Sigma and Pi Bonds and Hybridisation

Dot-and-cross diagrams tell you how many electrons are shared, but not where they are. This note describes covalent bonds in terms of overlapping atomic orbitals: the head-on overlap that makes a sigma ($\sigma$) bond and the sideways overlap that makes a pi ($\pi$) bond. It then explains hybridisation, the mixing of s and p orbitals into sp, sp² and sp³ orbitals, which accounts for the shapes of methane, ethene and ethyne. These ideas are tested in Paper 2 and are the foundation for explaining why alkenes react as they do in organic chemistry.

Bonds as overlapping orbitals

An atomic orbital is a region of space where there is a high probability of finding an electron, and it holds up to two electrons. A covalent bond forms when an orbital on one atom, containing one electron, overlaps with an orbital on another atom, also containing one electron. The two electrons pair up in the overlap region between the nuclei, and both nuclei are attracted to this concentration of negative charge. That is exactly the "shared pair" in the definition of covalent bonding.

The greater the overlap, the greater the electron density between the nuclei and the stronger the bond. There are two ways orbitals can overlap.

Sigma bonds

"Direct" means head-on or end-on: the orbitals point straight at each other along the line joining the two nuclei (the internuclear axis). The electron density of a $\sigma$ bond is concentrated on that axis, between the nuclei.

Sigma bonds can form from several kinds of overlap:

• s with s, as in $\ce{H2}$: two spherical 1s orbitals overlap.
• s with p, as in $\ce{HCl}$: the hydrogen 1s orbital overlaps end-on with a chlorine 3p orbital.
• p with p end-on, as in $\ce{Cl2}$: one 3p orbital from each chlorine points along the axis and they overlap.
• hybrid orbitals with s or other hybrid orbitals, as in $\ce{CH4}$ and $\ce{C2H6}$ (see below).

Every single covalent bond is a $\sigma$ bond. Because the electron density is symmetrical around the axis, one atom can rotate relative to the other without breaking a $\sigma$ bond. That is why the two $\ce{CH3}$ groups in ethane rotate freely.

Pi bonds

A $\pi$ bond can only form between two atoms that are already joined by a $\sigma$ bond. Each atom must have a p orbital at right angles to the $\sigma$ bond, and the two p orbitals must be parallel. They then overlap side by side, producing two regions of electron density: one above and one below the internuclear axis. These two regions together are one $\pi$ bond containing two electrons.

Three consequences follow, and all are examinable.

1. A $\pi$ bond is weaker than a $\sigma$ bond. Sideways overlap is less effective than head-on overlap, and the $\pi$ electrons are further from the nuclei. That is why the $\ce{C=C}$ bond energy ($610\ \text{kJ mol}^{-1}$) is less than twice the $\ce{C-C}$ bond energy ($2 \times 350 = 700\ \text{kJ mol}^{-1}$): the extra $\pi$ bond adds only about $260\ \text{kJ mol}^{-1}$.
2. $\pi$ electrons are exposed. They sit above and below the plane of the molecule, away from the nuclei, so they are attacked by electrophiles (electron-pair acceptors). This is why alkenes are much more reactive than alkanes and undergo addition reactions.
3. A $\pi$ bond prevents rotation. Rotating one end would twist the p orbitals out of line and destroy the overlap. This restricted rotation about $\ce{C=C}$ causes geometrical (cis/trans) isomerism in organic chemistry.

In a triple bond the two $\pi$ bonds are at right angles to each other: one from a pair of p orbitals pointing up and down, the other from a pair pointing in and out of the page. Together they make a cylinder of electron density around the $\sigma$ bond.

Hybridisation

The problem hybridisation solves

Carbon's ground-state configuration is $1s^2\,2s^2\,2p^2$. Only the two 2p electrons are unpaired, which suggests carbon should form two bonds. But carbon forms four bonds in methane, and all four C–H bonds are identical, pointing to the corners of a regular tetrahedron at $109.5^\circ$. Pure s and p orbitals cannot explain this: p orbitals are at $90^\circ$ to each other, and an s orbital has no direction at all.

The model that fixes this has two steps.

1. Promotion. One 2s electron is promoted to the empty 2p orbital: $1s^2\,2s^1\,2p^3$. Now there are four unpaired electrons. The energy needed is more than repaid by forming two extra bonds.
2. Hybridisation. The 2s orbital and some or all of the 2p orbitals mix to form new orbitals of equal energy, called hybrid orbitals. They point in directions that keep them as far apart as possible.

The number of hybrid orbitals always equals the number of atomic orbitals mixed. Any p orbitals that are not mixed stay as pure p orbitals and are available to form $\pi$ bonds.

sp³ hybridisation

The 2s and all three 2p orbitals mix to give four sp³ orbitals, each with one electron, directed to the corners of a tetrahedron at $109.5^\circ$.

• Methane: each sp³ orbital overlaps head-on with a hydrogen 1s orbital, giving four identical C–H $\sigma$ bonds.
• Ethane, $\ce{C2H6}$: each carbon is sp³. The C–C bond is a $\sigma$ bond formed by overlap of one sp³ orbital from each carbon; each C–H bond is an sp³–1s $\sigma$ bond. There are seven $\sigma$ bonds and no $\pi$ bonds.
• Ammonia and water: nitrogen and oxygen are also roughly sp³ hybridised, with lone pairs occupying some of the sp³ orbitals. This is consistent with their bond angles ($107^\circ$, $104.5^\circ$) being close to the tetrahedral angle.

sp² hybridisation

The 2s and two of the 2p orbitals mix to give three sp² orbitals, in one plane at $120^\circ$ (trigonal planar). The third 2p orbital is left unhybridised, at right angles to that plane.

Ethene, $\ce{C2H4}$: each carbon is sp².

• Two sp² orbitals on each carbon overlap with hydrogen 1s orbitals: four C–H $\sigma$ bonds.
• The third sp² orbital on each carbon overlaps head-on with the one on the other carbon: the C–C $\sigma$ bond.
• The unhybridised p orbitals, one on each carbon, are parallel and overlap sideways: the $\pi$ bond.

So the double bond is one $\sigma$ bond plus one $\pi$ bond. For the p orbitals to stay parallel, all six atoms must lie in one plane, with bond angles of about $120^\circ$. Ethene is a planar molecule.

sp hybridisation

The 2s and one 2p orbital mix to give two sp orbitals, pointing in opposite directions at $180^\circ$ (linear). Two 2p orbitals are left unhybridised, at right angles to each other and to the sp axis.

• Ethyne, $\ce{C2H2}$: each carbon uses one sp orbital for a C–H $\sigma$ bond and one for the C–C $\sigma$ bond. The two pairs of leftover p orbitals overlap sideways to give two $\pi$ bonds. The molecule is linear.
• Hydrogen cyanide, $\ce{HCN}$: carbon is sp. One sp orbital forms the C–H $\sigma$ bond; the other overlaps with an orbital on nitrogen to form the C–N $\sigma$ bond. Two p orbitals on carbon overlap sideways with two p orbitals on nitrogen to form two $\pi$ bonds. So $\ce{C#N}$ is one $\sigma$ plus two $\pi$, and the molecule is linear ($180^\circ$). Nitrogen keeps one lone pair, pointing away from carbon along the axis.
• Nitrogen, $\ce{N2}$: the $\sigma$ bond forms from head-on overlap of one orbital on each nitrogen along the axis (a p orbital, or an sp hybrid). The two remaining p orbitals on each atom overlap sideways in pairs to give two $\pi$ bonds. Each nitrogen keeps one lone pair. One $\sigma$ plus two $\pi$ bonds make the $\ce{N#N}$ bond very strong ($944\ \text{kJ mol}^{-1}$).

For example, in methanal, $\ce{H2C=O}$, carbon has three regions (two C–H and the C=O), so it is sp², with the $\pi$ bond formed from its unhybridised p orbital.

Worked examples

Practice

Answers

1. A $\sigma$ bond is formed by direct (head-on) overlap of orbitals between the bonding atoms, with electron density on the line joining the nuclei. A $\pi$ bond is formed by sideways overlap of adjacent p orbitals, with electron density above and below the $\sigma$ bond.
2. (a) $1\sigma$, $2\pi$. (b) Two C=O: $2\sigma$, $2\pi$. (c) Seven single bonds: $7\sigma$, $0\pi$. (d) C–H $1\sigma$; C≡N $1\sigma + 2\pi$: $2\sigma$, $2\pi$.
3. The 1s orbital of each hydrogen atom, each containing one electron, overlaps directly (head-on) with the other. The two electrons pair in the overlap region between the nuclei, forming a $\sigma$ bond; both nuclei are attracted to this shared pair.
4. In ethane the C–C bond is only a $\sigma$ bond; its electron density is symmetrical about the axis, so rotation does not reduce the overlap. In ethene the $\pi$ bond needs the two p orbitals to stay parallel; rotating one carbon would twist the p orbitals apart and break the $\pi$ bond, which needs a large amount of energy.
5. (a) sp³ (four $\sigma$ bonds). (b) sp (two regions: two C=O; carbon uses two unhybridised p orbitals for the two $\pi$ bonds). (c) $\ce{CH3}$ carbon: sp³ (four $\sigma$ bonds); $\ce{CHO}$ carbon: sp² (three regions: C–C, C–H, C=O).
6. One 2s electron is promoted to 2p, giving $2s^1\,2p^3$. The 2s orbital mixes with two of the 2p orbitals to form three equivalent sp² hybrid orbitals, each holding one electron, in one plane at $120^\circ$. The third 2p orbital is unchanged, perpendicular to the plane.
7. A double bond is one $\sigma$ bond plus one $\pi$ bond, not two $\sigma$ bonds. The $\pi$ bond forms by sideways overlap, which is less effective than head-on overlap, and its electrons are further from the nuclei, so it is weaker than a $\sigma$ bond ($610 - 350 = 260$ compared with $350\ \text{kJ mol}^{-1}$).
8. Bonds: six C–H ($6\sigma$), one C–C ($1\sigma$), two C=C ($2\sigma + 2\pi$): 9 $\sigma$, 2 $\pi$. Every carbon has three regions of electron density, so all four are sp². Each sp² carbon and its attached atoms are planar, and the $\pi$ overlap keeps each double bond planar, so all four carbons can lie in one plane (the molecule's most stable form is planar).
9. The $\ce{CH3}$ carbon is sp³ and the nitrile carbon is sp. The C–C bond is a $\sigma$ bond from head-on overlap of an sp³ orbital with an sp orbital. The C≡N bond is one $\sigma$ bond (head-on overlap of the other carbon sp orbital with an orbital on nitrogen) and two $\pi$ bonds (sideways overlap of two pairs of p orbitals at right angles). Because the nitrile carbon is sp, the $\ce{C-C-N}$ angle is $180^\circ$.
10. The central carbon forms two $\sigma$ bonds and no lone pairs: two regions, so sp. It has two unhybridised p orbitals, at right angles to each other. One overlaps sideways with the p orbital on the left carbon, the other with the p orbital on the right carbon, so the two $\pi$ bonds are perpendicular. Each terminal carbon is sp², and its two H atoms lie in the plane perpendicular to its p orbital. Since the two p orbitals involved are at right angles, the two $\ce{CH2}$ groups lie in perpendicular planes, not the same plane.