Question

Which of the following ligands are capable of linkage isomerism? Explain your answer. $$\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-}, \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{OCN}^{-}, \mathrm{I}^{-}$$

Short answer

The ligands capable of linkage isomerism are \(\mathrm{SCN}^{-}\) (through S or N), \(\mathrm{N}_{3}^{-}\) (through the terminal N), \(\mathrm{NO}_{2}^{-}\) (through N or O), and \(\mathrm{OCN}^{-}\) (through O or N).

Step-by-step solution

Ligand 1: \(\mathrm{SCN}^{-}\)

The ligand can connect to the metal through the sulfur (S) or nitrogen (N) atom.

Ligand 2: \(\mathrm{N}_{3}^{-}\)

The ligand can connect to the metal through the terminal nitrogen (N) atoms.

Ligand 3: \(\mathrm{NO}_{2}^{-}\)

The ligand can connect to the metal through the nitrogen (N) or one of the oxygen (O) atoms.

Ligand 4: \(\mathrm{NH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\)

This is an example of a bidentate ligand, with two nitrogen (N) atoms as potential binding sites but they will bind simultaneously to the metal center, so there is no linkage isomerism.

Ligand 5: \(\mathrm{OCN}^{-}\)

The ligand can connect to the metal through the oxygen (O) or nitrogen (N) atom.

Ligand 6: \(\mathrm{I}^{-}\)

This ligand has only one atom, iodine (I), which can bind to the metal center, so it cannot exhibit linkage isomerism. #Step 2: Determine which ligands can exhibit linkage isomerism#

Linkage isomerism-capable ligands

The ligands capable of linkage isomerism are: 1. \(\mathrm{SCN}^{-}\) (through S or N) 2. \(\mathrm{N}_{3}^{-}\) (through the terminal N) 3. \(\mathrm{NO}_{2}^{-}\) (through N or O) 4. \(\mathrm{OCN}^{-}\) (through O or N) Thus, the ligands \(\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{OCN}^{-}\) are capable of linkage isomerism.