JEE MAIN - Chemistry (2025 - 22nd January Evening Shift - No. 4)
Explanation
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.
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