Concept:Hybridization is determined by the steric number of the central atom (number of sigma bonds + number of lone pairs).
Formula:$$ \text{Steric Number (SN)} = 3 \implies \text{sp}^2 $$
Solution:- In Boron trifluoride (\( \text{BF}_3 \)), Boron is the central atom in Group IIIA, having 3 valence electrons.
- It forms 3 single sigma bonds with 3 Fluorine atoms. It has 0 lone pairs left.
- Steric Number = 3 + 0 = 3.
- This corresponds perfectly to \( \text{sp}^2 \) hybridization, giving the molecule a flat, trigonal planar geometry.
Why other options are incorrect:- Option A & Option C: Ammonia and Water have a steric number of 4 (bonds + lone pairs), making them \( \text{sp}^3 \) hybridized.
- Option D: Beryllium chloride has a steric number of 2, making it \( \text{sp} \) hybridized.
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