The percentage of s-character in the hybridised orbitals of B in `BF_(3)` is

Options

A. 25
B. 30
C. 75
D. 33.3

Explanation

To determine the percentage of s-character in the hybridized orbitals of boron in BF3, we can follow these steps:

### Step 1: Identify the central atom and its electronic configuration
- The central atom in BF3 is boron (B).
- The atomic number of boron is 5, which gives it the electronic configuration of 1s² 2s² 2p¹.

### Step 2: Determine the valence electrons
- The valence electrons of boron are in the 2s and 2p orbitals. Boron has 3 valence electrons (2 from 2s and 1 from 2p).

### Step 3: Determine the hybridization
- In BF3, boron forms three bonds with fluorine atoms. To accommodate these three bonds, boron undergoes hybridization.
- The hybridization of boron in BF3 is sp² because it uses one 2s orbital and two 2p orbitals to form three equivalent sp² hybrid orbitals.

### Step 4: Calculate the percentage of s-character
- In sp² hybridization, there is one s orbital and two p orbitals involved.
- The formula to calculate the percentage of s-character is:
\[ \text{Percentage of s-character} = \left( \frac{\text{Number of s orbitals used}}{\text{Total number of hybrid orbitals}} \right) \times 100 \]
- Here, the number of s orbitals used is 1, and the total number of hybrid orbitals is 3 (1 s + 2 p).
- Thus, the calculation is:
\[ \text{Percentage of s-character} = \left( \frac{1}{3} \right) \times 100 = 33.33\% \]

### Conclusion
- Therefore, the percentage of s-character in the hybridized orbitals of boron in BF3 is **33.3%**.