The correct combination is :

Options

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

Answer

To solve the problem, we need to analyze the coordination compounds mentioned in the question. We will determine the oxidation states, hybridization, geometry, and magnetic properties of each compound.

### Step-by-Step Solution

1. **Identify the Compounds**:
We are given three coordination compounds: NiCl4^2-, Ni(CN)4^2-, and Na2[NiCl4].

2. **Determine the Oxidation State of Nickel in NiCl4^2-**:
- Let the oxidation state of Ni be \( x \).
- The oxidation state of Cl is -1.
- The overall charge of the complex is -2.
- The equation is:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]

3. **Determine the Hybridization and Geometry of NiCl4^2-**:
- Nickel in this state is Ni^2+ (3d^8).
- Chlorine is a weak field ligand and does not cause pairing of electrons.
- The hybridization is \( sp^3 \) (due to 4 ligands).
- The geometry is tetrahedral.

4. **Determine the Oxidation State of Nickel in Ni(CN)4^2-**:
- Let the oxidation state of Ni be \( x \).
- The oxidation state of CN is -1.
- The equation is:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]

5. **Determine the Hybridization and Geometry of Ni(CN)4^2-**:
- Nickel is again in the +2 state (3d^8).
- Cyanide (CN) is a strong field ligand and causes pairing of electrons.
- The hybridization is \( dsp^2 \) (due to 4 ligands).
- The geometry is square planar.

6. **Determine the Oxidation State of Nickel in Na2[NiCl4]**:
- The oxidation state of Ni is still +2 as determined previously.
- The overall charge of the complex is neutral.

7. **Determine the Hybridization and Geometry of Na2[NiCl4]**:
- The hybridization remains \( sp^3 \) (due to 4 ligands).
- The geometry is tetrahedral.

8. **Check for Paramagnetism or Diamagnetism**:
- NiCl4^2- has 2 unpaired electrons (paramagnetic).
- Ni(CN)4^2- has no unpaired electrons (diamagnetic).
- Na2[NiCl4] also has 2 unpaired electrons (paramagnetic).

### Summary of Findings:
- **NiCl4^2-**: Tetrahedral, Paramagnetic
- **Ni(CN)4^2-**: Square planar, Diamagnetic
- **Na2[NiCl4]**: Tetrahedral, Paramagnetic

### Correct Combination:
Based on the analysis, the correct combination is found in option C, which states that NiCl4^2- is tetrahedral and paramagnetic, while Ni(CN)4^2- is square planar and diamagnetic.