The total number of isomers for the square pl | vedclass

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(C) The complex is $[PdClBr_2(SCN)]^{2-}$,which is of the type $Ma_2bc$ where $M = Pd^{2+}$,$a = Br^-$,$b = Cl^-$,and $c = SCN^-$.For a square planar

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$4$

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(C) The complex is $[PdClBr_2(SCN)]^{2-}$,which is of the type $Ma_2bc$ where $M = Pd^{2+}$,$a = Br^-$,$b = Cl^-$,and $c = SCN^-$.For a square planar complex of type $Ma_2bc$,there are $2$ geometrical isomers:$1$. $cis$-isomer: where the two $Br^-$ ligands are adjacent to each other.$2$. $trans$-isomer: where the two $Br^-$ ligands are opposite to each other.Additionally,the $SCN^-$ ligand is an ambidentate ligand,which can coordinate through either the $S$ atom (thiocyanate) or the $N$ atom (isothiocyanate),leading to linkage isomerism.Each of the $2$ geometrical isomers can exist in $2$ linkage forms ($Pd-SCN$ and $Pd-NCS$).Therefore,the total number of isomers = $2 	ext{ (geometrical)} 	imes 2 	ext{ (linkage)} = 4$.

The total number of isomers for the square planar complex $K_2[PdClBr_2(SCN)]$ is/are:

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