Question

Hybridization of Cu2+ ion in complex ion [Cu(NH3)4]2+ is

Answer 1

Answer:

So, the 4 electron pairs from N would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the $$e^-$$ from $$3d$$ would take place. So, you can say the hybridisation here would be $$dsp^2$$. Square planar with one unpaired electron.

Explanation:

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