Concept:In \( \text{sp}^2 \) hybridization, three hybrid orbitals form a planar trigonal structure, and the remaining unhybridized p-orbital is situated perpendicular to this plane.
Formula:$$ \text{sp}^2 \text{ plane} \perp \text{unhybridized } p_z \text{ orbital} $$
Solution:- The three \( \text{sp}^2 \) hybrid orbitals arrange themselves in a flat plane with 120° angles between them.
- The unhybridized p-orbital (which will be used to form a pi bond) stands straight up and down, piercing the center of this plane.
- Thus, the angle between the flat plane (the hybrid orbitals) and the unhybridized p-orbital is exactly 90°.
Why other options are incorrect:- Option A: 120° is the angle between the \( \text{sp}^2 \) hybrid orbitals themselves.
- Option B & Option D: Correspond to tetrahedral and linear geometries, respectively.
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