The number of atoms which are typically bonded to a given atom is termed the valence of that atom. Thus, in the examples shown below, hydrogen would have a valence of one, oxygen would have a valence of two, nitrogen and boron would have a valence of three, and carbon would have a valence of four.

As shown above, chemical analysis of simple hydrocarbons clearly demonstrates that carbon has a valence of four; that is, carbon forms four bonds and the simplest hydrocarbon (methane) has a molecular formula of CH₄. The arrangement of these atoms in space, however, is not immediately apparent, and at least three possibilities should be considered:

• square planar, in which the four hydrogens occupy the corners of a square, centered about the carbon.
• pyramidal, in which the carbon and three hydrogens occupy a plane with carbon in the middle and hydrogens at the vertices of a triangle, and,
• tetrahedral, in which the carbon is in the center of a regular tetrahedron and hydrogens are at each vertex.

2s 2px 2py 2pz

Recalling the geometry of the three p-orbitals, the pyramidal geometry might seem to be favored, since the three p-orbitals would form bonds at 90 angles, with the s-orbital on the opposite face.

This can easily be shown to be incorrect, however, by simply examining the number of molecules having the molecular formula CH₂Cl₂. As shown below, if carbon was square-planar, there should be two forms of the molecule CH₂Cl₂ (molecules having the same molecular formula, but different structures are termed isomers), one in which the two chlorines are on the same face of the square, and one in which they are in opposite corners.

If carbon was pyramidal, again two isomers would be predicted, one in which both chlorines are in the trigonal plane (with an angle of 120) and one in which one chlorine is in the apical position, with an angle of 90.

For a tetrahedron, however, since all bond angles are equal at 109.5, both chlorines would always occupy one adjacent face of the tetrahedron (no matter which face you choose, as shown below) and only one isomer would be predicted.

In fact, there is only one isomer with the molecular formula CH₂Cl₂, and carbon has been confirmed to be tetrahedral, using modern x-ray diffraction methods. The observed tetrahedral geometry, however, does not agree with the prediction based on the orbital description, which would seem to predict a pyramidal structure. The explanation which is commonly given for this is that the four orbitals around carbon hybridize to form four equivalent orbitals having 75% p-character and 25% s-character. The resulting geometry is predicted to be tetrahedral and the driving force is electronic repulsion; placing the four orbitals in tetrahedral geometry provides the maximum separation between the electron pairs and minimizes electronic repulsion (the VESPER model cited in the previous section). A hybrid consisting of one s- and three p-orbitals is termed an sp³ hybrid (shown below) and an sp³ center should always be considered to approximate tetrahedral geometry, with the overriding factor being the driving force for the molecule to assume the lowest energy geometry, which is readily accessible. Hydrocarbons containing only sp³ carbons are called alkanes and represent the simplest and least reactive class of organic compounds.

You should note that "hybridization" is a simple mathematical process which is useful in modifying electron densities, as predicted from the classical hydrogen atom analysis, to allow bonding geometries and electron densities in more complex molecules to be described. Again, the driving force for "hybridization" is the formation of a bonding geometry with the lowest net potential energy, which is accessible by energetically feasible mEANS. The tendency of molecules to seek to form the lowest energy structures (i.e., the most stable reaction intermediate) is a commonly occurring theme in organic reactions and a knowledge of intermediate stability will often allow the direction of a reaction pathway to be accurately predicted.

Two additional orbital hybrids are also available for carbon; one in which the 2s-orbital combines with two of the available p-orbitals (sp²) and a second in which the 2s combines with one p (sp). The geometry predicted for the sp² hybrid is trigonal with the unused p-orbital perpendicular to the plane. Carbons which are sp² must be connected to at least one additional sp² atom and organic molecules containing one or more pairs of sp² carbons are called alkenes. A model for the simplest alkene, ethene (or ethylene) is shown below.

As the drawing shows, the hydrogens and the second carbon are attached to the sp² carbon by 120^o bond angles. The p-orbital on each of the sp² carbons each contain one electron ("split" between the top and bottom lobes) and the two adjacent p-orbitals are in a position to overlap with each other to form a p-bond in which electron density is concentrated between the atoms, in a diffuse orbital above and below the plane. The combination of the single bond, formed from the sp² orbitals, and the p-bond, are represented as a carbon-carbon double bond. In this representation, the valence of carbon is still four, with the double bond counting for two. The final hybrid which is common in organic chemistry is constructed from one s- and one p-orbital, termed an sp-hybrid, and has linear geometry, as shown below. The two remaining p-orbitals on each sp atom each participate in p-bond formation with an adjacent sp center and form a "triple bond"; one sigma bond and two p-bonds.

The three types of hybridization which are encountered in carbon compounds are shown below for the molecules ethane (sp³, an alkane), ethene (sp², an alkene) and ethyne (sp, an alkyne) in both ball-and-stick and space-filling format.