The concept of atomic orbital overlap should apply also to polyatomic molecules.

The concept of hybridization is the overlap or blending of s, p and d orbitals to explain bond formation.

In applying the valence bond method to the ground state electronic configuration of carbon you can assume that the filled 1s orbital will not be involved in the bonding and focus your attention on the valence-shell orbitals.

From its ground state configuration, two unpaired electrons in the 2p subshell are observed. One can predict the simplest hydrocarbon molecule to be CH₂, by overlapping the two unpaired electrons from two H atoms with the two unpaired electrons of carbon.

sp hybrid orbitals

When two electron sets surround the central atom, you observe a linear molecular shape. VB theory explains this by proposing that mixing two non-equivalent orbitals of a central atom one s and one p, gives rise to two equivalent sp hybrid orbitals that lie 180° apart. Combination (mixing) of one s and one p orbital is called sp hybridization and the resultant orbitals are called sp hybrid orbitals.

sp² hybrid orbitals

Now turn your attention to boron, a Group IIIA element. The boron atom has four orbitals but only three electrons in its valence shell. In most boron compounds the hybridization scheme combines one 2s and two 2p orbitals into three sp² hybrid orbitals.

sp² hybrization for B

The atom in its ground state has only one unpaired electron, so that it can form only one covalent bond, but in the excited state there are three unpaired electrons, hence three bonds can be formed.

This indeed is what is observed experimentally. By far the most common examples of sp² hybridization are found in organic molecules with double bonds.

sp³ hybridization

Carbon, the central atom in a molecule of methane CH₄ has only two unpaired electrons in the ground state. The two electrons in the 2p level are not paired, i.e., put into the same box, in accordance with Hund’s rule.

sp³ hybridization for carbon

We might also expect to use sp³ hybridization not only for structures of the type AX₄ type (as in CH₄), but also for AX₃E type (as in NH₃) and AX₂E₂ type (as in H₂O). Nitrogen has three unpaired electrons in its ground state, sufficient to form three bonds; so it is not necessary to excite the atom.

sp³ hybridization of the central atom N in NH₃:

Accounts for the formation of three N–H bonds and a lone-pair of electrons on the N atom. The predicted H–N–H bond angles of 109.5° are close to the experimentally observed angles of 107°.

A similar scheme for H₂O accounts for the formation of two O–H bonds and two lone-pairs of electrons on the oxygen atom.

The predicted H–O–H bond angle of 109.5° is also reasonably close to the observed angle of 104.5°.