Look at methane, CH₄. Its carbon holds four hydrogens. The four C–H bonds are identical, and every pair of them meets at 109.5°. The electron configuration from the Orbitals section cannot explain that: it gives carbon only two unpaired electrons.
Hybridization is the idea that settles it. An atom mixes its own orbitals into new ones before it bonds. Once you can name an atom’s hybridization, that one label tells you its geometry and bond angles, whether it is flat or pyramidal, whether its lone pair is free to be shared, and how acidic its hydrogens are.
Resolving the carbon paradox
Ground-state carbon is 1s² 2s² 2p². Two of its electrons are unpaired, so it should form two bonds. It forms four. Two steps close the gap.
Promotion comes first. One of the paired 2s electrons moves up into the empty third 2p orbital. That gives 2s¹ 2p³: four unpaired electrons, one in each orbital. The move costs roughly 96 kcal/mol. Each C–H bond formed releases roughly 100 kcal/mol, so the two extra bonds bring in about 200 against a bill of 96. Bonding is so profitable that promotion happens essentially every time carbon bonds.
Hybridization comes second. Carbon now has four half-filled orbitals, but they are not alike. One is a round s orbital. Three are dumbbell-shaped p orbitals at right angles to each other. If carbon bonded with them as they are, methane would have three bonds at 90° and a fourth pointing some other way, and that fourth bond, made from the s orbital, would differ measurably from the other three. Methane shows neither.
So the atom mathematically mixes the four orbitals into four new, identical ones. Each new orbital is a hybrid orbital: a blend of the orbitals that went in. The four point as far apart as they can, toward the corners of a tetrahedron (a pyramid with four triangular faces), 109.5° from each other. Four orbitals go in and four come out; mixing changes their shape, never their number. These are sp³ orbitals: one part s, three parts p.
The three hybridizations, and what gets left over
An atom does not have to mix all three p orbitals. What it leaves unmixed matters as much as what it mixes.
- Mix 1 s with 3 p → four sp³ orbitals, 109.5° apart, tetrahedral. 0 p orbitals left over.
- Mix 1 s with 2 p → three sp² orbitals, 120° apart, all in one plane (trigonal planar). 1 p orbital left over, standing perpendicular to that plane.
- Mix 1 s with 1 p → two sp orbitals, 180° apart, linear. 2 p orbitals left over, both perpendicular to the axis and to each other.
Ethene, CH₂=CH₂, shows what each kind of orbital does. Both carbons are sp². Each carbon aims its three hybrids at two hydrogens and at the other carbon. Where two orbitals meet head-on, along the line between two nuclei, they form a sigma (σ) bond. Every single bond is a σ bond.
That leaves one p orbital on each carbon, standing at right angles to the flat molecule. The two p orbitals are parallel, so they overlap side by side, above and below the plane. That sideways overlap is a pi (π) bond. A double bond is one σ bond plus one π bond; the next section, Bonding, takes the two apart.
So the leftover p orbitals are what π bonds are made of. A double bond needs at least one leftover p on each atom, so an atom in a double bond is sp² or sp. It is sp when it has a second π bond as well, like the carbon of O=C=O below. A triple bond needs two leftover p orbitals on each atom, so both its atoms are sp.
A leftover p orbital can also hold a lone pair. When that p orbital sits next to a π bond, the pair overlaps sideways with the π bond instead of staying on its own atom. That arrangement makes amides flat, and it is the main exception to the shortcut below.
The shortcut you will actually use
Take one carbon of ethene again. It has two bonds to hydrogen and one double bond to the other carbon. The double bond points at a single neighbor, so it takes up one direction in space. That makes three directions, and three means sp².
That count is the shortcut. You do not redo promotion and mixing for every atom. You count its electron groups and read off the answer. An electron group is either a bond to a neighboring atom or a lone pair. A double or triple bond counts as one group, because all of it points at the same neighbor.
Methane, CH₄. The carbon has 4 bonds and 0 lone pairs → 4 groups → sp³, tetrahedral, 109.5°.
Carbon dioxide, O=C=O. The carbon has two double bonds, each counting once → 2 groups → sp, linear, 180°.
Ammonia, NH₃. Nitrogen has 3 bonds plus 1 lone pair → 4 groups → sp³. The lone pair counts. So ammonia is pyramidal, not flat, and its lone pair points in one definite direction. Because it points one way, that pair can reach out and bond to something else, as later chapters show.
Water, H₂O. Oxygen has 2 bonds plus 2 lone pairs → 4 groups → sp³, bent. The measured H–O–H angle is 104.5°, a little under the ideal 109.5°, because two lone pairs take more room than two bonds would. Molecular Geometry, later in this chapter, explains why the lone pairs squeeze the angles.
Keep hybridization and shape apart. Methane, ammonia and water are all sp³, because all three have four groups. Their shapes differ, tetrahedral, trigonal pyramidal and bent, because a shape describes where the atoms are and ignores the lone pairs. Molecular Geometry is about that second question.
A carbocation, CH₃⁺. A carbocation is a carbon with only three bonds and a positive charge, as Atomic Structure described. The carbon has 3 bonds and no lone pair → 3 groups → sp², trigonal planar, 120°. Its leftover p orbital is empty. That empty orbital, standing perpendicular to the flat carbon, is what makes a carbocation so reactive. Because it sticks out on both sides of the plane, an incoming electron pair can reach it from either face.
A carbanion, CH₃⁻. A carbanion is a carbon with three bonds, a lone pair and a negative charge. Three bonds plus 1 lone pair → 4 groups → sp³, pyramidal, with the lone pair in the fourth corner. It has the same three bonds as the cation; the difference is the pair, and the pair fills a corner.
A radical, CH₃•. A radical has a single unpaired electron. Here the carbon has three bonds plus that one electron. A lone electron needs much less room than a pair, so a methyl radical sits close to sp² and nearly flat, with its electron in the p orbital. The radical needs a caveat rather than an answer.
When the count misleads: a lone pair beside a π bond
Try the count on the nitrogen of an amide, R–C(=O)–NH₂. The nitrogen has 3 bonds and 1 lone pair, which is 4 groups. That predicts sp³ and a pyramid. Measured, the nitrogen is flat, and it is sp².
The cause is the C=O next door. Its π bond is made of two parallel p orbitals. The nitrogen puts its lone pair in a p orbital parallel to those two, so the pair overlaps sideways with the π bond. The pair is more stable spread over that overlap than kept in a hybrid on the nitrogen. When it does, the atom holding it has only three hybrid orbitals left, so it is sp².
The same thing happens wherever an atom that the count makes sp³ has a lone pair and is bonded to an atom with a π bond:
- the single-bonded oxygen of an ester, R–C(=O)–O–R′ (R and R′ are any two carbon groups);
- the OH oxygen of a carboxylic acid, R–C(=O)–O–H, worked through for acetic acid below;
- the oxygen of an enolate, C=C–O⁻, an oxygen with a negative charge on a carbon of a C=C;
- the nitrogen of pyrrole, a flat ring of four carbons and one N–H, whose nitrogen sits between two C=C bonds;
- the nitrogen of aniline, an NH₂ on a benzene ring (a flat ring of six carbons with alternating double bonds). Courses label it sp². Measured, it is only partly flattened, so it sits between the two pictures.
Every atom in one molecule
Exams rarely ask about one atom. They give a molecule and ask for the hybridization of every atom other than hydrogen. The method does not change; you run the count once per atom.
Acetonitrile, CH₃C≡N. The methyl carbon: 3 bonds to H plus 1 to carbon = 4 groups → sp³. The nitrile carbon: 1 bond to the methyl carbon, plus the triple bond counting once = 2 groups → sp. The nitrogen: the triple bond counts once, plus 1 lone pair = 2 groups → sp. So the C–C≡N unit is linear and the methyl end is tetrahedral, in one molecule.
Acetic acid, CH₃COOH. The methyl carbon: 4 groups → sp³. The carbonyl carbon: bond to CH₃, bond to OH, and the C=O counting once = 3 groups → sp², so that carbon and its three neighbors are flat. The C=O oxygen: 1 group from the double bond plus 2 lone pairs = 3 groups → sp². The OH oxygen: 2 bonds (to the carbonyl carbon and to H) plus 2 lone pairs = 4 groups, so the plain count says sp³. But this oxygen is bonded to the carbonyl carbon, right beside the C=O π bond. By the rule above, one of its pairs moves into a p orbital, and the OH oxygen is sp² as well. Its shape is still bent, because it has only two neighbors.
Propenal, CH₂=CH–CHO. Both alkene carbons: 3 groups each → sp². The aldehyde carbon: bond to the neighboring carbon, bond to H, C=O counting once = 3 groups → sp². The oxygen: 3 groups → sp². Every heavy atom is sp², so each one is flat with its neighbors. That alone does not make the whole molecule flat, because it could still twist about the single bond in the middle. The molecule stays flat because that lines up all four leftover p orbitals, and lined-up p orbitals let its two π bonds share electrons with each other, which Drawing Molecules & Moving Electrons calls resonance.
What hybridization buys you: s-character
An sp orbital is made from one s orbital and one p orbital, so half of it is s. Chemists say it has 50% s-character: the share of the hybrid that came from the s orbital. An sp² orbital is one s out of three orbitals, 33%. An sp³ orbital is one s out of four, 25%.
An s orbital holds its electrons closer to the nucleus than a p orbital does. So the more s-character a hybrid has, the closer in it holds its electrons. Three consequences follow, and all three are examinable.
Bond length and strength. More s-character gives shorter, stronger bonds: sp > sp² > sp³. The bonding electrons sit closer to the carbon, so the two atoms sit closer together. A C–H bond on an sp carbon is about 106 pm long; on an sp³ carbon, about 109 pm.
Electronegativity. An sp carbon pulls harder on the electrons in its bonds than an sp³ carbon does, because it holds them closer in. That pull has a name, electronegativity: how strongly an atom draws bonding electrons toward itself. The Electronegativity section, later in this chapter, treats it as a fixed property of each element. This is the exception: the same element changes its pull with its hybridization.
Acidity is a labeled preview here, because the scale itself belongs to Acids & Bases. An acid gives up a hydrogen ion, H⁺. When H⁺ leaves a carbon, the bonding pair stays behind on that carbon, in the same hybrid orbital that held the bond. So “how acidic is this C–H?” really asks “how comfortably does this carbon hold the leftover pair?” A hybrid with more s-character holds the pair closer to the nucleus, and so holds it more willingly.
Acidity is measured as pKa. A lower pKa means a stronger acid, and each unit is a factor of ten. Take three C–H bonds that differ in nothing but hybridization: alkyne (sp) ≈ 25, alkene (sp²) ≈ 44, alkane (sp³) ≈ 50. That is twenty-five powers of ten from the mixing alone.
Beyond four groups
Carbon, nitrogen and oxygen never have more than four electron groups. The reason is the one Orbitals gave: the n = 2 shell has only four orbitals to mix, so there is no fifth hybrid to make.
Atoms from period 3 and beyond sometimes manage five or six. Phosphorus in PCl₅ has five groups, arranged as a trigonal bipyramid: three around the middle, one above and one below. Sulfur in SF₆ has six, arranged as an octahedron: four around the middle, one above and one below.
Older books label those atoms “sp³d” and “sp³d²”, as if d orbitals joined the mix. That label has been retired, because calculations show the d orbitals take almost no part. The Lewis Structures section sets out what does the work instead, under the octet-rule exceptions. While you are assigning hybridization, the practical rule is all you need: count groups, and if the count exceeds four you are not looking at a carbon, nitrogen or oxygen.
What carries forward
Naming hybridization from an electron-group count is a reflex you will use on nearly every page from here on. It gives you geometry in Molecular Geometry, it explains π bonding in the next section, it decides whether a lone pair can join a neighboring π bond in Drawing Molecules & Moving Electrons, it sets alkyne acidity in Acids & Bases, and it is the test for whether a ring can be aromatic in Aromatic Chemistry.