Concept:A molecule has a zero dipole moment if it is highly symmetrical and lacks lone pairs on its central atom, causing all individual bond dipoles to cancel each other out.
Formula:$$ \mu = 0 \text{ for perfectly symmetrical molecules.} $$
Solution:- Boron trifluoride (\( \text{BF}_3 \)) has \( \text{sp}^2 \) hybridization, giving it a flat, trigonal planar geometry.
- Boron has no lone pairs to distort the symmetry.
- The three polar B-F bonds pull equally at 120° angles, resulting in a net vector sum (dipole moment) of zero.
Why other options are incorrect:- Option A, Option C, Option D: \( \text{PCl}_3 \), \( \text{NH}_3 \), and \( \text{H}_2\text{O} \) all possess lone pairs on their central atoms, which creates asymmetry and results in a net, non-zero dipole moment.
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