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

Solution

Chemistry ma’am · 4 steps

Rank by three levers: geometry first, then the charge on cobalt, then how strong the ligand field is.

1. Geometry. [Co(NH₃)₄]²⁺ is tetrahedral, and a tetrahedral split is small: Δₜ ≈ 4/9Δₒ. With fewer ligands and a smaller gap, it has the least stabilisation of the four.
2. Oxidation state. Co²⁺ is d⁷. With NH₃ it stays high spin (t_(2g)⁵ e_g²), so CFSE is only -0.8 Δₒ. Co³⁺ is d⁶, and its higher charge widens Δₒ enough to pair all six electrons in t_(2g), giving -2.4 Δₒ with a larger Δₒ too.

   Complex	Shape, configuration	CFSE	Rank
   [Co(NH₃)₄]²⁺	tetrahedral, e⁴ t₂³	-1.2 Δₜ ≈ -0.53 Δₒ	smallest
   [Co(NH₃)₆]²⁺	octahedral, t_(2g)⁵ e_g²	-0.8 Δₒ	second
   [Co(NH₃)₆]³⁺	octahedral, t_(2g)⁶ e_g⁰	-2.4 Δₒ	third
   [Co(en)₃]³⁺	octahedral, t_(2g)⁶ e_g⁰	-2.4 Δₒ, larger Δₒ	largest

3. Ligand strength. Both Co³⁺ complexes are t_(2g)⁶. The tie is broken by the ligand: en is higher than NH₃ in the spectrochemical series and binds as a chelate, so Δₒ for [Co(en)₃]³⁺ is larger.
4. So. The order of stabilisation is [Co(NH₃)₄]²⁺ < [Co(NH₃)₆]²⁺ < [Co(NH₃)₆]³⁺ < [Co(en)₃]³⁺, which is option (a).

The idea to keep. Tetrahedral complexes have the least CFSE, since Δₜ ≈ 4/9Δₒ. Among octahedral ones, a higher metal charge and a stronger-field ligand (en over NH₃) both widen the split.

Answer from the NTA final answer key. Solution by PrepWiser.