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

The given complex, \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+}\), has a central chromium atom (Cr) which is bonded to an ethylenediamine (en) ligand, two ammonia (NH3) ligands, a bromine (Br) atom, and a chlorine (Cl) atom. The coordination sphere of the metal atom (Cr) contains six coordination sites, making it an octahedral complex.

Considering the octahedral arrangement, there are two possible geometrical isomers for this complex: 1. Cis-isomer: In this case, the two NH3 ligands and the Br and Cl ligands occupy adjacent coordination sites (90° apart) on the octahedral sphere. \[ \begin{matrix} \text{Cl} & - & \text{Cr} & - & \text{Br} \\ & \diagup & | & \diagdown& \\ \text{NH}_3 & & \text{en}& & \text{NH}_3 \end{matrix} \] 2. Trans-isomer: In this case, the two NH3 ligands are on opposite sides of the octahedral sphere (180° apart), and the Br and Cl ligands are also on opposite sides. \[ \begin{matrix} \text{NH}_3 & - & \text{Cr} & - & \text{Br} \\ & \diagup & | & \diagdown& \\ \text{NH}_3 & & \text{en}& & \text{Cl} \end{matrix} \]

Now, we need to determine whether the geometrical isomers also exhibit optical isomerism. Optical isomers are non-superimposable mirror images of each other, known as enantiomers. In order for a complex to have optical isomers, it must lack a center of symmetry or a plane of symmetry. For the cis-isomer, we can identify a plane of symmetry. The presence of a plane of symmetry indicates that this geometrical isomer does not exhibit optical isomerism. However, the trans-isomer does not have a center of symmetry or a plane of symmetry, which means that it exhibits optical isomerism. Below are the two enantiomers of the trans-isomer: Enantiomer 1: \[ \begin{matrix} \text{NH}_3 & - & \text{Cr} & - & \text{Br} \\ & \diagup & | & \diagdown& \\ \text{NH}_3 & & \text{en}^{S}& & \text{Cl} \end{matrix} \] Enantiomer 2: \[ \begin{matrix} \text{NH}_3 & - & \text{Cr} & - & \text{Br} \\ & \diagup & | & \diagdown& \\ \text{NH}_3 & & \text{en}^{R}& & \text{Cl} \end{matrix} \] In conclusion, the given complex, \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+}\), has two geometrical isomers, cis and trans. Only the trans-isomer exhibits optical isomerism, with two enantiomers: \( [\text{en}^{S}] \) and \( [\text{en}^{R}] \).