Which one of the following has a square planar geometry ?

Options

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

Explanation

To determine which of the given complexes has a square planar geometry, we need to analyze each option based on their oxidation states, electronic configurations, and the type of ligands involved. Square planar geometry typically arises from a dsp² hybridization.

### Step-by-Step Solution:

1. **Identify the Complexes**: We have four complexes to analyze:
- (a) CoCl₄²⁻
- (b) FeCl₄²⁻
- (c) NiCl₄²⁻
- (d) PtCl₄²⁻

2. **Determine the Oxidation State**:
- For CoCl₄²⁻: Let x be the oxidation state of Co. The equation is x + 4(-1) = -2, leading to x = +2.
- For FeCl₄²⁻: Let x be the oxidation state of Fe. The equation is x + 4(-1) = -2, leading to x = +2.
- For NiCl₄²⁻: Let x be the oxidation state of Ni. The equation is x + 4(-1) = -2, leading to x = +2.
- For PtCl₄²⁻: Let x be the oxidation state of Pt. The equation is x + 4(-1) = -2, leading to x = +2.

3. **Write the Electronic Configurations**:
- Co²⁺: Argon [Ar] 3d⁷
- Fe²⁺: Argon [Ar] 3d⁶
- Ni²⁺: Argon [Ar] 3d⁸
- Pt²⁺: Xenon [Xe] 4f¹⁴ 5d⁸

4. **Analyze the Ligands**:
- Cl⁻ is a weak field ligand, which means it does not cause pairing of electrons in the d-orbitals for Co²⁺, Fe²⁺, and Ni²⁺.

5. **Determine Hybridization**:
- **CoCl₄²⁻**: 3d⁷ does not pair, leading to sp³ hybridization (tetrahedral).
- **FeCl₄²⁻**: 3d⁶ does not pair, leading to sp³ hybridization (tetrahedral).
- **NiCl₄²⁻**: 3d⁸ does not pair, leading to sp³ hybridization (tetrahedral).
- **PtCl₄²⁻**: 5d⁸ undergoes pairing due to the strong electronic repulsion and results in dsp² hybridization, leading to square planar geometry.

6. **Conclusion**: The complex that has a square planar geometry is **PtCl₄²⁻**.

### Final Answer:
**The complex with square planar geometry is PtCl₄²⁻.**