[Cu(NH_3)_4]^{2+} shows the following hybridi | vedclass

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Question

(A) The atomic number of $Cu$ is $29$. The electronic configuration is $[Ar] 3d^{10} 4s^1$.In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$.

Vedclass · Accepted Answer

$dsp^2$

Vedclass · Answer

(A) The atomic number of $Cu$ is $29$. The electronic configuration is $[Ar] 3d^{10} 4s^1$.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 $4$ $NH_3$ ligands,one electron from the $3d$ orbital is promoted to the $4p$ orbital to facilitate $dsp^2$ hybridization.Thus,the hybridization is $dsp^2$.

$[Cu(NH_3)_4]^{2+}$ shows the following hybridization:

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