Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: AB01 on February 27, 2011, 03:02:43 PM
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Hello everyone,
I was going through some research articles and they say palladium (II) complex is more stable than Palladium IV so the Palladium (IV) complex undergoes reductive elimination to give more stable Palladium (II) complex.
So please can anyone explain me why Pd (II) is more stable than Pd(IV)?
Thanks
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Pd(IV) has six d electrons and favors octahedral coordination.
Pd(II) has eight d electrons and forms square planar complexes.
Look up the ligand field diagrams for the energy levels of octahedral versus square planar complexes.