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Two peanut-shaped orbitals are shown, lying vertically and parallel with one another. They overlap one another along the top and bottom of the orbital.
Pi (π) bonds form from the side-by-side overlap of two p orbitals. The dots indicate the location of the nuclei.

While all single bonds are σ bonds, multiple bonds consist of both σ and π bonds. As the Lewis structures in suggest, O 2 contains a double bond, and N 2 contains a triple bond. The double bond consists of one σ bond and one π bond, and the triple bond consists of one σ bond and two π bonds. Between any two atoms, the first bond formed will always be a σ bond, but there can only be one σ bond in any one location. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. These bonds are described in more detail later in this chapter.

A diagram contains three Lewis structures. The left most structure shows an H atom bonded to a C l atom by a single bond. The C l atom has three lone pairs of electrons. The phrase “One sigma bond No pi bonds” is written below the drawing. The center structure shows two O atoms bonded by a double bond. The O atoms each have two lone pairs of electrons. The phrase “One sigma bond One pi bond” is written below the drawing. The right most structure shows two N atoms bonded by a triple bond. Each N atom has a lone pairs of electrons. The phrase “One sigma bond Two pi bonds” is written below the drawing.

As seen in [link] , an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. Adding an additional π bond causes a further increase of 225 kJ/mol. We can see a similar pattern when we compare other σ and π bonds. Thus, each individual π bond is generally weaker than a corresponding σ bond between the same two atoms. In a σ bond, there is a greater degree of orbital overlap than in a π bond.

Counting σ and π bonds

This figure shows a molecule composed of four carbon atoms. There is a double bond between carbons one and two and three and four, while a single bond holds carbon two and three together. Carbons one and four are also bonded to two hydrogens with a single bond while carbons two and three are each bonded to one hydrogen each by a single bond.

Butadiene, C 6 H 6 , is used to make synthetic rubber. Identify the number of σ and π bonds contained in this molecule.

Solution

There are six σ C–H bonds and one σ C–C bond, for a total of seven from the single bonds. There are two double bonds that each have a π bond in addition to the σ bond. This gives a total nine σ and two π bonds overall.

Check your learning

Identify each illustration as depicting a σ or π bond:

(a) side-by-side overlap of a 4 p and a 2 p orbital

(b) end-to-end overlap of a 4 p and 4 p orbital

(c) end-to-end overlap of a 4 p and a 2 p orbital

Three diagrams are shown and labeled “a,” “b,” and “c.” Diagram a depicts two peanut-shaped orbitals lying vertically side-by-side and overlapping. One orbital is smaller than the other. Diagram b shows two peanut-shaped orbitals lying end-to-end and overlapping. Diagram c shows two unequally sized peanut-shaped orbitals lying end-to-end and overlapping.

Answer:

(a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis.

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Key concepts and summary

Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.

Chemistry end of chapter exercises

Explain how σ and π bonds are similar and how they are different.

Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).

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Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.

(a) Use the bond energy found in [link] to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)

(b) Use the enthalpy of reaction and the bond energies for H 2 and Cl 2 to solve for the energy of one mole of HCl bonds.

H 2 ( g ) + Cl 2 ( g ) 2 HCl ( g ) Δ H rxn ° = −184.7 kJ/mol
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Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.

The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.

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Use valence bond theory to explain the bonding in F 2 , HF, and ClBr. Sketch the overlap of the atomic orbitals involved in the bonds.

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Use valence bond theory to explain the bonding in O 2 . Sketch the overlap of the atomic orbitals involved in the bonds in O 2 .

Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side-by-side to form the π bond.
This figure shows the orbitals of two atoms being added together as they form bonds. The two atoms are shown separately on the right, each having two peanut-shaped orbitals lying perpendicularly to one another. A right-facing arrow shows that the two have moved closer together and now the upper and lower portions of the vertical peanut-shaped orbitals are shown as merging together above and below the plane of the molecule while the horizontal peanut-shaped orbitals are overlapping between the two nuclei.

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How many σ and π bonds are present in the molecule HCN?

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A friend tells you N 2 has three π bonds due to overlap of the three p -orbitals on each N atom. Do you agree?

No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.
Two nitrogen atoms are shown both in a Lewis structure and as a diagram. The Lewis structure depicts the two nitrogen atoms bonded by a triple bond. The diagram shows two nitrogen atoms and their three peanut-shaped p-orbitals. One of the orbitals lies horizontally and overlaps between the two nuclei. It is labeled, “sigma bond.” The other two lie vertically and in the z-plane of the page. They overlap above and below and into and out of the page in relation to the nuclei. They are each labeled, “pi bond.”

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Draw the Lewis structures for CO 2 and CO, and predict the number of σ and π bonds for each molecule.

(a) CO 2

(b) CO

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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