Revise Shapes of organic molecules; σ and π bonds for Cambridge 9701 (13.3)

A $\sigma$ bond is formed by the head-on overlap of atomic orbitals. This overlap can occur between two s-orbitals, an s-orbital and a p-orbital, or two p-orbitals. The crucial feature is that the region of electron overlap, and therefore the highest electron density, lies directly on the axis connecting the two nuclei. This creates a strong, single covalent bond.

Consider ethane, $C_2H_6$. It contains seven $\sigma$ bonds in total: one C-C bond and six C-H bonds. Because the electron density is symmetrical around the internuclear axis of the C-C bond, there is free rotation of the methyl ($-CH_3$) groups relative to one another. This rotation does not disrupt the orbital overlap.

A $\pi$ bond is formed by the sideways overlap of two parallel p-orbitals. This can only happen after a $\sigma$ bond has already formed, bringing the atoms close enough together. The p-orbitals overlap in two regions: one above and one below the plane of the $\sigma$ bond. A $\pi$ bond is weaker than a $\sigma$ bond because the orbital overlap is less effective.

The presence of a $\pi$ bond restricts rotation. To rotate the atoms, the sideways p-orbital overlap must be broken, which requires a substantial input of energy. This restricted rotation is a key feature of alkenes and leads to the possibility of E/Z isomerism.

Single Bond = 1 $\sigma$ bond Double Bond = 1 $\sigma$ bond + 1 $\pi$ bond Triple Bond = 1 $\sigma$ bond + 2 $\pi$ bonds

We use VSEPR theory, treating each bonding region (single, double, or triple bond) as one area of electron density. The shape is determined by the number of bonding regions and lone pairs around the central atom.

Ethane ($C_2H_6$): Each carbon atom is the centre. It forms four single bonds (four $\sigma$ bonds) and has no lone pairs. These four regions of electron density repel each other equally, resulting in a tetrahedral arrangement with bond angles of 109.5°.

Ethene ($C_2H_4$): Each carbon atom is the centre. It forms two single bonds and one double bond. The double bond counts as one region of electron density. With three regions of electron density and no lone pairs, the arrangement is trigonal planar with bond angles of approximately 120°. The entire molecule lies in a single plane.

Benzene, $C_6H_6$, presents a unique bonding scenario. It is a planar, hexagonal ring of six carbon atoms. Each carbon atom forms three $\sigma$ bonds: one to a hydrogen atom and one to each of its two neighbouring carbon atoms. Since there are three bonding regions around each carbon, the geometry is trigonal planar, and all bond angles are 120°.

This leaves one p-orbital on each carbon atom, perpendicular to the plane of the ring. Instead of forming three distinct $\pi$ bonds, these six p-orbitals overlap sideways all around the ring. This creates a continuous, delocalised $\pi$ system – two doughnut-shaped clouds of electron density, one above and one below the plane of the ring. The electrons are shared equally among all six carbon atoms.