Draw all the geometrical isomers of \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+1}\) . Which of these isomers also have an optical isomer? Draw the various isomers.

To draw the geometrical isomers of the complex, we will consider different arrangements of the monodentate ligands (\(\mathrm{NH_3, Br, Cl}\)) around the metal ion Cr. Isomer 1 (cis): In the cis isomer, the monodentate ligands are adjacent to each other: \( \begin{array}{r} \mathrm{Cr} (\mathrm{en})\left(\mathrm{NH}_{3}\right)\left(\mathrm{Br}\right)\\ \hspace{1.3cm} |\\ \hspace{1.3cm}\mathrm{Cl} \end{array} \) Isomer 2 (trans): In the trans isomer, the monodentate ligands are opposite to each other: \( \begin{array}{r} \mathrm{Cr} (\mathrm{en})\left(\mathrm{NH}_{3}\right)\left(\mathrm{Br}\right)\\ \hspace{1.7cm} | \\ \hspace{1.7cm} \mathrm{NH_3} \\ \hspace{1.7cm} | \\ \mathrm{Cl} \end{array} \)

Now, we need to determine which of these geometrical isomers also have optical isomers. Optical isomers are non-superimposable mirror images of each other, which means they have a chiral carbon atom and a plane of symmetry that cannot be overlapped when rotated or flipped. Isomer 1 (cis): For the cis isomer, there is a plane of symmetry that divides the complex into two symmetrical halves. Since this isomer has a plane of symmetry, it does not have optical isomers. Isomer 2 (trans): For the trans isomer, there is no plane of symmetry which means it could possibly have an optical isomer. However, since there are no chiral carbon centers in the ethylenediamine ligand or elsewhere, this complex does not have optical isomers either. In conclusion, there are two geometrical isomers of the complex \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+1}\), namely cis and trans, but neither of them have optical isomers.