Ammonia acts as a very good ligand, but the ammonium ion (NH4+) does not form complexes. Which of the following is the correct reason?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The nitrogen atom in NH4+ has no lone pair of electrons.
A ligand bonds to a metal ion by donating an electron pair into an empty orbital of the metal. The nitrogen of \( \ce{NH3} \) carries an unshared pair of electrons, so ammonia is a good ligand.
In \( \ce{NH4^+} \) that pair has already been used to bond an \( \ce{H^+} \):
\( \displaystyle \ce{NH3 + H^+ -> NH4^+} \)
All four orbitals of nitrogen in \( \ce{NH4^+} \) now hold N–H bonds, so no lone pair is left to donate to a metal ion.
Why the other options are incorrect
Option (1) NH4+ is a larger species than NH3 and cannot approach the metal ion. — Incorrect. Size is not the reason: ligands may be much larger molecules, such as \( \ce{N(CH2CH2NH2)3} \) or even proteins.
Option (2) The nitrogen atom in NH4+ is sp2 hybridised, while in NH3 it is sp3 hybridised. — Incorrect. Nitrogen is sp3 hybridised in both: pyramidal in \( \ce{NH3} \) and tetrahedral in \( \ce{NH4^+} \).
Option (4) The nitrogen atom in NH4+ has an unpaired electron. — Incorrect. All eight valence electrons around nitrogen are paired in the four N–H bonds; \( \ce{NH4^+} \) has no unpaired electron.
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “… These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding.”
9.1 Structure of Amines – “Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons.”