Which type of hybridisation is found in ammonia ?

The correct answer is Option (1) → sp³ hybridisation

The hybridisation of a central atom is decided by the number of electron pairs around it, counting both bond pairs (σ bonds) and lone pairs:

• 2 pairs: sp, linear (e.g. \( \ce{BeCl2} \))

• 3 pairs: sp², trigonal planar (e.g. \( \ce{BF3} \))

• 4 pairs: sp³, tetrahedral arrangement (e.g. \( \ce{CH4} \), \( \ce{NH3} \), \( \ce{H2O} \))

• 5 pairs: sp³d, trigonal bipyramidal (e.g. \( \ce{PCl5} \))

• 6 pairs: sp³d², octahedral (e.g. \( \ce{SF6} \))

Nitrogen has five valence electrons. In \( \ce{NH3} \) it forms three N–H bonds and keeps one lone pair, so there are four electron pairs around it. The four pairs occupy four sp³ hybrid orbitals. Because the lone pair repels the bond pairs more strongly, the molecule is trigonal pyramidal with an H–N–H angle of about 107°, a little less than the tetrahedral 109.5°.

So the nitrogen in ammonia is sp³ hybridised.

Why the other options are incorrect

• Option (2) sp² hybridisation — Incorrect. It needs only three electron pairs, as in trigonal planar \( \ce{BF3} \); nitrogen in \( \ce{NH3} \) has four.

• Option (3) sp hybridisation — Incorrect. It needs two electron pairs and gives a linear molecule such as \( \ce{BeCl2} \).

• Option (4) sp³d² hybridisation — Incorrect. It needs six electron pairs, as in \( \ce{SF6} \); nitrogen has no d orbitals in its valence shell.

NCERT (2026-27 edition):
Structure of Amines