The correct statement is

Options

A. `[NiCl_(4)]^(2-)` is `sp^(3)` hybridized and paramagnetic in nature.
B. `[PtCl_(4)]^(2-)` is `dsp^(2)` hybridized and paramagnetic in nature.
C. `[Ni(CO)_(4)]` is `dsp^(2)` hybridized and diamagnetic in nature.
D. `[Co(H_(2)O)_(6)]^(3+)` is `sp^(3)d^(2)` hybridized and diamangnetic in nature.

Answer

To determine the correct statement regarding coordination compounds, we need to analyze each of the given statements step by step.

### Step 1: Analyze NiCl4^2-
1. **Oxidation State of Nickel**:
- The formula is NiCl4^2-.
- Let the oxidation state of Ni be \( x \).
- The equation becomes: \( x + 4(-1) = -2 \).
- Solving gives: \( x - 4 = -2 \) → \( x = +2 \).

2. **Electron Configuration**:
- Nickel in +2 oxidation state has the electron configuration of Argon (Ar) 3d^8.

3. **Hybridization**:
- NiCl4^2- has 4 chloride ligands which are weak field ligands.
- Therefore, there is no pairing of electrons in the 3d orbital.
- The hybridization is sp^3 (4 pairs of electrons).

4. **Magnetic Nature**:
- Since there are unpaired electrons, NiCl4^2- is paramagnetic.

**Conclusion for NiCl4^2-**: The statement about NiCl4^2- being sp^3 hybridized and paramagnetic is correct.

### Step 2: Analyze PtCl4^2-
1. **Oxidation State of Platinum**:
- The formula is PtCl4^2-.
- Let the oxidation state of Pt be \( y \).
- The equation becomes: \( y + 4(-1) = -2 \).
- Solving gives: \( y - 4 = -2 \) → \( y = +2 \).

2. **Hybridization**:
- PtCl4^2- involves strong field ligands (Cl^- is a weak field ligand, but in this case, we consider the overall field).
- The hybridization is dsp^2 (square planar).

3. **Magnetic Nature**:
- In this case, all electrons pair up, making it diamagnetic.

**Conclusion for PtCl4^2-**: The statement about PtCl4^2- being dsp^2 hybridized and paramagnetic is incorrect; it is diamagnetic.

### Step 3: Analyze Ni(CO)4
1. **Oxidation State of Nickel**:
- The formula is Ni(CO)4.
- The oxidation state of Ni is 0 (CO is a neutral ligand).

2. **Hybridization**:
- For Ni(CO)4, CO is a strong field ligand, leading to pairing of electrons.
- The hybridization is sp^3 (tetrahedral).

3. **Magnetic Nature**:
- Since all electrons are paired, Ni(CO)4 is diamagnetic.

**Conclusion for Ni(CO)4**: The statement about Ni(CO)4 being dsp^2 hybridized is incorrect; it is sp^3 hybridized.

### Step 4: Analyze Co(H2O)6^3+
1. **Oxidation State of Cobalt**:
- The formula is Co(H2O)6^3+.
- The oxidation state of Co is +3.

2. **Electron Configuration**:
- Cobalt in +3 oxidation state has the electron configuration of 3d^6.

3. **Hybridization**:
- H2O is a weak field ligand, leading to no pairing in the 3d orbital.
- The hybridization is sp^3d^2 (octahedral).

4. **Magnetic Nature**:
- Since there are unpaired electrons, Co(H2O)6^3+ is paramagnetic.

**Conclusion for Co(H2O)6^3+**: The statement about Co(H2O)6^3+ being sp^3d^2 is incorrect; it is octahedral and paramagnetic.

### Final Conclusion
The only correct statement is regarding NiCl4^2- being sp^3 hybridized and paramagnetic.