The central atom of a molecule has `33%` s-character in its hybrid orbitals. Its geometry will be :

Options

A. Trigonal Planar
B. Pyramidal
C. Linear
D. T-shaped

Answer

To determine the geometry of a molecule with a central atom that has 33% s-character in its hybrid orbitals, we can follow these steps:

### Step 1: Understand the concept of hybridization
Hybridization is the mixing of atomic orbitals to form new hybrid orbitals, which can then be used to form bonds in a molecule. The type of hybridization depends on the number of sigma bonds and lone pairs around the central atom.

### Step 2: Determine the hybridization based on s-character
The percentage of s-character in hybrid orbitals can be calculated using the formula:
\[ \text{Percentage of s-character} = \left( \frac{\text{Number of s orbitals}}{\text{Total number of hybrid orbitals}} \right) \times 100 \]
Given that the central atom has 33% s-character, we can deduce the hybridization:
- If there is 33% s-character, it implies that there is 1 s orbital and 2 p orbitals involved in hybridization (since 1 s and 2 p gives a total of 3 orbitals).

### Step 3: Identify the hybridization type
From the above deduction, we can conclude that the hybridization is **sp²** (1 s orbital and 2 p orbitals).

### Step 4: Determine the geometry associated with sp² hybridization
The geometry associated with sp² hybridization is **trigonal planar**. This means that the atoms bonded to the central atom will be arranged in a plane at 120-degree angles from each other.

### Conclusion
Thus, the geometry of the molecule with a central atom that has 33% s-character in its hybrid orbitals is **trigonal planar**.

### Summary of Steps:
1. Understand hybridization and its significance.
2. Calculate the percentage of s-character to determine the hybridization type.
3. Identify that 33% s-character corresponds to sp² hybridization.
4. Conclude that the geometry is trigonal planar.