Concept:A perfect bond angle of 109.5° occurs in molecules with a regular tetrahedral geometry, which requires \( \text{sp}^3 \) hybridization and zero lone pairs on the central atom.
Formula:$$ \text{Geometry} = \text{Tetrahedral } (\text{AB}_4) $$
Solution:- \( \text{SiCl}_4 \), \( \text{NH}_4^+ \), and \( \text{CH}_4 \) all feature a central atom with exactly 4 bond pairs and 0 lone pairs.
- Because there are no lone pairs to compress the bond angles, the geometry is a perfectly symmetrical tetrahedron.
- The resulting bond angle in all three species is exactly 109.5°.
Why other options are incorrect:- Option B: \( \text{H}_2\text{O} \) has an angle of 104.5°, and \( \text{BeCl}_2 \) is 180°.
- Option C: \( \text{NH}_3 \) has one lone pair, compressing its angle to 107.5°.
- Option D: \( \text{PH}_3 \) has one lone pair, making its angle even smaller than ammonia.
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