Chemical Bonding and Molecular Structure · Lesson 7 of 12
Valence Bond Theory and Orbital Overlap
“Explain covalent-bond formation through forces, energy changes, orbital overlap, and the distinction between sigma and pi bonds.”
• Describe the attractive and repulsive interactions involved in forming a hydrogen molecule. • Interpret a potential-energy curve and identify equilibrium bond length and dissociation energy. • Explain orbital overlap, phase, and the role of electron pairing. • Distinguish sigma and pi bonds by overlap direction and electron-density distribution.
From an electron-pair drawing to a physical explanation
A Lewis structure tells us that H₂ contains one shared electron pair. It does not explain how the attraction develops or why the nuclei settle about 74 pm apart. Valence bond theory connects covalent bonding with atomic orbitals, electron pairing, and a lowering of energy.
An atomic orbital is a wave-based description of an electron’s spatial distribution. Its boundary drawing represents a region where the electron is likely to be found, rather than a solid container. When two atoms approach, their electron distributions can overlap. A favourable overlap allows a paired-electron distribution that helps stabilise the bonded system.
Attraction and repulsion as hydrogen atoms approach
Here “bonding in hydrogen” means formation of the covalent H–H bond, not the intermolecular interaction called hydrogen bonding. Imagine two H atoms far apart. Each has one nucleus and one electron. At a great separation their mutual interactions are negligible, so we can use that state as an energy reference.
As the atoms approach, each electron is attracted to both nuclei. These new cross-attractions lower the energy. At the same time, the two electrons repel one another and the two positively charged nuclei repel one another. Bond formation depends on the balance, not on attraction acting alone.
Problem
Identify whether each new interaction attracts or repels as two H atoms approach.
- 1.An electron on one atom and the nucleus of the other have opposite charges, so they attract.
- 2.The two electrons have the same negative charge, so they repel. The two nuclei have the same positive charge, so they also repel.
- 3.The net energy can decrease over a range of distances when stabilising effects outweigh repulsive ones. The final bond is a balance among the interactions.
Why there is a definite bond length
At first, bringing the atoms closer lowers the potential energy. Eventually the energy reaches a minimum. At still smaller separation, strong repulsion raises the energy steeply. The equilibrium bond length is the separation at the minimum, where the net force is zero.
For H₂, the energy minimum occurs near 74 pm and is roughly 436 kJ per mole below the separated-atom level. Forming the bond from separated gaseous H atoms releases approximately that energy. Dissociating a mole of H₂ back into gaseous atoms requires the corresponding positive energy input.
Problem
A qualitative curve has a minimum at 74 pm. What happens if the nuclei are pushed closer or pulled farther apart?
- 1.At the minimum, a small displacement in either direction raises the energy of the system.
- 2.Pushing the nuclei too close increases strong repulsive effects. Pulling them apart weakens the stabilising interactions until the separated-atom energy is approached.
- 3.The preferred separation is not zero distance. A chemical bond has an equilibrium length because attraction and repulsion compete.
Overlap and the pairing of electrons
In the valence bond picture of a simple two-electron bond, two half-filled atomic orbitals overlap and their electrons pair with opposite spins. Increased electron density in the region between the nuclei helps hold the atoms together. Appropriate orbital orientation is therefore important, not merely the presence of unpaired electrons somewhere in the atom.
The interpenetration of orbital distributions on approaching atoms, allowing the electron wavefunctions to interact in a bonding region.
Greater effective overlap generally gives a stronger bond when comparing otherwise similar situations. Hydrogen uses overlapping 1s orbitals, which are spherical. A p orbital has two lobes pointing in particular directions, so its overlap depends on how the atoms approach. This directional behaviour connects atomic orbitals with molecular geometry.
Orbital drawings often mark lobes with plus and minus signs or different colours. These represent the phase, or sign, of the wavefunction. They do not mean that one lobe carries positive electric charge and the other negative charge. Both lobes describe the same negatively charged electron distribution through the wavefunction.
Same-phase regions can reinforce in a favourable overlap. Opposite-phase overlap produces destructive interference. An orientation may also make the net overlap zero when positive and negative contributions cancel by symmetry. Orbital phases and orientation must therefore be considered together.
Problem
Let the internuclear axis be x. Why does overlap between an s orbital and a pᵧ orbital give no net sigma overlap?
- 1.The s orbital is symmetric about the x-axis, but the pᵧ orbital has opposite-sign lobes on the two sides of that axis.
- 2.The s orbital overlaps both pᵧ lobes equally. One contribution has the opposite phase to the other.
- 3.The contributions cancel, giving zero net overlap by symmetry. By contrast, a pₓ orbital points along the internuclear axis and can overlap head-on with s.
Sigma and pi bonds
The direction of overlap distinguishes two important kinds of covalent bond. Head-on overlap occurs along the line joining the nuclei, called the internuclear axis. Sideways overlap occurs between parallel orbitals on neighbouring atoms, producing electron density on opposite sides of that axis.
A covalent bond formed by head-on overlap along the internuclear axis, with electron density distributed around that axis.
A covalent bond formed by sideways overlap of parallel p orbitals, with electron density on either side of a plane containing the internuclear axis.
| Overlap | Possible bond type | Reason |
|---|---|---|
| s–s, head-on | Sigma | Spherical orbitals overlap along the nuclear axis. |
| s–p, appropriately directed head-on | Sigma | One p lobe points along the nuclear axis. |
| p–p, head-on | Sigma | Both p orbitals point along the nuclear axis. |
| p–p, parallel and sideways | Pi | Matching lobes overlap on both sides of the axis. |
In otherwise comparable overlaps, a sigma interaction is generally stronger because head-on overlap is more extensive than sideways overlap. An ordinary localised single bond contains one sigma bond. A double bond contains one sigma and one pi bond; a triple bond contains one sigma and two pi bonds. Adding pi bonding strengthens the total multiple bond even though an individual pi interaction is generally weaker than the corresponding sigma interaction.
A double bond includes both a sigma and a pi contribution. Comparing the whole double bond with a single bond is different from comparing its two components. Also, one pi bond contains two connected regions of density; the upper and lower regions are not two separate pi bonds.
Why overlap alone does not explain methane
The carbon ground-state valence arrangement has two unpaired electrons. A promoted arrangement can provide four, but using one pure s orbital and three pure p orbitals would not naturally give four identical bonds pointing at tetrahedral angles. Pure p orbitals are mutually perpendicular, which would suggest 90° directions for three bonds.
Likewise, simple use of unmodified p orbitals does not reproduce the observed ammonia and water angles. Hybridisation, developed next, provides a valence bond description using equivalent combinations of orbitals directed towards the observed bonding regions. The energy cost of promotion is considered within the overall energy gained through bonding, not as an isolated atom deciding to rearrange first.
Check your understanding
Use charge for the force questions, the energy minimum for equilibrium, and phase and geometry for overlap. These are related explanations, but they describe different aspects of the same bonding event.
Quiz
At equilibrium bond length on a potential-energy curve, the system has:
What do opposite signs on two lobes of a p orbital represent?
Which overlap forms a pi bond?
An ordinary localised triple bond consists of:
Why does the energy rise when bonded nuclei are pushed much closer than equilibrium?
Practice Problems
- Draw the attractive and repulsive interactions between the particles of two approaching H atoms.
- Sketch a potential-energy curve for H₂, mark the separated-atom reference, the equilibrium distance, and the dissociation-energy difference.
- Explain why opposite electron spins and favourable orbital overlap are both relevant to the simple valence bond description.
- Draw s–s, s–p, head-on p–p, and sideways p–p overlap, identifying sigma and pi cases.
- Using x as the internuclear axis, compare s–pₓ and s–pᵧ overlap and explain the different outcomes.
- A molecule has one ordinary double bond and four ordinary single bonds. Count the sigma and pi bonds.
- Explain why four unpaired electrons in pure carbon orbitals are insufficient by themselves to explain methane’s four equivalent tetrahedral bonds.
Key Takeaways
• Covalent bonding reflects a balance of attractive and repulsive interactions. • The potential-energy minimum gives the equilibrium bond length; dissociation requires energy. • Favourable orbital overlap supports an opposite-spin shared pair in the simple valence bond picture. • Orbital plus and minus signs indicate phase, not electric charge. • Head-on overlap gives sigma bonding; sideways overlap gives pi bonding.