9.3 Hybridization
PRE-HEALTH PREP
Hybridization
Combining Atomic Orbitals for the Perfect Bond Angles
General Chemistry 9.3
Hybridization is the process of combining atomic orbitals into new hybrid orbitals that account for experimentally observed bond angles in molecules, arising from the framework of valence bond theory. Five hybridization types — sp, sp2, sp3, sp3d, and sp3d2 — correspond to two through six electron domains and produce characteristic geometries ranging from linear (180°) to octahedral (90°). The hybridization of any atom can be determined by counting electron domains on its Lewis structure, where each bond (regardless of bond order) and each lone pair counts as one domain.
In this lesson you will learn:
- What hybrid orbitals are and the types of hybridization (sp, sp2, sp3 hybridization, etc.)
- How to identify an atom's hybridization from the Lewis structure
- The relationship between an atom's hybridization and its bond angles
Valence Bond Theory
Core Idea: Atomic Orbitals Overlap to Form Bonds
Before diving into hybridization itself, it helps to understand the theoretical framework that motivates it: valence bond theory. Valence bond theory makes two key claims. First, atoms use their unpaired electrons to form bonds — each of the two bonding atoms contributes one unpaired electron, and that pair of electrons is shared between them to create the bond. Second, the atomic orbitals from both atoms physically overlap during bond formation. The bond, in this picture, lives in the region where those two orbitals overlap. This overlap-based picture forms the foundation for understanding why hybrid orbitals — and the sigma bonds they form — have the geometries they do.
3 Examples: H2, HF, and F2
Three simple molecules illustrate this idea well.
In H2, each hydrogen atom has the electron configuration 1s1 — one electron in a spherical 1s orbital, and that electron is unpaired. When the two hydrogen atoms bond, their 1s orbitals overlap, placing both electrons in the shared region between the nuclei. This is the simplest possible picture of a covalent bond according to valence bond theory.
In HF, things are slightly more involved. Hydrogen still brings its 1s orbital, but fluorine has the configuration 1s2 2s2 2p5. Looking at the valence shell, the single unpaired electron resides in one of the 2p orbitals. The bond in HF therefore forms from the overlap of hydrogen's 1s orbital with fluorine's 2p orbital — two different orbital types combining to create one sigma bond.
In F2, both fluorine atoms have an unpaired electron in a 2p orbital, so the bond arises from the overlap of two 2p orbitals — one from each fluorine atom.
Looking Ahead — Molecular Orbital Theory: Valence bond theory is a useful and intuitive model, and it handles most of what we need in this chapter. However, a more complete (and more accurate) treatment is molecular orbital theory, which we will cover in the final lesson of this chapter. Whether or not your general chemistry course requires molecular orbital theory is roughly a coin flip — but you will definitely need valence bond theory and hybridization.
Why Hybridization Is Necessary
Here is where the story gets interesting. Valence bond theory works cleanly for diatomic molecules, but the moment we consider a polyatomic molecule — one where a central atom forms multiple bonds — the raw, unhybridized atomic orbitals run into a serious problem: they predict the wrong bond angles.
The molecular geometries we identified using VSEPR theory (linear, trigonal planar, tetrahedral, and so on) each have characteristic bond angles: 180°, 120°, 109.5°, etc. These angles are experimentally confirmed. But if we simply use the original s and p orbitals of an atom to form its bonds, the predicted angles come out wrong. Something else must be going on.
The answer is hybridization — the mixing of an atom's original atomic orbitals into a new set of equivalent hybrid orbitals whose spatial orientations match the observed bond angles. Hybridization is what bridges VSEPR theory and valence bond theory: VSEPR predicts the geometry, and hybridization explains why the bonds actually point in those directions. For atoms in the second period, only s and p orbitals are involved. For atoms in the third period and beyond that form expanded octets, d orbitals join the mix as well.
The Problem with Methane (CH4): Why sp3 Hybridization Is Needed
Methane, CH4, is the canonical example used to expose the flaw in using bare atomic orbitals for bonding. Carbon has the configuration 1s2 2s2 2p2. In its ground state, only two of its four valence electrons are unpaired. Valence bond theory says an atom uses its unpaired electrons to form bonds — which would imply carbon can only make two bonds. Yet we know carbon routinely forms four bonds, and with good reason: it has four valence electrons and needs four more to complete an octet.
Valence bond theory resolves this with the concept of promotion: one electron from the filled 2s orbital is promoted into the empty 2p orbital, producing four unpaired valence electrons and allowing four bonds to form. The analogy is a job promotion — the electron moves to a higher-energy position.
The Bond Angle Problem: 90° vs. 109.5°
Promotion gets us four unpaired electrons, but it introduces a second problem. After promotion, carbon's four valence electrons occupy one 2s orbital and three 2p orbitals. The three p orbitals are oriented along the x, y, and z axes — they are mutually perpendicular, separated by exactly 90°. If these p orbitals were the ones overlapping with the 1s orbitals of the four hydrogen atoms, the H–C–H bond angles would be 90°.
But they are not 90°. Experiment — and VSEPR theory — tells us methane has bond angles of exactly 109.5°. That is a 19.5-degree discrepancy that cannot be ignored. The p orbitals, as they stand, simply cannot account for the geometry of methane. Therefore, they cannot be the orbitals involved in forming the bonds. Something new must be created: sp3 hybrid orbitals.
The Key Insight: The mismatch between the 90° orientation of raw p orbitals and the experimentally measured 109.5° bond angles in methane is the entire reason hybridization exists as a concept. If p orbitals naturally pointed 109.5° apart, there would be no need to invoke hybrid orbitals at all.
What Is Hybridization?
Atomic orbitals — the familiar s, p, and d shapes — are not arbitrary drawings. They are mathematical solutions to the Schrödinger equation, expressed as three-dimensional wave functions. Because they are mathematical functions, they can be combined mathematically. The technique is called a linear combination of atomic orbitals: the wave functions are added together (in various proportions and with various signs) to generate brand-new wave functions.
Each new wave function describes a new orbital — a hybrid orbital. The critical rule is this: the number of hybrid orbitals produced always equals the number of atomic orbitals mixed. Mix four atomic orbitals, get four hybrid orbitals. Mix three, get three. Mix two, get two.
The energy of a hybrid orbital falls between the energies of the atomic orbitals used to create it. An sp3 hybrid, built from one 2s and three 2p orbitals, has energy higher than the 2s but lower than the 2p. The spatial orientations of the resulting hybrids are what give us the correct bond angles — that is precisely what makes hybridization so powerful as a model in general chemistry and organic chemistry alike.
Naming Convention: Hybrid orbital names are simply a list of the atomic orbitals combined to make them. One s + three p's = sp3. One s + two p's = sp2. One s + one p = sp. One s + three p's + one d = sp3d. No creativity required — the name tells you exactly what went into the mix.
How to Determine Hybridization: Count Electron Domains
The universal method for determining hybridization is to count the electron domains on the atom of interest in its Lewis structure. Each electron domain — a single bond, double bond, triple bond, or lone pair — requires exactly one hybrid orbital. The total number of electron domains equals the number of hybrid orbitals needed, which directly tells you the hybridization type. This method works for every hybridization type from sp through sp3d2.
sp3 Hybridization (4 Electron Domains, 109.5°)
When a central atom has four electron domains, it requires four hybrid orbitals — one to accommodate each domain. To produce four hybrid orbitals, four atomic orbitals must be combined: the 2s and all three 2p orbitals (px, py, pz). The result is a set of four equivalent sp3 hybrid orbitals.
Each sp3 hybrid orbital looks like an asymmetric dumbbell: one large, prominent lobe on one side and a much smaller lobe on the other. It is fatter and more directional than a pure p orbital. The four sp3 hybrids point toward the corners of a tetrahedron — separated by 109.5° — which is exactly the bond angle measured in molecules like methane. These sp3 hybrid orbitals are the ones that form sigma bonds with neighboring atoms.
Example: Methane (CH4)
Carbon in methane has four bonding pairs and zero lone pairs — four electron domains in total. Carbon hybridizes its 2s, 2px, 2py, and 2pz orbitals to produce four sp3 hybrids pointing 109.5° apart. Each sp3 hybrid overlaps with a hydrogen 1s orbital to form one C–H sigma bond. The result is the perfectly tetrahedral geometry of methane.
Example: Water (H2O)
Oxygen in water also has four electron domains: two O–H bonds and two lone pairs. It is also sp3 hybridized. Two of its four sp3 hybrid orbitals hold lone pairs; the other two overlap with hydrogen 1s orbitals to form sigma bonds. The lone pairs compress the H–O–H angle slightly below the ideal 109.5° (to about 104.5°), but the hybridization is still sp3.
Identifying sp3 Hybridization from Electron Domains
The shortcut for identifying sp3 hybridization directly from a Lewis structure is straightforward: count the electron domains on the central atom. Each electron domain — whether a bonding pair, lone pair, single bond, double bond, or triple bond — requires one hybrid orbital. Therefore, the number of electron domains equals the number of hybrid orbitals needed, which equals the number of atomic orbitals to mix.
Four electron domains → four hybrid orbitals needed → mix 1s + 3p → sp3 hybridization.
sp2 Hybridization (3 Electron Domains, 120°)
When a central atom has three electron domains, only three hybrid orbitals are needed. The atom mixes the 2s orbital with two of the three 2p orbitals, leaving one 2p orbital entirely unhybridized. The result is three sp2 hybrid orbitals and one leftover p orbital.
The three sp2 hybrids lie in a plane and point 120° apart — consistent with trigonal planar electron domain geometry. The unhybridized p orbital stands perpendicular to that plane, above and below it. This perpendicular p orbital has an important role to play: it is the orbital involved in forming the second bond of a double bond — a pi bond — as you will see in the next lesson on sigma and pi bonding. The sp2 hybrid orbitals themselves form the sigma bond framework.
Example: Formaldehyde (CH2O)
The central carbon in formaldehyde is bonded to two hydrogen atoms and one oxygen atom via a double bond, giving it exactly three electron domains (each bond — single or double — counts as one domain). Carbon is therefore sp2 hybridized: three sp2 hybrid orbitals point 120° apart in a plane, and one unhybridized p orbital stands perpendicular. The three sp2 hybrids form the framework of sigma bonds; the unhybridized p orbital participates in the pi bond of the C=O double bond.
In the molecule above the carbon atom has 3 electron domains as it is bonded to 3 other atoms but has no lone pairs. Therefore, it is sp2 hybridized and has bond angles of 120°. Incidentally, the oxygen atom also has 3 electrons domains (bonded to 1 atom and has 2 lone pairs) and is also sp2 hybridized.
Double Bonds Count as One Domain: A common mistake is to count the two bonds of a double bond as two separate electron domains. They are not — a double bond is a single electron domain. Always count groups of electrons (bonds, lone pairs), not individual bond lines. A carbon with one double bond and two single bonds has three domains and is sp2 hybridized.
sp Hybridization (2 Electron Domains, 180°)
When a central atom has only two electron domains, it needs only two hybrid orbitals. The 2s orbital mixes with just one 2p orbital, leaving two 2p orbitals entirely unhybridized. The result is two sp hybrid orbitals and two leftover p orbitals.
The two sp hybrids point in exactly opposite directions — 180° apart — which is the linear geometry. The two unhybridized p orbitals are each perpendicular to the sp axis and perpendicular to each other. These unhybridized p orbitals go on to form the two pi bonds of a triple bond, while the sp hybrids form the sigma bonds and house lone pairs.
Example: Carbon Monoxide (CO)
Carbon in carbon monoxide has two electron domains: the triple bond to oxygen counts as one domain, and the lone pair on carbon counts as the other. Carbon is therefore sp hybridized. One sp hybrid orbital participates in the sigma bond with oxygen; the other sp hybrid orbital houses the lone pair on the far side of carbon. The two remaining unhybridized p orbitals on carbon overlap with p orbitals on oxygen to form the two pi bonds of the triple bond — this will be fully explored in the next lesson.
Triple Bonds Count as One Domain: Just as a double bond counts as one electron domain, so does a triple bond. All three bonds in a triple bond — one sigma and two pi — collectively occupy one electron domain. An atom with a triple bond and nothing else (like the carbon in CO, which also has a lone pair) has only two domains and is sp hybridized.
2 Expanded-Octet Hybridization Types: sp3d and sp3d2
For atoms in the second period (carbon, nitrogen, oxygen, etc.), the s and p orbitals of the valence shell are the only options — and there are exactly four of them (one s + three p). This places a ceiling of four electron domains for second-period atoms, which is why they obey the octet rule.
Atoms in the third period and beyond have accessible d orbitals in their valence shell. When these atoms form five or six electron domains (expanded octets), the d orbitals are recruited into the hybridization scheme, producing two additional hybridization types.
- Five electron domains: Mix 1s + 3p + 1d → sp3d hybridization → five hybrid orbitals pointing toward the vertices of a trigonal bipyramid (90° and 120° bond angles).
- Six electron domains: Mix 1s + 3p + 2d → sp3d2 hybridization → six hybrid orbitals pointing toward the vertices of an octahedron (90° bond angles).
Example: Phosphorus Pentachloride (PCl5)
Phosphorus in PCl5 forms five bonds and has no lone pairs — five electron domains. It is sp3d hybridized, with five sp3d hybrid orbitals arranged in a trigonal bipyramidal geometry.
Example: Sulfur Hexafluoride (SF6)
Sulfur in SF6 forms six bonds with no lone pairs — six electron domains. It is sp3d2 hybridized, with six sp3d2 hybrid orbitals arranged in an octahedral geometry.
Curriculum Note: Many general chemistry courses omit sp3d and sp3d2 hybridization, and there is genuine theoretical debate about whether d-orbital participation accurately describes expanded-octet bonding. Check your syllabus. If your course covers expanded octets in VSEPR, it likely covers sp3d and sp3d2 hybridization as well — but not all courses do.
5 Hybridization Types: Summary Table
The table below consolidates all five hybridization types. Once you can count electron domains on a Lewis structure, you can instantly read off the hybridization, the electron domain geometry, the ideal bond angles, and which orbitals were mixed. This is the most efficient way to answer hybridization questions on a general chemistry exam.
| Hybridization | Electron Domains | Electron Domain Geometry | Bond Angles | Orbitals Mixed | Hybrid Orbitals Formed |
|---|---|---|---|---|---|
| sp | 2 | Linear | 180° | 1s + 1p | 2 sp hybrids + 2 unhybridized p |
| sp2 | 3 | Trigonal planar | 120° | 1s + 2p | 3 sp2 hybrids + 1 unhybridized p |
| sp3 | 4 | Tetrahedral | 109.5° | 1s + 3p | 4 sp3 hybrids |
| sp3d | 5 | Trigonal bipyramidal | 90°, 120° | 1s + 3p + 1d | 5 sp3d hybrids |
| sp3d2 | 6 | Octahedral | 90° | 1s + 3p + 2d | 6 sp3d2 hybrids |
Quick-Count Method: To find the hybridization of any atom in a Lewis structure: (1) count all electron domains on that atom — each single bond, double bond, triple bond, and lone pair each count as one domain; (2) look up that number in the table above. That is all there is to it. You do not need to understand the underlying wave-function mathematics to apply hybridization correctly on an exam.
8 Key Takeaways
- Valence bond theory describes bonds as regions where atomic orbitals from two atoms overlap, sharing a pair of electrons — one contributed by each atom.
- Hybridization is necessary because raw atomic orbitals predict the wrong bond angles. The p orbitals of carbon, for example, are 90° apart, but the measured bond angles in methane are 109.5°.
- Hybrid orbitals are formed by combining (taking linear combinations of) atomic orbitals. The number of hybrid orbitals produced always equals the number of atomic orbitals mixed.
- The number of electron domains determines hybridization: 2 domains → sp; 3 domains → sp2; 4 domains → sp3; 5 domains → sp3d; 6 domains → sp3d2.
- Bond angles arise naturally from the geometry of the hybrid orbitals: sp hybrids point 180° apart (linear), sp2 hybrids 120° apart (trigonal planar), sp3 hybrids 109.5° apart (tetrahedral).
- Leftover unhybridized p orbitals remain after mixing: sp hybridization leaves two unhybridized p orbitals; sp2 leaves one. These unhybridized p orbitals are the orbitals involved in forming pi bonds in double and triple bonds.
- Expanded-octet hybridization (sp3d and sp3d2) applies to third-period atoms and beyond that form five or six electron domains, where d orbitals are recruited into the hybrid set.
- The shortcut: count electron domains on any atom in a Lewis structure — each bond (regardless of order) and each lone pair counts as one domain — and you can immediately identify the hybridization.