The shape of ammonia molecules is

Analysis

The question asks for the molecular geometry (shape) of an ammonia molecule ($NH_3$). To determine this, we must look at the electron distribution around the central nitrogen atom.

Concept: VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory helps us predict the shape of molecules based on the number of bonding pairs and lone pairs of electrons around the central atom.

1. Electronic Configuration: Nitrogen ($N$) has an atomic number of 7. Its electron configuration is $1s^2 2s^2 2p^3$. It has 5 valence electrons.
2. Bonding: In $NH_3$, the nitrogen atom forms three covalent bonds with three hydrogen atoms. This uses 3 of its valence electrons.
3. Lone Pairs: After bonding with the three hydrogens, nitrogen has 2 electrons remaining, which form one lone pair.
4. Steric Number: The total number of electron domains (bonding pairs + lone pairs) is $3 + 1 = 4$. • A molecule with 4 electron domains typically adopts a tetrahedral electron geometry. • However, because one of these domains is a lone pair, it exerts a stronger repulsive force on the bonding pairs, pushing the hydrogen atoms closer together.

Solution

• Electron Geometry: Tetrahedral (due to 4 electron domains).
• Molecular Geometry: Because one of the four positions is occupied by a lone pair rather than an atom, the shape formed by the atoms is trigonal pyramidal. The nitrogen sits at the apex, and the three hydrogen atoms form the base of a pyramid.

Conclusion

While the electron arrangement is tetrahedral, the actual shape of the molecule (the positions of the atoms) is trigonal pyramidal.

Exam Tip: Always remember the difference between electron geometry and molecular geometry. If a central atom has lone pairs, the molecular geometry will be different from the electron geometry. For $NH_3$, the lone pair "squashes" the bond angles from the ideal $109.5^\circ$ (tetrahedral) down to approximately $107^\circ$.