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 s-character and hybridization
- The percentage of s-character in hybrid orbitals helps us identify the type of hybridization.
- If the central atom has 33% s-character, we can calculate the hybridization using the formula:
\[ \text{s-character} = \frac{\text{number of s orbitals}}{\text{total number of orbitals}} \times 100 \]
- Rearranging this, we find that if there is 33% s-character, it corresponds to 1 s orbital and 2 p orbitals, leading to the hybridization of sp².

### Step 2: Determine the hybridization
- From the above calculation, since we have 1 s orbital and 2 p orbitals, the hybridization is sp².
- The sp² hybridization indicates that there are three hybrid orbitals formed.

### Step 3: Identify the geometry associated with sp² hybridization
- The geometry associated with sp² hybridization is trigonal planar.
- In a trigonal planar arrangement, the three hybrid orbitals are arranged in a plane at 120° angles to each other.

### Step 4: Conclusion
- Therefore, the geometry of the molecule with a central atom that has 33% s-character in its hybrid orbitals is **trigonal planar**.

### Summary
The central atom with 33% s-character has sp² hybridization, leading to a trigonal planar geometry.

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