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.
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
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.
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.
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.
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.
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.
The number of shared bonding pairs between two atoms, averaged over resonance structures if necessary.
As bond order rises, bond length falls and bond dissociation energy rises.
Rotation about isolated single bonds often interconverts conformers readily, with finite barriers; π bonding and delocalization can substantially restrict rotation.
Bond order 1.5 gives two identical O-O bonds intermediate in length between single and double.
bond order = total bonds between the pair / number of resonance structures
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.
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.
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.
| Mistake | Why it's wrong | Check 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. |
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After the lesson · Question bank
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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.
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