Which of the following compounds has square planar geometry ?

Options

A. `[Ni(CN)_4]^(2-)`
B. `[NiCl_4]^(2-)`
C. `[Ni(CO)_4]`
D. Both (1) and (3)

Answer

To determine which of the given compounds has a square planar geometry, we will analyze the coordination number, oxidation state, and the type of ligands involved in each compound.

### Step-by-Step Solution:

1. **Identify the Coordination Compound**:
We need to analyze the compound provided in the question, which is Nickel (Ni) complexed with cyanide ions (CN⁻). The specific compound we are considering is Ni(CN)₄²⁻.

2. **Determine the Oxidation State of Nickel**:
In the complex Ni(CN)₄²⁻, the cyanide ion (CN⁻) has a charge of -1. Since there are four cyanide ligands, the total negative charge contributed by the ligands is -4. To balance this with the overall charge of the complex (-2), the oxidation state of nickel must be +2.
\[ \text{Oxidation state of Ni} = +2 \]

3. **Electron Configuration of Nickel**:
The atomic number of nickel is 28. The electron configuration of neutral nickel is:
\[ \text{Ni: } [Ar] 3d^8 4s^2 \]
For Ni²⁺, we remove two electrons from the 4s orbital:
\[ \text{Ni}^{2+}: [Ar] 3d^8 \]

4. **Identify the Type of Ligands**:
Cyanide (CN⁻) is a strong field ligand. Strong field ligands cause pairing of electrons in the d-orbitals.

5. **Hybridization**:
With Ni²⁺ having the electron configuration of 3d⁸, and considering that CN⁻ is a strong field ligand, the 3d electrons will pair up. The hybridization can be determined as follows:
- The 3d orbitals will pair up, and we will use one d orbital, two p orbitals, and one s orbital for hybridization.
- This results in DSP² hybridization.

6. **Geometry**:
The DSP² hybridization leads to a square planar geometry. Therefore, Ni(CN)₄²⁻ has a square planar geometry.

7. **Conclusion**:
Based on the analysis, the compound that exhibits square planar geometry is Ni(CN)₄²⁻.

### Final Answer:
The compound that has square planar geometry is **Ni(CN)₄²⁻**.