7.2 · MO & VB theory

Valence bond theory and hybridization

sp, sp² and sp³ orbitals, σ and π bonds.

By the end you should be able to:

Assign hybridization from the steric number
Count σ and π bonds and describe the orbital overlap in each

Key idea

Hybrid orbitals follow the steric number

Valence bond (VB) theory describes each bond as the overlap of two orbitals, one on each atom, sharing a pair of electrons. To match the VSEPR shapes, a central atom mixes its valence s and p (and d) orbitals into hybrid orbitals. The number of hybrids equals the number of atomic orbitals mixed, which equals the steric number.

Steric number	Hybridization	Electron geometry	Angles	Examples
2	$sp$	linear	180°	$\ce{BeCl2}$ , $\ce{CO2}$ , $\ce{C2H2}$
3	$sp^2$	trigonal planar	120°	$\ce{BF3}$ , $\ce{C2H4}$
4	$sp^3$	tetrahedral	109.5°	$\ce{CH4}$ , $\ce{NH3}$ , $\ce{H2O}$
5	$sp^3d$	trigonal bipyramidal	90°, 120°, 180°	$\ce{PCl5}$ , $\ce{SF4}$
6	$sp^3d^2$	octahedral	90°, 180°	$\ce{SF6}$ , $\ce{XeF4}$

Steric number = atoms bonded to the central atom + lone pairs on it; a double or triple bond counts once. Lone pairs occupy hybrid orbitals too, so N in NH₃ is $sp^3$ .

Key idea

sp³, sp² and sp carbon

Methane: C is $sp^3$ . Four equivalent hybrids point to the corners of a tetrahedron (109.5°); each C–H σ bond is a C $sp^3$ orbital overlapping an H $1s$ orbital.
Ethene: each C is $sp^2$ . Three hybrids in a plane (120°) form σ bonds (two C–H, one C–C). The leftover unhybridized $2p$ orbital on each C, perpendicular to the plane, overlaps side-on with its neighbour's to form the π bond. In BF₃ the boron is also $sp^2$ , but its unhybridized $2p$ orbital is empty.
Ethyne: each C is $sp$ . Two hybrids at 180° form the σ bonds (C–H and C–C), and the two unhybridized $2p$ orbitals on each carbon form two perpendicular π bonds. In CO₂ the $sp$ carbon forms one σ and one π bond to each oxygen, with the two π bonds at right angles.

Key idea

σ and π bonds

	σ bond	π bond
Overlap	head-on, along the internuclear axis	side-on, between parallel p orbitals
Electron density	on the axis	above and below the axis (a nodal plane contains the axis)
Rotation	free rotation about the bond	rotation would break the overlap, so cis and trans isomers exist
Strength	stronger	weaker: C–C ≈ 348 and C=C ≈ 614 kJ/mol, so the π part is ≈ 266 kJ/mol

Every bond contains exactly one σ bond:

single bond = 1σ
double bond = 1σ + 1π
triple bond = 1σ + 2π

Method

Assigning hybridization and counting bonds

Draw the Lewis structure.
For each atom of interest, steric number = bonded atoms + lone pairs.
Hybridization: 2 → $sp$ , 3 → $sp^2$ , 4 → $sp^3$ , 5 → $sp^3d$ , 6 → $sp^3d^2$ .
Count bonds: each bond (single, double or triple) gives one σ; each double adds one π and each triple adds two π.
Describe the overlap: σ bonds use hybrid orbitals (or H $1s$ ); π bonds use unhybridized p orbitals. So an atom in one π bond is $sp^2$ , and an atom in two π bonds is $sp$ .

Common mistake

Hybridization traps

Wrong: a double bond counts as two electron domains. Right: a multiple bond is one domain; C in CO₂ has steric number 2, so it is $sp$ .
Wrong: ignoring lone pairs. Right: N in NH₃ (3 bonds + 1 lone pair) and O in H₂O (2 bonds + 2 lone pairs) are both $sp^3$ .
Wrong: a double bond is two identical bonds. Right: it is one σ plus one weaker π bond.
Wrong: π bonds are made from hybrid orbitals. Right: π bonds use the unhybridized p orbitals left over after hybridization.
Wrong: hybridization decides the shape. Right: the shape (the steric number from the Lewis structure) decides which hybridization we assign.

Worked example

Worked example: HCN and acrylonitrile

HCN (H–C≡N): C has 2 electron domains, so it is $sp$ ; N has 1 bond + 1 lone pair, also $sp$ . Bonds: C–H (1σ) + C≡N (1σ + 2π) = 2σ + 2π. The C–H σ bond is C $sp$ + H $1s$ ; the two π bonds come from side-on overlap of C and N $2p$ orbitals.

Acrylonitrile, $\ce{CH2=CH-C#N}$ :

Bonds	σ	π
three C–H	3	0
C=C	1	1
C–C	1	0
C≡N	1	2
Total	6	3

The CH₂ and CH carbons are $sp^2$ (three domains each); the nitrile carbon and the nitrogen are $sp$ .

Key idea

Valence bond versus molecular orbital theory

	Valence bond (VB)	Molecular orbital (MO)
Electrons	localized: in a bond between two atoms, or in a lone pair	delocalized over the whole molecule
Built from	overlap of (hybrid) atomic orbitals on neighbouring atoms	combinations of the atomic orbitals of all the atoms
Best for	shapes, hybridization, σ/π description of organic molecules	bond order, magnetism, ions, electronic spectra
Weakness	predicts O₂ is diamagnetic; needs resonance for delocalized electrons	diagrams become complicated for large molecules

In practice chemists combine them: a VB framework of localized σ bonds plus delocalized, MO-style π systems. That is exactly the picture used for conjugated molecules in Module 8.

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