Question

Write shape, hybridization and no of lone pair of electrons for following: ${\mathrm{NH}}_3$

Answer 1

Lone pair of ${\mathbf{NH}}_{\mathbf{3}}$ molecule

• The atomic number of Nitrogen $\left(\mathrm{N}\right)$ is $7$.
• The electronic configuration of oxygen atom is as follows:

$\mathrm{N}=1s^{2}2s^{2}2p^3$

• The number of valence electrons of nitrogen is $5$.
• Nitrogen form three sigma bonds with three hydrogen $\left(\mathrm{H}\right)$ atoms to form a ammonia $\left({\mathrm{NH}}_3\right)$ molecule.
• ${\mathrm{NH}}_3$ molecule has one lone pairs.

Hybridization of ${\mathbf{NH}}_{\mathbf{3}}$ molecule

• In ${\mathrm{NH}}_3$, nitrogen is the central atom and its one $2s$ and three $2p$ orbitals.
• It will form an $s{p}^3$ hybridized orbital.

Shape of ${\mathbf{NH}}_{\mathbf{3}}$ molecule

• According to VSEPR theory, the order of repulsion between electron pairs is as follows:
• lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair repulsions
• It has tetrahedral geometry, but due to bond pair-lone pair repulsion, ${\mathrm{NH}}_3$ will form a trigonal pyramidal shape.
• Structure of ammonia: