Which of the following complex is homoleptic ?

Options

A. `H_(2)[PtCl_(6)]`
B. `Li[AlH_(4)]`
C. `[Ni(CO)_(4)]`
D. All of these

Answer

To determine which of the following complexes is homoleptic, we first need to understand what a homoleptic complex is. A homoleptic complex is defined as a coordination compound in which all the ligands attached to the central metal atom are the same or identical.

Let's analyze the given options step by step:

1. **Understanding Homoleptic Complexes**:
- A homoleptic complex contains only one type of ligand.
- For example, if a complex has ligands like Cl, NH3, or CO, and all are the same, it is homoleptic.

2. **Analyzing the Given Complexes**:
- **H2PtCl6**: This complex can be represented as [PtCl6]²⁻. Here, all six ligands are chloride ions (Cl⁻). Since all ligands are identical, this is a homoleptic complex.
- **LiAlH4**: This complex dissociates to form AlH4⁻ and Li⁺. In AlH4⁻, there are four identical hydride ions (H⁻) attached to aluminum. Thus, this is also a homoleptic complex.
- **NiCl4²⁻**: In this complex, all four ligands are chloride ions (Cl⁻). Therefore, this is a homoleptic complex as well.
- **Ni(CO)4**: In this complex, all four ligands are carbon monoxide (CO) molecules. Since all ligands are identical, this is also a homoleptic complex.

3. **Conclusion**:
- All the complexes listed (H2PtCl6, LiAlH4, NiCl4²⁻, and Ni(CO)4) are homoleptic complexes because they each contain only one type of ligand.

Thus, the answer is that all of the given complexes are homoleptic.