Valence Bond Theory

Lewis structures tell you that atoms share electrons; valence bond theory (VBT) tells you why and in what direction. A covalent bond, in this picture, is not a symbol on paper but a real overlap of two atomic orbitals in space — each holding one unpaired electron — and the strength and direction of the bond follow directly from how completely those orbitals overlap. VBT also supplies hybridisation, the mixing of an atom's own orbitals before bonding, which is what finally explains why methane is a perfect tetrahedron and not some arbitrary tangle of four bonds. Together, orbital overlap and hybridisation turn the qualitative shapes of VSEPR theory into a mechanism grounded in orbitals — the bridge this article builds, from $\sigma$ / $\pi$ bonds through sp, $sp^2$ , $sp^3$ , $sp^3d$ and $sp^3d^2$ hybridisation, to a from-scratch derivation of the 109.5 degree tetrahedral angle.

The core insight

A covalent bond forms along the direction of maximum overlap between two singly-occupied atomic orbitals, and mixing an atom's own orbitals into equivalent hybrid orbitals (sp, $sp^2$ , $sp^3$ , $sp^3d$ , $sp^3d^2$ ) is what fixes the bond angles and molecular shape that VSEPR only predicts qualitatively.

The core idea

Orbital overlap is the bond. When two atomic orbitals, each carrying one unpaired electron, approach and overlap in space, the shared region of high electron density between the two nuclei is the covalent bond. Overlap is not symmetric in strength: the greater the extent of overlap, the stronger and shorter the resulting bond. Because overlap can only be large along a specific line joining the two orbitals, covalent bonds are inherently directional — unlike ionic bonds, which are non-directional electrostatic attractions that act equally in every direction around an ion.

Sigma and pi bonds are two different overlap geometries. A $\sigma$ bond comes from head-on (axial) overlap along the internuclear axis — s–s, s–p, or end-on p–p overlap — and every bonded pair of atoms has exactly one $\sigma$ bond, formed first. A $\pi$ bond comes from sideways (lateral) overlap of two parallel, unhybridised p orbitals, one lobe above and one below the internuclear axis, and it can only form in addition to an existing $\sigma$ bond, never alone. A single bond is one $\sigma$ ; a double bond is one $\sigma$ plus one $\pi$ ; a triple bond is one $\sigma$ plus two $\pi$ . Sideways overlap is inherently less effective than head-on overlap, so $\pi$ bonds are weaker than $\sigma$ bonds — which is why double and triple bonds break more easily at the $\pi$ component first (as in addition reactions).

Hybridisation mixes orbitals on one atom before bonding. Atomic orbitals of comparable energy on the same atom — one s and some number of p (and, for the third row onward, d) orbitals — combine mathematically into an equal number of new, equivalent hybrid orbitals, oriented to keep mutual electron-pair repulsion at a minimum. This is what fixes geometry before any bond is even drawn:

Hybridisation	Orbitals mixed	Geometry	Bond angle	Example
sp	one $s$ + one $p$	Linear	$180°$	$\text{BeCl}_2$
$sp^2$	one $s$ + two $p$	Trigonal planar	$120°$	$\text{BF}_3$
$sp^3$	one $s$ + three $p$	Tetrahedral	$109.5°$	$\text{CH}_4$
$sp^3d$	one $s$ + three $p$ + one $d$	Trigonal bipyramidal	$90°/120°$	$\text{PCl}_5$
$sp^3d^2$	one $s$ + three $p$ + two $d$	Octahedral	$90°$	$\text{SF}_6$

The hybridisation-predicted geometry matches the VSEPR electron-pair geometry exactly for every row of this table — VSEPR reasons from electron-pair repulsion, VBT reasons from orbital mixing, and both land on the same shape because they are describing the same physical minimum-repulsion arrangement.

Figure

Four sp3 hybrid orbitals pointing from a central carbon atom to the four corners of a tetrahedron with bond angle 109.5 degrees marked between two of the lobes

Derivation AThe 109.5 degree tetrahedral angle between two sp3 hybrid orbitals

The goal is to show, using nothing beyond three-dimensional vectors and the dot-product formula for the angle between two vectors, that four equivalent orbitals arranged to point as far apart from each other as possible must sit at exactly $109.47°$ — the bond angle quoted for every $sp^3$ -hybridised centre such as $\text{CH}_4$ , and the number that VBT and VSEPR both converge on for a regular tetrahedron.

Step 1. Place the central atom at the origin and put the four hybrid orbitals along the four directions that reach the alternating corners of a cube — this is the standard way to generate a regular tetrahedron with simple coordinates. The four direction vectors are $\vec{a}=(1,1,1)$ , $\vec{b}=(1,-1,-1)$ , $\vec{c}=(-1,1,-1)$ , and $\vec{d}=(-1,-1,1)$ . By symmetry all four vectors have the same length and each pair makes the same angle with every other pair, so it suffices to compute the angle between any two, say $\vec{a}$ and $\vec{b}$ .

Step 2. Compute the dot product. $\vec{a}\cdot\vec{b} = (1)(1) + (1)(-1) + (1)(-1) = 1 - 1 - 1 = -1$ .

Step 3. Compute the magnitudes. $|\vec{a}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$ , and likewise $|\vec{b}| = \sqrt{3}$ , since every coordinate of every vector is $\pm 1$ .

Step 4. Apply the angle formula and solve.

\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{-1}{\sqrt{3}\cdot\sqrt{3}} = \frac{-1}{3}.

where: $\theta$ is the angle between any two of the four hybrid orbitals. Taking the inverse cosine, $\theta = \cos^{-1}(-1/3) = 109.47°$ , which rounds to the familiar $109.5°$ . Because the four chosen vectors are related by the full symmetry of the tetrahedron, this single calculation gives the angle between every pair of the four $sp^3$ lobes at once — there is no separate case to check, and the same $109.5°$ appears at the C–H–H angle in $\text{CH}_4$ predicted in Worked example 1 below.

Worked example

Use valence bond theory to explain the bonding and shape of methane, $\text{CH}_4$ , and confirm every H–C–H angle is 109.5 degrees.

Step 1. Ground-state carbon is $1s^2\,2s^2\,2p^2$ , with only 2 unpaired electrons in the $2p$ orbitals — this alone cannot explain 4 identical C–H bonds, so the simple ground-state configuration is not the whole story.

Step 2. One $2s$ electron is promoted to the empty $2p_z$ orbital, giving the excited configuration $2s^1\,2p_x^1\,2p_y^1\,2p_z^1$ , now with 4 unpaired electrons, one in each of the four orbitals.

Step 3. The one $2s$ and three $2p$ orbitals mix to form four identical $sp^3$ hybrid orbitals. Each hybrid orbital is one part $s$ -character and three parts $p$ -character, and being identical in energy and shape, they arrange themselves to minimise mutual repulsion.

Step 4. As derived above, four equivalent directions repelling each other maximally point to the corners of a regular tetrahedron, with $\cos\theta = -1/3$ giving $\theta = 109.5°$ between every pair.

Step 5. Each $sp^3$ orbital overlaps head-on with a hydrogen $1s$ orbital, forming 4 equivalent C–H $\sigma$ bonds. Answer: methane is tetrahedral with all four H–C–H bond angles equal to $109.5°$ , exactly the angle fixed by the geometry of four equivalent sp3 orbitals.

Competitive stretch

Ground-state phosphorus in $\text{PCl}_5$ is $[\text{Ne}]\,3s^2\,3p^3$ . Use VBT to justify its trigonal bipyramidal shape with two distinct bond angles. One $3s$ electron is promoted into an empty $3d$ orbital, giving $3s^1\,3p_x^1\,3p_y^1\,3p_z^1\,3d^1$ — five unpaired electrons. These five orbitals mix into five $sp^3d$ hybrids, which VSEPR-style repulsion minimisation arranges as a trigonal bipyramid: three equatorial hybrids at $120°$ to each other, and two axial hybrids at $90°$ to the equatorial plane, one above and one below. Each hybrid overlaps head-on with a chlorine orbital, giving 5 equivalent P–Cl $\sigma$ bonds arranged with bond angles of $90°$ (axial–equatorial) and $120°$ (equatorial–equatorial), matching the $sp^3d$ row of the hybridisation table.

Common traps

Confusing hybridisation (mixing orbitals on the same atom, before any bond forms) with orbital overlap (the interaction between orbitals on two different atoms during bond formation) — VBT needs both, in that order, not one instead of the other.
Assuming a $\pi$ bond can exist by itself; a $\pi$ bond only ever accompanies a $\sigma$ bond between the same two atoms, so a double bond is one $\sigma$ plus one $\pi$ , never two $\pi$ bonds alone.
Forgetting that hybrid orbitals hold only $\sigma$ bonds and lone pairs — unhybridised p orbitals are what participate in $\pi$ bonding, so in ethene each carbon keeps one unhybridised $2p$ orbital outside its three $sp^2$ hybrids.
Mismatching hybridisation with molecular shape when lone pairs are present — e.g. in $\text{NH}_3$ and $\text{H}_2\text{O}$ the central atom is still $sp^3$ hybridised, but the molecular shape (pyramidal, bent) excludes the lone pairs from the description, unlike the electron-pair geometry, which stays tetrahedral.

Ask the tutor: Want to work through the hybridisation and shape of a specific molecule, such as $\text{SF}_6$ or $\text{NH}_3$ , step by step?

Why does carbon need to promote an electron before it can form four bonds in methane?

Ground-state carbon, $1s^2\,2s^2\,2p^2$ , has only two unpaired electrons, in two separate $2p$ orbitals — by itself that predicts just two bonds, not four. Promoting one $2s$ electron into the empty $2p_z$ orbital costs a small amount of energy but gives four unpaired electrons instead of two. That energy cost is repaid many times over because forming four C–H bonds instead of two releases far more energy overall, so the promoted state is the one actually adopted. The four singly-occupied orbitals then mix into four equivalent $sp^3$ hybrids before any bonding overlap happens.

If sp2 carbons in ethene each have three hybrid orbitals for three sigma bonds, what happens to the fourth orbital?

Carbon has four valence orbitals total (one $2s$ , three $2p$ ), but $sp^2$ hybridisation mixes only three of them — one $s$ and two $p$ — leaving the third $2p$ orbital completely unhybridised and perpendicular to the plane of the three $sp^2$ hybrids. That leftover p orbital is exactly what forms the $\pi$ bond by sideways overlap with the matching p orbital on the other carbon, giving the C=C double bond its $\sigma$ plus $\pi$ character and forcing the whole molecule to stay planar.

Do VSEPR theory and valence bond theory ever disagree on molecular shape?

No — for every hybridisation scheme on this syllabus (sp, $sp^2$ , $sp^3$ , $sp^3d$ , $sp^3d^2$ ), the hybridisation-predicted geometry matches the VSEPR electron-pair geometry exactly, because both are really describing the same underlying arrangement: orbitals or electron pairs spaced out to minimise mutual repulsion. The two theories differ only in why — VSEPR argues from electron-pair repulsion directly, VBT argues from orbital mixing and overlap — which is why questions often ask you to explain the same shape "using VBT" or "using VSEPR" as two separate skills testing the same answer.

Valence Bond Theory

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads