This result can be used to infer the hybridization scheme. Trigonal planar electron-pair geometry requires three sp2 hybrid orbitals, formed from the 2s, 2px, and 2py orbitals. Each of these hybrid orbitals requires an unpaired electron in order to be able to form three bonds. Hybridization begins with the lowest electrons in the lowest energy state. This means the carbon atom’s paired 2s electrons and the 2p electron combine to form the three sp2 orbitals using the px and py orbitals. These sp2 orbitals overlap end to end with the s orbitals of the hydrogen atoms and sp2 orbital of the other carbon atom, forming bonds.But carbon has four valence electrons. This means each carbon atom has one more electron in an unhybridized pz orbital. These two pz orbitals from the carbon atoms overlap along the x-axis in the side-by-side manner of a bond and are observed as a density of electrons above and below the plane of the molecule. This combination of and bonds produces the double bond. Triple bonds are modeled in a similar way. Consider the structure of ethyne (C2H2). Its Lewis structure is linear with triple-bonded carbon atoms. A linear geometry requires sp hybridization, which requires a minimum of two electrons, one for each sp orbital. In the case of ethyne, the two electrons in the s orbital of the carbon atom hybridize with the empty p orbital to form the two sp orbitals. The sp hybrids form bonds with the hydrogen atom and the other carbon atom. The two leftover electrons in the p orbitals form pi bonds with their counterparts on the other carbon atom. Thus, the triple bond is formed by one bond and two bonds. The two bonds form a sort of tube of electron density around the straight bond.
The bonds do not affect the shape of the molecule, but they introduce rigidity. Atoms are not able to easily rotate around a double bond in the way they can rotate around a single bond, and a triple bond is more rigid still.