Draw geometrical isomers of each of the following complex ions. a. \(\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{-}\) c. \(\left[\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]\) b. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]^{2+}\) d. \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{I}_{2}\right]^{+}\)

For each complex ion, we should first determine the coordination geometry of the metal ion, this will be based on the coordination number, which is the number of ligands bonded to the metal ion. a. In \(\left[\mathrm{Co}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2}\mathrm{O}\right)_{2}\right]^{-}\), the coordination number of Co is 6 (2 oxalate ions and 2 water molecules). This means the geometry is octahedral. b. In \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{I}_{2}\right]^{2+}\), the coordination number of Pt is also 6 (4 ammonia molecules and 2 iodide ions). The geometry is also octahedral. c. In \(\left[\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{3}\mathrm{Cl}_{3}\right]\), the coordination number of Ir is 6 (3 ammonia molecules and 3 chloride ions). The geometry is octahedral. d. In \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\mathrm{I}_{2}\right]^{+}\), the coordination number of Cr is 6 (2 ethylenediamine ligands, which are bidentate, 2 ammonia molecules and 2 iodide ions). The geometry is also octahedral.

In octahedral complexes, we can have two types of geometrical isomers: trans and cis. Cis: The ligands are adjacent to each other. Trans: The ligands are opposite to each other.

For each complex ion, we will draw the trans and cis geometrical isomers of the ligands. a. For \(\left[\mathrm{Co}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2}\mathrm{O}\right)_{2}\right]^{-}\), only 1 isomer is possible where both water molecules are adjacent and in cis positions. b. For \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{I}_{2}\right]^{2+}\), we have 2 geometrical isomers: - Cis: 2 ammonia molecules are adjacent to one iodide ion, making it cis. Remaining two ammonia molecules form a cis pair on the opposite side of the iodide ions. - Trans: All ammonia molecules form pairs on opposite sides of the iodides, making it trans. c. For \(\left[\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{3}\mathrm{Cl}_{3}\right]\), we have 2 geometrical isomers: - Cis: two ammonia molecules adjacent to one chloride ion, making it cis. Remaining ammonia molecule forms a cis pair on an opposite side of chloride ions. - Trans: All ammonia molecules form pairs on opposite sides of the chloride ions, making it trans. d. For \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\mathrm{I}_{2}\right]^{+}\), we have 2 geometrical isomers: - Cis: both ammonia molecules are adjacent to the same ethylenediamine ligand, making it cis. - Trans: ammonia molecules are opposite and adjacent to different ethylenediamine ligands, making it trans. In conclusion, we have drawn the geometrical isomers for each complex ion, identifying cis and trans isomers.