Total number of stereoisomers of [Co(acac)_2B | vedclass

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(B) The complex $[Co(acac)_2BrCl]^{\ominus}$ is of the type $[M(AA)_2ab]$,where $AA$ is a symmetrical bidentate ligand (acetylacetonate,$acac^-$) and

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The complex $[Co(acac)_2BrCl]^{\ominus}$ is of the type $[M(AA)_2ab]$,where $AA$ is a symmetrical bidentate ligand (acetylacetonate,$acac^-$) and $a, b$ are monodentate ligands ($Br^-$ and $Cl^-$).This complex exhibits geometrical isomerism,having two forms: $cis$ and $trans$.$1$. The $trans$-form has a plane of symmetry and is optically inactive (achiral).$2$. The $cis$-form does not have a plane of symmetry and exists as a pair of enantiomers (dextro and laevo forms),making it optically active.Therefore,the total number of stereoisomers is $1$ $(trans)$ + $2$ ($cis$ enantiomers) = $3$.

Total number of stereoisomers of $[Co(acac)_2BrCl]^{\ominus}$ are

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