Bond Energy, Length, and Order
Think of bond order like the number of ropes tying two boats together. One rope (single bond) is easy to cut and gives the boats lots of room to drift apart - the boats sit far from each other because the single rope is not pulling them very tight. Two ropes (double bond) are harder to cut and pull the boats closer together. Three ropes (triple bond) are the hardest to cut and keep the boats closest together. More ropes mean a tighter, shorter, stronger connection between the two boats.
That analogy captures the three-way relationship that the MCAT loves to test: bond order, bond energy, and bond length are all connected, and once you know one, you can predict the other two.
Bond Order
Bond order is simply the number of bonding electron pairs shared between two atoms. For straightforward molecules:
- Single bond = bond order 1
- Double bond = bond order 2
- Triple bond = bond order 3
When resonance structures exist, bond order becomes fractional. You calculate it by dividing the total number of bonds across all resonance structures by the number of bond positions.
The Key Relationships
Here is the central pattern. Memorize the direction of each arrow:
Hybridisation, sigma and pi, and what bond order does
Scroll sideways to see the whole map.
Counting to the hybridisationCount the groups attached to the atom, exactly as for VSEPR: four means sp³, three means sp², two means sp. The number of orbitals you mix always equals the number of groups you need to point at.
Sigma and pi, and what rotatesThe first bond between two atoms is always sigma, formed head-on. Every bond after that is pi, formed from p orbitals overlapping side-on above and below. A sigma bond alone lets the ends spin freely; a pi bond locks them, which is exactly why alkenes have cis and trans forms and alkanes do not.
One chain, three consequencesMore s character pulls the bonding electrons closer to the nucleus, so sp bonds are shorter and stronger than sp³ bonds, and sp carbons are more electronegative. Bond order, bond length and bond energy are three readings of the same quantity.
Higher bond order → shorter bond length → greater bond energy (stronger bond)
And the inverse:
Lower bond order → longer bond length → lower bond energy (weaker bond)
Notice that bond energy and bond length are inversely related to each other but both track with bond order in predictable ways. A triple bond is the shortest and strongest. A single bond is the longest and weakest.
| Bond | Bond Order | Bond Length (pm) | Bond Energy (kJ/mol) |
|:---|:---:|:---:|:---:|
| C - C | 1 | 154 | 347 |
| C = C | 2 | 134 | 614 |
| C ≡ C | 3 | 120 | 839 |
Notice that doubling the bond order does not double the energy. Going from a single to a double bond adds about 267 kJ/mol, but going from a double to a triple adds only about 225 kJ/mol. Each additional pi bond contributes less than the sigma bond did, because lateral overlap is less effective than head-on overlap.
Bond Dissociation Energy
Bond dissociation energy (BDE) is the energy required to break one mole of a specific bond in the gas phase, producing two radical fragments. It is always endothermic and always reported as a positive value.
For example, the BDE of the O-H bond in water is about 463 kJ/mol. That means you must put in 463 kJ of energy to break one mole of O-H bonds in water molecules (in the gas phase). Breaking bonds always costs energy.
The reverse process - forming a bond - always releases energy. When two atoms come together and form a bond, the system drops to a lower energy state and gives off exactly the same amount of energy that would be needed to break that bond.
- Breaking bonds = endothermic = requires energy input (positive)
- Forming bonds = exothermic = releases energy (negative)
Estimating Enthalpy of Reaction from Bond Energies
You can use tabulated average bond energies to estimate the enthalpy change of a reaction. The logic is simple: break all the bonds in the reactants (costs energy), then form all the bonds in the products (releases energy). The difference tells you whether the overall reaction absorbed or released energy.
Here is how to apply this formula step by step:
- Draw out the full Lewis structures of all reactants and products.
- Identify every bond that is broken in the reactants.
- Identify every bond that is formed in the products.
- Look up the average bond energy for each type of bond.
- Sum the energies of all bonds broken (this is a positive number).
- Sum the energies of all bonds formed (this is also entered as a positive number in the formula).
- Subtract: ΔH = (energy in) - (energy out).
If the result is negative, the reaction is exothermic (more energy released forming bonds than consumed breaking them). If positive, the reaction is endothermic.
Bond Order in Resonance Structures
When a molecule has resonance structures, the actual bond order is the average across all contributing structures. This gives fractional bond orders.
Benzene (C6H6): Each carbon-carbon bond alternates between single and double in the two main resonance structures. That gives each C-C bond an order of (1 + 2) / 2 = 1.5. As a result, every C-C bond in benzene has the same length (140 pm), which falls between a typical C-C single bond (154 pm) and a C=C double bond (134 pm). The energy of each bond also falls between single and double bond values.
Carbonate ion (CO3 2-): Three resonance structures each place the double bond on a different oxygen. Bond order = 4 total bonds / 3 positions = 1.33 for each C-O bond. Every C-O bond in carbonate is identical in length and strength.
Common MCAT Traps
- Confusing bond energy with bond length trends. They are inversely related. Stronger bonds (higher energy) are shorter, not longer.
- Using the wrong sign convention for ΔH calculations. Always subtract bonds formed from bonds broken when using positive bond energy values.
- Treating resonance bonds as alternating. In benzene, all six C-C bonds are identical at bond order 1.5. They do not flip back and forth between single and double.
- Forgetting that BDE values are averages. The C-H bond energy is slightly different in methane vs. ethane vs. benzene. Tables give average values, so enthalpy estimates from bond energies are approximations.
C≡C (120 pm) < C=C (134 pm) < C-C (154 pm). Higher bond order means shorter bond length. The triple bond pulls the two carbons closest together because three shared electron pairs create the strongest attraction between nuclei, resulting in the shortest internuclear distance.
Apply ΔH ≈ Σ(bonds broken) - Σ(bonds formed). Sum the bond energies of all bonds broken in the reactants and subtract the sum of bond energies of all bonds formed in the products. If ΔH is negative, the reaction is exothermic (more energy released forming new bonds than consumed breaking old ones). If ΔH is positive, the reaction is endothermic.
Bond order = 1.5. Benzene has two equivalent resonance structures that alternate single and double bonds. Averaging gives (1 + 2) / 2 = 1.5 for each C-C bond. The actual bond length is 140 pm, which falls between a C-C single bond (154 pm) and a C=C double bond (134 pm). All six C-C bonds in benzene are identical.