The compound having a tetrahedral geometry is .

Options

A. `[Ni(CN_(4))]^(2-)`
B. `[Pd(CN)_(4)]^(2-)`
C. `[PdCI_(4)]^(2-)`
D. `[NiCI_(4)]^(2-)`

Explanation

To determine which compound has a tetrahedral geometry, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ligands**:
- We need to consider the ligands attached to the metal ion. In this case, we have two types of ligands mentioned: cyanide ion (CN⁻) which is a strong field ligand, and chloride ion (Cl⁻) which is a weak field ligand.

2. **Determine the Metal Ion and its Oxidation State**:
- The metal ion in question is Nickel (Ni). In the context provided, we are considering Ni²⁺, which has the electronic configuration of 3d⁸.

3. **Analyze the Hybridization**:
- For tetrahedral geometry, the hybridization is sp³. This occurs when a central atom forms four equivalent hybrid orbitals. In the case of Ni²⁺ with weak field ligands (like Cl⁻), the electrons do not pair up, allowing for the formation of four unpaired electrons.

4. **Count the Coordination Number**:
- A tetrahedral complex typically has a coordination number of 4. Since Cl⁻ is a weak field ligand, it can surround the Ni²⁺ ion in a tetrahedral arrangement.

5. **Conclusion**:
- Based on the analysis, the compound that exhibits tetrahedral geometry is Nickel(II) chloride (NiCl₂), which is formed by the combination of Ni²⁺ and four chloride ions.

### Final Answer:
The compound having a tetrahedral geometry is Nickel(II) chloride (NiCl₂).