Concept:In \( \text{sp}^2 \) hybridization, three hybrid orbitals form a flat geometric plane to form sigma bonds, while the remaining unhybridized p-orbital is reserved for forming pi bonds.
Formula:$$ \text{sp}^2 \text{ Plane (XY)} \perp \text{p}_z \text{ orbital (Z)} $$
Solution:- The three \( \text{sp}^2 \) hybrid orbitals arrange themselves at 120° angles in a flat plane (e.g., the XY plane) to minimize repulsion.
- The one remaining unhybridized p-orbital (e.g., \( \text{p}_z \)) stands straight up and down, piercing directly through the center of this plane.
- Therefore, it is strictly perpendicular (90°) to the plane of the hybrid orbitals.
Why other options are incorrect:- Option A, Option B, Option C: If the p-orbital were in the same plane or parallel, it could not form the sideways overlap necessary for a pi-bond without catastrophic spatial interference with the sigma bonds.
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