Chemistry: Atoms First 2e · Advanced Theories of Bonding
Hybrid Atomic Orbitals
On this page 9 sections
In 30 seconds
Valence bond theory explains bonds as orbital overlap, but it fails for real molecules: carbon's 2s and three 2p orbitals should give three bonds at 90° plus a different fourth bond, yet methane has four identical C–H bonds at 109.5°. Hybridization fixes this by mathematically mixing atomic orbitals into new hybrid orbitals aimed at the bonds. The number of hybrids equals the number of orbitals mixed: one s + three p → four sp³; one s + two p → three sp²; one s + one p → two sp. Hybridization is a bookkeeping model assigned after the geometry is known.
Why this matters
Molecular shape controls what molecules can do. Water's bent 104.5° shape, from sp³ hybridization One s + three p → four equivalent tetrahedral orbitals (109.5°) of oxygen, makes it a hydrogen-bonding solvent — essential to life. The flat geometry of ethene, aromatic rings, and DNA bases comes from sp² carbons whose leftover p orbitals form π bonds; that planarity shapes how drugs fit receptors. Hybridization also explains a measurable trend: more s character Fraction of s in a hybrid (sp³ 25%, sp² 33%, sp 50%) Full entry → (sp³ → sp² → sp) means shorter, stronger bonds. Exam tip: count electron groups and the label follows.
The college version
Core Concepts
The problem hybridization solves
Pure atomic orbitals cannot reproduce observed geometries: the 2p orbitals are mutually perpendicular (90°) and the 2s orbital is spherical, predicting right-angle, unequal bonds. Hybridization mixes the orbitals so each new hybrid has one large, directional lobe aimed at a bonding partner — and because all four sp³ hybrids are identical, methane's bonds come out equivalent.
sp³ hybridization: four electron groups
Mixing one s with all three p orbitals gives four equivalent sp³ orbitals arranged tetrahedrally, 109.5° apart. Any atom with four electron groups (σ bonds + lone pairs) uses sp³: carbon in methane, nitrogen in ammonia, oxygen in water. Lone pairs occupy hybrid orbitals too (107° for NH₃, 104.5° for H₂O).
sp² hybridization: three electron groups
Mixing one s with two p orbitals gives three equivalent sp² orbitals in one plane, 120° apart, leaving one unhybridized p orbital p orbital left out of the mixing Full entry → perpendicular to that plane. Atoms with three electron groups use sp² (boron in BF₃, carbon in ethene and carbonate). The leftover p orbital overlaps side-by-side with a neighbor's to form a π bond — exactly how the C=C double bond works.
sp hybridization: two electron groups
Mixing one s with just one p orbital gives two sp orbitals pointing in opposite directions, 180° apart, leaving two unhybridized p orbitals perpendicular to the axis. Atoms with two electron groups use sp: carbon in CO₂ and ethyne, beryllium in BeCl₂. The leftover p orbitals give ethyne its second and third bonds (one σ + two π) and CO₂ its two C=O double bonds. sp systems are linear.
Lone pairs count as electron groups
Hybridization depends on the total number of electron groups — σ bonds and lone pairs — not bond count alone. Ammonia has three N–H bonds but four electron groups (one lone pair), so nitrogen is sp³.
Expanded octets: d-orbital hybrids (with a caveat)
Period-3 and heavier atoms can host five or six electron groups, requiring d-orbital mixing: sp³d gives trigonal bipyramidal geometry (PCl₅), sp³d² an octahedron (SF₆). Model limitation: modern computational chemistry often describes these "hypervalent" molecules without d-orbital participation; for OpenStax-style problems, use sp³d/sp³d² for five/six groups.
How It Works / Step-by-Step Process
- Draw the Lewis structure.
- Count electron groups on the central atom: each σ bond = one, each lone pair = one (π bonds don't count here).
- Match: 2 → sp, 3 → sp², 4 → sp³, 5 → sp³d, 6 → sp³d².
- Note p orbitals left unmixed: 4 groups → none; 3 → one; 2 → two — these make π bonds, while hybrids hold σ bonds and lone pairs.
- Check angles: 180°, 120°, or ~109.5° (compressed by lone pairs).
Common Confusions
| Do Not Confuse | With | The Difference |
|---|---|---|
| Hybridization (a model) | A real event before bonding | We assign hybridization to explain geometry, not the reverse |
| sp² carbon | sp³ carbon | 3 groups → sp² (planar, one p for π); 4 groups → sp³ (tetrahedral) |
| Number of bonds | Number of electron groups | NH₃: 3 bonds but 4 groups (one lone pair) → sp³ |
| Hybrid orbitals | Molecular orbitals (next topic) | Hybrids belong to one atom; MOs spread over the molecule |
| π bonds in hybrid orbitals | π bonds from unhybridized p orbitals | Hybrids make σ bonds and hold lone pairs; π bonds need leftover p |
| "sp³ always means 109.5°" | Lone-pair-compressed angles | NH₃ (107°) and H₂O (104.5°) are sp³ but bent; expect deviations |
| sp³d hybridization | Proven fact for expanded octets | A textbook model with known limitations; advanced treatments differ |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine an artist melting one round brush and three flat ones into four identical brushes pointing in four directions. Atoms do the same: they mix their round s orbital and dumbbell-shaped p orbitals into "hybrid" orbitals that all look alike and point where the bonds go.
Worked example
Example 1: Percent s character across hybrid types
S character is the fraction of mixed orbitals that came from an s orbital:
% s character = number of s orbitals mixedtotal number of orbitals mixed × 100%
Substitute for each hybrid type:
sp3: 14 × 100% = 25% sp2: 13 × 100% ≈ 33% sp: 12 × 100% = 50%
This matches real measurements: C–H bonds shorten as s character rises — ethane (sp³) ≈ 1.096 Å, ethene (sp²) ≈ 1.087 Å, ethyne (sp) ≈ 1.060 Å.
Example 2: Assigning hybridization by electron-group count
Count electron groups for each central atom, then name hybridization and geometry.
| Molecule | Electron groups | Hybridization | Geometry (angle) |
|---|---|---|---|
| CO₂ | 2 double bonds = 2 | sp | linear (180°) |
| BF₃ | 3 single bonds = 3 | sp² | trigonal planar (120°) |
| CH₄ | 4 single bonds = 4 | sp³ | tetrahedral (109.5°) |
| H₂O | 2 bonds + 2 lone pairs = 4 | sp³ | bent (104.5°) |
| SF₆ | 6 single bonds = 6 | sp³d² | octahedral (90°) |
Water is the classic trap: only two bonds but four electron groups — oxygen is sp³, not sp²; the lone pairs' repulsion squeezes the angle from 109.5° to 104.5°.
Example 3: Converting an sp C–H bond length (dimensional analysis)
The C–H bond in ethyne (sp carbon) is ≈1.060 Å. Express it in pm and nm using 1 Å = 100 pm and 1 nm = 1000 pm, setting up so units cancel:
1.060 Å × 100 pm1 Å = 106.0 pm
106.0 pm × 1 nm1000 pm = 0.1060 nm
For comparison, ethane's sp³ C–H bond (≈1.096 Å) is longer than ethyne's — the same shortening predicted in Example 1.
Key takeaways
- Hybrid orbitals = mathematical mixtures of atomic orbitals; number made = number mixed (s+3p → 4 sp³; s+2p → 3 sp²; s+1p → 2 sp)
- Hybridization is assigned after geometry is known — a model, not a physical event.
- 2 electron groups → sp (180°); 3 → sp² (120°); 4 → sp³ (109.5°); 5 → sp³d; 6 → sp³d².
- Lone pairs count as electron groups: NH₃ and H₂O are sp³, not sp².
- Unhybridized p orbitals form π bonds: sp³ none, sp² one, sp two.
- More s character (sp³ 25% → sp² 33% → sp 50%) → shorter, stronger bonds; lone pairs compress angles (NH₃ ≈ 107°, H₂O ≈ 104.5°).
- sp³d/sp³d² (expanded octets) is a model with known limits.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Why can't pure 2s and 2p orbitals explain methane's four identical C–H bonds?
Show answer
Pure s and p orbitals aren't directional enough — they'd give three 90° bonds plus one unequal bond; four equivalent 109.5° bonds need four identical directional orbitals.
How many hybrid orbitals form when one s and two p orbitals are mixed, and what geometry results?
Show answer
Three sp² orbitals, trigonal planar at 120°, one p orbital left unmixed.
Which hybridization leaves two unhybridized p orbitals, and what do they do?
Show answer
sp hybridization leaves two unhybridized p orbitals that overlap side-by-side to form the two π bonds of a triple bond or double bonds (ethyne, CO₂).
Water has only two bonds — why is oxygen sp³?
Show answer
Hybridization depends on electron groups, not bond count: two O–H bonds + two lone pairs = four groups → sp³, and lone-pair repulsion compresses the angle to 104.5°.
What happens to C–H bond length as s character increases sp³ → sp² → sp?
Show answer
Bond length decreases: ≈1.096 Å (sp³) → ≈1.087 Å (sp²) → ≈1.060 Å (sp).
What are the hybridization and geometry of the carbon in CO₂?
Show answer
Two electron groups (each C=O contributes one σ bond) → sp, linear at 180°.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- hybrid orbital
- Orbital made by mathematically mixing an atom's s and p (sometimes d) orbitals
- sp³ hybridization
- One s + three p → four equivalent tetrahedral orbitals (109.5°)
- sp² hybridization
- One s + two p → three planar orbitals (120°) plus one leftover p
- sp hybridization
- One s + one p → two linear orbitals (180°) plus two leftover p's
- unhybridized p orbital
- p orbital left out of the mixing
- electron group
- Any σ bond or lone pair around a central atom
- s character
- Fraction of s in a hybrid (sp³ 25%, sp² 33%, sp 50%)
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.