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. `[Ni(Cl)_(4)]^(2-)`tetrahedral, paramagnetic

Explanation

To determine which statement is incorrect regarding the coordination compound Ni(CO)4, we will analyze the oxidation state, electronic configuration, hybridization, geometry, and magnetic nature of the compound step by step.

### Step-by-Step Solution:

1. **Identify the Coordination Compound**:
The compound in question is Ni(CO)4.

2. **Calculate the Oxidation State of Nickel (Ni)**:
- Carbonyl (CO) is a neutral ligand, contributing 0 charge.
- Therefore, the oxidation state of Ni in Ni(CO)4 is 0.

3. **Determine the Electronic Configuration of Nickel**:
- Nickel has an atomic number of 28.
- The ground state electronic configuration is:
\[ \text{Ni: } [\text{Ar}] 4s^2 3d^8 \]

4. **Draw the Orbital Diagram**:
- The 3d subshell has 8 electrons, and the 4s subshell has 2 electrons:
- 3d: ↑↓ ↑↓ ↑↓ ↑↓ ↑
- 4s: ↑↓

5. **Analyze the Ligand**:
- Carbon monoxide (CO) is a strong field ligand, which causes pairing of electrons in the d-orbitals.
- The 3d orbitals will rearrange to accommodate the ligands:
- After pairing, the configuration becomes:
\[ 3d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ \]
- The 4s and 4p orbitals remain empty.

6. **Determine the Hybridization**:
- With 4 ligands (CO) bonded to Ni, we need 4 hybrid orbitals.
- The hybridization will be \( sp^3 \).

7. **Determine the Geometry**:
- The geometry corresponding to \( sp^3 \) hybridization is tetrahedral.

8. **Determine the Magnetic Nature**:
- Since all the electrons in the d-orbitals are paired, Ni(CO)4 is diamagnetic.

9. **Evaluate the Statements**:
- The question provides a statement that Ni(CO)4 is tetrahedral and paramagnetic.
- However, we have established that Ni(CO)4 is tetrahedral but **diamagnetic**, not paramagnetic.

### Conclusion:
The incorrect statement is that Ni(CO)4 is tetrahedral and paramagnetic. The correct nature is tetrahedral and diamagnetic.