Hybridisation is the mixing of atomic orbital of comparable energy
and the number of hybrid orbitals formed is equal to the number of pure atomic orbitals mixed up and hybrid are occupied by `sigma`-bond
pair and lone pair.
Which of the following geometry is most likely to not form from
`sp^(3)d` hybridisation of the central atom.

Options

A. Linear
B. Tetrahedral
C. T-shaped
D. See-saw

Answer

To determine which geometry is least likely to form from sp³d hybridization of a central atom, we need to analyze the hybridization and the geometries associated with it.

### Step-by-Step Solution:

1. **Understanding sp³d Hybridization**:
- The sp³d hybridization involves the mixing of one s orbital, three p orbitals, and one d orbital. This results in five hybrid orbitals.

2. **Identifying Possible Geometries**:
- The geometries that can arise from sp³d hybridization include:
- **Trigonal Bipyramidal**: This occurs when there are five bond pairs and no lone pairs (AX5).
- **Seesaw**: This occurs when there are four bond pairs and one lone pair (AX4E).
- **T-shaped**: This occurs when there are three bond pairs and two lone pairs (AX3E2).
- **Linear**: This occurs when there are two bond pairs and three lone pairs (AX2E3).

3. **Identifying the Geometry that Cannot Form**:
- The tetrahedral geometry (AX4) is not possible with sp³d hybridization. Tetrahedral geometry arises from sp³ hybridization, which involves four hybrid orbitals formed from one s and three p orbitals, without any d orbitals.

4. **Conclusion**:
- Therefore, the geometry that is most likely to not form from sp³d hybridization is **tetrahedral**.

### Final Answer:
- The geometry that is least likely to form from sp³d hybridization is **tetrahedral**.

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