Question

What is the hybridization of $N{{H}_{3}}$ ?

Answer 1

Hint : We know that the $N{{H}_{3}}$ is known as Ammonia. It has Nitrogen as the atom with three Hydrogens around it attached with a sigma bond and a lone pair on Nitrogen. If the three hydrogen atoms would bond with nitrogen using the available p-orbitals, the bond angles would be ${{90}^{0}}.$

Complete Step By Step Answer:
The atom in Ammonia, i.e. Nitrogen is $s{{p}^{3}}$ hybridized. It is a gas at normal conditions and is known as Azane. The molecular weight of Ammonia is $17\text{ }g/mol.$ It is a colourless alkaline gas. Starting with the Lewis dot structure of Ammonia, Nitrogen has $5$ valence electrons and each hydrogen has one valence electron.
Ammonia has pyramidal or distorted tetrahedral structure due to the repulsive lone pair – bond pair interaction. So, the total valence electrons are eight. Hydrogen always goes on the outside, so Nitrogen is the atom. After the three valence electrons of Nitrogen have bonded with three Hydrogens, we still have two valence electrons left, which make up one lone pair of Nitrogen
$H-\ddot{N}(H)-H$
To understand the hybridization of Nitrogen in ammonia, we need to look into the configuration; $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$ . During hybridization, one s orbital and three p orbitals of nitrogen hybridize to form four hybrid orbitals having equal energy levels, thus making its hybridization $s{{p}^{3}}.$ Half three $s{{p}^{3}}$ orbitals of nitrogen form a bond with three hydrogens. The fourth fully filled hybridized orbital holds the lone pair of nitrogen.

Note :
Note that a student might not consider the lone pair of nitrogen in the hybridization of the central atom of Ammonia. The answer in that case would come out to be $s{{p}^{2}}.$ Also, the bond angle in ammonia is less than standard ${{109}^{0}}73'$ due to the same reason. The bond angle is ${{107}^{0}}.$