Which of the following compound has tetrahedral geometry?

Options

A. `[Ni(CN)_(4)]^(2-)`
B. `[Pd(CN)_(4)]^(2-)`
C. `[PdCl_(4)]^(2-)`
D. `[NiCl_(4)]^(2-)`

Answer

To determine which of the given compounds has a tetrahedral geometry, we can analyze the coordination compounds based on their ligands and the oxidation states of the central metal ions. Here's a step-by-step solution:

### Step 1: Identify the Compounds
We need to evaluate the coordination compounds mentioned in the question. The compounds we will analyze are:
1. Nickel(CN)₄²⁻
2. Palladium(CN)₄²⁻
3. Palladium(Cl)₂⁻
4. Nickel(Cl)₂⁻

### Step 2: Determine the Oxidation State of the Metal
- For Nickel(CN)₄²⁻, the oxidation state of nickel is +2.
- For Palladium(CN)₄²⁻, the oxidation state of palladium is also +2.
- For Palladium(Cl)₂⁻, the oxidation state of palladium is +2.
- For Nickel(Cl)₂⁻, the oxidation state of nickel is +2.

### Step 3: Identify the Nature of the Ligands
- CN⁻ (cyanide) is a strong field ligand.
- Cl⁻ (chloride) is a weak field ligand.

### Step 4: Determine the Electron Configuration of the Metal Ions
- Nickel in +2 oxidation state (Ni²⁺) has the configuration [Ar] 3d⁸.
- Palladium in +2 oxidation state (Pd²⁺) has the configuration [Kr] 4d⁸.

### Step 5: Analyze Hybridization and Geometry
- **Nickel(CN)₄²⁻**: The strong field ligand CN⁻ will lead to pairing of electrons in the 3d orbitals. The hybridization is dsp², which corresponds to square planar geometry.

- **Palladium(CN)₄²⁻**: Similar to nickel, the strong field ligand CN⁻ will lead to pairing of electrons in the 4d orbitals. The hybridization is also dsp², leading to square planar geometry.

- **Palladium(Cl)₂⁻**: The weak field ligand Cl⁻ will not cause pairing of electrons. The hybridization is likely to be sp³, which corresponds to tetrahedral geometry.

- **Nickel(Cl)₂⁻**: The weak field ligand Cl⁻ will also not cause pairing of electrons. The hybridization is likely to be sp³, which corresponds to tetrahedral geometry.

### Step 6: Conclusion
From the analysis, both Nickel(Cl)₂⁻ and Palladium(Cl)₂⁻ can exhibit tetrahedral geometry due to the weak field nature of the chloride ligands. However, since the question asks for which compound has tetrahedral geometry, we can conclude that:

**Answer: Nickel(Cl)₂⁻ has tetrahedral geometry.**