Which one of the following has a square planar geometry?
`(Co=27, Ni=28, Fe=26, Pt=78)`

Options

A. `[NiCl_(4)]^(2-)`
B. `[PtCl_(4)]^(2-)`
C. `[CoCl_(4)]^(2-)`
D. `[FeCl_(4)]^(2-)`

Answer

To determine which of the given coordination compounds has a square planar geometry, we will analyze the oxidation states, electronic configurations, and hybridization of the central metal ions in each complex.

### Step-by-Step Solution:

1. **Identify the Complexes**: The question provides four metal ions: Co (Cobalt), Ni (Nickel), Fe (Iron), and Pt (Platinum). We need to analyze the complexes formed with Cl⁻ ligands.

2. **Determine the Oxidation States**:
- For each complex, the oxidation state of the metal can be calculated. Since all complexes are in the form of MCl₄²⁻, we can set up the equation:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]
- Thus, all metals (Co, Ni, Fe, Pt) are in the +2 oxidation state.

3. **Electronic Configuration**:
- **Cobalt (Co)**: Atomic number 27, electronic configuration: [Ar] 3d⁷ 4s². In +2 state: 3d⁷.
- **Nickel (Ni)**: Atomic number 28, electronic configuration: [Ar] 3d⁸ 4s². In +2 state: 3d⁸.
- **Iron (Fe)**: Atomic number 26, electronic configuration: [Ar] 3d⁶ 4s². In +2 state: 3d⁶.
- **Platinum (Pt)**: Atomic number 78, electronic configuration: [Xe] 4f¹⁴ 5d⁹ 6s². In +2 state: 4d⁸.

4. **Determine the Hybridization**:
- **Nickel (NiCl₄²⁻)**: With 3d⁸ and Cl⁻ being a weak field ligand, it forms a high-spin complex. The hybridization is sp³, leading to a tetrahedral geometry.
- **Platinum (PtCl₄²⁻)**: With 4d⁸ and Cl⁻ being a strong field ligand, it forms a low-spin complex. The hybridization is dsp², leading to a square planar geometry.
- **Cobalt (CoCl₄²⁻)**: With 3d⁷ and Cl⁻ being a weak field ligand, it forms a high-spin complex. The hybridization is sp³, leading to a tetrahedral geometry.
- **Iron (FeCl₄²⁻)**: With 3d⁶ and Cl⁻ being a weak field ligand, it forms a high-spin complex. The hybridization is sp³, leading to a tetrahedral geometry.

5. **Conclusion**: Among the complexes analyzed, only **PtCl₄²⁻** exhibits square planar geometry due to the strong field ligand causing electron pairing and resulting in dsp² hybridization.

### Final Answer:
The complex that has a square planar geometry is **PtCl₄²⁻**.

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