There is a problem with basic valence bond theory We would expect that an atom could form a bond for every unpaired electron that is has in its valence orbitals. This is basically true, but consider an atom like beryllium. Using Hunds rule, we can draw out the electron configuration as

Be: 1s() 2s() 2p()()()

In order to form more bonds, the beryllium atom will hybridize, or mix, the 2s orbital and one of its 2p orbitals to form two 2sp hybrid orbitals. These hybrid orbitals have the same energy: less than a 2p but more than a 2s, and so according to Hunds rule the electrons in the beryllium atom will spread out into these hybrid orbitals:

Be: 1s() 2s() 2p()()() -> 1s() 2sp()( ) 2p()()

Similar things happen in atoms with more electrons. In the case of boron, the 2s and two 2p orbitals combine to form three 2sp² orbitals, and in carbon the 2s and all three 2p orbitals combine to four 2sp³ orbitals.

B: 1s() 2s() 2p()()() -> B: 1s() 2sp²() ()() 2p()

C: 1s() 2s() 2p()()() -> C: 1s() 2sp³() ()()()

Atoms beyond the second row like sulfur and phosphorus can combine 3s, 3p and 3d orbitals to form even more hybrids: this explains why they can violate the octet rule. The shape of the hybrid orbitals determines the geometry of the molecule: the shapes derived from theory are in exact accord with the predictions of VSEPR theory. For example, sp³ hybrids, as found around the carbon atom in methane, CH₄, have a tetrahedral geometry, thus the bond to the hydrogens are also in a tetrahedral geometry.

The table below summarizes the basic hybridization cases:

Number of electron pairs	Atomic orbitals	Hybrid orbitals formed	Geometry
2	one s + one p	two sp	AX2: linear
3	one s + two p	three sp2	AX3: trigonal planar
4	one s + three p	four sp3	AX4: tetrahedral
5	s + three p + one d	five sp3d	AX5: trigonal bipyramidal
6	one s + three p + two d	six sp3d2	AX6: octahedral

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