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 complexes \([Ni(CO)_{4}]\) and \([PdCl_{4}]^{2-}\), we will follow a systematic approach for each complex.

### Step 1: Determine the oxidation state of Nickel in \([Ni(CO)_{4}]\)

- The ligand CO is neutral, so it does not contribute to the charge.
- The overall charge of the complex is 0.
- Therefore, the oxidation state of Ni is 0.

### Step 2: Determine the electronic configuration of Nickel

- Nickel (Ni) has an atomic number of 28.
- The ground state electronic configuration of Ni is \([Ar] 4s^2 3d^8\).

### Step 3: Analyze the effect of the ligand (CO)

- CO is a strong field ligand, which causes pairing of electrons.
- The 4s electrons will be vacant after pairing occurs in the 3d orbitals.

### Step 4: Determine the hybridization of \([Ni(CO)_{4}]\)

- With the pairing of electrons, the configuration can be represented as \(3d^{10}\).
- Since there are four ligands, the hybridization will be \(sp^3\).

### Step 5: Determine the geometry of \([Ni(CO)_{4}]\)

- The \(sp^3\) hybridization leads to a tetrahedral geometry.

### Step 6: Determine the oxidation state of Palladium in \([PdCl_{4}]^{2-}\)

- Each Cl ligand has a charge of -1, and there are four Cl ligands.
- Therefore, the total contribution from Cl is -4.
- Let the oxidation state of Pd be \(x\). Thus, \(x - 4 = -2\).
- Solving for \(x\), we find \(x = +2\).

### Step 7: Determine the electronic configuration of Palladium

- Palladium (Pd) has an atomic number of 46.
- The ground state electronic configuration of Pd is \([Kr] 4d^{10}\).
- For the oxidation state of +2, we remove two electrons, typically from the 4d subshell, resulting in \(4d^8\).

### Step 8: Analyze the effect of the ligand (Cl)

- Chloride is a weak field ligand, which does not cause significant pairing.
- The \(4d^8\) configuration remains as is.

### Step 9: Determine the hybridization of \([PdCl_{4}]^{2-}\)

- With four ligands, the hybridization will be \(dsp^2\).

### Step 10: Determine the geometry of \([PdCl_{4}]^{2-}\)

- The \(dsp^2\) hybridization leads to a square planar geometry.

### Final Answer

- The geometry of \([Ni(CO)_{4}]\) is **tetrahedral**.
- The geometry of \([PdCl_{4}]^{2-}\) is **square planar**.

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