Concept:The geometry of a molecule is determined by the number of electron domains around its central atom (VSEPR theory).
Formula:$$ \text{Hybridization of B in BF}_3 = \text{sp}^2 $$
Solution:- Boron trifluoride (\( \text{BF}_3 \)) belongs to the \( \text{AB}_3 \) type molecule.
- The central boron atom has 3 bond pairs and 0 lone pairs.
- This results in a perfect trigonal planar geometry, distributing the domains evenly to minimize repulsion.
- The circle (360°) divided evenly among 3 domains yields exactly 120°.
Why other options are incorrect:- Option A: Associated with octahedral or square planar geometries.
- Option B: Bond angle for water (\( \text{H}_2\text{O} \)) due to 2 lone pairs.
- Option C: Represents slight compression seen in structures with double bonds (like alkenes), not a symmetrical \( \text{AB}_3 \) molecule.
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