Question

How and why does hybridization occur $ ? $ For example, why does $ N{H_3} $ form an $ s{p^3} $ orbital $ ? $

Answer 1

Hint: Hybridization is the concept of mixing atomic orbitals into new hybrid orbitals with different energies, shapes, etc., than the component atomic orbitals, suitable for the pairing of electrons to form chemical bonds in valence bond theory. Write the electronic configuration for the central atom in $ N{H_3} $ and check for the hybridization.

Complete step by step solution:
The idea of hybridization was developed because of the fact that all the $ C - H $ bonds in methane are similar. Hybridization is a theory which starts with the consideration that when atoms combine to form a molecule combination of atomic orbitals to produce new orbitals of varying energies and shapes than the initial orbitals occur. The new orbital will contain the same total electron number as the old electrons. This new hybrid molecular orbital belongs to the molecule as a whole, but its geometry is determined by the types of atomic orbitals in the central atom that were involved in the bonding of the molecule. Hybrid orbitals are useful in the explanation of molecular geometry and atomic bonding properties and are symmetrically disposed in space. For example, in a carbon atom which forms four single bonds the valence-shell $ s $ orbital combines with three valence-shell $ p $ orbitals to form four equivalent $ s{p^3} $ hybrid orbitals which are arranged in a tetrahedral arrangement around the carbon to bond to four different atoms.
In case of $ N{H_3} $ , $ N $ is the central atom having an electronic configuration, $ \left[ {He} \right]2{s^2}2{p^3} $ which can be represented as:
 $
  \underline { \uparrow \downarrow } \,\,\,\,\,\underline \uparrow \,\,\,\,\,\,\underline \uparrow \,\,\,\,\,\,\underline \uparrow \\
  2s\,\,\,2{p_x}\,\,2{p_y}\,\,2{p_z} \\
  $ , which is the ground state as well as the excited state configuration. The one $ s $ orbital combines with three valence-shell $ p $ orbitals to form four equivalent $ s{p^3} $ hybrid orbitals where the paired electrons form a lone pair on $ N $ while the three unpaired electrons form sigma bond with $ H $ and hence has a distorted tetrahedral geometry.

Note:
You can also use Valence Shell Electron Pair Repulsion theory to know the structure or geometry of the compound by using electron count of the central atom of the molecule and then by using the structure or geometry you can determine the hybridization of the compound.