Chemistry 2e · Advanced Theories of Covalent Bonding
Hybrid Atomic Orbitals
On this page 8 sections
In 30 seconds
Topic 1 left a loose end: valence bond theory explains bonds as orbital overlaps, but carbon in its ground state (2s2 2p2) should form only two bonds at 90°, yet methane has four identical bonds at 109.5°. Hybridization resolves this by mathematically mixing the atomic orbitals on an atom to create new, equivalent hybrid orbitals before bonding occurs.
The key idea: the number of hybrid orbitals equals the number of atomic orbitals mixed. An s orbital plus three p orbitals gives four sp3 hybrids; an s plus two p's gives three sp2 hybrids; an s plus one p gives two sp hybrids. The hybrid orbitals point in specific directions, and those directions are exactly the electron-pair geometries VSEPR predicts: tetrahedral (109.5°), trigonal planar (120°), and linear (180°). Adding d orbitals produces sp3d (trigonal bipyramidal) and sp3d2 (octahedral) hybrids for expanded octets.
Hybridization is not a claim that atoms "really" have these orbitals before bonding. It is a mathematical model — a way of re-describing the atom's orbitals so that VB theory's overlap picture matches observed geometry and bond equivalence.
Why this matters
- One theory, all geometries: Hybridization lets a single framework (orbital overlap) explain linear CO2, trigonal planar BF3, tetrahedral CH4, bent H2O, and octahedral SF6 — the shapes that determine how molecules interact.
- Biology and medicine: The shape of a molecule controls what it can bind to. The planar sp2 geometry of the peptide bond and the tetrahedral sp3 carbons in amino acids shape proteins; drug designers think in hybridization terms because sp3 carbons add 3-D "bulk" while sp2 regions stay flat.
- Reactivity: Hybridization affects bond angles, bond strengths, and acidity (an sp-hybridized C–H bond is more acidic than an sp3 one because the sp hybrid holds electrons closer to carbon, stabilizing the anion).
- Exams: Hybridization questions are among the most common in general chemistry — identify the hybridization, count σ/π bonds, and predict geometry.
The college version
Core Concepts
How mixing works
Hybridization combines one s orbital with n orbitals of similar energy (usually p, sometimes d) to make n+1 equivalent hybrid orbitals. The hybrids keep the total number of orbitals and electrons — nothing is created or destroyed; the atom's orbitals are just re-packaged so they point in the directions where bonds actually form.
sp3: four orbitals, tetrahedral
Mixing one s + three p gives four sp3 hybrids pointing to the corners of a tetrahedron, 109.5° apart. Examples:
- CH4 — four σ bonds to H, 109.5°.
- NH3 — three sp3 hybrids bond to H; the fourth holds the lone pair Valence electron pair not involved in bonding Full entry →. The lone pair pushes the N–H bonds down to about 107°.
- H2O — two sp3 hybrids bond to H; two hold lone pairs. Lone-pair repulsion squeezes the bond angle to about 104.5°.
A common error is to think that only the bonding orbitals count. Lone pairs occupy hybrids too: the hybridization of the central atom is decided by the total number of electron domains (bonding pairs + lone pairs), exactly as in VSEPR.
sp2: three orbitals, trigonal planar
One s + two p → three sp2 hybrids in a plane at 120°. The remaining unhybridized p orbital sticks out above and below the plane and is available for a π bond. Example: BF3 (three σ bonds, no lone pairs, 120°) and the carbon atoms in ethene, C2H4 (three σ domains each: two C–H and one C–C).
sp: two orbitals, linear
One s + one p → two sp hybrids 180° apart. Two unhybridized p orbitals remain, both perpendicular to the axis, ready to form two π bonds. Examples: BeCl2 and the carbon in CO2 or HCCH (ethyne).
Adding d orbitals: expanded octets
Atoms in period 3 and beyond can use d orbitals:
- sp3d (one s + three p + one d) → five hybrids, trigonal bipyramidal (e.g., PCl5).
- sp3d2 (one s + three p + two d) → six hybrids, octahedral (e.g., SF6).
The rules are the same: total electron domains = number of hybrids = geometry.
A summary shortcut table
| Hybridization | Orbitals mixed | Number of hybrids | Electron-domain geometry | Bond angle |
|---|---|---|---|---|
| sp | one s + one p | 2 | linear | 180° |
| sp2 | one s + two p | 3 | trigonal planar | 120° |
| sp3 | one s + three p | 4 | tetrahedral | 109.5° |
| sp3d | one s + three p + one d | 5 | trigonal bipyramidal | 90°/120° |
| sp3d2 | one s + three p + two d | 6 | octahedral | 90° |
How It Works / Step-by-Step Process
Worked example 1: assign hybridization and geometry to CH4, NH3, and H2O
Problem. Determine the hybridization of the central atom in each molecule.
Solution.
- Count electron domains around the central atom (bonding pairs + lone pairs):
- CH4: 4 bonds, 0 lone pairs → 4 domains → sp3, tetrahedral, 109.5°.
- NH3: 3 bonds + 1 lone pair → 4 domains → sp3; observed angle ~107° (lone pair compresses it).
- H2O: 2 bonds + 2 lone pairs → 4 domains → sp3; observed angle ~104.5°.
- Rule: the hybridization depends on domains, not on bonds alone — all three are sp3.
Worked example 2: predict hybridization of carbon in CO2 and C2H4
Problem. Find the hybridization and bond angles at each carbon.
Solution.
- CO2: carbon has 2 double bonds → 2 electron domains → sp, linear, 180°. The two unhybridized p orbitals form the two π bonds of the double bonds (one per C=O).
- C2H4: each carbon has 3 domains (two C–H single bonds + one C=C double bond) → sp2, trigonal planar, ~120° around each carbon. Each carbon's leftover p orbital overlaps sideways with the other carbon's → the π half of the double bond. All six atoms lie in one plane.
Worked example 3: account for the geometry of SF6 with an expanded octet
Problem. Sulfur in SF6 has six bonds. Assign hybridization and geometry.
Solution.
- Count domains: 6 S–F bonds, 0 lone pairs → 6 domains.
- Six hybrids require one s + three p + two d → sp3d2.
- Geometry: octahedral, all F–S–F angles 90°. The d orbitals make expanded octets possible — a reminder that hybridization can extend beyond s and p.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Number of bonds | Number of electron domains | Lone pairs count as domains; H2O has only 2 bonds but 4 domains → sp3, not sp |
| Hybridization "happening" physically | Hybridization as a mathematical model | No experiment separates "before" and "after" mixing; hybrids are a re-description that matches geometry and equivalence |
| sp2 carbon with a double bond | sp3 carbon with a single bond | sp2 has 3 domains and a leftover p orbital for a π bond; sp3 has 4 domains and no leftover p |
| Bond angle exactly 109.5° | Actual measured angle | Lone pairs compress angles (NH3 ~107°, H2O ~104.5°); 109.5° is the lone-pair-free ideal |
| Hybridization of the atom | Geometry of the molecule | Hybridization describes the central atom's orbitals; molecular shape also depends on which domains are lone pairs |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine you have a bag of round and squished balloons. Before making a model, you blow them all up into identical long balloons so they point in different directions — that's mixing orbitals into hybrids. Mix four of them and they naturally point to the four corners of a pyramid (tetrahedral). Mix three and they lie flat like a peace sign (trigonal planar). Mix two and they point straight out (linear). The direction the balloons point is the shape of the molecule.
Key takeaways
- Hybridization = mathematical mixing of atomic orbitals on one atom into equivalent hybrid orbitals; count is conserved.
- Total electron domains (bonds + lone pairs) determine hybridization, matching VSEPR.
- sp → 2 domains → linear (180°); sp2 → 3 domains → trigonal planar (120°); sp3 → 4 domains → tetrahedral (109.5°); sp3d → 5 domains → trigonal bipyramidal; sp3d2 → 6 domains → octahedral.
- Lone pairs occupy hybrids: NH3 (107°) and H2O (104.5°) are sp3 with lone-pair compression.
- Unhybridized p orbitals on sp2/sp atoms are reserved for π bonds (topic 3).
- CH4, NH3, H2O are all sp3; BF3, C2H4 carbons are sp2; CO2, C2H2 carbons are sp.
- Hybridization is a model, not a claim about a pre-existing physical state.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Why was hybridization introduced in VB theory?
Show answer
Because pure atomic orbitals cannot explain equivalent bonds and correct geometries — e.g., carbon's 2s2 2p2 predicts two 90° bonds instead of four 109.5° bonds in CH4.
How many hybrid orbitals result from mixing one s with three p? What geometry?
Show answer
Four sp3 hybrids → tetrahedral geometry, 109.5°.
What is the hybridization of oxygen in H2O, and why is the angle less than 109.5°?
Show answer
sp3 (2 bonds + 2 lone pairs = 4 domains); the two lone pairs repel the bonding pairs, compressing the angle to ~104.5°.
Which atoms in C2H4 are sp2, and what orbital makes the π bond?
Show answer
Both carbons are sp2; each carbon's leftover unhybridized 2p orbital overlaps sideways to form the π bond of the C=C double bond.
What hybridization does PCl5 (five bonds, no lone pairs) use?
Show answer
sp3d (one s + three p + one d) — five hybrids → trigonal bipyramidal.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- hybrid orbital
- New orbital made by mixing atomic orbitals on the same atom
- sp³ hybridization
- One s + three p → four equivalent hybrids
- sp² hybridization
- One s + two p → three hybrids + one p left over
- sp hybridization
- One s + one p → two hybrids + two p's left over
- electron domain
- Any bonding pair or lone pair around an atom
- lone pair
- Valence electron pair not involved in bonding
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.