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) $= 3$.

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

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