In [Cu(NH_3)_4]SO_4,Cu has the following hybr | vedclass

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(A) In the complex $[Cu(NH_3)_4]SO_4$,the central metal ion is $Cu^{2+}$.The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.Therefor

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$dsp^2$

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(A) In the complex $[Cu(NH_3)_4]SO_4$,the central metal ion is $Cu^{2+}$.The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.Therefore,the electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.$NH_3$ is a strong field ligand,but in the case of $Cu^{2+}$,the $3d^9$ configuration leaves one unpaired electron in the $3d$ orbital.To accommodate four $NH_3$ ligands,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridization,utilizing one $3d$,one $4s$,and two $4p$ orbitals.Thus,the hybridization is $dsp^2$.

In $[Cu(NH_3)_4]SO_4$,$Cu$ has the following hybridization:

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