Problem 88 - Chemistry The Central Science

Question

CC(=O)[O-] can act a… # Many trace metal ions exist in the blood complexed with amino acids or small peptides. The anion of the amino acid glycine (gly), N#CC(=O)[O-] can act as a bidentate ligand, coordinating to the metal through nitrogen and oxygen atoms. How many isomers are possible for (a) \(\left[\mathrm{Zn}(\mathrm{gly})_{2}\right]\) (tetrahedral), (b) \(\left[\mathrm{Pt}(\mathrm{gly})_{2}\right]\) (square planar), (c) \(\left[\mathrm{Co}(\mathrm{gly})_{3}\right]\) (octahedral)? Sketch all possible isomers. Use the symbol \(\mathrm{N}\) O to represent the ligand.

Step-by-Step Solution

Answer

For the given metal complexes with glycine as the ligand, there is only one possible isomer for each: (a) Zn(gly)\(_2\): 1 isomer (tetrahedral) (b) Pt(gly)\(_2\): 1 isomer (square planar) (c) Co(gly)\(_3\): 1 isomer (octahedral)

Step 1: Determine the coordination geometries of the complexes

(a) Zn(gly)\(_2\): Coordination geometry is tetrahedral with a central metal ion and two bidentate glycine ligands. (b) Pt(gly)\(_2\): Coordination geometry is square planar with a central metal ion and two bidentate glycine ligands. (c) Co(gly)\(_3\): Coordination geometry is octahedral with a central metal ion and three bidentate glycine ligands.

Step 2: Find the isomers for the tetrahedral complex Zn(gly)\(_2\)

Since the coordination geometry is tetrahedral and we have two bidentate ligands, only one isomer is possible. The metal binds to the oxygen atom from the carboxylate group and the nitrogen atom from the amino group in each ligand. Sketch: ``` O \ O-Zn-N / \ (NH3)C \ || \ C N // (NH3)C N==C /\. \ O O ```

Step 3: Find the isomers for the square planar complex Pt(gly)\(_2\)

For the square planar complex, we have two bidentate ligands. We only have one possible geometrical isomer. Each ligand will bind to the metal through nitrogen and oxygen atoms. Sketch: ``` O \ O-Pt-N \ (NH3)C \ || \ C N // (NH3)C N==C /\. \ O O ```

Step 4: Find the isomers for the octahedral complex Co(gly)\(_3\)

For the octahedral complex, we have three bidentate ligands. Since each ligand will bind through nitrogen and oxygen atoms, there is only one possible geometrical isomer. Sketch: ``` O \ O-Co-N / \ (NH3)C / \ || / \ C / N //\/ (NH3)C N=\O \ / \.\ N-Co-O O O / \ N O (NH3)C / || / C / // / N==C /^\ \ O O ``` To summarize, we found that there is only one possible isomer for each of the complexes: - (a) Zn(gly)\(_2\): 1 isomer (tetrahedral) - (b) Pt(gly)\(_2\): 1 isomer (square planar) - (c) Co(gly)\(_3\): 1 isomer (octahedral)