The correct order of the following complexes in terms of their crystal field stabilization energies is :

To determine the correct order of crystal field stabilization energies (CFSE) for the given cobalt complexes, we need to analyze each complex based on the oxidation state of cobalt, the geometry, the nature of the ligands, and the electron configuration. CFSE is the energy stabilization due to the splitting of d-orbitals in a ligand field, and it is more negative for greater stabilization.

- $$[Co(NH_3)_4]^{2+}$$: Cobalt oxidation state is +2 (since NH₃ is neutral), so Co(II) with d⁷ electrons. This complex is tetrahedral because Co(II) with strong field ligands like NH₃ forms tetrahedral complexes.

- $$[Co(NH_3)_6]^{2+}$$: Cobalt oxidation state is +2, so Co(II) d⁷. This complex is octahedral.

- $$[Co(NH_3)_6]^{3+}$$: Cobalt oxidation state is +3, so Co(III) d⁶. This complex is octahedral.

- $$[Co(en)_3]^{3+}$$: Cobalt oxidation state is +3, so Co(III) d⁶ (en is ethylenediamine, a neutral ligand). This complex is octahedral, and en is a stronger field ligand than NH₃ due to chelation.

The CFSE depends on the crystal field splitting parameter (Δ) and the number of electrons in t₂g and e_g orbitals. The formulas are:

- For octahedral complexes: $$CFSE_{oct} = [-0.4 \times n(t_{2g}) + 0.6 \times n(e_g)] \Delta_o$$

- For tetrahedral complexes: $$CFSE_{tet} = [-0.6 \times n(e) + 0.4 \times n(t_2)] \Delta_t$$ where $$\Delta_t = \frac{4}{9} \Delta_o$$ for the same metal and ligand.

Additionally, Δ_o increases with higher oxidation state and stronger field ligands. The spectrochemical series shows that en > NH₃, so Δ_o(en) > Δ_o(NH₃). Also, Δ_o for Co(III) > Δ_o for Co(II).

Use Δ_o(1) as the reference splitting for octahedral Co(II)-NH₃.

Complex 1: $$[Co(NH_3)_4]^{2+}$$ (tetrahedral, Co(II) d⁷ high-spin)

Electron configuration: e⁴ t₂³

$$CFSE_{tet} = [-0.6 \times 4 + 0.4 \times 3] \Delta_t = [-2.4 + 1.2] \Delta_t = -1.2 \Delta_t$$

Since $$\Delta_t = \frac{4}{9} \Delta_o(1)$$,

$$CFSE = -1.2 \times \frac{4}{9} \Delta_o(1) = -\frac{4.8}{9} \Delta_o(1) \approx -0.533 \Delta_o(1)$$

Complex 2: $$[Co(NH_3)_6]^{2+}$$ (octahedral, Co(II) d⁷ high-spin)

Electron configuration: t₂g⁵ e_g² (high-spin, as pairing energy is high for Co(II))

$$CFSE_{oct} = [-0.4 \times 5 + 0.6 \times 2] \Delta_o(1) = [-2.0 + 1.2] \Delta_o(1) = -0.8 \Delta_o(1)$$

Complex 3: $$[Co(NH_3)_6]^{3+}$$ (octahedral, Co(III) d⁶ low-spin)

Electron configuration: t₂g⁶ e_g⁰ (low-spin, as NH₃ is strong field)

Δ_o for Co(III)-NH₃ (denoted Δ_o(2)) is larger than for Co(II)-NH₃. Typically, Δ_o(2) ≈ 1.5 Δ_o(1).

$$CFSE_{oct} = [-0.4 \times 6 + 0.6 \times 0] \Delta_o(2) = -2.4 \Delta_o(2) \approx -2.4 \times 1.5 \Delta_o(1) = -3.6 \Delta_o(1)$$

Complex 4: $$[Co(en)_3]^{3+}$$ (octahedral, Co(III) d⁶ low-spin)

Electron configuration: t₂g⁶ e_g⁰

en is a stronger field ligand than NH₃, so Δ_o for Co(III)-en (denoted Δ_o(3)) is larger than Δ_o(2). Typically, Δ_o(3) ≈ 1.25 Δ_o(2).

$$CFSE_{oct} = [-0.4 \times 6 + 0.6 \times 0] \Delta_o(3) = -2.4 \Delta_o(3) \approx -2.4 \times 1.25 \Delta_o(2) = -3.0 \Delta_o(2)$$

Substituting Δ_o(2) ≈ 1.5 Δ_o(1),

$$CFSE \approx -3.0 \times 1.5 \Delta_o(1) = -4.5 \Delta_o(1)$$

Summarizing the CFSE in terms of Δ_o(1):

- $$[Co(NH_3)_4]^{2+}$$: ≈ -0.533 Δ_o(1)

- $$[Co(NH_3)_6]^{2+}$$: ≈ -0.8 Δ_o(1)

- $$[Co(NH_3)_6]^{3+}$$: ≈ -3.6 Δ_o(1)

- $$[Co(en)_3]^{3+}$$: ≈ -4.5 Δ_o(1)

Since CFSE is more negative for greater stabilization, the order of increasing CFSE (from least to greatest stabilization) is:

$$[Co(NH_3)_4]^{2+} < [Co(NH_3)_6]^{2+} < [Co(NH_3)_6]^{3+} < [Co(en)_3]^{3+}$$

Option D matches this order:

D. $$[Co(NH_{3})_{4}]^{2+} < [Co(NH_{3})_{6}]^{2+} < [Co(NH_{3})_{6}]^{3+} < [Co(en)_{3}]^{3+}$$

Thus, the correct answer is D.