Question

Acetylacetone, abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac2, as shown below, where M is a transition metal:

Which of the following complexes are optically active:$\mathbf{cis}\mathbf{-}\mathbf{Cr}{\left(\mathrm{acac}\right)}_{\mathbf{2}}{\left(H_{2}O\right)}_{\mathbf{2}}\mathbf{,}\mathbf{trans}\mathbf{-}\mathbf{Cr}{\left(\mathrm{acac}\right)}_{\mathbf{2}}{\left(H_{2}O\right)}_{\mathbf{2}}\mathbf{,}\mathbf{andCr}{\left(\mathrm{acac}\right)}_{\mathbf{3}}$ ?

Short answer

Optically active: $\mathrm{cis}-\mathrm{Cr}{\left(\mathrm{acac}\right)}_2{\left({\mathrm{H}}_2\mathrm{O}\right)}_2,\mathrm{Cr}{\left(\mathrm{acac}\right)}_3$

Step-by-step solution

Check for complexes are optically active 

$\mathbf{cis}\mathbf{-}\mathbf{Cr}{\left(\mathrm{acac}\right)}_{\mathbf{2}}{{\left(H_{2}O\right)}_{\mathbf{2}}}_{}$Cis isomer of octahedral complexes of type$\left[\mathrm{M}{\left(\mathrm{aa}\right)}_2{\mathrm{b}}_2\right]$ is found to exhibit optical isomerism whereas trans form$\mathrm{trans}-\mathrm{Cr}{\left(\mathrm{acac}\right)}_2{\left({\mathrm{H}}_2\mathrm{O}\right)}_2$ does not exhibit optical isomerism because they are symmetrical.

Check for complexes are optically active 

$\mathbf{Cr}{{\left(\mathrm{acac}\right)}_{\mathbf{3}}}_{\mathbf{}\mathbf{}}$is octahedral complex of type$\left[\mathrm{M}{\left(\mathrm{aa}\right)}_3\right]$ whereas (aa) is symmetrical bidentade ligand exhibit optical isomerism. The d and l form of$\mathrm{Cr}{{\left(\mathrm{acac}\right)}_3}_{}$ isomers are there.