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

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(A) The complex $[PdCl_2(PMe_3)_2]$ is a square planar,diamagnetic complex of $Pd(II)$.The analogous complex of $Ni(II)$ is $[NiCl_2(PMe_3)_2]$.$Ni(II

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(A) The complex $[PdCl_2(PMe_3)_2]$ is a square planar,diamagnetic complex of $Pd(II)$.The analogous complex of $Ni(II)$ is $[NiCl_2(PMe_3)_2]$.$Ni(II)$ has a $d^8$ configuration. In the presence of weak field ligands like $Cl^-$,it forms a tetrahedral complex with $sp^3$ hybridization,which is paramagnetic.For a tetrahedral complex of the type $[Ma_2b_2]$,all positions are equivalent relative to each other.Therefore,it does not exhibit geometric isomerism.Since it is a tetrahedral complex,it also does not exhibit optical isomerism.Thus,the total number of isomers possible is $0$.

Compound	Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$	$P$. $2$
$B$. $[Co(en)_3]Cl_3$	$Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$	$R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$	$S$. $5$

$[PdCl_2(PMe_3)_2]$ is a diamagnetic complex of $Pd(II)$. How many total isomers are possible for the analogous paramagnetic complex of $Ni(II)$?

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Compound Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$ $P$. $2$
$B$. $[Co(en)_3]Cl_3$ $Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$ $R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$ $S$. $5$

Match the compounds with their total number of stereoisomers and select the correct code:

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