The geometry of `[Ni(CO)_(4)]` and `[PdCl_(4)]^(2-)`respectively

Options

A. Both are tetrahedral
B. Both are square planar
C. Square planar and tetrahedral
D. Tetrahedral and square planar

Answer

To determine the geometry of the coordination compounds `[Ni(CO)_(4)]` and `[PdCl_(4)]^(2-)`, we will follow these steps:

### Step 1: Analyze the compound `[Ni(CO)_(4)]`
1. **Identify the oxidation state of Nickel (Ni)**:
- Carbon monoxide (CO) is a neutral ligand, which means it does not contribute any charge. Therefore, the oxidation state of Ni in `[Ni(CO)_(4)]` is 0.

2. **Determine the electronic configuration of Ni**:
- The atomic number of Ni is 28. Its electronic configuration is:
\[ [Ar] 3d^{8} 4s^{2} \]
- In the case of `[Ni(CO)_(4)]`, Ni is in the zero oxidation state, so it retains its full configuration.

3. **Determine the hybridization**:
- Since there are 4 CO ligands, which are strong field ligands, they will cause pairing of electrons in the 3d orbitals. The hybridization can be determined as follows:
- 4 ligands require 4 orbitals for bonding, which leads to sp³ hybridization.

4. **Determine the geometry**:
- The geometry corresponding to sp³ hybridization is tetrahedral.

### Step 2: Analyze the compound `[PdCl_(4)]^(2-)`
1. **Identify the oxidation state of Palladium (Pd)**:
- Chloride (Cl) is a negatively charged ligand (-1). In `[PdCl_(4)]^(2-)`, the total charge contributed by 4 Cl ligands is -4. Therefore, the oxidation state of Pd can be calculated as:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]
- Thus, Pd is in the +2 oxidation state.

2. **Determine the electronic configuration of Pd**:
- The atomic number of Pd is 46. Its electronic configuration is:
\[ [Kr] 4d^{10} 5s^{2} \]
- For Pd in the +2 oxidation state, it loses 2 electrons, which are from the 5s orbital:
\[ [Kr] 4d^{8} \]

3. **Determine the hybridization**:
- With 4 chloride ligands, the hybridization can be determined as follows:
- 4 ligands require 4 orbitals for bonding, which leads to dsp² hybridization (since Pd is a transition metal).

4. **Determine the geometry**:
- The geometry corresponding to dsp² hybridization is square planar.

### Final Answer:
- The geometry of `[Ni(CO)_(4)]` is **tetrahedral**.
- The geometry of `[PdCl_(4)]^(2-)` is **square planar**.

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