Sigma and Pi Bonding
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You should be comfortable with the concept of atomic orbitals (s, p, d) and their shapes, and have a basic understanding of how hybridisation determines molecular geometry as established in the previous lecture.
Why this matters
Understanding the distinction between sigma and pi bonds explains why double and triple bonds are shorter and stronger than single bonds, yet less flexible.
While hybridisation theory successfully predicts the three-dimensional arrangement of atoms in molecules, it does not fully describe the nature of the bonds themselves. In reality, a double bond is not simply two identical single bonds; it is a composite of two distinct types of electron density distributions. Recognising the difference between sigma (\(\sigma\)) and pi (\(\pi\)) bonding is fundamental to understanding molecular reactivity, stereochemistry, and the electronic structure of organic and inorganic compounds. This lecture bridges the gap between simple Lewis structures and the more rigorous molecular orbital theory that follows.
The Geometry of Overlap
Imagine two atoms approaching each other. The strength of the bond depends on how much their electron clouds overlap. There are two fundamental ways this overlap can occur. The first is head-on overlap, where the orbitals align along the axis connecting the two nuclei. This creates a cylindrical symmetry around the bond axis; if you were to rotate the molecule around this axis, the electron density would look exactly the same. This type of bond is called a sigma bond. It is the strongest type of covalent bond because the electron density is concentrated directly between the nuclei, maximising the attractive force between the positive nuclei and the negative electrons.
The second type of overlap is sideways. If two parallel p-orbitals overlap laterally, the electron density is concentrated above and below the internuclear axis, not directly between the nuclei. This creates a nodal plane that contains the bond axis. This type of bond is called a pi bond. Because the electron density is not directly between the nuclei, the pi bond is generally weaker than a sigma bond. Crucially, a pi bond cannot exist in isolation; it must always be accompanied by a sigma bond. This is why a double bond consists of one sigma and one pi bond, and a triple bond consists of one sigma and two pi bonds.
Sigma Bonds: The Foundation
A sigma bond is formed by the constructive interference of atomic orbitals along the internuclear axis. This can occur through several combinations: s-s overlap, s-p overlap, or p-p overlap (specifically the end-to-end overlap of p-orbitals). The key characteristic of a sigma bond is its cylindrical symmetry. The electron density is highest at the point midway between the two nuclei and decreases as you move away from the axis.
In the context of hybridisation, sigma bonds are typically formed using hybrid orbitals. For example, in methane (\(\mathrm{CH_4}\)), the carbon atom uses four \(sp^3\) hybrid orbitals to form four sigma bonds with hydrogen 1s orbitals. In ethene (\(\mathrm{C_2H_4}\)), each carbon uses three \(sp^2\) hybrid orbitals to form sigma bonds with two hydrogens and the other carbon. The remaining unhybridised p-orbital on each carbon is perpendicular to the plane of the sigma bonds. The sigma framework defines the skeleton of the molecule and allows for free rotation around the bond axis, which is why single bonds in alkanes are flexible.
Pi Bonds: The Sidekick
A pi bond arises from the lateral overlap of unhybridised p-orbitals. For a pi bond to form, the p-orbitals must be parallel to each other. This parallel alignment is only possible if the sigma bond framework has already fixed the relative orientation of the atoms. In ethene, the \(sp^2\) hybridisation forces the molecule to be planar, aligning the unhybridised p-orbitals on both carbons parallel to each other. Their sideways overlap creates the pi bond.
Unlike sigma bonds, pi bonds lack cylindrical symmetry. They possess a nodal plane that contains the internuclear axis. This structural feature has significant consequences. First, pi bonds are weaker than sigma bonds because the electron density is further from the nuclei. Second, and more importantly, pi bonds restrict rotation. If you attempt to rotate one carbon of ethene relative to the other, the parallel alignment of the p-orbitals is broken, and the pi bond is severed. This is why alkenes exhibit cis-trans isomerism, whereas alkanes do not. The pi bond acts as a "lock" that holds the molecular geometry rigid.
Multiple Bonds and Bond Order
When atoms form multiple bonds, the first bond is always a sigma bond, and any additional bonds are pi bonds. A double bond consists of one sigma and one pi bond. A triple bond consists of one sigma and two pi bonds. In a triple bond, such as in ethyne (\(\mathrm{C_2H_2}\)), the carbon atoms are \(sp\) hybridised. The two \(sp\) orbitals on each carbon form a sigma bond between the carbons and sigma bonds with the hydrogens. The two remaining unhybridised p-orbitals on each carbon are perpendicular to each other and to the bond axis. They overlap sideways in two mutually perpendicular planes to form two pi bonds.
The presence of pi bonds affects bond length and strength. Multiple bonds are shorter than single bonds because the increased electron density between the nuclei pulls them closer together. They are also stronger, but not by a simple additive factor. For example, the bond dissociation energy of a C=C double bond is less than twice that of a C-C single bond, and the C\(\equiv\)C triple bond is less than three times the strength of a C-C single bond. This is because the pi bonds are inherently weaker than the sigma bond. The bond order, defined as the number of bonding pairs minus anti-bonding pairs (or simply the number of shared pairs in simple Lewis terms), correlates with bond strength and inversely with bond length.
The result
Orbital hybridisation
One \(s\) orbital and \((n-1)\) \(p\) orbitals on a central atom can be mixed into \(n\) equivalent, directional hybrid orbitals. Choosing the mixing so all \(n\) hybrids are equally weighted and mutually orthogonal forces the angle \(\theta\) between any two of them to satisfy \(\cos\theta=-\tfrac{1}{n-1}\) — for \(n=2,3,4\) this reproduces the linear, trigonal planar, and tetrahedral angles exactly, giving a quantum-mechanical, orbital-level account of the geometries VSEPR already predicts.
A common misconception is that the pi bond in a double bond is "weaker" simply because it is the second bond formed. While it is true that the pi bond is weaker than the sigma bond, the total bond energy of a double bond is significantly higher than that of a single bond. Furthermore, the reactivity of alkenes is largely due to the high electron density of the pi bond, which is exposed above and below the molecular plane, making it susceptible to electrophilic attack. The pi bond is not just a "weak add-on"; it is the primary site of chemical reactivity in unsaturated compounds. C
Worked example
Problem. Consider the molecule propene (\(\mathrm{CH_3-CH=CH_2}\)). Determine the number of sigma (\(\sigma\)) and pi (\(\pi\)) bonds in the molecule. Given that the average bond dissociation energy for a C-C single bond is \(348\ \mathrm{kJ\,mol^{-1}}\) and for a C=C double bond is \(614\ \mathrm{kJ\,mol^{-1}}\), calculate the approximate energy contribution of the pi bond component in the double bond, assuming the sigma bond component in the double bond is similar in strength to a standard C-C sigma bond.
- Identify the bonding framework. Propene has three carbon atoms and six hydrogen atoms. The carbon skeleton is C-C=C. Each carbon forms four bonds in total.
- Count the sigma bonds. There is one C-C sigma bond between C1 and C2, one C-C sigma bond between C2 and C3 (part of the double bond), three C-H sigma bonds on C1, one C-H sigma bond on C2, and two C-H sigma bonds on C3. Total sigma bonds = \(1 + 1 + 3 + 1 + 2 = 8\).
- Count the pi bonds. There is one C=C double bond, which contains one pi bond. Total pi bonds = 1.
- Calculate the pi bond energy. The total energy of the C=C double bond is \(614\ \mathrm{kJ\,mol^{-1}}\). The energy of the C-C sigma bond is approximated by the single bond energy, \(348\ \mathrm{kJ\,mol^{-1}}\). The pi bond energy is the difference: \(614 - 348 = 266\ \mathrm{kJ\,mol^{-1}}\).
Units check. The calculation involves subtracting two energies in \(\mathrm{kJ\,mol^{-1}}\), resulting in a final value in \(\mathrm{kJ\,mol^{-1}}\), which is consistent with the definition of bond dissociation energy.
Three quick tests
- Why can a pi bond not exist without a sigma bond?
Answer
A pi bond requires the parallel alignment of p-orbitals, which is only possible if the sigma bond framework has already fixed the relative positions and orientations of the atoms. - How many sigma and pi bonds are present in the nitrile group (\(-\mathrm{C\equiv N}\))?
Answer
There is one sigma bond and two pi bonds in the carbon-nitrogen triple bond. - Explain why rotation around a C=C double bond is restricted, but rotation around a C-C single bond is free.
Answer
Rotation around a C-C single bond (sigma only) does not break any bonds. Rotation around a C=C double bond would break the pi bond because the parallel alignment of the p-orbitals required for pi overlap would be lost.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.