Which of the following complexes has a square | vedclass

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(D) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization,resu

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(D) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.$2$. $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$. The unpaired electron is in the $4p_z$ orbital,and the four ligands form bonds using $dsp^2$ hybrid orbitals,resulting in a square planar geometry.$3$. $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d^8$ ion. $Pt$ is a $5d$ series element,so it always forms square planar complexes with $dsp^2$ hybridization regardless of the ligand strength.Since all three complexes exhibit $dsp^2$ hybridization and square planar geometry,the correct answer is $D$.

Which of the following complexes has a square planar geometry and $dsp^2$ hybridization?

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