Hybridisation and Multiple Bonds
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You should be comfortable with the concept of atomic orbitals (s, p, d) and have reviewed the previous lecture on sp, sp², and sp³ hybridisation geometries.
Why this matters
Hybridisation provides the orbital framework that explains how atoms form multiple bonds and achieve specific molecular geometries.
While VSEPR theory predicts molecular shapes based on electron pair repulsion, it does not explain the electronic structure underlying these shapes. Hybridisation theory bridges this gap by showing how atomic orbitals mix to form new, directional hybrid orbitals that overlap to create sigma (\(\sigma\)) and pi (\(\pi\)) bonds. Understanding this is crucial for predicting bond lengths, bond strengths, and the reactivity of molecules containing double or triple bonds, such as alkenes and alkynes.
The Intuitive Picture
Imagine you have a set of pure atomic orbitals: one spherical \(s\) orbital and three dumbbell-shaped \(p\) orbitals. In their pure state, these orbitals are not particularly good at forming strong, directional bonds with other atoms because their shapes and orientations are fixed by the atom's internal symmetry. However, when an atom prepares to bond, it can "mix" these orbitals mathematically to create new hybrid orbitals. Think of it like blending primary colours to get secondary colours; you are not creating new fundamental particles, but you are creating new shapes that are better suited for a specific task. For example, mixing one \(s\) and two \(p\) orbitals creates three \(sp^2\) hybrids that point towards the corners of a triangle. This allows the atom to form three strong bonds in a plane, leaving one unhybridised \(p\) orbital perpendicular to that plane. This remaining \(p\) orbital is key: it is the "spare" orbital that allows for the formation of a pi bond, which is the second component of a double bond. Thus, hybridisation is not just about geometry; it is about creating the correct orbital architecture to support both sigma and pi bonding.
Orbital Mixing and Geometry
The mathematical mixing of atomic orbitals is constrained by the requirement that the resulting hybrid orbitals must be orthogonal (mutually perpendicular in terms of their wavefunction overlap) and normalised. This constraint dictates the angles between the hybrid orbitals. For a central atom undergoing hybridisation, if we mix one \(s\) orbital and \((n-1)\) \(p\) orbitals, we generate \(n\) equivalent hybrid orbitals. The angle \(\theta\) between any two of these hybrids is determined by the orthogonality condition, which leads to the relationship \(\cos\theta = -\frac{1}{n-1}\). This result provides a rigorous quantum-mechanical justification for the geometries observed in molecules. For instance, in \(sp^3\) hybridisation (\(n=4\)), the angle is \(\arccos(-1/3) \approx 109.47^\circ\), which matches the tetrahedral geometry of methane. In \(sp^2\) hybridisation (\(n=3\)), the angle is \(\arccos(-1/2) = 120^\circ\), corresponding to trigonal planar geometry. This consistency between the orbital mixing rules and the observed molecular shapes validates the hybridisation model as a useful descriptive tool.
Formation of Multiple Bonds
A single covalent bond is always a sigma (\(\sigma\)) bond, formed by the head-on overlap of two hybrid orbitals (or one hybrid and one pure orbital). However, a double bond consists of one \(\sigma\) bond and one \(\pi\) bond. The \(\pi\) bond arises from the sideways overlap of unhybridised \(p\) orbitals that are parallel to each other. For example, in ethene (\(\mathrm{C_2H_4}\)), each carbon atom is \(sp^2\) hybridised. The three \(sp^2\) orbitals on each carbon form \(\sigma\) bonds with the other carbon and two hydrogen atoms. The remaining unhybridised \(p\) orbital on each carbon is perpendicular to the plane of the \(\sigma\) bonds. These two \(p\) orbitals overlap sideways to form a \(\pi\) bond. This \(\pi\) bond restricts rotation around the carbon-carbon axis, leading to cis-trans isomerism. A triple bond, as in ethyne (\(\mathrm{C_2H_2}\)), consists of one \(\sigma\) bond and two \(\pi\) bonds. Here, each carbon is \(sp\) hybridised, leaving two unhybridised \(p\) orbitals per carbon. These two pairs of \(p\) orbitals overlap sideways to form two orthogonal \(\pi\) bonds, resulting in a linear molecule with a very strong, short carbon-carbon bond.
Bond Strength and Length
The presence of multiple bonds significantly affects the physical properties of the bond. A triple bond is shorter and stronger than a double bond, which in turn is shorter and stronger than a single bond. This is because the \(\pi\) bonds add additional electron density between the nuclei, pulling them closer together. However, \(\pi\) bonds are generally weaker than \(\sigma\) bonds because the sideways overlap of \(p\) orbitals is less effective than the head-on overlap of hybrid orbitals. For example, the bond dissociation energy of a C-C single bond is approximately \(348\ \mathrm{kJ\,mol^{-1}}\), a C=C double bond is approximately \(614\ \mathrm{kJ\,mol^{-1}}\), and a C\(\equiv\)C triple bond is approximately \(839\ \mathrm{kJ\,mol^{-1}}\). Note that the increase in energy from single to double is not double the single bond energy, reflecting the weaker nature of the \(\pi\) component. This difference in bond strength and length is critical for understanding reaction mechanisms, where the breaking of a \(\pi\) bond is often the rate-determining step in addition reactions.
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 hybridisation is a physical process that occurs in time, as if the atom "decides" to hybridise before bonding. In reality, hybridisation is a mathematical model used to describe the ground state of the molecule. The orbitals do not mix and then unmix; the hybrid orbitals are simply a convenient basis for describing localised bonds, and the same electron density can be described equally well by delocalised molecular orbitals. Furthermore, hybridisation is not a universal requirement; some molecules, such as those with significant ionic character or where d-orbitals participate, may not fit neatly into the simple sp, sp², or sp³ models. Always check if the hybridisation model is appropriate for the specific chemical system you are analysing. C
Worked example
Problem. Calculate the bond length difference between a C-C single bond and a C=C double bond, given that the C-C single bond length is \(154\ \mathrm{pm}\) and the C=C double bond length is \(134\ \mathrm{pm}\). Determine the percentage reduction in bond length due to the formation of the \(\pi\) bond.
- Identify the given bond lengths: \(L_{\sigma} = 154\ \mathrm{pm}\) and \(L_{\sigma+\pi} = 134\ \mathrm{pm}\).
- Calculate the absolute difference in length: \(\Delta L = L_{\sigma} - L_{\sigma+\pi}\).
- Calculate the percentage reduction: \(\% \text{ reduction} = \frac{\Delta L}{L_{\sigma}} \times 100\%.\)
Units check. The units of length (pm) cancel out in the percentage calculation, leaving a dimensionless percentage, which is correct.
Three quick tests
- What is the hybridisation of the carbon atoms in ethyne (\(\mathrm{C_2H_2}\))?
Answer
sp hybridisation, resulting in a linear geometry with a bond angle of \(180^\circ\). - How many \(\pi\) bonds are present in a carbon-carbon triple bond?
Answer
Two \(\pi\) bonds, formed by the sideways overlap of two pairs of unhybridised \(p\) orbitals. - Why is a \(\pi\) bond generally weaker than a \(\sigma\) bond?
Answer
Because \(\pi\) bonds result from sideways overlap of \(p\) orbitals, which is less effective at concentrating electron density between the nuclei compared to the head-on overlap of hybrid orbitals in \(\sigma\) bonds.
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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.