JEE Main 2025 — Coordination Compounds Question with Solution

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:

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

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.