How many isomers are possible for the complex | vedclass

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(D) The complex ion is $[Cr(NH_3)(OH)_2Cl_3]^{2-}$.This is an octahedral complex of the type $[M(a)(b)_2(c)_3]$.Here,$a = NH_3$,$b = OH^-$,and $c = Cl

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$5$

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(D) The complex ion is $[Cr(NH_3)(OH)_2Cl_3]^{2-}$.This is an octahedral complex of the type $[M(a)(b)_2(c)_3]$.Here,$a = NH_3$,$b = OH^-$,and $c = Cl^-$.For a complex of the type $[M(a)(b)_2(c)_3]$,the possible geometric isomers are:$1$. $mer-fac$ isomerism is not applicable here due to the presence of different ligands.$2$. The possible arrangements for the ligands are:- $fac$ arrangement of $Cl$ ligands with $NH_3$ trans to $OH$.- $mer$ arrangement of $Cl$ ligands with $NH_3$ trans to $Cl$.- Other variations based on the relative positions of $OH$ and $Cl$ ligands.For the coordination number $6$,the total number of geometric isomers for $[M(a)(b)_2(c)_3]$ is $5$.

List-$I$ (Complex)	List-$II$ (Type of isomerism)
$A$. $[Co(NH_3)_5(NO_2)]Cl_2$	$I$. Solvate isomerism
$B$. $[Co(NH_3)_5(SO_4)]Br$	$II$. Linkage isomerism
$C$. $[Co(NH_3)_6][Cr(CN)_6]$	$III$. Ionization isomerism
$D$. $[Co(H_2O)_6]Cl_3$	$IV$. Coordination isomerism

How many isomers are possible for the complex ion $[Cr(NH_3)(OH)_2Cl_3]^{2-}$?

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