We need to arrange the given cobalt complexes in the increasing order of their Crystal Field Stabilization Energy (CFSE) magnitude by calculating their values in terms of octrahedral splitting ($$\Delta_o$$) and tetrahedral splitting ($$\Delta_t$$).
CFSE Calculations:
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1. $$[\text{Co}(\text{NH}_3)_4]^{2+}$$ (Tetrahedral, weak-field ligand):
- $$\text{Co}^{2+} \implies d^7$$ system.
- Tetrahedral configuration: $$e^4 t_2^3$$
- $$\text{CFSE} = [-0.6 \times 4 + 0.4 \times 3]\Delta_t = -1.2\Delta_t$$
- Since $$\Delta_t \approx \frac{4}{9}\Delta_o$$, this corresponds to a very small energy splitting:
$$\text{CFSE} \approx -1.2 \times \frac{4}{9}\Delta_o \approx -0.53\Delta_o$$
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2. $$[\text{Co}(\text{NH}_3)_6]^{2+}$$ (Octahedral, weak-to-moderate field ligand):
- $$\text{Co}^{2+} \implies d^7$$ system (high-spin).
- Configuration: $$t_{2g}^5 e_g^2$$
- $$\text{CFSE} = [-0.4 \times 5 + 0.6 \times 2]\Delta_o = -0.8\Delta_o$$
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3. $$[\text{Co}(\text{NH}_3)_6]^{3+}$$ (Octahedral, strong-field behavior with $$\text{Co}^{3+}$$):
- $$\text{Co}^{3+} \implies d^6$$ system (low-spin due to high charge on cobalt pairing the electrons).
- Configuration: $$t_{2g}^6 e_g^0$$
- $$\text{CFSE} = [-0.4 \times 6]\Delta_o + 2P = -2.4\Delta_o + 2P$$
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4. $$[\text{Co}(\text{en})_3]^{3+}$$ (Octahedral, strong chelating ligand):
- $$\text{Co}^{3+} \implies d^6$$ system (low-spin).
- Configuration: $$t_{2g}^6 e_g^0$$
- $$\text{CFSE} = [-0.4 \times 6]\Delta_o + 2P = -2.4\Delta_o + 2P$$
- Because ethylenediamine ($$\text{en}$$) is a stronger chelating ligand than ammonia ($$\text{NH}_3$$), its splitting parameter ($$\Delta_o$$) is significantly larger. Consequently, its absolute CFSE magnitude is the highest.
Conclusion:
Comparing the calculated values and ligand field strengths, the increasing order of CFSE magnitude is:
$$\text{[Co(NH}_3)_4]^{2+} < \text{[Co(NH}_3)_6]^{2+} < \text{[Co(NH}_3)_6]^{3+} < \text{[Co(en)}_3]^{3+}$$
Answer: Option D