sp3 hybridization is the mixing of one s orbital and three p orbitals on an atom to form four equivalent hybrid orbitals that point toward the corners of a tetrahedron (109.5° apart), which happens whenever a central atom has four regions of electron density.
sp3 hybridization is what happens when an atom mixes one s orbital and three p orbitals into four new, identical hybrid orbitals. Each one is 25% s character and 75% p character, and the four of them spread out as far apart as possible, pointing at the corners of a tetrahedron with 109.5° angles between them. The classic example is carbon in methane (CH₄), where four equal C-H bonds at 109.5° would be impossible with plain s and p orbitals.
Here's the shortcut that makes it click: hybridization is just VSEPR counted from the orbital side. Count the regions of electron density around the central atom (bonds and lone pairs both count, and a double or triple bond counts as one region). Four regions means sp3. That's it. Water's oxygen is sp3 too, even though its molecular geometry is bent, because two bonding pairs plus two lone pairs still equals four electron domains.
This term lives in Unit 2 (Compound Structure and Properties), Topic 2.7, under learning objective 2.7.A, which asks you to connect Lewis diagrams and VSEPR theory to explain a molecule's structural and electronic properties. Per essential knowledge 2.7.A.1, VSEPR is built on Coulombic repulsion between electron pairs, and sp3 hybridization is the orbital-level story of that repulsion. Four electron domains repel each other into a tetrahedral arrangement, and sp3 orbitals are the orbitals that actually point in those directions. The AP exam keeps hybridization deliberately simple. You only need sp, sp2, and sp3, and the whole game is counting electron domains and matching them to the right hybridization, geometry, and bond angle.
Keep studying AP Chemistry Unit 2
Molecular Geometry (Unit 2)
sp3 hybridization sets the electron geometry (tetrahedral), but lone pairs can change the molecular geometry you actually name. CH₄ is tetrahedral, NH₃ is trigonal pyramidal, and H₂O is bent, yet all three central atoms are sp3.
sp2 hybridization (Unit 2)
Drop one electron domain and you drop one p orbital from the mix. sp2 means three domains, trigonal planar, 120° angles, and a leftover unhybridized p orbital that forms the pi bond in a double bond. sp3 atoms have no leftover p orbital, which is why they only make single (sigma) bonds.
sp Hybridization (Unit 2)
Two electron domains gives sp hybridization, linear geometry, and 180° angles, like the carbon in CO₂. Together sp, sp2, and sp3 are the complete hybridization menu for the AP exam, mapped directly to 2, 3, and 4 electron domains.
Covalent Bonding (Unit 2)
Hybrid orbitals are how covalent bonds actually form in this model. Each sp3 orbital overlaps head-on with an orbital from another atom to make a sigma bond, or holds a lone pair, which is the orbital-level version of the shared electron pairs in your Lewis diagram.
Hybridization shows up almost entirely as a counting task. Multiple-choice stems give you a central atom's bonding situation and ask you to name the hybridization, the geometry, or the bond angle. For example, a carbon with four single bonds at 109.5° (like methane) should instantly read as sp3 and tetrahedral, while two bonds at 180° should read as sp. The reverse direction gets tested too, where the question names the hybridization and you supply the angle (sp3 means about 109.5°). On FRQs, hybridization typically appears as one part of a larger structure question, where you draw a Lewis diagram, then identify the hybridization of a specific atom and justify it by counting electron domains. The justification matters; write "four regions of electron density, therefore sp3" rather than just the label.
The difference is one electron domain. sp3 means four domains, tetrahedral electron geometry, and 109.5° angles; sp2 means three domains, trigonal planar geometry, and 120° angles. The structural giveaway is double bonds. An sp2 atom has an unhybridized p orbital available for a pi bond, so a carbon in a C=C double bond is sp2, while a carbon with four single bonds is sp3. If you mix these up, you'll also get the geometry and bond angle wrong, so the errors snowball.
sp3 hybridization mixes one s orbital and three p orbitals into four equivalent hybrid orbitals arranged tetrahedrally at 109.5°.
An atom is sp3 hybridized whenever it has four regions of electron density, counting both bonds and lone pairs, with multiple bonds counting as one region.
Lone pairs don't change the hybridization, so the oxygen in bent H₂O and the nitrogen in trigonal pyramidal NH₃ are both sp3 just like the carbon in tetrahedral CH₄.
sp3 atoms have no leftover unhybridized p orbitals, so they form only single (sigma) bonds, never pi bonds.
On the exam, justify a hybridization answer by counting electron domains from the Lewis diagram, not by memorizing molecules.
It's the mixing of one s and three p orbitals into four equivalent hybrid orbitals that point toward the corners of a tetrahedron, 109.5° apart. Any atom with four electron domains (bonds plus lone pairs) is sp3 hybridized.
No. sp3 guarantees a tetrahedral electron geometry, but lone pairs can shrink the molecular shape you name. NH₃ (one lone pair) is trigonal pyramidal and H₂O (two lone pairs) is bent, yet both central atoms are sp3.
sp3 comes from four electron domains and gives 109.5° tetrahedral angles; sp2 comes from three domains and gives 120° trigonal planar angles. sp2 atoms also keep one unhybridized p orbital for a pi bond, which is why double-bonded carbons are sp2 and four-single-bond carbons are sp3.
Approximately 109.5°, the tetrahedral angle. Methane (CH₄) is the textbook example, with four C-H bonds at exactly 109.5°; lone pairs squeeze that angle slightly smaller, like the roughly 104.5° angle in water.
Yes. Hybridization is determined by total electron domains, so a lone pair counts the same as a bond. That's why oxygen in water (two bonds plus two lone pairs) is sp3 even though it only forms two bonds.