The number of possible isomers for the square | vedclass

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(D) The complex is $[PdClBr_2(SCN)]^{2-}$.This is a square planar complex of the type $[Mabcd]$ where $a=Cl$,$b=Br$,$c=Br$,and $d=SCN$.However,the lig

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$6$

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(D) The complex is $[PdClBr_2(SCN)]^{2-}$.This is a square planar complex of the type $[Mabcd]$ where $a=Cl$,$b=Br$,$c=Br$,and $d=SCN$.However,the ligand $SCN^-$ is an ambidentate ligand,which can coordinate through $S$ $(SCN^-)$ or $N$ $(NCS^-)$.For a square planar complex of the type $[M(a)(b)(c)(d)]$,the number of geometrical isomers is $3$.Here,the ligands are $Cl$,$Br$,$Br$,and $SCN$. Since there are two identical $Br$ ligands,the formula is $[M(a)(b_2)(d)]$.For $[M(a)(b_2)(d)]$,the possible geometrical isomers are $3$: $(1)$ $Cl$ and $SCN$ are trans,$(2)$ $Cl$ and $Br$ are trans,$(3)$ $Br$ and $Br$ are trans.Additionally,$SCN$ can link via $S$ or $N$ (linkage isomerism).For each geometrical isomer,there are $2$ linkage isomers.Total isomers = $3 	imes 2 = 6$.

The number of possible isomers for the square planar complex $K_2[PdClBr_2SCN]$ is .................

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