Concept:Removing a lone pair by converting it into a bonding pair removes the strong LP-BP repulsion, allowing the molecule's bond angles to expand to their ideal geometric state.
Formula:$$ \text{Angle}_{(\text{NH}_3)} = 107^{\circ} \longrightarrow \text{Angle}_{(\text{NH}_4^+)} = 109.5^{\circ} $$
Solution:- In ammonia (\( \text{NH}_3 \)), the nitrogen has 3 bond pairs and 1 lone pair. The lone pair repels the bonds strongly, compressing the angle to 107°.
- When a proton (\( \text{H}^+ \)) attaches, the lone pair becomes a bond pair, forming \( \text{NH}_4^+ \).
- The ion now has 4 identical bond pairs and 0 lone pairs. It assumes a perfect tetrahedral geometry with angles of exactly 109.5°.
Why other options are incorrect:- Option A, Option B, Option D: These state incorrect initial angles or falsely claim the angle decreases. The angle must increase because the compressing force of the lone pair is neutralized.
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