The number of stereoisomers possible for [Co( | vedclass

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(D) The complex $[Co(ox)_2(Br)(NH_3)]^{2-}$ is of the type $[M(AA)_2ab]$,where $M = Co^{3+}$,$AA = ox^{2-}$,$a = Br^-$,and $b = NH_3$.$1$. For the $[M

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$3$

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(D) The complex $[Co(ox)_2(Br)(NH_3)]^{2-}$ is of the type $[M(AA)_2ab]$,where $M = Co^{3+}$,$AA = ox^{2-}$,$a = Br^-$,and $b = NH_3$.$1$. For the $[M(AA)_2ab]$ type complex,there are two geometrical isomers: $cis$ and $trans$.$2$. The $trans$ isomer has a plane of symmetry and is optically inactive (achiral).$3$. The $cis$ isomer does not have a plane of symmetry and exists as a pair of enantiomers ($d$ and $l$ forms),making it optically active.$4$. Therefore,the total number of stereoisomers is $1$ $(trans)$ $+ 2$ ($cis$ enantiomers) $= 3$.

List-$I$	List-$II$
$(a).$ $[Co(NH_3)_6][Cr(CN)_6]$	$(i).$ Linkage isomerism
$(b).$ $[Co(NH_3)_3(NO_2)_3]$	$(ii).$ Solvate isomerism
$(c).$ $[Cr(H_2O)_6]Cl_3$	$(iii).$ Coordination isomerism
$(d).$ $cis-[CrCl_2(ox)_2]^{3-}$	$(iv).$ Optical isomerism

The number of stereoisomers possible for $[Co(ox)_2(Br)(NH_3)]^{2-}$ is $...$. $[ox = \text{oxalate}]$

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