The complex ion [Cu(NH_3)_4]^{2+} has: | vedclass

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(B) In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. In the presence of $NH_3$ (a

Vedclass · Accepted Answer

Square planar geometry with one unpaired electron

Vedclass · Answer

(B) In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. In the presence of $NH_3$ (a strong field ligand),the $3d$ electrons are arranged such that one electron remains unpaired in the $3d$ orbital. The hybridization involved is $dsp^2$,which corresponds to a square planar geometry. Thus,the complex has a square planar geometry with one unpaired electron.

The complex ion $[Cu(NH_3)_4]^{2+}$ has:

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