JEE MAIN - Chemistry (2020 - 4th September Evening Slot - No. 2)

The Crystal Field Stabilization Energy
(CFSE) of [CoF3(H2O)3] ($$\Delta $$0 < P) is :
-0.8 $$\Delta $$0
-0.4 $$\Delta $$0
-0.8 $$\Delta $$0 + 2P
-0.4 $$\Delta $$0 + P

Explanation



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

For octahedral

Crystal field stabilization energy (CFSE)

= (-0.4$$\Delta $$0) $$ \times $$ nt2g + (+0.6$$\Delta $$0) $$ \times $$ neg + np

nt2g = number of electrons in t2g orbital

neg = number of electrons in eg orbital

n = number of extra pairs

p = Pairing energy

Here nt2g = 1

neg = 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

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