Which of the following has a square planar geometry? .

Options

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

Answer

To determine which of the given coordination compounds has a square planar geometry, we will analyze each option step by step.

### Step 1: Analyze PtCl4²⁻
1. **Identify the central atom**: Platinum (Pt).
2. **Determine oxidation state**: For PtCl4²⁻, the oxidation state of Pt can be calculated as follows:
\[ x - 1 \times 4 = -2 \implies x - 4 = -2 \implies x = +2 \]
Thus, Pt is in the +2 oxidation state.
3. **Electronic configuration**: The electronic configuration of Pt (atomic number 78) is:
\[ [Xe] 4f^{14} 5d^9 6s^1 \]
In the +2 state, it becomes:
\[ [Xe] 4f^{14} 5d^8 \]
4. **Hybridization**: The presence of strong field ligands (Cl⁻) causes pairing of electrons. The hybridization is dsp².
5. **Geometry**: dsp² hybridization corresponds to square planar geometry.

### Step 2: Analyze CoCl4²⁻
1. **Identify the central atom**: Cobalt (Co).
2. **Determine oxidation state**: For CoCl4²⁻, the oxidation state of Co is:
\[ x - 1 \times 4 = -2 \implies x - 4 = -2 \implies x = +2 \]
3. **Electronic configuration**: The electronic configuration of Co (atomic number 27) is:
\[ [Ar] 4s^2 3d^7 \]
In the +2 state, it becomes:
\[ [Ar] 4s^0 3d^7 \]
4. **Hybridization**: Cl⁻ is a weak field ligand, so no pairing occurs. The hybridization is sp³.
5. **Geometry**: sp³ hybridization corresponds to tetrahedral geometry.

### Step 3: Analyze FeCl4⁻
1. **Identify the central atom**: Iron (Fe).
2. **Determine oxidation state**: For FeCl4⁻, the oxidation state of Fe is:
\[ x - 1 \times 4 = -1 \implies x - 4 = -1 \implies x = +3 \]
3. **Electronic configuration**: The electronic configuration of Fe (atomic number 26) is:
\[ [Ar] 4s^2 3d^6 \]
In the +3 state, it becomes:
\[ [Ar] 4s^0 3d^5 \]
4. **Hybridization**: Cl⁻ is a weak field ligand, so no pairing occurs. The hybridization is sp³.
5. **Geometry**: sp³ hybridization corresponds to tetrahedral geometry.

### Step 4: Analyze Ni(CN)4²⁻
1. **Identify the central atom**: Nickel (Ni).
2. **Determine oxidation state**: For Ni(CN)4²⁻, the oxidation state of Ni is:
\[ x - 1 \times 4 = -2 \implies x - 4 = -2 \implies x = +2 \]
3. **Electronic configuration**: The electronic configuration of Ni (atomic number 28) is:
\[ [Ar] 4s^2 3d^8 \]
In the +2 state, it becomes:
\[ [Ar] 4s^0 3d^8 \]
4. **Hybridization**: CN⁻ is a strong field ligand, so pairing occurs. The hybridization is dsp².
5. **Geometry**: dsp² hybridization corresponds to square planar geometry.

### Conclusion
The compounds with square planar geometry are:
- PtCl4²⁻
- Ni(CN)4²⁻

### Final Answer
The compounds that have square planar geometry are **PtCl4²⁻ and Ni(CN)4²⁻**.

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