\( sp^3 \) hybridization leads to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → tetrahedral geometry with bond angles 109.5° each
In \( sp^3 \) hybridisation one s and three p orbitals mix to give four equivalent hybrid orbitals. Four electron pairs keep as far apart as possible, so the orbitals point to the corners of a regular tetrahedron, with an angle of 109.5° (109°28′) between any two. Examples are \( \ce{CH4} \) and the complexes \( \ce{[Ni(CO)4]} \) and \( \ce{[NiCl4]^2-} \).
Why the other options are incorrect
Option (1) tetrahedral geometry with bond angles 90° each — Incorrect. Four bonds at 90° to one another cannot point to the corners of a tetrahedron; 90° angles belong to square planar and octahedral shapes.
Option (3) trigonal geometry with bond angles 120° each — Incorrect. This is the trigonal planar shape of \( sp^2 \) hybridisation (three hybrid orbitals), as in \( \ce{BF3} \).
Option (4) square planar geometry with bond angles 90° each — Incorrect. This shape comes from \( dsp^2 \) hybridisation, which uses one d orbital, as in \( \ce{[Ni(CN)4]^2-} \).
NCERT (2026-27 edition):
Valence Bond Theory – …
