Mixing orbitals into hybrids.

One s and up to three p (then d) orbitals blend into equivalent hybrids; lobes rearrange to linear, trigonal, tetrahedral, TBP or octahedral with % s character and angle readouts, and left-over p orbitals are drawn for π bonds.

For C2H2, set the mix that matches each carbon (sp, 2 σ regions) and confirm 2 unhybridised p orbitals remain for the two π bonds.

Counting π bonds when deciding hybridisation; assuming all five P–Cl bonds of PCl5 are equal.

• Hybridisation is the mixing of atomic orbitals of similar energy on the same atom to form an equal number of equivalent hybrid orbitals.
• Only orbitals of similar energy hybridise; hybrid orbitals form σ bonds or hold lone pairs, never π bonds.
• sp: one s + one p, linear, 180°, 50% s character (BeCl₂; each carbon in ethyne).
• sp²: one s + two p, trigonal planar, 120°, 33% s character (BCl₃; each carbon in ethene).
• sp³: one s + three p, tetrahedral, 109.5° (109°28′), 25% s character (CH₄); NH₃ and H₂O are also sp³, with angles squeezed by lone pairs.
• In ethene each carbon uses sp² orbitals for σ bonds and the unhybridised p orbital for the π bond; in ethyne each carbon forms 2 σ bonds and 2 π bonds.
• sp³d (dz² orbital used) gives a trigonal bipyramid, as in PCl₅; the two axial P–Cl bonds are longer and weaker than the three equatorial bonds, which makes PCl₅ reactive.
• sp³d² (dz² and dx²−y² used) gives an octahedron, as in SF₆, with six equal S–F bonds at 90°.
• Quick rule: count σ bonds plus lone pairs on the central atom; 2, 3, 4, 5 and 6 correspond to sp, sp², sp³, sp³d and sp³d².

Teaching this idea? Put the simulation on your own page or LMS with this code.