Sigma and Pi bondsCourse level: AP Chemistry

Bond Order, Length, Strength and Rotation

What you'll be able to do: Use bond order to compare bond length and strength, and explain why pi bonds block rotation.

Best after: Sigma and Pi Bonds

Introduction

Bond order is a single number that predicts length, strength and even whether a molecule can twist.

These are recommended, not required. You can start this lesson at any time.

Learning objectives

  • Calculate bond order including fractional values from resonance
  • Rank bonds by length and strength using bond order
  • Explain free rotation about sigma bonds
  • Explain restricted rotation about double bonds and cis-trans isomerism
  • Relate delocalization to intermediate bond lengths

Bond order and its consequences

Bond order is the number of bonding pairs shared between two atoms. C-C is 1, C=C is 2 and a carbon-carbon triple bond is 3. Length falls and strength rises as bond order rises: roughly 154, 134 and 120 picometres respectively for those three. More shared electron density between the nuclei means a stronger electrostatic pull holding them closer together.

The comparison is strongest within the same atom pair and broadly comparable environments. A C≡C bond is normally shorter and stronger than C=C, which is shorter and stronger than C–C. Atomic size and electronic environment also affect lengths and bond enthalpies, so do not compare unrelated atom pairs using bond order alone. A double bond is not exactly twice the energy of a single bond because its σ and π components are not equivalent contributions.

Shorter almost always means stronger for bonds between the same pair of elements.

Average bond orders only when the contributors justify it

For equivalent resonance contributors, follow one fixed pair of atoms through every contributor and average equally. An O–O pair in ozone is single once and double once, giving (1 + 2)/2 = 1.5. A C–O pair in carbonate is double once and single twice, giving 4/3. Equivalent positions therefore have equivalent intermediate bond lengths in the hybrid.

If contributors are not equivalent, their weights need not be equal; an arithmetic mean of every drawable structure is not generally valid. Fractional Lewis bond order summarizes delocalization in this model, not a literal mixture of separately measured single and double bonds. Molecular-orbital bond order is a related concept obtained by a different counting scheme, so state the model being used.

Equivalent resonance contributors explain equal measured bonds and fractional average bond order; this is one delocalized species.

Distinguish rotation from geometric isomerism

Rotation about an isolated C–C single bond preserves its σ overlap, so ordinary conformations of ethane and butane interconvert readily at room temperature. Rotation still has finite torsional barriers from electron interactions and steric effects; it is not literally energy free. Rings and bulky substituents can restrict it, and resonance gives some nominal single bonds, such as an amide C–N bond, partial double-bond character.

Twisting a C=C bond disrupts the parallel p overlap, producing a much larger barrier. Distinct alkene geometric isomers require each double-bond carbon to have two different substituents. 2-butene satisfies that condition: each alkene carbon has H and CH₃, giving cis and trans arrangements. Propene does not, because its terminal CH₂ carbon has two identical H atoms. A double bond alone is therefore not sufficient to guarantee cis/trans isomerism. More general substituent patterns use E/Z nomenclature.

But-2-ene has two different groups at each double-bonded carbon, allowing distinct cis/trans configurations with the same connectivity.
Restricted rotation is a property of pi bonds, not of bond strength in general.

Use bond measurements as evidence of delocalization

If a symmetric ion is represented by equivalent single/double contributors but has equal measured bond lengths, the intermediate lengths support a delocalized hybrid. Do not claim the instrument merely averages rapidly alternating long and short bonds. In nitrate or carbonate, the equivalent bonds belong to one electronic description at the given geometry.

The same logic can explain restricted rotation in a bond drawn as single in one contributor. In an amide, overlap of the nitrogen lone pair with the carbonyl π system gives the C–N link partial double-bond character and favors a planar arrangement. A local bond line is therefore not sufficient on its own to establish completely free rotation. Check resonance as well as the σ/π labels before interpreting the observed behavior.

Average bond order follows the shared bonding pattern across equivalent positions.

Key ideas

Definition
Bond order

The number of shared bonding pairs between two atoms, averaged over resonance structures if necessary.

Rule
Length and strength trend

As bond order rises, bond length falls and bond dissociation energy rises.

Rule
Rotation

Rotation about isolated single bonds often interconverts conformers readily, with finite barriers; π bonding and delocalization can substantially restrict rotation.

Key concept
Ozone

Bond order 1.5 gives two identical O-O bonds intermediate in length between single and double.

Equation
Resonance bond order

bond order = total bonds between the pair / number of resonance structures

    Worked examples

    Worked example 1

    Rank the carbon-carbon bonds in ethane, ethene and ethyne by length and explain the order.

    Try it first: Write the bond order for each molecule.

      0 of 4 steps revealed.

      Worked example 2

      Why does 2-butene have cis and trans isomers while butane does not have separable isomers of the same kind?

      Try it first: Identify which bond would need to rotate.

        0 of 4 steps revealed.

        Worked example 3

        What is the average C-O bond order in carbonate CO₃²⁻?

        Try it first: Draw three equivalent contributors, each with one C=O and two C-O bonds.

          0 of 3 steps revealed.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck instead
          Assuming ozone has one short and one long O-O bond.Resonance delocalizes the pi electrons, so both bonds are identical with order 1.5.Average the bond orders across all resonance structures.
          Saying a double bond is simply twice as strong as a single bond.The pi component is weaker than the sigma component, so the total is less than double.Compare measured bond enthalpies rather than multiplying.
          Thinking single bonds cannot rotate.Sigma overlap is unchanged by rotation about the axis.Ask whether a pi bond is present before claiming restricted rotation.

          Scroll horizontally if needed to read all columns.

          After the lesson · Question bank

          Independent practice

          Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.

          22 topic questions across 2 difficulty levels.

          Loading practice questions…

          What you should now know

          Within comparable bonds, higher bond order generally means shorter and stronger bonding. Equivalent resonance contributors can give equal intermediate bonds. Single-bond rotation often has a modest barrier, whereas twisting a π system costs much more; alkene geometric isomers additionally require different substituents at each double-bond carbon.

          • Compare bond order within similar atom pairs and environments.
          • Average equally only when resonance contributors are equivalent.
          • Isolated σ-bond rotation is often accessible but has finite barriers.
          • π overlap gives a large twisting barrier and conjugation can restrict nominal single bonds.
          • Cis/trans alkene isomers require two different substituents on each double-bond carbon.

          Your next step

          Lesson: Not marked completed

          Marking a lesson completed records that you finished reading it. Practice establishes mastery separately. Saved on this device.

          Try a short set of different questions to check what you can do independently.

          Watch and Learn

          An optional video to support this lesson.

          Loading video resources…

          Supplementary learning resources

          More resources for this topic →

          External resources belong to their respective creators. Chem Help is not affiliated with or endorsed by these providers. Links open the original provider; their terms apply.

          Sources and further reading

          Source attribution for this lesson has not been verified. Missing source records do not establish originality or reuse permission.