Which of the following are tetrahedral structures ? .

Options

A. `[Ni(CN)_(4)]^(2-)`
B. `[Ni(CO)_(4)]`
C. `[NiCI_(4)]^(2-)`
D. `CrO_(4)^(2-)`

Explanation

To determine which of the given complexes have tetrahedral structures, we need to analyze each complex based on their hybridization and geometry.

### Step-by-Step Solution:

1. **Identify the Complexes**: The complexes to analyze are:
- A) Ni(CN)₄²⁻
- B) Ni(CO)₄
- C) NiCl₄²⁻
- D) Chromate ion (CrO₄²⁻)

2. **Analyze Complex A: Ni(CN)₄²⁻**:
- **Oxidation State**: Let x be the oxidation state of Ni. The equation is:
\[ x + 4(-1) = -2 \implies x = +2 \]
- **Electronic Configuration**: For Ni²⁺, the configuration is [Ar] 3d⁸.
- **Hybridization**: CN⁻ is a strong field ligand, leading to pairing of electrons. The hybridization is dsp², which corresponds to a square planar geometry, not tetrahedral.

3. **Analyze Complex B: Ni(CO)₄**:
- **Oxidation State**: Here, the oxidation state of Ni is 0.
- **Electronic Configuration**: The configuration remains [Ar] 3d⁸ 4s².
- **Hybridization**: CO is also a strong field ligand, leading to pairing. The hybridization is sp³, which corresponds to a tetrahedral geometry.

4. **Analyze Complex C: NiCl₄²⁻**:
- **Oxidation State**: Let x be the oxidation state of Ni. The equation is:
\[ x + 4(-1) = -2 \implies x = +2 \]
- **Electronic Configuration**: For Ni²⁺, the configuration is [Ar] 3d⁸.
- **Hybridization**: Cl⁻ is a weak field ligand, so no pairing occurs. The hybridization is sp³, which corresponds to a tetrahedral geometry.

5. **Analyze Complex D: Chromate Ion (CrO₄²⁻)**:
- **Oxidation State**: Let x be the oxidation state of Cr. The equation is:
\[ x + 4(-2) = -2 \implies x = +6 \]
- **Hybridization**: The hybridization is sp³, which corresponds to a tetrahedral geometry.

### Conclusion:
Based on the analysis:
- **Tetrahedral Structures**:
- B) Ni(CO)₄
- C) NiCl₄²⁻
- D) Chromate ion (CrO₄²⁻)

Thus, the complexes that have tetrahedral structures are **B, C, and D**.