Which of the following involves `dsp^(2)`-hybridisation and are square planar ?

Options

A. `SF_(6)`
B. `BrF_(5)`
C. `PCl_(5)`
D. `[Ni(CN)_(4)]^(2-)`

Answer

To determine which of the given compounds involves `dsp²` hybridization and exhibits a square planar geometry, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding dsp² Hybridization**:
- `dsp²` hybridization occurs when one d orbital, one s orbital, and two p orbitals combine to form four equivalent hybrid orbitals. This type of hybridization typically leads to a square planar geometry.

2. **Analyzing the Options**:
- **Option 1: Sulfur Hexafluoride (SF₆)**:
- SF₆ has six fluorine atoms bonded to sulfur. This corresponds to `sp³d²` hybridization, not `dsp²`, and has an octahedral geometry. Therefore, this option is incorrect.

- **Option 2: Bromine Pentafluoride (BrF₅)**:
- BrF₅ has five fluorine atoms and one lone pair on bromine. This corresponds to `sp³d` hybridization, which results in a square pyramidal geometry. Hence, this option is also incorrect.

- **Option 3: Phosphorus Pentachloride (PCl₅)**:
- PCl₅ has five chlorine atoms bonded to phosphorus. This corresponds to `sp³d` hybridization, leading to a trigonal bipyramidal geometry. Therefore, this option is incorrect as well.

- **Option 4: Tetra Cyanonickelate Ion ([Ni(CN)₄]²⁻)**:
- The tetra cyanonickelate ion consists of nickel coordinated with four cyanide ligands. Nickel in this ion has a +2 oxidation state, leading to an electronic configuration of 3d⁸. The strong field cyanide ligands cause pairing of electrons, allowing for `dsp²` hybridization. The resulting geometry is square planar, as the four cyanide ligands occupy the corners of a square. Thus, this option is correct.

3. **Conclusion**:
- The only compound that involves `dsp²` hybridization and exhibits a square planar geometry is the tetra cyanonickelate ion ([Ni(CN)₄]²⁻).

### Final Answer:
The correct answer is **Option 4: Tetra Cyanonickelate Ion ([Ni(CN)₄]²⁻)**.