4.4 Valence Bond Theory, Hybridization & Sigma/Pi Bonding

Key Takeaways

  • Valence Bond theory models covalent bonds through the quantum mechanical overlap of atomic orbitals sharing paired spins, where greater overlap produces stronger bonds.
  • Hybridization mathematically combines atomic orbitals (s, p, d) to produce degenerate directional hybrid orbitals whose geometries match observed molecular shapes.
  • Sigma (σ) bonds form by head-on overlap along the internuclear axis with cylindrical symmetry, permitting free rotation, whereas pi (π) bonds form by lateral overlap of parallel unhybridized p orbitals.
  • Multiple bonds consist of one σ bond accompanied by one π bond (double bond) or two orthogonal π bonds (triple bond), with restricted π bond rotation producing geometric (cis/trans) isomerism.
  • Increasing s-character in hybrid orbitals (sp > sp² > sp³) draws electron density closer to the nucleus, yielding shorter and stronger bonds along with enhanced effective electronegativity.
Last updated: September 2026

Valence Bond Theory, Hybridization & Sigma/Pi Bonding

Quick Summary: Valence Bond (VB) theory explains covalent bonding through localized orbital overlap sharing paired electron spins. To account for observed molecular shapes that pure ss and pp orbitals cannot explain, atomic wavefunctions mix mathematically into directional hybrid orbitals (sp,sp2,sp3,sp3d,sp3d2sp, sp^2, sp^3, sp^3d, sp^3d^2). Direct head-on overlap along the internuclear axis creates cylindrical sigma (σ\sigma) bonds with free rotation, whereas lateral overlap of unhybridized pp orbitals creates pi (π\pi) bonds whose rotational rigidity gives rise to geometric (cis/trans) isomerism.


1. Orbital Overlap & Foundations of Valence Bond Theory

Valence Bond (VB) theory models covalent bonding as the spatial overlap of two atomic orbitals, each containing an unpaired electron:

  1. Orbital Overlap: Covalent bonds form when an orbital on one atom shares space with an orbital on an adjacent atom.
  2. Spin Pairing: Under the Pauli exclusion principle, the two electrons in the overlap region must have opposite (antiparallel) spins (↑↓\uparrow\downarrow).
  3. Maximum Overlap Condition: Bond strength correlates directly with spatial overlap; greater overlap concentrates negative charge between nuclei, lowering potential energy.

The Limitation of Pure Atomic Orbitals

Ground-state carbon has configuration 1s22s22px12py11s^2 2s^2 2p_x^1 2p_y^1. With only two unpaired pp electrons at 90∘90^\circ, pure orbital theory predicts carbon would form only two bonds at 90∘90^\circ. Promoting a 2s2s electron yields 2s12px12py12pz12s^1 2p_x^1 2p_y^1 2p_z^1, predicting three bonds from pp orbitals at 90∘90^\circ and one non-directional bond from ss overlap. This contradicts experimental reality: methane (CH4\text{CH}_4) possesses four identical C−H\text{C}-\text{H} bonds of equal length (109 pm109\text{ pm}) and strength (435 kJ/mol435\text{ kJ/mol}) oriented at regular tetrahedral angles of 109.5∘109.5^\circ.


2. Hybridization of Atomic Orbitals

Linus Pauling proposed hybridization: the mathematical mixing of atomic wavefunctions on a central atom to generate directional hybrid orbitals matching VSEPR geometries:

  • spsp (Steric Number 2): One ss + one pp orbital yield two equivalent spsp hybrid orbitals directed 180∘180^\circ apart (linear). Two unhybridized pp orbitals remain perpendicular. Examples: BeCl2,CO2,C2H2\text{BeCl}_2, \text{CO}_2, \text{C}_2\text{H}_2 (ethyne).
  • sp2sp^2 (Steric Number 3): One ss + two pp orbitals yield three sp2sp^2 hybrids directed 120∘120^\circ apart in a plane (trigonal planar). One unhybridized pp orbital remains perpendicular to the plane. Examples: BF3,C2H4\text{BF}_3, \text{C}_2\text{H}_4 (ethene), CO32−\text{CO}_3^{2-}.
  • sp3sp^3 (Steric Number 4): One ss + three pp orbitals yield four sp3sp^3 hybrids directed 109.5∘109.5^\circ apart toward tetrahedral vertices. Zero unhybridized pp orbitals remain. Examples: CH4,NH3,H2O,C2H6\text{CH}_4, \text{NH}_3, \text{H}_2\text{O}, \text{C}_2\text{H}_6.
  • sp3dsp^3d (Steric Number 5): One ss + three pp + one dd orbital yield five sp3dsp^3d hybrids in a trigonal bipyramidal geometry (90∘,120∘,180∘90^\circ, 120^\circ, 180^\circ). Examples: PCl5,SF4\text{PCl}_5, \text{SF}_4.
  • sp3d2sp^3d^2 (Steric Number 6): One ss + three pp + two dd orbitals yield six equivalent sp3d2sp^3d^2 hybrids in an octahedral geometry (90∘,180∘90^\circ, 180^\circ). Examples: SF6,XeF4\text{SF}_6, \text{XeF}_4.

3. Sigma (σ\sigma) vs. Pi (π\pi) Bonding & Multiple Bonds

Covalent bonds are classified by their spatial overlap geometry:

Sigma (σ\sigma) Bonds

  • Overlap: Head-on (end-to-end) overlap directly along the internuclear axis (s−s,s−p,p−ps-s, s-p, p-p, or hybrid-hybrid).
  • Symmetry & Strength: Exhibiting cylindrical symmetry around the bond axis, σ\sigma bonds concentrate electron density between nuclei and are stronger than π\pi bonds.
  • Free Rotation: Rotating atoms around a single σ\sigma bond maintains constant orbital overlap, permitting free rotation at room temperature.

Pi (π\pi) Bonds

  • Overlap: Lateral (sideways) overlap of parallel unhybridized pp orbitals above and below the internuclear axis.
  • Nodal Plane: The internuclear axis lies in a nodal plane of zero electron density. Due to less effective lateral overlap, π\pi bonds are weaker than σ\sigma bonds and only form after a primary σ\sigma bond is established.

Multiple Bond Composition

  • Single Bond: 1 σ1\,\sigma bond.
  • Double Bond: 1 σ+1 π1\,\sigma + 1\,\pi bond (e.g., ethene, H2C=CH2\text{H}_2\text{C}=\text{CH}_2).
  • Triple Bond: 1 σ+2 π1\,\sigma + 2\,\pi bonds oriented mutually perpendicular (e.g., ethyne, HC≡CH\text{HC}\equiv\text{CH}, and N2\text{N}_2).

4. Rotational Rigidity & Geometric Isomerism

Unlike σ\sigma bonds, double bonds cannot rotate at ambient temperature. Rotating one sp2sp^2 carbon twists its unhybridized 2p2p orbital out of parallel coplanarity, destroying the π\pi bond. Overcoming this barrier requires ≈260 kJ/mol\approx 260\text{ kJ/mol}, locking substituents into fixed positions and creating geometric (cis/trans) isomerism:

  • In cis-1,2-dichloroethene, chlorine atoms reside on the same side of the double bond (net polar).
  • In trans-1,2-dichloroethene, chlorines reside on opposite sides across the inversion center (nonpolar).

5. Delocalized π\pi Systems & Conjugation

When adjacent atoms possess parallel unhybridized pp orbitals, π\pi electrons delocalize across multiple nuclei:

  • Benzene (C6H6\text{C}_6\text{H}_6): Six sp2sp^2 carbons form a planar hexagonal ring. Six perpendicular 2pz2p_z orbitals overlap continuously, forming toroidal π\pi clouds above and below the ring. This equalizes all C−C\text{C}-\text{C} bonds to 139.7 pm139.7\text{ pm} (between single 154 pm154\text{ pm} and double 134 pm134\text{ pm}) and provides ≈150 kJ/mol\approx 150\text{ kJ/mol} of aromatic stabilization.
  • Carbonate & Nitrate: Continuous π\pi overlap across four atoms yields fractional bond orders of 4/34/3.

6. Hybrid Orbital ss-Character & Structural Trends

The fraction of ss-character (sp=50%,sp2=33.3%,sp3=25%sp = 50\%, sp^2 = 33.3\%, sp^3 = 25\%) determines orbital proximity to the positive nucleus:

  • Bond Length & Strength: Greater ss-character holds electrons tighter, producing shorter, stronger bonds: C−C\text{C}-\text{C} in ethane (154 pm,347 kJ/mol154\text{ pm}, 347\text{ kJ/mol}) vs. C=C\text{C}=\text{C} in ethene (134 pm,614 kJ/mol134\text{ pm}, 614\text{ kJ/mol}) vs. C≡C\text{C}\equiv\text{C} in ethyne (120 pm,839 kJ/mol120\text{ pm}, 839\text{ kJ/mol}).
  • Acidity of Hydrocarbons: The spsp carbon in ethyne holds the conjugate base lone pair close to the nucleus, stabilizing the acetylide anion (HC≡C:−\text{HC}\equiv\text{C}:^-) and giving ethyne mild acidity (pKa≈25pK_a \approx 25), whereas ethane (sp3,pKa≈50sp^3, pK_a \approx 50) is non-acidic.

7. Hybridization Master Summary Table

SNHybridizationPure OrbitalsUnhybridized ppGeometryAnglesMultiple Bond CapacityKey Examples
2spsps+ps + p2Linear180∘180^\circ1 σ+2 π1\,\sigma + 2\,\piBeCl2,CO2,C2H2\text{BeCl}_2, \text{CO}_2, \text{C}_2\text{H}_2
3sp2sp^2s+2ps + 2p1Trigonal Planar120∘120^\circ1 σ+1 π1\,\sigma + 1\,\piBF3,C2H4,CO32−\text{BF}_3, \text{C}_2\text{H}_4, \text{CO}_3^{2-}
4sp3sp^3s+3ps + 3p0Tetrahedral109.5∘109.5^\circσ\sigma onlyCH4,NH3,H2O\text{CH}_4, \text{NH}_3, \text{H}_2\text{O}
5sp3dsp^3ds+3p+ds + 3p + d0Trigonal Bipyramidal90∘,120∘,180∘90^\circ, 120^\circ, 180^\circσ\sigma onlyPCl5,SF4,XeF2\text{PCl}_5, \text{SF}_4, \text{XeF}_2
6sp3d2sp^3d^2s+3p+2ds + 3p + 2d0Octahedral90∘,180∘90^\circ, 180^\circσ\sigma onlySF6,BrF5,XeF4\text{SF}_6, \text{BrF}_5, \text{XeF}_4
Test Your Knowledge

How many total sigma (σ) bonds and pi (π) bonds are present in a molecule of acrylonitrile, CH₂=CH-C≡N?

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Test Your Knowledge

What is the hybridization state of the central atom in the carbonate ion (CO₃²⁻) and in the ozone molecule (O₃)?

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Test Your Knowledge

Why is free rotation permitted around the carbon-carbon bond in ethane (C₂H₆), whereas rotation around the central carbon-carbon bond in 2-butene (CH₃-CH=CH-CH₃) is strictly restricted at room temperature?

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Test Your Knowledge

Which of the following statements correctly explains why terminal alkynes, such as ethyne (HC≡CH), are significantly more acidic (pKa ≈ 25) than alkanes, such as ethane (CH₃CH₃, pKa ≈ 50)?

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