The geometry of `Ni(CO)_4` and `Ni(PPh_3)Cl_2` are

Options

A. both square planar
B. tetrahedral and square planar respectively
C. both tetrahedral
D. square planar and tetrahedral respectively.

Explanation

To determine the geometry of the complexes Ni(CO)₄ and Ni(PPh₃)Cl₂, we can follow these steps:

### Step 1: Analyze Ni(CO)₄
1. **Identify the oxidation state of Ni**: In Ni(CO)₄, the oxidation state of Ni is 0 because CO is a neutral ligand.
2. **Determine the electron configuration of Ni**: Nickel has the electron configuration of [Ar] 3d⁸ 4s².
3. **Hybridization**: Since CO is a strong field ligand, it causes the 4s electrons to be promoted to the 3d orbital, resulting in the configuration 3d¹⁰. The hybridization for four ligands (4 CO) is sp³.
4. **Geometry**: The geometry corresponding to sp³ hybridization is tetrahedral.

### Step 2: Analyze Ni(PPh₃)Cl₂
1. **Identify the oxidation state of Ni**: In Ni(PPh₃)Cl₂, let the oxidation state of Ni be X. The oxidation state can be calculated as follows:
\[ X + 2(-1) = 0 \quad \Rightarrow \quad X = +2 \]
2. **Determine the electron configuration of Ni in +2 state**: The electron configuration for Ni in the +2 state is [Ar] 3d⁸ (the 4s electrons are removed).
3. **Hybridization**: With two PPh₃ (which are strong field ligands) and two Cl⁻ (which are weak field ligands), the hybridization is also sp³.
4. **Geometry**: The geometry corresponding to sp³ hybridization is tetrahedral.

### Final Answer
Both Ni(CO)₄ and Ni(PPh₃)Cl₂ have a tetrahedral geometry.

### Summary
- **Ni(CO)₄**: Tetrahedral geometry (sp³ hybridization)
- **Ni(PPh₃)Cl₂**: Tetrahedral geometry (sp³ hybridization)