Δ_o itself differs between en and NH₃ — that's how the question separates C and D.

$\boxed{\text{Answer: (3) A < B < C < D}}$