JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Coordination Compounds

Crystal field stabilization energy rises with both the metal's effective charge and the ligand field strength, and drops sharply for a lower coordination number.

${\left[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_4\right]}^{2+}$ is a 4-coordinate (tetrahedral) $\mathrm{C}{\mathrm{o}}^{2+}$ complex; tetrahedral geometry splits the d-orbitals into a much smaller gap than octahedral geometry does, so this complex has the lowest CFSE of the set.

${\left[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6\right]}^{2+}$ is octahedral $\mathrm{C}{\mathrm{o}}^{2+}$ ($d^7$) with $\mathrm{N}{\mathrm{H}}_3$ — a bigger splitting than the tetrahedral case, but smaller than the $\mathrm{C}{\mathrm{o}}^{3+}$ complexes below because of the lower metal charge.

${\left[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6\right]}^{3+}$ is octahedral $\mathrm{C}{\mathrm{o}}^{3+}$ ($d^6$); the higher metal oxidation state pulls in the ligand electron density more strongly and increases ${\Delta }_o$, giving a larger CFSE than either $\mathrm{C}{\mathrm{o}}^{2+}$ complex.

${\left[\mathrm{Co}(\mathrm{en}{)}_3\right]}^{3+}$ is also octahedral $\mathrm{C}{\mathrm{o}}^{3+}$, but ethylenediamine sits higher than $\mathrm{N}{\mathrm{H}}_3$ in the spectrochemical series (a stronger-field, chelating amine donor), producing the largest splitting and the largest CFSE of all four.

Order of increasing CFSE: ${\left[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_4\right]}^{2+}<{\left[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6\right]}^{2+}<{\left[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6\right]}^{3+}<{\left[\mathrm{Co}(\mathrm{en}{)}_3\right]}^{3+}$.