sp, sp², and sp³ Hybridisation
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the shapes of atomic s and p orbitals and the basic principles of Valence Bond Theory, specifically how overlapping orbitals form sigma bonds.
Why this matters
Hybridisation resolves the geometric mismatch between atomic orbitals and observed molecular shapes.
Atomic orbitals are fixed in orientation: s orbitals are spherical, and p orbitals are mutually perpendicular. If carbon atoms used pure p orbitals to form bonds, we would expect bond angles of 90 degrees. However, methane has a perfect tetrahedral geometry with angles of 109.5 degrees. Hybridisation is the mathematical mixing of atomic orbitals to create new hybrid orbitals that point in the specific directions required by the molecule's geometry, allowing for stronger, more directional bonding.
The Concept of Mixing
Imagine you have a set of tools: one round hammer (the s orbital) and three square chisels (the p orbitals). If you try to build a tetrahedron using only these, the fit is poor. Hybridisation is like reshaping these tools into four identical, pointed wedges that fit perfectly into the corners of a tetrahedron. In quantum mechanics, this "reshaping" is a linear combination of the original wavefunctions. The s orbital contributes to the electron density along the bond axis, while the p orbitals provide the directional lobes. By mixing them, we create orbitals that are no longer purely s or p, but hybrids with specific directional properties. This process conserves the total number of orbitals: one s and three p orbitals mix to form four new sp3 orbitals. The key insight is that the shape of the orbital is dictated by the need to maximise overlap with neighbouring atoms and minimise electron-electron repulsion, leading to the specific geometries we observe in nature.
sp3 Hybridisation: The Tetrahedral Case
Carbon in methane (CH4) is the classic example. The ground state electron configuration of carbon is 1s2 2s2 2p2. To form four bonds, one electron from the 2s orbital is promoted to the empty 2p orbital, creating four unpaired electrons. These four orbitals (one 2s and three 2p) then hybridise to form four equivalent sp3 orbitals. Each sp3 orbital consists of 25% s character and 75% p character. Because the s orbital is spherical and the p orbitals are directional, the resulting hybrid orbitals are asymmetric, with a larger lobe on one side and a smaller lobe on the other. The four orbitals arrange themselves in space to be as far apart as possible, resulting in a tetrahedral geometry. The angle between any two sp3 orbitals is 109.47 degrees, which matches the experimental bond angle in methane perfectly.
sp2 and sp Hybridisation: Planar and Linear
When a central atom forms three sigma bonds and has no lone pairs, it uses sp2 hybridisation. One s orbital mixes with two p orbitals to form three sp2 orbitals, leaving one unhybridised p orbital perpendicular to the plane of the hybrids. The three sp2 orbitals are arranged in a trigonal planar geometry with 120-degree angles. Each sp2 orbital has 33.3% s character and 66.7% p character. This geometry is seen in molecules like ethene (C2H4), where the unhybridised p orbitals overlap sideways to form the pi bond of the double bond. For sp hybridisation, one s orbital mixes with one p orbital to form two sp orbitals. These two orbitals are linear, pointing in opposite directions at 180 degrees. Each sp orbital has 50% s character and 50% p character. Two unhybridised p orbitals remain perpendicular to the axis, allowing for the formation of two pi bonds, as seen in acetylene (C2H2). The higher the s character, the shorter and stronger the bond, because s orbitals are closer to the nucleus.
Quantitative Relationship Between Geometry and Mixing
The specific angles of hybridisation are not arbitrary; they are derived from the requirement that the hybrid orbitals be orthogonal (mutually perpendicular in the mathematical sense of their wavefunctions). For a set of \(n\) equivalent hybrid orbitals formed from one s and \(n-1\) p orbitals, the angle \(\theta\) between any two hybrids is determined by the mixing coefficients. The general relationship is given by the cosine of the angle. For sp3 (\(n=4\)), the angle is 109.47 degrees. For sp2 (\(n=3\)), it is 120 degrees. For sp (\(n=2\)), it is 180 degrees. This mathematical consistency provides a robust framework for predicting molecular geometry based on the number of electron domains around the central atom. The s-character percentage can be calculated as \(100/n\%\). Thus, sp3 is 25% s, sp2 is 33.3% s, and sp is 50% s. This variation in s-character explains trends in bond length and acidity; for instance, terminal alkynes (sp C-H) are more acidic than alkenes (sp2 C-H) because the sp hybrid orbital holds the bonding electrons closer to the nucleus, stabilising the conjugate base.
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 happens in time, like atoms "deciding" to mix their orbitals. In reality, hybridisation is a mathematical model used to describe the stationary states of the molecule. The orbitals do not change shape dynamically; rather, the hybrid orbitals are the correct basis set to describe the electron distribution in the bonded state. Furthermore, hybridisation is not limited to carbon; it applies to any atom where the geometry requires mixing of s and p orbitals, such as nitrogen in ammonia or oxygen in water, although the presence of lone pairs can distort the ideal angles slightly due to repulsion differences. C
Worked example
Problem. Calculate the percentage of s-character in the hybrid orbitals of a carbon atom in a molecule where the bond angle between two C-H bonds is measured to be 109.5 degrees. Determine the hybridisation type.
- Identify the relationship between bond angle and hybridisation. The angle 109.5 degrees corresponds to the tetrahedral geometry.
- Recall that tetrahedral geometry arises from sp3 hybridisation, which involves mixing one s orbital and three p orbitals.
- Calculate the s-character. The total number of hybrid orbitals is 4. The contribution of the s orbital is 1 out of 4.
- Compute the percentage: \(\frac{1}{4} \times 100\% = 25\% \).
Units check. The percentage is a dimensionless ratio, which is correct for orbital character composition.
Three quick tests
- What is the bond angle associated with sp hybridisation?
Answer
180 degrees. - If a carbon atom forms three sigma bonds and one pi bond, what is its hybridisation?
Answer
sp2. - Which hybrid orbital has the highest s-character?
Answer
sp (50% s-character).
Back to CU-102 →
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.