Hybridization

The central atom of nitrogen has 5 valence electrons as per the electronic configuraion. During the formation of ${\mathrm{N}\mathrm{H}}_3$ , 3 valence electrons forms three sigma bonds with hydrogen and one lone electron pair is left.

• The steric number  of the ammonia molecule: SN=3 bonds +1 lone pair $=4$ total electron domains.
• As per the steric number, there is ${\mathrm{s}\mathrm{p}}^3$ hybridisation in ammonia.
• The lone pair causes repulsion, which leads to a trigonal pyramidal geometry with bond angles of ${107}^{?}$ .

Xenon has 8 valence electrons. In ${\mathrm{X}\mathrm{e}\mathrm{F}}_2$ , it forms two bonds with fluorine, leaving three lone pairs. The steric number =2 bonds +3 lone pairs $=5$ .
As per the VSEPR Theory  the electron? pair geometry due to SN = 5 will be trigonal bipyramidal. The hybridization type will be ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ hybridisation. However, the three lone pairs occupy equatorial positions, resulting in a linear molecular geometry.

Following have the sp³d² hybridisation