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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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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 ```
03

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 ```
04

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)

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Most popular questions from this chapter

Suppose that a transition-metal ion was in a lattice in which it was in contact with just two nearby anions, located on opposite sides of the metal. Diagram the splitting of the metal \(\bar{d}\) orbitals that would result from such a crystal field. Assuming a strong field, how many unpaired electrons would you expect for a metal ion with six \(d\) electrons? (Hint: Consider the linear axis to be the \(z\) -axis)

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Indicate the coordination number of the metal and the oxidation number of the metal as well as the number and type of each donor atom of the ligands for each of the following complexes: (a) \(\mathrm{Na}_{2}\left[\mathrm{CdCl}_{4}\right]\) (b) \(\mathrm{K}_{2}\left[\mathrm{MoOCl}_{4}\right]\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (d) \(\left[\mathrm{Ni}(\mathrm{CN})_{5}\right]^{3-}\) (e) \(\mathrm{K}_{3}\left[\mathrm{~V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) (f) \(\left[\mathrm{Zn}(\mathrm{en})_{2}\right] \mathrm{Br}_{2}\)

The +2 oxidation state is common for almost all the transition metals. Suggest an explanation.

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