The formation of molecular orbitals from atomic orbitals, sigma and pi bonds, sp, sp2 and sp3 hybridisation and the shapes they give, and how conjugation and chromophores lead to the absorption of visible light and colour in organic molecules.

What this key area is asking

The SQA wants you to explain how molecular orbitals form from atomic orbitals, to describe sigma and pi bonds, to use sp, sp2 and sp3 hybridisation to predict shape, and to explain how conjugation and chromophores let organic molecules absorb visible light and appear coloured. The sigma-versus-pi distinction, hybridisation-to-shape link and the conjugation explanation of colour are reliable exam earners.

Molecular orbitals from atomic orbitals

Sigma and pi bonds

A sigma bond allows free rotation, whereas a pi bond locks the atoms in place, which is why double bonds give rise to geometric (cis-trans) isomerism.

Hybridisation and shape

To explain the observed shapes of molecules, the s and p atomic orbitals on carbon mix to form equivalent hybrid orbitals:

• sp3 hybridisation mixes one s and three p orbitals into four equivalent orbitals at $109.5^{\circ}$, giving a tetrahedral shape and four single bonds (as in methane and the alkanes).
• sp2 hybridisation mixes one s and two p orbitals into three orbitals at $120^{\circ}$ in a plane, leaving one unhybridised p orbital for a pi bond, giving a planar arrangement (as in alkenes).
• sp hybridisation mixes one s and one p orbital into two orbitals at $180^{\circ}$, leaving two unhybridised p orbitals for two pi bonds, giving a linear shape (as in alkynes).

Conjugation, chromophores and colour

This is why short-chain molecules are colourless (they absorb only ultraviolet light) while long conjugated molecules such as beta-carotene are strongly coloured.

Examples in context

Molecular orbitals and conjugation explain colour throughout chemistry and biology. Beta-carotene, the orange pigment in carrots, has eleven conjugated double bonds, giving a chromophore long enough to absorb blue light and reflect orange. The same idea explains the colours of dyes, indicators and the visual pigment retinal in the eye, where absorbing a photon changes the shape of a conjugated molecule and triggers a nerve signal. Hybridisation accounts for the shapes that determine reactivity: the planar sp2 carbon of a carbonyl group is open to attack by nucleophiles, while the tetrahedral sp3 carbons of an alkane are unreactive. The sigma-pi picture also underpins why double bonds cannot rotate, the basis of stereochemistry covered next.

Try this

Q1. State how a sigma bond differs from a pi bond in terms of orbital overlap. [2 marks]

• Cue. A sigma bond is end-on overlap along the bond axis; a pi bond is side-on overlap of p orbitals above and below the axis.

Q2. State the hybridisation and shape around each carbon in ethane, $\text{C}_2\text{H}_6$. [2 marks]

• Cue. sp3 hybridised, tetrahedral ($109.5^{\circ}$).

Q3. Explain why a longer conjugated system absorbs light of lower energy. [1 mark]

• Cue. More delocalisation lowers the gap between the occupied and unoccupied molecular orbitals, so lower-energy (longer-wavelength) light is absorbed.