Question

Sketch all the possible stereoisomers of (a) tetrahedral $\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right],(\mathbf{b})\( square-planar \)\left[\operatorname{Ir} \mathrm{Cl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-},$ (c) octahedral $\left[\mathrm{Fe}(\sigma \text { -phen })_{2} \mathrm{Cl}_{2}\right]^{+}$

Short answer

In summary, the possible stereoisomers for the given metal complexes are as follows: a) Tetrahedral complex: Only one stereoisomer (no need for cis/trans consideration due to different ligands) b) Square-planar complex: Two stereoisomers - cis and trans configurations c) Octahedral complex: Two stereoisomers - cis and trans configurations

Step-by-step solution

a) Tetrahedral Complex

For the tetrahedral complex \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\mathrm{Cl}_{2}\right]\), we have a central atom (Cd) and four ligands around it (two water molecules and two chloride ions). In a tetrahedral geometry, all the bond angles are approximately 109.5°. Since all the ligands are different, there is no need to consider any stereoisomers in this case. Therefore, the only possible stereoisomer of this complex is the given structure itself.

b) Square-planar Complex

For the square-planar complex \(\left[\operatorname{Ir}\mathrm{Cl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-}\), we have a central atom (Ir) and four ligands around it (two chloride ions and two triphenylphosphine molecules). In a square-planar geometry, all the bond angles are 90°, and the ligands occupy positions around the central atom in a square configuration. In this case, there are two possible arrangements for the ligands: 1. Cis - the two chloride ions and the two triphenylphosphine molecules are adjacent (next) to each other, forming a cis configuration 2. Trans - the two chloride ions and the two triphenylphosphine molecules are opposite to each other, forming a trans configuration Therefore, the two possible stereoisomers of this complex are the cis and trans forms.

c) Octahedral Complex

For the octahedral complex \(\left[\mathrm{Fe}(\sigma \text { -phen })_{2}\mathrm{Cl}_{2}\right]^{+}\), we have a central atom (Fe) and six ligands around it (two sigma-bonded phen ligands and two chloride ions). In an octahedral geometry, the bond angles are 90°, and the ligands are arranged in a regular octahedron around the central atom. In this case, since the sigma-phen ligands are bidentate, each of them can bind to the iron center in two different ways – via either ring of the bidentate ligand. The possible stereoisomers can be determined by the arrangement of the bidentate ligands and chloride ions in the coordination sphere. There are two possible arrangements for the ligands: 1. Cis - the two sigma-bonded phen ligands are bound to the iron center via the same face, and the two chloride ions are adjacent to each other 2. Trans - the two sigma-bonded phen ligands are bound to the iron center via opposite faces, and the two chloride ions are opposite to each other Therefore, the two possible stereoisomers of this complex are the cis and trans forms.