QJEEM25Chd6cd: JEE Main 2025 Chemistry - Class 12 JEEM25Chd6cd Solution: The correct order of the following complexes in terms of their…

The crystal field stabilization energy (CFSE) of a complex depends on two main factors:

Charge/Oxidation State of the Central Metal Atom: Higher oxidation states generally lead to greater crystal field splitting energy (Δ).

Field Strength of the Ligand: Stronger field ligands also result in greater splitting. Additionally, chelating ligands typically increase the field strength.

For octahedral complexes, the CFSE is calculated as:

$\text{CFSE} = [-0.4 \, t_{2g} + 0.6 \, e_g ] \, \Delta_\circ$

Considering the complexes:

$[\text{Co(en)}_3]^{3+}$ :

Co exists as Co $^{3+}$ with electronic configuration $t_{2g}^6 e_g^0$ .

CFSE = $-2.4 \, (\Delta_0)_1$ .

$[\text{Co(NH}_3)_6]^{3+}$ :

Co also as Co $^{3+}$ with $t_{2g}^6 e_g^0$ .

CFSE = $-2.4 \, (\Delta_0)_2$ .

$[\text{Co(NH}_3)_6]^{2+}$ :

Co with Co $^{2+}$ configuration $t_{2g}^5 e_g^2$ .

CFSE = $-0.8 \, (\Delta_0)_3$ .

$[\text{Co(NH}_3)_4]^{2+}$ :

Co as Co $^{2+}$ with different configuration $e^4 t_2^3$ .

CFSE = $-1.2 \, \Delta_t$ , where $\Delta_t = \frac{4}{9} (\Delta_0)_3$ .

The order of crystal field splitting energies, based on our criteria, is:

\Delta_t < (\Delta_0)_3 < (\Delta_0)_2 < (\Delta_0)_1

Therefore, the CFSEs of these complexes compare as follows:

$[\text{Co(en)}_3]^{3+}$ has the highest CFSE due to the $\Delta_0$ of the strongest field ligand and oxidation state.

$[\text{Co(NH}_3)_6]^{3+}$ follows with similarly high CFSE due to Co $^{3+}$ .

$[\text{Co(NH}_3)_6]^{2+}$ has a moderate CFSE.

$[\text{Co(NH}_3)_4]^{2+}$ has the lowest CFSE with $\Delta_t$ .

This hierarchy of CFSE is determined by both the oxidation state of the cobalt metal and the field strength of the ligands involved.