Which statement is incorrect -

Options

A. `Ni(CO_(4))`- Tetrahedral, paramagnetic
B. `[Ni(CN)_(4)]^(2-)` - Square planar, diamagnetic
C. `Ni(CO)_(4)-` Tetrahedral, diamagnetic
D. `[NiCl_(4)]^(-2)`- Tetrahedral, paramagnetic

Answer

To solve the question, we need to analyze the given complexes and determine the correct statements regarding their hybridization and magnetic properties. Let's break this down step by step.

### Step 1: Analyze the first complex (Ni(CO)₄)
1. **Identify the oxidation state of Nickel (Ni)**: In Ni(CO)₄, Nickel is in the zero oxidation state.
2. **Determine the electronic configuration of Nickel**: The electronic configuration of Ni in the zero oxidation state is Argon 3d⁸ 4s².
3. **Hybridization**: CO is a strong field ligand, which means it will cause pairing of electrons. The 3d orbitals will fill as follows:
- 3d: ↑↓ ↑↓ ↑↓ ↑↓ (4 pairs)
- 4s: ↑↓ (2 electrons)
- 4p: (empty)
4. **Shape and magnetic property**: Since all electrons are paired, Ni(CO)₄ is tetrahedral and **diamagnetic**.

### Step 2: Analyze the second complex (Ni(CN)₄)²⁻
1. **Identify the oxidation state of Nickel (Ni)**: In Ni(CN)₄²⁻, Nickel is in the +2 oxidation state.
2. **Determine the electronic configuration of Nickel**: The electronic configuration of Ni in the +2 oxidation state is 3d⁸ 4s⁰.
3. **Hybridization**: CN is also a strong field ligand, leading to pairing of electrons:
- 3d: ↑↓ ↑↓ ↑↓ ↑↓ (4 pairs)
- 4s: (empty)
- 4p: (empty)
4. **Shape and magnetic property**: The hybridization is dsp², leading to a square planar shape. Since all electrons are paired, Ni(CN)₄²⁻ is **diamagnetic**.

### Step 3: Analyze the third complex (Ni(Cl)₄)²⁻
1. **Identify the oxidation state of Nickel (Ni)**: In Ni(Cl)₄²⁻, Nickel is also in the +2 oxidation state.
2. **Determine the electronic configuration of Nickel**: The electronic configuration of Ni in the +2 oxidation state is 3d⁸ 4s⁰.
3. **Hybridization**: Cl is a weak field ligand, which does not cause pairing:
- 3d: ↑ ↑ ↑ ↑ (4 unpaired electrons)
- 4s: (empty)
- 4p: (empty)
4. **Shape and magnetic property**: The hybridization is sp³, leading to a tetrahedral shape. Since there are unpaired electrons, Ni(Cl)₄²⁻ is **paramagnetic**.

### Step 4: Evaluate the statements
1. **Option 1**: Ni(CO)₄ is tetrahedral and paramagnetic. **(Incorrect, it is tetrahedral and diamagnetic)**
2. **Option 2**: Ni(CN)₄²⁻ is square planar and diamagnetic. **(Correct)**
3. **Option 3**: Ni(Cl)₄²⁻ is tetrahedral and paramagnetic. **(Correct)**
4. **Option 4**: Ni(Cl)₄²⁻ is tetrahedral and paramagnetic. **(Correct)**

### Conclusion
The incorrect statement is **Option 1**: Ni(CO)₄ is tetrahedral and paramagnetic.