Structure of Double Bond

Fig. 9.4 is a schematic orbital diagram of ethene (C₂H₄) that shows only the σ-bond framework — the skeleton of the molecule formed by head-on orbital overlaps. The figure deliberately omits the π bond (which is shown separately in Fig. 9.5) so you can focus on the geometry and hybridisation of the sigma bonds.

Each carbon atom in ethene is $sp^2$ hybridised. This means one $2s$ orbital mixes with two $2p$ orbitals to produce three equivalent $sp^2$ hybrid orbitals that lie in a plane, pointing toward the vertices of an equilateral triangle — separated by $120^\circ$. The remaining unhybridised $2p$ orbital on each carbon sticks out perpendicular to this plane (it will form the π bond later).

In the figure, you see the three $sp^2$ hybrid orbitals on each carbon. Two of them overlap head-on with the $1s$ orbitals of two hydrogen atoms, forming two C–H σ bonds. The third $sp^2$ orbital on each carbon overlaps head-on with the corresponding $sp^2$ orbital of the other carbon, forming the C–C σ bond. All six atoms — two carbons and four hydrogens — lie in the same plane. The bond angles are close to $120^\circ$, and the C–C σ bond length is about $154\ \text{pm}$ (the same as a typical single bond in ethane).

The key physical idea the figure teaches is that the σ bond is strong and localised along the internuclear axis. Because it is formed by direct, head-on overlap, it has high electron density between the nuclei and a bond enthalpy of about $397\ \text{kJ mol}^{-1}$ for the C–C σ component. The planar geometry is a direct consequence of $sp^2$ hybridisation: the three σ bonds around each carbon repel each other to maximise separation, giving the $120^\circ$ angles.

The textbook develops the following central relationship from this figure:

The figure also reinforces that the π bond is weaker than the σ bond because sideways overlap is less effective than head-on overlap. This weakness is why alkenes are reactive toward electrophiles: the π electrons are loosely held and easily attacked. …

The figure shows two carbon atoms, each with three $sp^2$ hybrid orbitals lying in a plane at $120^\circ$ angles. On each carbon, one unhybridised $2p$ orbital sticks out perpendicular to that plane — one lobe above and one below. The two $2p$ orbitals, one from each carbon, are parallel to each other and aligned so that their lobes overlap sideways, both above and below the plane of the molecule. This sideways overlap creates the $\pi$ bond.

The key physical idea is that a $\pi$ bond is not concentrated along the internuclear axis. Instead, its electron density is distributed in two regions — one above and one below the molecular plane — with a node (zero electron density) exactly in the plane, right along the line joining the two carbon nuclei. This is why the $\pi$ bond is weaker than a $\sigma$ bond: sideways overlap is less effective than head-on overlap. The electrons in a $\pi$ bond are also more exposed and more mobile, which is why alkenes are attacked by reagents that seek electrons (electrophiles).

The textbook uses this figure to explain the bond enthalpy data for the double bond. The total bond enthalpy of the C=C double bond in ethene is $681\ \text{kJ mol}^{-1}$. This is the sum of the $\sigma$ bond enthalpy (about $397\ \text{kJ mol}^{-1}$) and the $\pi$ bond enthalpy (about $284\ \text{kJ mol}^{-1}$). For comparison, a single C–C $\sigma$ bond in ethane has a bond enthalpy of only $348\ \text{kJ mol}^{-1}$.

The $\pi$ bond alone is weaker than a single $\sigma$ bond, but the double bond as a whole is stronger than a single bond. The bond length of the double bond ($134\ \text{pm}$) is also shorter than that of a single bond ($154\ \text{pm}$), because the sideways overlap pulls the two carbons closer together. …