How to Find Hybridization: A Step-by-Step Guide

Hybridization is the bookkeeping that connects a Lewis structure to a three-dimensional shape. Once you know an atom is $sp$, $sp^2$ or $sp^3$, you also know its ideal bond angles, its electron geometry, and how many unhybridized p orbitals it has left over for pi bonding. The good news: for almost every question you will be asked, finding hybridization is a three-step mechanical procedure built around a single number.

What hybridization actually is

An isolated carbon atom has one 2s orbital and three 2p orbitals. Those four orbitals have different energies and different shapes, which cannot explain why methane's four C–H bonds are identical and point at the corners of a tetrahedron. Hybridization is the model that resolves the mismatch: the atomic orbitals are mathematically combined into a new set of equivalent hybrid orbitals, one for every direction in which the atom needs to point a sigma bond or a lone pair.

The arithmetic is conservative — mix $n$ atomic orbitals, get $n$ hybrid orbitals:

• 1 s + 1 p gives two $sp$ orbitals
• 1 s + 2 p gives three $sp^2$ orbitals
• 1 s + 3 p gives four $sp^3$ orbitals

Any p orbitals left over stay unhybridized and remain available for pi bonds.

The one number you need: the steric number

Steric number (SN) = number of sigma bonds on the atom + number of lone pairs on the atom.

Notice what is not in that count: pi bonds. A double bond contributes one sigma bond and one pi bond; a triple bond contributes one sigma and two pi. Only the sigma bond counts toward SN. This is the single largest source of wrong answers.

Once you have SN, the hybridization is a lookup:

Steric number	Hybridization	Electron geometry	Ideal bond angle
2	sp	Linear	180°
3	sp2	Trigonal planar	120°
4	sp3	Tetrahedral	109.5°
5	sp3d	Trigonal bipyramidal	120° and 90°
6	sp3d2	Octahedral	90°

The four steps

Step 1 — Draw the Lewis structure. Count the total valence electrons, connect the atoms, complete the octets, and place any electrons left over as lone pairs on the central atom. You cannot skip this step, because the lone pairs are half of the answer and they are invisible in the molecular formula.

Step 2 — Pick the atom. Hybridization is a property of an atom, not of a molecule. "The hybridization of ethanol" is not a well-posed question; each carbon and the oxygen has its own. When a problem says "the hybridization of SF4", it means the central sulfur.

Step 3 — Count sigma bonds and lone pairs. Every bonded neighbour contributes exactly one sigma bond, whether the connection is a single, double or triple bond. Add the lone pairs sitting on that same atom.

Step 4 — Read the table. SN 2 gives $sp$, SN 3 gives $sp^2$, SN 4 gives $sp^3$, and so on.

Worked examples

CH4 (methane). Carbon has four bonded neighbours and no lone pairs. SN = 4 + 0 = 4, so carbon is $sp^3$: tetrahedral, with 109.5° bond angles.

NH3 (ammonia). Nitrogen has three sigma bonds and one lone pair, so SN = 4 and nitrogen is $sp^3$. The electron geometry is tetrahedral, but the molecular shape is trigonal pyramidal, because shape names describe only where the atoms are. The lone pair repels the bonding pairs more strongly than they repel each other, compressing the H–N–H angle to about 107°.

H2O (water). Oxygen has two sigma bonds and two lone pairs: SN = 4, so oxygen is $sp^3$. The shape is bent, with an angle near 104.5° — two lone pairs squeeze harder than one.

BF3 (boron trifluoride). Boron has three sigma bonds and no lone pairs; it is one of the classic electron-deficient exceptions to the octet rule. SN = 3, so boron is $sp^2$: trigonal planar, 120°.

CO2 (carbon dioxide). Carbon carries two double bonds. Count sigma bonds only: two sigma bonds, zero lone pairs, so SN = 2 and carbon is $sp$ — linear, 180°. Carbon keeps two unhybridized p orbitals, and those form the two pi bonds. Each oxygen, meanwhile, has one sigma bond and two lone pairs, so SN = 3 and each oxygen is $sp^2$. Same molecule, two different hybridizations — which is exactly why you must name the atom.

C2H4 (ethene). Each carbon is bonded to two hydrogens and one carbon: three sigma bonds, no lone pairs, SN = 3, so each carbon is $sp^2$. The leftover p orbital on each carbon overlaps side-on to form the pi bond. That pi bond is why ethene is planar and why the C=C cannot freely rotate — rotation would break the side-on overlap.

C2H2 (ethyne). Each carbon has one sigma bond to hydrogen and one sigma bond to the other carbon: SN = 2, so each carbon is $sp$ and the molecule is linear. Two unhybridized p orbitals per carbon build the two pi bonds of the triple bond.

SF4. Sulfur has four sigma bonds and one lone pair, so SN = 5 and sulfur is $sp^3d$. The electron geometry is trigonal bipyramidal; the molecular shape is a seesaw, because the lone pair takes an equatorial position.

XeF4. Xenon has four sigma bonds and two lone pairs: SN = 6, so xenon is $sp^3d^2$. The electron geometry is octahedral and the shape is square planar, with the two lone pairs opposite each other.

A shortcut when the outer atoms are monovalent

For a central atom A surrounded only by monovalent atoms (H, F, Cl, Br, I), you can get the steric number without drawing anything:

$SN = \frac{V + M - C + A}{2}$

where $V$ is the number of valence electrons on the central atom, $M$ is the number of monovalent atoms attached, $C$ is the charge if the species is a cation, and $A$ is the charge if it is an anion. Doubly bonded oxygens contribute nothing to $M$.

• NH3: $(5 + 3)/2 = 4$, so $sp^3$
• NH4+: $(5 + 4 - 1)/2 = 4$, so $sp^3$
• ClF3: $(7 + 3)/2 = 5$, so $sp^3d$
• SO4 2-: $(6 + 0 + 2)/2 = 4$, so $sp^3$
• SO2: $(6 + 0)/2 = 3$, so $sp^2$

Treat it as a fast cross-check rather than a replacement. It breaks down when the outer atoms are not monovalent, and it tells you nothing about the molecular shape.

Common mistakes

• Counting pi bonds in the steric number. Only sigma bonds and lone pairs count. Carbon in CO2 has four bonds drawn but only two sigma bonds, so it is $sp$, not $sp^3$.
• Forgetting the lone pairs. If you count only bonds, water comes out as SN = 2 and you land on $sp$ — wrong. Its two lone pairs push it to SN = 4 and $sp^3$.
• Confusing electron geometry with molecular shape. Hybridization tracks the electron geometry, which counts bonds and lone pairs together. The shape name reports only the atoms. Water is $sp^3$ and tetrahedral in electron geometry, but bent in shape.
• Expecting exactly ideal angles. $sp^3$ predicts 109.5°, but lone pairs repel more strongly than bonding pairs, so measured angles are usually a little smaller (ammonia about 107°, water about 104.5°).
• Asking about a molecule instead of an atom. Always identify which atom the question means.

One caveat worth knowing: $sp^3d$ and $sp^3d^2$ are still standard in general chemistry courses, but modern computational work indicates that d-orbital participation in main-group elements is minimal, and hypervalent bonding is better described by three-centre four-electron bonds. Use the $sp^3d$ labels because your course expects them, and keep in mind that they are a teaching model rather than the final word.

Quick self-check

Work these out before reading on: PCl5, SO3, the carbon in HCN, OF2, and BeCl2.

• PCl5: phosphorus has 5 sigma bonds, 0 lone pairs, SN = 5, so $sp^3d$.
• SO3: sulfur has 3 sigma bonds, 0 lone pairs, SN = 3, so $sp^2$.
• HCN: the carbon has one sigma bond to H and one sigma bond to N (the triple bond is 1 sigma + 2 pi), SN = 2, so $sp$.
• OF2: oxygen has 2 sigma bonds and 2 lone pairs, SN = 4, so $sp^3$.
• BeCl2: beryllium has 2 sigma bonds and 0 lone pairs, SN = 2, so $sp$.

If all five came out right, you have the method. The Lewis structure supplies the sigma bonds and the lone pairs; the steric number turns them into a hybridization label; the label hands you the geometry and the bond angles.