Question

The compounds \( [\text{PtCl}_2(\text{NH}_3)_4]\text{Br}_2 \) and \( [\text{PtBr}_2(\text{NH}_3)_4]\text{Cl}_2 \) constitute a pair of:

• A. \( \text{Coordination isomers} \)
• B. \( \text{Linkage isomers} \)
• C. \( \text{Ionization isomers} \)
• D. \( \text{Optical isomers} \)

Hint:

To identify ionization isomers, compare the ions present in the outer ionic sphere and the coordination sphere. These isomers dissociate into different ions when dissolved in solution, making their identification straightforward.

Answer 1

The Correct Option is C

Step 1: Understanding the structures of the compounds.
The given compounds, \( [\text{PtCl}_2(\text{NH}_3)_4]\text{Br}_2 \) and \( [\text{PtBr}_2(\text{NH}_3)_4]\text{Cl}_2 \), are coordination complexes containing \( \text{Pt(NH}_3\text{)}_4 \) as the central unit. The primary difference lies in the arrangement of \( \text{Cl}^- \) and \( \text{Br}^- \) ions between the inner coordination sphere and the outer ionic sphere. 

Step 2: Analyzing ionization isomerism. Ionization isomers are formed when the ions in the coordination sphere and the ionic sphere exchange positions. These isomers have the same molecular formula but release different ions in solution. For example: \[ [\text{PtCl}_2(\text{NH}_3)_4]\text{Br}_2 \xrightarrow[]{\text{Dissociation}} 2\text{Br}^- + [\text{PtCl}_2(\text{NH}_3)_4] \] \[ [\text{PtBr}_2(\text{NH}_3)_4]\text{Cl}_2 \xrightarrow[]{\text{Dissociation}} 2\text{Cl}^- + [\text{PtBr}_2(\text{NH}_3)_4] \] Here: - In \( [\text{PtCl}_2(\text{NH}_3)_4]\text{Br}_2 \), the bromide ions are in the ionic sphere, and the chloride ions are directly bonded to the platinum center. - In \( [\text{PtBr}_2(\text{NH}_3)_4]\text{Cl}_2 \), the chloride ions occupy the ionic sphere, while bromide ions are in the coordination sphere. 

Step 3: Distinguishing from other isomerism types.
- Coordination isomerism: This occurs when ligands are exchanged between two different metal centers, which is not relevant here as there is only one metal center. - Linkage isomerism: This arises when a ligand can bind through different donor atoms, such as \( \text{NO}_2^- \) through \( \text{N} \) or \( \text{O} \). This is not applicable here as the ligands involved do not show linkage isomerism. - Optical isomerism: This requires non-superimposable mirror images, which is not observed in these square-planar complexes. Thus, the two given compounds exhibit ionization isomerism. 

Final Answer: \[ \boxed{\text{Ionization isomers}} \]