


Geometry of Ammonia

We can determine the hybridisation of ammonia using the simple formula:


Interpretation:
3 σ bonds + 1 lone pair → 4 regions of electron density → sp³ hybridisation

Ground state and hybridised state of Nitrogen and Hydrogen

For hybridisation:
Orbital overlap in NH3

Number of electron domains around nitrogen: 4 (3 bonds + 1 lone pair)
Electron domain geometry: Tetrahedral
Molecular geometry: Trigonal pyramidal
The lone pair is responsible for the pyramidal shape and bond angle (~107° instead of 109.5°).
| Property | NH3 Value / Feature |
|---|---|
| Central Atom | Nitrogen (N) |
| Hybridisation | sp³ |
| Bond Formation | 3 σ bonds (N–H) |
| Lone Pairs on Central Atom | 1 |
| Electron Domain Geometry | Tetrahedral |
| Molecular Shape | Trigonal pyramidal |
| Bond Angle | ~107° |
| Polarity | Yes, a Polar molecule |

The formal charge (FC) of an atom in a molecule is given by:
Formal Charge = Valence electrons − (Non-bonding electrons + ½ Bonding electrons)

Valence electrons of nitrogen = 5
Non-bonding electrons = 2 (one lone pair)
Bonding electrons = 6 (three N–H bonds)
Half of the bonding electrons = 3
Formal Charge = 5 − (2 + 3)
Formal Charge = 0
This means nitrogen has a formal charge of 0, and the molecule is neutral overall.

Valence electrons = 1
Non-bonding electrons = 0
Bonding electrons (N–H single bond) = 2
Half of the bonding electrons = 1
Formal charge = 1 − (0 + 1)
Formal Charge = 0
Each hydrogen in ammonia has a formal charge of 0 (all three H atoms are identical).



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