[PdCl_2(PMe_3)_2] is a diamagnetic complex of | vedclass

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(C) $Pd(II)$ is a $4d$ transition metal, and due to the high crystal field splitting energy $(\Delta_o)$, it forms square planar complexes with strong

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$2$

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(C) $Pd(II)$ is a $4d$ transition metal, and due to the high crystal field splitting energy $(\Delta_o)$, it forms square planar complexes with strong field ligands, resulting in $0$ unpaired electrons.$Ni(II)$ is a $3d$ transition metal with a $d^8$ configuration.In the complex $[NiCl_2(PMe_3)_2]$, $Cl^-$ is a weak field ligand.For $Ni(II)$ $(d^8)$ with weak field ligands, the complex typically adopts a tetrahedral geometry ($sp^3$ hybridization).In a tetrahedral field, the $d$-orbitals split as $e The $8$ electrons fill as $e^4 t_2^4$, resulting in $2$ unpaired electrons.

$[PdCl_2(PMe_3)_2]$ is a diamagnetic complex of $Pd(II)$. How many unpaired electrons are present in the analogous complex of $Ni(II)$?

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