Two hybrid orbitals have a bond angle of `120^(@)`. The percentage of s-character in the hybrid orbitals is nealy:

Options

A. 0.25
B. 0.33
C. 0.5
D. 0.66

Answer

To solve the problem of determining the percentage of s-character in hybrid orbitals with a bond angle of 120 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Hybridization**:
- The bond angle of 120 degrees indicates that the hybridization is sp². This is characteristic of molecules with trigonal planar geometry.

2. **Determine the Hybrid Orbital Composition**:
- In sp² hybridization, one s orbital and two p orbitals combine to form three equivalent sp² hybrid orbitals.

3. **Calculate the Percentage of s-Character**:
- The formula to calculate the percentage of s-character in hybrid orbitals is given by:
\[ \text{Percentage of s-character} = \left( \frac{\text{Number of s orbitals}}{\text{Total number of hybrid orbitals}} \right) \times 100 \]
- For sp² hybridization:
- Number of s orbitals = 1
- Total number of hybrid orbitals = 3 (1 s + 2 p)
- Therefore, the calculation becomes:
\[ \text{Percentage of s-character} = \left( \frac{1}{3} \right) \times 100 = 33.33\% \]

4. **Final Result**:
- The percentage of s-character in the sp² hybrid orbitals is approximately 33%.

### Conclusion:
Thus, the percentage of s-character in the hybrid orbitals with a bond angle of 120 degrees is nearly **33%**.

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