$sp^3$ hybridization leads to

tetrahedral geometry with bond angles 109.5° each

The correct answer is Option (2) → tetrahedral geometry with bond angles 109.5° each

sp³ hybridization occurs when one s orbital and three p orbitals mix to form four equivalent sp³ hybrid orbitals. These orbitals arrange themselves to minimize electron repulsion (VSEPR theory), resulting in a tetrahedral geometry.

• Ideal bond angles in a perfect tetrahedron: 109.5° (approximately 109°28').
• Example: Methane (CH₄) — carbon is sp³ hybridized with four identical C–H bonds at 109.5°.

Why not the others?

• 90° bond angles: Typical for unhybridized p orbitals (not sp³).
• Trigonal geometry with 120°: From sp² hybridization (e.g., ethene, boron trifluoride).
• Square planar with 90°: Typically for dsp² hybridization in d⁸ metal complexes (e.g., [PtCl₄]²⁻), not sp³.