8.2 Hybridization, VSEPR & Molecular Geometry

Key Takeaways

  • A single bond is always one sigma bond; a double bond is one sigma plus one pi bond; a triple bond is one sigma plus two pi bonds — pi bonds form from unhybridized parallel p orbitals and restrict rotation around the bond axis.
  • Hybridization is determined by the number of electron domains around an atom: four domains give sp3 (tetrahedral, 109.5°), three give sp2 (trigonal planar, 120°), and two give sp (linear, 180°).
  • VSEPR theory predicts molecular geometry from the repulsion between electron domains, and lone pairs occupy more space than bonding pairs — this converts the tetrahedral electron geometry of ammonia and water into trigonal pyramidal (~107°) and bent (~104.5°) molecular geometries, respectively.
  • As bond order increases from single to double to triple, bond length decreases and bond dissociation energy increases, because more shared electron pairs pull the bonded nuclei closer together with greater total binding energy.
Last updated: July 2026

Sigma and Pi Bonds

Every covalent bond, whether single, double, or triple, contains exactly one sigma (σ) bond — formed by the head-on, end-to-end overlap of two atomic or hybrid orbitals along the axis connecting the two nuclei. A pi (π) bond forms only in double and triple bonds, from the sideways overlap of two unhybridized, parallel p orbitals above and below (or in front of and behind) the sigma-bond axis.

  • A single bond = 1 sigma bond only.
  • A double bond = 1 sigma bond + 1 pi bond.
  • A triple bond = 1 sigma bond + 2 pi bonds (the two pi bonds occupy mutually perpendicular planes).

This distinction matters mechanically and geometrically. Because sigma-bond overlap is symmetric around the internuclear axis, the atoms joined by a sigma bond alone (a single bond) can rotate freely relative to each other without breaking any orbital overlap — this free rotation is the basis of conformational isomerism, covered in the next section. Pi-bond overlap, in contrast, depends on the parallel alignment of two p orbitals; rotating one end of a double bond by 90° relative to the other would break the pi overlap entirely, so double bonds are rotationally rigid and planar. This rigidity is what makes cis/trans (E/Z) isomerism possible around double bonds, and it is the same kind of resonance-driven rigidity that locks the protein backbone's peptide bond into a planar arrangement.

Hybrid Orbitals: sp3, sp2, and sp

An atom's hybridization is determined by its number of electron domains — the count of sigma bonds plus lone pairs surrounding it (a domain counts as one region regardless of whether it is a single, double, or triple bond). Mixing (hybridizing) the atom's available s and p orbitals produces a set of new, equivalent hybrid orbitals used for sigma bonding and lone pairs, while any leftover unhybridized p orbitals are reserved for pi bonding.

HybridizationElectron domainsOrbitals mixedIdealized geometryBond angleExample
sp341 s + 3 pTetrahedral109.5°Methane (CH4), ethane
sp231 s + 2 pTrigonal planar120°Ethylene (C2H4), formaldehyde, benzene carbons
sp21 s + 1 pLinear180°Acetylene (C2H2), the central carbon of CO2

An sp3 carbon has four equivalent hybrid orbitals and no leftover p orbitals, so it can only form sigma bonds (never pi bonds) — this is why every bond to an sp3 carbon is a single bond. An sp2 carbon has three hybrid orbitals for sigma bonds arranged in a plane, plus one unhybridized p orbital perpendicular to that plane available for exactly one pi bond. An sp carbon has two hybrid orbitals for sigma bonds arranged linearly, plus two unhybridized, mutually perpendicular p orbitals available for up to two pi bonds — enough for either two separate double bonds (as in the O=C=O of carbon dioxide) or one triple bond (as in the C≡C of acetylene, alongside one more sigma bond to another atom).

Test Your Knowledge

In formaldehyde (H2C=O), what is the hybridization of the carbon atom, and how many pi bonds does that carbon form?

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VSEPR and Molecular Geometry

Valence shell electron pair repulsion (VSEPR) theory predicts a molecule's three-dimensional shape by assuming that electron domains around a central atom arrange themselves as far apart as possible to minimize electrostatic repulsion. Critically, lone pairs are held closer to the nucleus and spread into more space than bonding pairs, so repulsion strength follows this order: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. This means lone pairs compress the bond angles of the surrounding bonding pairs below the idealized value predicted by electron-domain geometry alone.

Three worked examples show this progression clearly, each with four electron domains and sp3 hybridization at the central atom:

MoleculeBonding pairsLone pairsElectron geometryMolecular geometryBond angle
CH4 (methane)40TetrahedralTetrahedral109.5°
NH3 (ammonia)31TetrahedralTrigonal pyramidal~107°
H2O (water)22TetrahedralBent~104.5°

Methane has no lone pairs, so its molecular geometry matches its electron geometry exactly. Ammonia's single lone pair compresses the H-N-H angle from the ideal 109.5° down to about 107°. Water's two lone pairs compress the H-O-H angle even further, to about 104.5°, because two lone pairs now compete for space around the central oxygen. Carbon dioxide (CO2) illustrates a different case: the central carbon has only two electron domains (each C=O double bond counts as a single domain), giving a linear electron geometry and molecular geometry with a 180° bond angle and sp hybridization at carbon — this is also why CO2, despite having two polar bonds, has zero net dipole moment, as discussed in the previous section.

Exam strategy: always distinguish electron geometry (based on all domains, including lone pairs) from molecular geometry (based only on the positions of atoms, ignoring lone pairs) — the MCAT frequently asks for one or the other specifically.

Test Your Knowledge

Methane (CH4), ammonia (NH3), and water (H2O) all have four electron domains around the central atom, yet their bond angles decrease across that series from about 109.5° to about 107° to about 104.5°. What causes this trend?

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Typical Bonding Patterns for Common Elements

Quickly recognizing how many bonds and lone pairs a neutral atom typically forms lets you draw and check structural formulas without recounting electrons from scratch every time. For second-row elements, a useful shortcut is that the number of bonds a neutral atom forms tends to equal 8 minus its group number (assuming zero formal charge).

ElementTypical bondsTypical lone pairsNotes
H10Always terminal; never a central atom
C, Si40No lone pairs in typical neutral structures
N, P31P can expand its octet (e.g., PCl5, phosphate)
O, S22S can expand its octet (e.g., SO2, SO4^2-, SF6)
F, Cl13Halogens are always terminal in typical structures

Silicon behaves like carbon (group 14, four bonds, no lone pairs) and appears in inorganic and geological chemistry passages. Phosphorus and sulfur both belong to period 3 and, unlike their lighter second-row analogs nitrogen and oxygen, can expand beyond an octet — phosphorus reaches five bonds in the biologically essential phosphate groups of ATP and the DNA/RNA backbone, and sulfur reaches six bonds in sulfate and related oxyanions. Recognizing when an atom is exhibiting its typical bonding pattern versus an expanded or charged state is a fast way to verify a drawn Lewis structure or to spot when a nitrogen or oxygen must be carrying a formal charge.

Test Your Knowledge

Based on typical (lowest formal charge) bonding patterns for main-group atoms, how many covalent bonds and lone pairs does a neutral nitrogen atom typically have in an uncharged organic molecule?

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Delocalization, Resonance, and the Effects of Multiple Bonding

Resonance (introduced in the previous section as an electron-counting concept) has a direct structural consequence: when p orbitals on three or more adjacent atoms are all aligned in parallel, their electrons delocalize across the whole system rather than staying confined to one pi bond. In the carbonate ion (CO3^2-), the central sp2 carbon and all three oxygens contribute a p orbital to a delocalized pi system, making all three carbon-oxygen bonds identical in length (bond order 4/3) rather than two single bonds and one double bond. The guanidinium group of the arginine side chain shows the same phenomenon across three nitrogens, delocalizing positive charge so effectively that arginine remains protonated and positively charged at physiological pH (pKa near 12.5) — a key feature exploited by DNA-binding proteins and enzyme active sites.

Multiple bonding directly affects bond length and bond energy. As bond order increases from single to double to triple, more electron pairs are shared between the same two nuclei, pulling them closer together (shorter bond length) and requiring more energy to separate them (higher bond dissociation energy):

BondApproximate lengthApproximate bond energy
C-C (single)154 pm83 kcal/mol
C=C (double)134 pm146 kcal/mol
C≡C (triple)120 pm200 kcal/mol

Multiple bonding also affects rigidity. As discussed above, the pi bond(s) present in double and triple bonds prevent free rotation around the bond axis, locking the atoms directly attached to a double bond into a single, planar arrangement (all atoms directly bonded to a simple alkene lie in one plane). This rigidity is the structural basis for cis/trans and E/Z isomerism, and it is also why resonance-stabilized bonds with partial double-bond character — such as the peptide bond or the carbon-nitrogen bonds in the guanidinium group — impose planarity even though they may be formally drawn as single bonds in one resonance contributor.

Test Your Knowledge

Ranking carbon-carbon bonds by increasing bond order (single, double, triple), what happens to bond length and bond dissociation energy?

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