QJEEM20Che9e6: JEE Main 2020 Chemistry - Class 12 JEEM20Che9e6 Solution: The Crystal Field Stabilization Energy (CFSE) of [CoF 3 (H 2 O) …

As $\Delta$ ₀ < P, so all ligands behaves as weak field ligands.

For octahedral

Crystal field stabilization energy (CFSE)

= (-0.4 $\Delta$ ₀) $\times$ n_t2g + (+0.6 $\Delta$ ₀) $\times$ n_eg + np

n_t2g = number of electrons in t_2g orbital

n_eg = number of electrons in e_g orbital

n = number of extra pairs

p = Pairing energy

Here n_t2g = 1

n_eg = 0

n = 0 ( For weak field ligands n always zero. Here H₂O is weak field ligand)

$\therefore$ Crystal field stabilization energy (CFSE)

= (-0.4 $\Delta$ ₀) $\times$ 4 + (+0.6 $\Delta$ ₀) $\times$ 2 + 0 $\times$ P

= -1.6 $\Delta$ ₀ + 1.2 $\Delta$ ₀

= -0.4 $\Delta$ ₀