How many maximum atoms are present in one plane of `[PtCl_(3)(n^(2) C_(2)H_(4))]^(-)` ?

Answer

To determine the maximum number of atoms present in one plane of the coordination complex \([PtCl_3(η^2C_2H_4)]^-\), we can break down the structure of the complex step by step.

### Step-by-Step Solution:

1. **Identify the Central Metal Atom**:
The central metal atom in the complex is platinum (Pt).

2. **Identify the Ligands**:
The complex contains three chloride ions (Cl) and one ethylene (C2H4) ligand, which is coordinated in an eta-squared (η²) manner.

3. **Understand the Coordination Geometry**:
In coordination complexes, the arrangement of ligands around the central metal atom can be visualized in a plane. For this complex, we have:
- Three chloride ligands (Cl)
- The ethylene ligand (C2H4) which contributes two carbon atoms (C) and four hydrogen atoms (H).

4. **Visualizing the Structure**:
- The three chloride ions will be arranged around the platinum.
- One chloride will be in the plane, while the other two will be positioned above and below the plane.
- The ethylene ligand will also be in the plane, with both carbon atoms contributing to the planar structure.

5. **Counting Atoms in the Plane**:
- In the plane, we have:
- 1 Chlorine atom from the chloride ligands.
- 2 Carbon atoms from the ethylene ligand.
- 2 Hydrogen atoms (one from each carbon) that are also in the plane.
- Therefore, the total number of atoms in the same plane is:
- 1 Cl + 2 C + 2 H = 5 atoms.

6. **Conclusion**:
The maximum number of atoms present in one plane of the complex \([PtCl_3(η^2C_2H_4)]^-\) is **5**.

### Final Answer:
The maximum number of atoms present in one plane of \([PtCl_3(η^2C_2H_4)]^-\) is **5**.