why cu show dsp2 hybridization in [Cu(NH3)4]2+
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Practically it is observed that the given compund is square planar.So it is either sp2d or dsp2. From chemical analysis it can be said that in [Cu(NH3)4]2+ Cu will lose its nature if the hybridization is dsp2 since during this formation the ligands will take over the place of the free electron and that electron will move to last shell. But it is observed that in the compound Cu2+ is already stable and it shows no tendency to donate electron.So its hybridization must be sp2d. It is exceptional.
Jaspreet26131:
thank you
Don't feel confused : in case of sp2d, one p orbital will remain empty and lone pair electrons of NH4 will fill up the next d orbital.
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