Valence Bond Theory

Core Concepts

A bond is an orbital overlap holding a shared electron pair

In VB theory, a covalent bond forms when two atomic orbitals overlap and the two bonding electrons (one from each atom, with opposite spins) occupy the merged region. Each bond uses one pair of electrons, exactly as in a Lewis structure — the difference is that VB theory says where those electrons sit: between the nuclei, in the overlap zone.

Sigma bonds: head-on overlap

When orbitals overlap along the line connecting the two nuclei (the internuclear axis), the result is a sigma bond. The electron density of a σ bond is cylindrically symmetric around the bond axis — you can spin the bond like a pencil and the electron cloud looks the same. Sigma bonds form from:

• s–s overlap (as in H₂),
• s–p overlap (as in H-F),
• p–p head-on overlap (as in F₂).

Every single bond is one σ bond.

Pi bonds: sideways overlap

Two parallel p orbitals can also overlap sideways (above and below the internuclear axis). This produces a pi bond, with electron density concentrated in two lobes on opposite sides of the bond axis. Pi bonds are weaker than sigma bonds for the same two atoms because sideways overlap is less extensive than head-on overlap. Pi bonds appear only in multiple bonds, which are covered in topic 3 — but knowing that π density sits off the axis explains a lot of chemistry (rotation around double bonds is restricted, for example).

Overlap extent and energy matching control bond strength

Two factors decide how strong an overlap bond is:

1. How much the orbitals overlap — more overlap means more time in the low-energy region between the nuclei, so the bond is stronger and shorter.
2. How well the orbital energies match — orbitals very different in energy (say, a 1s and a high-energy 3d) overlap poorly; similar-energy orbitals (like 1s and 2p in HF) mix effectively.

Bond formation is a tug-of-war between attractions (each electron pulled toward both nuclei) and repulsions (nucleus–nucleus and electron–electron); the bond length is the separation where net energy is lowest.

Where VB theory breaks down

VB theory with pure atomic orbitals fails to explain molecules like methane, CH₄. Carbon's valence configuration is 2s² 2p², which suggests two half-filled p orbitals — predicting only two bonds at 90° to each other. Methane actually has four equivalent bonds at 109.5°. The fix is hybridization (topic 2): atomic orbitals on the same atom are mathematically combined into new, equivalent hybrid orbitals before bonding. VB theory therefore survives as the core idea, with hybridization as its repair kit.

How It Works / Step-by-Step Process

Worked example 1: H₂ from two 1s orbitals

Problem. Use VB theory to describe the bond in H₂, and interpret its bond length (74 pm) and bond energy (436 kJ/mol).

Solution.

1. Each H atom contributes one electron in a 1s orbital.
2. As the atoms approach, the two 1s orbitals overlap head-on along the internuclear axis → one σ bond holding the shared pair.
3. Attraction (electrons to both nuclei) pulls the atoms together; repulsion (nucleus–nucleus) pushes them apart. The minimum-energy separation is 74 pm — the bond length.
4. The depth of that energy minimum, 436 kJ/mol, is the bond energy: the energy needed to separate one mole of H₂ into H atoms.

Worked example 2: HF — a 1s–2p sigma bond

Problem. Identify the orbitals that overlap in HF, the type of bond formed, and why the molecule is polar.

Solution.

1. H contributes its 1s orbital; F contributes a half-filled 2p orbital (F's configuration is 2s² 2p⁵).
2. The 1s orbital overlaps the 2p orbital head-on along the bond axis → one σ bond.
3. The energies of H's 1s and F's 2p are close enough to overlap effectively (bond energy 565 kJ/mol, bond length 92 pm).
4. F is far more electronegative than H, so the shared pair sits closer to F → a polar σ bond with a permanent dipole. Overlap explains the bond; electronegativity difference explains the polarity.

Worked example 3: why pure VB theory fails for methane

Problem. Predict the bonding in CH₄ using only carbon's ground-state valence orbitals, and explain the contradiction with experiment.

Solution.

1. Carbon's ground state is 2s² 2p²: the 2s orbital is full, and only two of the three 2p orbitals hold one electron each.
2. Pure VB logic → carbon can form only two bonds, using its two half-filled 2p orbitals, with a predicted 90° angle between them.
3. Experiment shows CH₄ has four equivalent C–H bonds at 109.5° (tetrahedral).
4. Resolution: the 2s orbital "mixes" with the three 2p orbitals to make four identical sp³ hybrid orbitals. This is the motivation for the hybridization topic that follows — VB theory's overlap picture is kept, but the orbitals themselves are rebuilt.

Common Confusions