Build a model with LEGO and every brick is a different shape, pointing a different way. A carbon atom starts out a bit like that: it owns one round orbital (the $s$) and three dumbbell-shaped orbitals (the $p$'s) that sit at right angles to each other. Yet methane, CH₄, has four identical bonds spread evenly in space, each one the same length and the same strength. How does a round orbital plus three perpendicular dumbbells turn into four matching bonds? The atom blends them. Think of mixing one can of white paint with three cans of blue: you end up with four cans of the exact same light-blue shade. You can't pull the white back out, and every can is identical. That blending of orbitals is hybridization.
The bookkeeping rule is simple: mix $N$ atomic orbitals and you get exactly $N$ hybrids back. One $s$ + one $p$ makes two $sp$ hybrids that point straight apart at $180°$ (linear, like BeCl₂). One $s$ + two $p$ makes three $sp^2$ hybrids in a flat triangle at $120°$ (like BH₃). One $s$ + three $p$ makes four $sp^3$ hybrids reaching to the corners of a tetrahedron at $109.5°$ (like CH₄). The more of the round $s$ orbital you pour into each hybrid, the wider the bonds open and the shorter they become. Since each $sp^3$ hybrid is built from one-quarter of one $s$ orbital, it carries 25% "s-character" — and that fixes methane's bond angle at $109.5°$.
Each hybrid is a weighted sum of the atomic wavefunctions, $\psi_{\text{hyb}} = c_s\,\psi_s + c_p\,\psi_p$, where the s-character is $\alpha = |c_s|^2$. Coulson's directionality formula ties the geometry straight to the mixing:
Plug in the s-character and the angle falls out: $sp$ has $\alpha=\tfrac12 \Rightarrow \theta=180°$; $sp^2$ has $\alpha=\tfrac13 \Rightarrow 120°$; $sp^3$ has $\alpha=\tfrac14 \Rightarrow 109.5°$. In the simulation, the Hybridization Type menu picks how many orbitals are mixed; the Mixing parameter t morphs the picture from the raw atomic orbitals ($t=0$) to the finished hybrids ($t=1$); and the π bonds slider adds the leftover, un-mixed $p$ orbitals that overlap sideways to make the second and third bonds of double and triple bonds.
First, set the type to sp and watch the two lobes line up at $180°$, then switch to sp³ and see them swing open into a tetrahedron. Next, drag the Mixing parameter t from 1 down to 0 to pull the hybrids back apart into a round $s$ sphere and a dumbbell $p$ — the blend in reverse. Finally, open the Multiple Bonds (π) tab and push the π slider to 2: the extra sideways $p$-orbital overlaps appear, building the second and third bonds of a triple bond like the one in acetylene.
Hybridization is the mathematical procedure of constructing new directional orbitals from a linear combination of atomic orbitals (LCAO) on the same atom. Pauling introduced this in 1931 to reconcile experimental molecular geometries (like methane's 109.5°) with the spherical/dumbbell shapes of pure $s$ and $p$ orbitals. Hybrid orbitals are not physically distinct — they are a particular orthonormal basis that gives maximum overlap with bonded atoms.
| Symbol | Meaning | Range / Unit |
|---|---|---|
| $\psi_{\text{sp}^n}$ | Hybrid orbital wavefunction | — |
| $\psi_s, \psi_{p_i}$ | Atomic $s$, $p_i$ orbital basis functions | — |
| $c_s, c_i$ | Mixing coefficients (orthonormal) | $|c|^2 \le 1$ |
| $\alpha = c_s^2$ | $s$-character fraction | 0 to 1 |
| $\theta_{ij}$ | Angle between hybrids $i$ and $j$ | degrees |
| $N$ | Total number of hybrids = atomic orbitals mixed | 2, 3, 4, 5, 6 |
| $E_h$ | Energy of hybrid orbital | eV (between $E_s$ and $E_p$) |
Each carbon uses sp hybridization (50% $s$, 50% $p$), forming a linear C–C–H–H arrangement. The two unused $p_y, p_z$ orbitals on each C overlap sideways to form two π bonds — giving the C≡C triple bond. Bond length: 120 pm (vs 154 pm for sp³ C–C single).