Hybridization and Sigma and Pi Bonds: Orbitals Behind Molecular Shape

Bonds as overlapping orbitals

Valence bond theory describes a covalent bond as two electrons with opposite spins in a region where orbitals on two atoms overlap. The overlap concentrates electron density between the nuclei, where it attracts both. More overlap generally means a stronger bond.

Overlap can happen in two ways. In a sigma (σ) bond, the orbitals meet end to end, along the line joining the nuclei, so the electron density is greatest on that axis. In a pi (π) bond, two parallel p orbitals overlap side by side, so the density lies above and below the axis, with a nodal plane containing the axis. Every single bond is a σ bond.

Why carbon's 2s and 2p orbitals must mix

Carbon's valence orbitals are one 2s and three 2p. The three p orbitals point along x, y and z, at 90° to each other. If methane's four C–H bonds used them directly, three bonds would be at 90° and one would be different. Instead, all four bonds in CH₄ are identical and 109.5° apart.

Hybridization resolves this. Mathematically combining the 2s orbital with all three 2p orbitals gives four equivalent sp³ hybrid orbitals pointing to the corners of a tetrahedron. The number of hybrid orbitals always equals the number of atomic orbitals mixed. Each sp³ orbital has one large lobe that overlaps well with a hydrogen 1s orbital, which is why hybrid orbitals form strong σ bonds.

Assigning hybridization from the steric number

Hybridization follows from the number of electron domains around an atom, its steric number from VSEPR. Lone pairs count as domains, and a multiple bond counts once. Once you know the steric number, the hybridization is fixed.

Lone pairs also occupy hybrid orbitals and push the bonds closer together. OpenStax Chemistry 2e (Section 8.2) gives 107.3° for the H–N–H angle in ammonia and 104.5° for H–O–H in water, both sp³ with bond angles below the ideal 109.5°.

• Steric number 2: sp, linear, 180°. One s and one p mix, leaving two unhybridized p orbitals. Each sp orbital is 50% s character.
• Steric number 3: sp², trigonal planar, 120°. One s and two p mix, leaving one p orbital perpendicular to the plane. 33% s character.
• Steric number 4: sp³, tetrahedral, 109.5°. One s and three p mix, leaving none. 25% s character.
• Steric numbers 5 and 6 are often labeled sp³d and sp³d², as in OpenStax. Treat these as geometry labels; see the limitations below.

Double and triple bonds as sigma plus pi

In ethylene, H₂C=CH₂, each carbon has three domains and is sp². The three sp² orbitals form σ bonds to two hydrogens and the other carbon. The leftover p orbitals on the two carbons are parallel and overlap side by side to form a π bond. A double bond is therefore one σ bond plus one π bond. A triple bond, as in acetylene, HC≡CH, is one σ bond plus two π bonds, made from two pairs of unhybridized p orbitals at right angles to each other on sp carbons.

The π bond explains why double bonds do not rotate freely. Twisting one end by 90° would make the p orbitals perpendicular and break the π overlap. That is why cis-2-butene and trans-2-butene are different compounds that do not interconvert at room temperature, while the two ends of a C–C single bond rotate easily.

Worked examples: counting bonds in acrylonitrile and aspirin

To count σ and π bonds, draw a complete Lewis structure, including every hydrogen. Then every bond line contributes exactly one σ bond, each double bond adds one π bond, and each triple bond adds two.

Worked example 1. Acrylonitrile, H₂C=CH–C≡N, used to make acrylic fibers. σ bonds: three C–H, one in C=C, one in C–C and one in C≡N, a total of 6. π bonds: one in C=C and two in C≡N, a total of 3. Hybridization: both carbons of the double bond have three domains and are sp²; the nitrile carbon has two domains and is sp; nitrogen, with one bond domain and one lone pair, is also sp. As a check, the valence electrons are 3(4) + 3(1) + 5 = 20: 9 bonds hold 18, and nitrogen's lone pair holds the other 2.

Worked example 2. Aspirin, C₉H₈O₄, drawn with alternating double bonds in the benzene ring. σ bonds: 6 C–C in the ring, 4 ring C–H, and 4 in the carboxylic acid group plus its link to the ring (ring C–C, C=O, C–O and O–H), and 7 in the ester group (ring C–O, O–C, C=O, C–CH₃ and three C–H). That gives 6 + 4 + 4 + 7 = 21 σ bonds. π bonds: 3 in the ring and 1 in each C=O, 5 in total. Check: aspirin has 21 atoms and one ring, and any connected structure has (atoms − 1 + rings) σ bonds, 21 − 1 + 1 = 21. The electron count also works: 26 bonds hold 52 electrons, the four oxygens hold eight lone pairs, or 16 electrons, and 9(4) + 8(1) + 4(6) = 68. Eight carbons are sp²; only the methyl carbon is sp³.

How strong is a pi bond compared with a sigma bond?

Average bond energies in OpenStax Chemistry 2e Table 7.3 show that adding bonds between two atoms gives diminishing returns. For carbon–carbon bonds: C–C, 1.54 Å and 345 kJ/mol; C=C, 1.34 Å and 611 kJ/mol; C≡C, 1.20 Å and 837 kJ/mol.

Worked example 3. Subtracting gives a rough energy for each added π bond. The first π bond adds 611 − 345 = 266 kJ/mol, and the second adds 837 − 611 = 226 kJ/mol. Both are smaller than the 345 kJ/mol of a C–C σ bond, because side-by-side overlap is weaker than end-to-end overlap. A double bond is only 1.77 times as strong as a single bond, not twice as strong. The weaker π bond is the part that breaks in addition reactions, which is why alkenes add Br₂ and HBr readily, and H₂ in the presence of a metal catalyst. Treat these differences as estimates; average bond energies combine values from many different molecules.

Delocalized pi electrons and the limits of hybrid orbitals

In benzene, C₆H₆, every carbon is sp² and has one unhybridized p orbital perpendicular to the ring. Six parallel p orbitals overlap all the way around, so the six π electrons are spread over the whole ring rather than fixed in three double bonds. All six C–C bonds have the same length, between the typical single and double bond lengths.

Resonance rearranges π electrons and lone pairs, never σ bonds, so for atoms like the carbons in benzene any valid structure gives the same hybridization; OpenStax makes this point explicitly for benzene. Take care with an atom whose lone pair becomes a π bond in another structure, such as the nitrogen in an amide: its domain count changes between structures, and its observed planar geometry shows that it is sp². Delocalization is an area where valence bond theory becomes awkward, and molecular orbital theory gives a cleaner description.

Valence bond theory pairs every electron in a bond or lone pair, so it predicts that O₂ has no unpaired electrons. Liquid oxygen is in fact attracted to a magnet, which shows that O₂ has unpaired electrons. Molecular orbital theory builds orbitals that belong to the whole molecule. Bond order is half the difference between electrons in bonding and antibonding orbitals. For O₂, the 12 valence electrons give 8 bonding and 4 antibonding, so bond order (8 − 4)/2 = 2. The last two electrons occupy two separate π* orbitals with parallel spins, explaining the paramagnetism. For N₂, (8 − 2)/2 = 3, matching its triple bond.

Hybridization errors and how to catch them

Most errors come from skipping the Lewis structure or from miscounting domains. Count σ bonds and lone pairs on the atom first; hybridization follows from that count alone.

• Counting π bonds toward the steric number. Only σ bonds and lone pairs count as domains.
• Forgetting lone pairs, which makes the nitrogen in NH₃ look sp² instead of sp³.
• Counting a double bond as two σ bonds. Every double bond is one σ and one π.
• Leaving out hydrogens when counting σ bonds in a skeletal or condensed formula.
• Assuming every carbon with a double bond is sp². The central carbon of CO₂, with two double bonds, has two domains and is sp.
• Believing that a resonance structure changes an atom's hybridization. Resonance moves π electrons, not σ bonds.