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 each statement regarding the nickel complexes and determine which one is incorrect.

### Step-by-Step Solution:

1. **Analyze Ni(CO)₄**:
- Nickel (Ni) has an electronic configuration of [Ar] 4s² 3d⁸.
- In Ni(CO)₄, CO is a strong field ligand and causes pairing of electrons.
- The hybridization is sp³, leading to a tetrahedral geometry.
- Since all electrons are paired, Ni(CO)₄ is diamagnetic, not paramagnetic.
- **Conclusion**: The statement "Ni(CO)₄ tetrahedral paramagnetic" is **incorrect**.

2. **Analyze Ni(CN)₄²⁻**:
- In this case, nickel is in the +2 oxidation state, so its electronic configuration is [Ar] 3d⁸.
- CN⁻ is a strong field ligand, which will cause pairing of electrons.
- The hybridization is dsp², leading to a square planar geometry.
- Since all electrons are paired, Ni(CN)₄²⁻ is diamagnetic.
- **Conclusion**: The statement "Ni(CN)₄²⁻ square planar diamagnetic" is **correct**.

3. **Analyze Ni(CO)₄ again**:
- As previously discussed, Ni(CO)₄ is tetrahedral and diamagnetic.
- **Conclusion**: The statement "Ni(CO)₄ tetrahedral diamagnetic" is **correct**.

4. **Analyze NiCl₄²⁻**:
- Here, nickel is again in the +2 oxidation state, with an electronic configuration of [Ar] 3d⁸.
- Cl⁻ is a weak field ligand, leading to no pairing of electrons.
- The hybridization is sp³, resulting in a tetrahedral geometry.
- There are unpaired electrons, making NiCl₄²⁻ paramagnetic.
- **Conclusion**: The statement "NiCl₄²⁻ tetrahedral paramagnetic" is **correct**.

### Final Answer:
The incorrect statement is: **Ni(CO)₄ tetrahedral paramagnetic**.