The total number of isomers for a square plan | vedclass

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(D) For a square planar complex of the type $[M(abcd)]$,there are $3$ geometrical isomers.In the given complex $[M(F)(Cl)(SCN)(NO_2)]$,the ligands $SC

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$12$

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(D) For a square planar complex of the type $[M(abcd)]$,there are $3$ geometrical isomers.In the given complex $[M(F)(Cl)(SCN)(NO_2)]$,the ligands $SCN^-$ and $NO_2^-$ are ambidentate ligands.$SCN^-$ can coordinate through $S$ $(SCN^-)$ or $N$ $(NCS^-)$.$NO_2^-$ can coordinate through $N$ $(NO_2^-)$ or $O$ $(ONO^-)$.This gives rise to linkage isomerism.There are $4$ possible combinations of these ambidentate ligands:$1$. $[M(F)(Cl)(SCN)(NO_2)]$$2$. $[M(F)(Cl)(NCS)(NO_2)]$$3$. $[M(F)(Cl)(SCN)(ONO)]$$4$. $[M(F)(Cl)(NCS)(ONO)]$Each of these $4$ combinations exhibits $3$ geometrical isomers.Therefore,the total number of isomers = $4 	imes 3 = 12$.

Complex	Structure and magnetic property
$i$. $NiCl_4^{2-}$	$p$. tetrahedral and diamagnetic
$ii$. $Ni(CO)_4$	$q$. tetrahedral and paramagnetic
$iii$. $PtCl_4^{2-}$	$r$. square planar and diamagnetic
$iv$. $Ni(CN)_4^{2-}$	$s$. square planar and paramagnetic

The total number of isomers for a square planar complex $[M(F)(Cl)(SCN)(NO_2)]$ is

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