Question

How do you predict the hybridization of an atom in a molecule?

Answer 1

To solve this question we should know about:
Hybridization: In valence bond theory, orbital hybridization (or hybridization) is the concept of combining atomic orbitals into new hybrid orbitals (with different energies, shapes, and other properties than the component atomic orbitals) suited for electron pairing to create chemical bonds.
So, we have to first calculate the number of electrons in given atoms. Then we will decide the nature of an electron like it is a valence electron or something else.

Complete answer:
You can always assign hybridization if you can assign total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR. Hybridization was created to improve the compatibility of quantum mechanical bonding theories with empirical geometries. You always know the other if you know one.
-Linear - $sp$ - two hybrid orbitals oriented 180 degrees apart result from the hybridization of one s and one p orbital.
-The hybridization of one s and two p orbitals produces a Trigonal planar - $s{p^2}$ .
-Tetrahedral - $s{p^3}$ - the hybridization of one s and three p orbitals results in four hybrid orbitals orientated toward the ${109.5^ \circ }$ points of a normal tetrahedron.
-The hybridization of ones, three p, and one d orbitals produces five hybrid orbitals arranged in this strange shape: three equatorial hybrid orbitals positioned ${120^ \circ }$ from each other in the same plane, and two axial orbitals orientated ${180^ \circ }$ from each other, orthogonal to the equatorial orbitals.
-Octahedral - ${d^2}s{p^3}$ or $s{p^3}{d^2}$ - produced by the hybridization of ones, three p, and two d orbitals, resulting in six hybrid orbitals positioned ${90^ \circ }$ degrees apart toward the points of a conventional octahedron.
I'm guessing you haven't learnt any of the geometries above steric number 6 (because of their rarity), yet each one corresponds to a distinct hybridization.
For example: $N{H_3}$
Which of the following categories does $N{H_3}$ belong in? When calculating total electron geometry, keep in mind that the lone pair counts as an electron domain. Because $N{H_3}$ is $s{p^3}$ , it must be tetrahedral, according to the sample question. Make sure you understand how $N{H_3}$ electron shape is tetrahedral.

Note:
In view of computational chemistry, a preferable approach would be to use sigma bond resonance in addition to hybridization, implying that each resonance structure has its own hybridization strategy. The octet rule must be followed by all resonance structures.
Although ideal hybrid orbitals can be advantageous in some cases, most bonds require intermediate orbitals. This necessitates an expansion to incorporate flexible weightings of atomic orbitals of each type (s, p, d), allowing for a quantitative representation of bond creation when molecular geometry deviates from ideal bond angles. The number of p-characters is not limited to integers.