JEE MAIN - Chemistry (2020 - 4th September Evening Slot - No. 2)
The Crystal Field Stabilization Energy
(CFSE) of [CoF3(H2O)3] ($$\Delta $$0 < P) is :
(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
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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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