Bond Energies and Bond Lengths

Last updated: 6 August 2026

Diagnose → Learn → Master: Every study-note section includes 15 free MCQs — start with 5 Diagnostic MCQs to identify gaps in your understanding, then complete 10 Mastery MCQs near the end to check what you have learned.

A structural formula tells us which atoms are joined, but it does not tell us how far apart those atoms are or how difficult their bond is to break. Yet these differences can have a major effect on how a molecule behaves.

For example, the bonds C–C, C=C and C≡C all join two carbon atoms, but they do not have the same length or strength. Similarly, the H–F and H–I bonds both join hydrogen to a halogen, yet one is much shorter and stronger than the other.

Understanding bond length and bond energy therefore gives us a way to connect microscopic bonding to chemical stability and reactivity. The important goal is not simply to memorise that “short bonds are strong”, but to understand why the trend occurs and when it can be used.

Diagnostic Quiz

What Do You Already Know About Covalent Bonds?

Try these five questions before reading the chapter. They check the ideas about attraction, bond breaking and covalent bonding that will make the discussion of bond length and bond energy easier to understand.

5 quick questions
Instant right-or-wrong feedback
Every answer option is explained

Which statement best describes the bond length of a covalent bond?

The diameter of the larger bonded atom

Bond length does not describe the size of either individual atom. It refers to the separation between the two bonded nuclei.

The equilibrium distance between the nuclei of the two bonded atoms

Correct. Bond length is measured between the centres of the two nuclei at their equilibrium separation.

The distance between the outermost electrons of the two atoms

Electron clouds do not have sharp outer edges. Bond length is defined using the positions of the nuclei.

The distance travelled by an electron while forming the bond

Bond length is a structural distance within the bonded molecule, not a distance travelled by an electron.

Which statement about breaking a covalent bond is correct?

Bond breaking always releases energy

Energy is released when bonds form. Breaking an attractive bond requires energy to be supplied.

Bond breaking has no energy change

The bonded atoms are in a lower-energy state than separated atoms, so energy must be supplied to separate them.

Bond breaking is exothermic only for weak bonds

Weak bonds require less energy to break, but breaking them is still endothermic.

Bond breaking is endothermic because energy must be supplied

Correct. Energy must be supplied to overcome the attractive interaction holding the bonded atoms together.

Which carbon–carbon bond would you normally expect to be shortest?

C≡C

Correct. The triple bond has the highest bond order and is normally the shortest of the three carbon–carbon bonds.

C=C

A C=C bond is shorter than C–C but longer than C≡C.

C–C

A carbon–carbon single bond has the lowest bond order and is normally the longest of these three bonds.

They should all have the same length because the same atoms are bonded

The bonded atoms are the same, but the bond order differs. Bond order has a major effect on bond length.

Compared with fluorine, iodine has a much larger atomic radius. What would you therefore expect when comparing H–I with H–F?

H–I should be shorter and stronger

The much larger iodine atom places the bonding region farther from its nucleus, so H–I is not expected to be shorter.

H–I should be shorter but weaker

Increasing atomic size generally increases the internuclear distance rather than decreasing it.

H–I should be longer and weaker

Correct. Iodine is larger than fluorine, so H–I is longer and its orbital overlap is less effective. The bond therefore has a lower bond energy.

H–I should have the same length and strength as H–F

The large difference in halogen atomic size produces substantial differences in both bond length and bond energy.

In a polar bond C–X where X is more electronegative than carbon, which charge distribution is expected?

Cδ⁻–Xδ⁺

This reverses the polarity. The more electronegative atom attracts the bonding electron density more strongly and becomes δ⁻.

Cδ⁺–Xδ⁻

Correct. X attracts the shared electron pair more strongly, giving X a partial negative charge and carbon a partial positive charge.

C⁺–X⁻

A polar covalent bond normally involves partial charges, δ⁺ and δ⁻, rather than complete ionic charges.

Neither atom can carry a partial charge in a covalent bond

Covalent bonds can certainly be polar when the bonded atoms have different electronegativities.

Defining bond length and bond energy

Bond length and bond energy describe two different aspects of a covalent bond. Bond length tells us about the distance between the bonded nuclei, while bond energy tells us about the energy needed to separate them.

Quantity	Meaning	Typical unit
Bond length	Equilibrium distance between the nuclei of two covalently bonded atoms	pm
Bond energy	Energy required to break one mole of a specified covalent bond in gaseous molecules	kJ mol⁻¹

The word equilibrium is important when defining bond length. The two nuclei do not sit at an arbitrary distance from one another. They settle at a separation where the attractive and repulsive interactions within the bonded system balance.

1 pm = 1 × 10⁻¹² m

Bond length is measured between nuclei, not between the “edges” of atoms.

Electron clouds do not have sharp boundaries. The positions of the nuclei provide a clear and measurable way to describe the separation between two bonded atoms.

Why does a covalent bond have an equilibrium length?

To understand bond length properly, imagine two atoms beginning far apart and then moving towards one another.

At a large separation, the interaction between them is weak. As they approach, each positively charged nucleus attracts electron density associated with the other atom. These attractive interactions lower the potential energy of the system.

If the atoms come too close together, however, repulsive interactions become very strong:

the two positively charged nuclei repel one another; and
the electrons repel other electrons.

The most stable arrangement therefore occurs at an intermediate separation where the potential energy is at a minimum.

The bond length corresponds to the minimum in the potential-energy curve.

At this equilibrium separation, the attractive and repulsive effects balance. Moving the nuclei either much closer together or much farther apart raises the potential energy.

Bonded atoms are not completely stationary. They vibrate about this equilibrium distance, so a quoted bond length represents the characteristic equilibrium separation of the nuclei.

Bond breaking requires energy

A covalent bond represents an attractive interaction between atoms. The bonded state is lower in potential energy than the completely separated atoms.

Energy must therefore be supplied to pull bonded atoms apart.

X–Y(g) → X(g) + Y(g) ΔH > 0

Bond breaking is therefore always endothermic.

The reverse process is bond formation. When two separated atoms form a stable covalent bond, the system moves to a lower-energy state and energy is released.

X(g) + Y(g) → X–Y(g) ΔH < 0

Bond breaking: energy is absorbed.

Bond formation: energy is released.

An overall reaction can still be exothermic because the energy released when new bonds form may be greater than the energy absorbed when the original bonds break.

Why is bond energy quoted per mole of bonds?

The unit kJ mol⁻¹ refers to one mole of the specified bonds. It does not automatically mean one mole of molecules.

For example, each methane molecule, CH₄, contains four C–H bonds. One mole of CH₄ therefore contains four moles of C–H bonds.

If a mean C–H bond enthalpy of 413 kJ mol⁻¹ is used, the approximate energy needed to break all four moles of C–H bonds is:

Energy = 4 × 413 = 1652 kJ mol⁻¹ of CH₄

This distinction becomes particularly important when bond-energy data are later used to estimate enthalpy changes of reactions.

The relationship between bond length and bond strength

Within a chemically sensible series, there is an important general relationship:

shorter bond → stronger bond → higher bond energy

A shorter bond generally allows the nuclei to interact more strongly with the bonding electron density between them. More energy is then needed to separate the bonded atoms completely.

Conversely:

longer bond → weaker bond → lower bond energy

“Shorter means stronger” is a trend, not an unrestricted law.

The comparison works especially well when the same pair of elements or a closely related series of bonds is being compared. Bond strength also depends on the identities of the atoms, orbital overlap and the electronic environment.

Bond order: single, double and triple bonds

One of the clearest demonstrations of the bond-length–bond-energy relationship comes from comparing bonds between the same two atoms but with different bond orders.

Bond	Bond order	Approximate bond length / pm	Approximate bond energy / kJ mol⁻¹
C–C	1	154	347
C=C	2	134	612
C≡C	3	120	837

As the bond order increases, more bonding interaction exists between the two carbon atoms. The nuclei are drawn closer together, so the bond becomes shorter.

At the same time, separating the carbon atoms becomes more difficult, so the bond energy increases.

Bond length: C≡C < C=C < C–C
Bond energy: C≡C > C=C > C–C

Is a double bond exactly twice as strong as a single bond?

No. A carbon–carbon single bond consists of one σ bond, whereas a carbon–carbon double bond consists of one σ bond and one π bond.

The π bond is formed by sideways overlap and is weaker than the σ bond formed by head-on overlap. Adding a π bond therefore increases the total bond energy, but not by exactly the energy of another σ bond.

Similarly, a triple bond contains one σ bond and two π bonds, so its bond energy is not simply three times the energy of the corresponding single bond.

Atomic size and bond length

Bond length is also affected by the sizes of the bonded atoms. When larger atoms form a covalent bond, their nuclei generally remain farther apart.

Consider the hydrogen halides:

Bond	Approximate bond length / pm	Approximate bond energy / kJ mol⁻¹
H–F	92	562
H–Cl	127	431
H–Br	141	366
H–I	161	299

Down Group 17, the halogen atom becomes larger. The valence orbital involved in bonding is increasingly far from the halogen nucleus.

As a result:

H–X bond length increases: H–F < H–Cl < H–Br < H–I

H–X bond energy decreases: H–F > H–Cl > H–Br > H–I

This trend is an important example of how atomic size influences both bond length and bond strength.

Why does orbital overlap matter?

A strong covalent bond requires effective overlap between atomic orbitals. Effective overlap places substantial bonding electron density between the nuclei, allowing that electron density to attract both nuclei strongly.

When the orbitals overlap less effectively, the bonding interaction is weaker. Less energy is then required to separate the atoms.

Atomic size and orbital overlap are connected.

As atoms become larger, the orbitals involved in bonding become more diffuse. In a series such as H–F to H–I, this contributes to poorer overlap, longer bonds and lower bond energies.

Why the same type of bond can have slightly different energies

A C–H bond does not necessarily have exactly the same bond energy in every molecule. The surrounding atoms affect the electron distribution and therefore influence the strength of a particular bond.

For this reason, data tables often contain mean bond enthalpies. A mean bond enthalpy is an average value for a particular type of bond taken from a range of gaseous compounds.

Specific bond dissociation enthalpy: refers to breaking a particular bond in a particular gaseous molecule.

Mean bond enthalpy: is an average value for that type of bond across different gaseous molecules.

This is why calculations using mean bond enthalpies usually give an approximate enthalpy change rather than an exact experimental value.

How bond length and bond energy affect reactivity

Bond energy becomes particularly useful when the reaction mechanism requires a particular bond to be weakened or broken.

If two closely related molecules react by breaking the same type of bond, the molecule containing the weaker bond will often react more readily because less energy is required to weaken that bond.

lower bond energy → bond easier to break

Since longer comparable bonds are often weaker, bond length can provide another clue:

longer bond → usually weaker bond → usually easier to break

However, this does not mean that bond energy alone determines reaction rate. A reaction must still follow a suitable pathway and overcome its activation energy.

Always connect bond strength to the bond that actually changes in the reaction.

A molecule does not become generally “more reactive” simply because it contains one weak bond. Ask whether that bond is involved in the reaction mechanism being considered.

Bond polarity also influences reactivity

Bond strength is not the only bonding factor that matters. A covalent bond can also be polar when the bonded atoms have different electronegativities.

If X is more electronegative than carbon:

Cδ⁺–Xδ⁻

X attracts the shared electron pair more strongly. Carbon therefore carries a partial positive charge, δ⁺, while X carries a partial negative charge, δ⁻.

An electron-rich species such as a nucleophile can be attracted towards the δ⁺ carbon atom. Bond polarity can therefore help determine where a reagent attacks.

Bond polarity can favour attack, while bond energy affects how difficult the bond is to break.

A good explanation of reactivity may therefore require both ideas. The most polar bond is not automatically the bond that reacts fastest.

Example: C–X bonds in halogenoalkanes

Halogenoalkanes provide a useful example because their hydrolysis involves attack at a δ⁺ carbon atom and breaking of a C–X bond.

Bond	Approximate bond length / pm	Approximate bond energy / kJ mol⁻¹
C–F	135	485
C–Cl	177	327
C–Br	194	285
C–I	214	213

Down Group 17, the halogen atom becomes larger, so the C–X bond becomes longer. The bond energy decreases:

C–F > C–Cl > C–Br > C–I in bond energy

The C–I bond is therefore much easier to break than the C–F bond.

There is an important subtlety here. The C–F bond is actually the most polar C–X bond because fluorine is the most electronegative halogen. This gives the carbon a relatively large δ⁺ charge, which might initially suggest strong attraction towards a nucleophile.

Nevertheless, the C–F bond is exceptionally strong. The effect of bond strength is so important that fluoroalkanes hydrolyse much more slowly than the corresponding iodoalkanes.

Polarity argument: C–F is highly polar, so its carbon atom is strongly δ⁺ and can attract a nucleophile.

Bond-strength argument: C–F has a very high bond energy, so breaking it is difficult.

Observed trend: the weaker C–I bond is broken more readily, so iodoalkanes generally hydrolyse faster.

Why is nitrogen gas relatively unreactive?

Nitrogen gas contains an N≡N triple bond. The bond is very short and exceptionally strong.

N≡N bond energy ≈ 945 kJ mol⁻¹

Many reactions involving N₂ require this bond to be substantially weakened before new bonds can form. This requires a large amount of energy and contributes to a high activation energy.

N₂ is therefore relatively unreactive under ordinary conditions. This does not mean that nitrogen can never react. Suitable conditions such as high temperature, high pressure or catalysts can provide a practical reaction pathway.

Strong bond does not mean “chemically impossible to react”.

It means that reactions requiring substantial weakening of that bond may have a high activation-energy barrier and therefore occur slowly under ordinary conditions.

Worked reasoning with bond length and bond energy

Example 1: comparing C–C, C=C and C≡C

Arrange C–C, C=C and C≡C in order of increasing bond length.

All three bonds involve the same atoms, so bond order provides the clearest comparison. Increasing bond order strengthens the bonding interaction and draws the carbon nuclei closer together.

Increasing bond length: C≡C < C=C < C–C

The corresponding bond-energy order is the reverse:

Increasing bond energy: C–C < C=C < C≡C

Example 2: comparing H–Cl and H–I

Explain why H–I has a lower bond energy than H–Cl.

Iodine has a larger atomic radius than chlorine. The H–I bond is therefore longer than H–Cl, and the overlap between the relevant orbitals is less effective.

The H–I bond is consequently weaker, so less energy is needed to break one mole of H–I bonds.

Example 3: interpreting reactivity using two factors

A student states:

“Fluoroalkanes should hydrolyse fastest because C–F is the most polar C–X bond.”

This argument considers only bond polarity. Although greater polarity increases the δ⁺ character of carbon, hydrolysis also requires the C–X bond to be broken.

The C–F bond is much stronger than C–Cl, C–Br or C–I. The high bond energy makes C–F difficult to break, so fluoroalkanes hydrolyse slowly.

reactivity depends on polarity + bond strength + reaction mechanism

Common mistakes with bond energies and bond lengths

Common mistake	Why it is wrong	Correct approach
Defining bond length as the distance between the edges of two atoms	Atoms do not have sharply defined outer surfaces	Use the equilibrium distance between the two nuclei
Saying that bond breaking releases energy	Energy must be supplied to overcome the attractive bonding interaction	Bond breaking is endothermic; bond formation is exothermic
Forgetting that bond energy is quoted per mole of bonds	A molecule may contain several bonds of the specified type	Count the number of moles of bonds involved
Assuming a double bond has exactly twice the energy of a single bond	The additional π bond is not equivalent in strength to the σ bond	Use actual bond-energy data or a qualitative comparison
Applying “shorter means stronger” to any two unrelated bonds	Bond strength also depends on the identities of the atoms and orbital overlap	Use the trend within a chemically sensible comparison
Saying larger atoms form stronger bonds because they contain more electrons	The bonding orbitals are generally farther from the nuclei and often overlap less effectively	Relate larger atomic size to longer bond length and usually lower bond energy
Assuming the most polar bond must react fastest	Polarity may favour attack, but the bond may still be very difficult to break	Consider both bond polarity and bond energy
Treating mean bond enthalpy as an exact value for every molecule	The strength of the same bond type varies with molecular environment	Recognise mean bond enthalpies as average values
Saying N₂ is completely inert	N₂ can react when suitable conditions or catalysts provide an accessible pathway	Describe N₂ as relatively unreactive because its N≡N bond is exceptionally strong

Bond energy and bond length checklist

Know the definitions

Bond length is an equilibrium internuclear distance.
Bond energy is quoted per mole of specified covalent bonds.
Bond breaking requires gaseous bonds to be separated.
Use the correct units: pm and kJ mol⁻¹.

Compare bond strength

Higher bond order generally gives a shorter bond.
Shorter comparable bonds are generally stronger.
Larger atoms generally form longer bonds.
More effective orbital overlap generally strengthens a bond.

Think about energy

Bond breaking is endothermic.
Bond formation is exothermic.
Count moles of bonds rather than only molecules.
Mean bond enthalpies give approximate values.

Explain reactivity

Identify which bond changes during the reaction.
Lower bond energy usually makes that bond easier to break.
Use bond polarity to identify likely sites of attack.
Do not use polarity or bond energy in isolation when both matter.

Mastery Quiz

Bond Energies and Bond Lengths Mastery Quiz

Now apply the ideas from the chapter. These questions test whether you can use bond order, atomic size, bond energy, bond length and bond polarity to explain unfamiliar bonding and reactivity trends.

10 application-focused questions
Bonding and reactivity reasoning
Every answer option is explained

Which order correctly shows increasing carbon–carbon bond length?

C–C < C=C < C≡C

This is the reverse of the bond-length trend. Increasing bond order draws the two carbon nuclei closer together.

C≡C < C=C < C–C

Correct. C≡C has the highest bond order and is shortest, while C–C has the lowest bond order and is longest.

C=C < C–C < C≡C

A triple bond is not the longest. The greater bonding interaction in C≡C draws the carbon nuclei closest together.

C–C = C=C = C≡C

Although the same atoms are bonded, the different bond orders produce different equilibrium internuclear distances.

Which trend is expected for the H–X bond as X changes from F to Cl to Br to I?

Bond length decreases and bond energy increases

The halogen atoms become larger down the group, so the H–X bond becomes longer rather than shorter.

Bond length increases and bond energy increases

Bond length does increase, but the increasingly long bond is generally weaker, so its bond energy decreases.

Bond length decreases and bond energy decreases

The bond energy decreases, but the increasing size of X causes the bond length to increase.

Bond length increases and bond energy decreases

Correct. The larger halogen atoms give longer H–X bonds with less effective overlap, so the bonds become weaker down the group.

Fluoroalkanes have a more polar C–X bond than iodoalkanes, yet iodoalkanes generally hydrolyse more rapidly. Which explanation is best?

The C–I bond is much weaker than C–F, so the bond that must be broken is easier to break

Correct. C–F is highly polar but exceptionally strong. The much lower C–I bond energy makes C–I easier to break during hydrolysis.

The C–I bond is more polar because iodine is more electronegative than fluorine

Fluorine is more electronegative than iodine, so C–F is the more polar bond.

The C–F bond is ionic and therefore cannot react with nucleophiles

C–F in a fluoroalkane is a polar covalent bond, not an ionic bond.

Bond energy has no influence on reactions involving covalent bonds

If a reaction requires a bond to be weakened or broken, its bond energy can be an important factor affecting the reaction.

The mean C–H bond enthalpy is 413 kJ mol⁻¹. Approximately how much energy is required to break all the C–H bonds in one mole of CH₄(g)?

103 kJ

Dividing by four is inappropriate. One methane molecule contains four C–H bonds, so one mole of methane contains four moles of C–H bonds.

413 kJ

413 kJ corresponds to approximately one mole of C–H bonds. One mole of CH₄ contains four moles of C–H bonds.

1652 kJ

Correct. One mole of CH₄ contains four moles of C–H bonds, so the estimated energy required is 4 × 413 = 1652 kJ.

2065 kJ

This would correspond to five lots of 413 kJ. CH₄ contains four C–H bonds per molecule, not five.

Two related molecules react by breaking an A–X bond. Their bond data are: molecule P, 180 pm and 350 kJ mol⁻¹; molecule Q, 210 pm and 230 kJ mol⁻¹. If other relevant factors are similar, which prediction is most reasonable?

P should react more readily because its bond is shorter

A shorter bond with the higher bond energy is generally more difficult to break, not easier.

Q should react more readily because its longer bond has the lower bond energy

Correct. If breaking A–X is important in the mechanism and other factors are similar, the weaker 230 kJ mol⁻¹ bond should be easier to break.

P and Q must react at exactly the same rate because both contain A–X

Bonds of the same general type can have different lengths and strengths, and these differences may influence reactivity.

No bond can break if its bond energy is greater than 200 kJ mol⁻¹

Bond energy is not a threshold above which reaction becomes impossible. Strong bonds can react if a suitable reaction pathway is available.

Why can the quoted mean bond enthalpy for C–H differ slightly from the energy required to break a particular C–H bond in a particular molecule?

The unit kJ mol⁻¹ changes from one molecule to another

The unit remains kJ mol⁻¹. The difference arises from the chemical environment of the bond, not from changing units.

Carbon and hydrogen have different proton numbers in different compounds

Their proton numbers do not change when they appear in different compounds.

A mean bond enthalpy refers only to bonds in liquids

Mean bond enthalpies are defined using gaseous species rather than being values specifically for liquids.

The atoms surrounding the C–H bond alter its electronic environment

Correct. A mean bond enthalpy averages the values for a bond type in different molecular environments, so a particular bond need not equal the average exactly.

At the equilibrium bond length between two atoms, which description is most appropriate?

All attractive forces have disappeared

Attractive interactions are still present. They are balanced by repulsive interactions at the equilibrium separation.

The nuclei have moved as close together as physically possible

If the nuclei were forced closer than the equilibrium separation, strong repulsive interactions would raise the potential energy.

Attractive and repulsive effects balance and the potential energy is at a minimum

Correct. The equilibrium bond length corresponds to the most stable separation and the minimum of the potential-energy curve.

The bond energy has become zero

A stable bond still requires energy to break. Its bond energy is therefore not zero at the equilibrium separation.

N₂ is relatively unreactive under ordinary conditions. Which bonding explanation is most appropriate?

The N≡N bond has a very high bond energy, contributing to a large activation barrier for reactions that weaken it

Correct. The exceptionally strong N≡N triple bond is difficult to weaken, so many reactions of N₂ have high activation energies.

Nitrogen atoms cannot form any bonds other than N≡N

Nitrogen forms many compounds containing N–H, N–C, N–O and other bonds.

N₂ has no electrons that can participate in reactions

N₂ certainly contains electrons. Its low reactivity is related to the high stability and strength of the N≡N bond.

Triple bonds always prevent molecules from reacting under any conditions

Triple bonds can react. N₂ itself can react under suitable conditions, particularly when catalysts and appropriate temperatures are used.

The C=C bond energy is greater than the C–C bond energy, but it is not twice as large. What best explains this?

A double bond contains only one shared electron

A covalent bond contains shared electron pairs. A C=C double bond contains both a σ bonding interaction and a π bonding interaction.

The second bond causes the first bond to disappear

The σ bond remains present in a double bond. The π bond is added to it.

Bond energy is unrelated to the number and type of bonding interactions

The number and nature of bonding interactions strongly affect the total strength of a bond.

The π bond added to the σ bond is weaker than the σ bond itself

Correct. Sideways p-orbital overlap forming the π bond is less effective than the head-on overlap forming the σ bond, so the total is not simply twice the single-bond energy.

A student compares two related reactions and says, “The molecule with the weaker bond must always react faster.” Why is this conclusion too strong?

Weak covalent bonds cannot take part in chemical reactions

Weak covalent bonds can certainly participate in reactions and are often easier to break.

Bond energies can be used only for ionic compounds

Bond energies describe covalent bonds and are particularly useful when discussing covalent bond breaking.

Reaction rate also depends on the reaction pathway, activation energy and other interactions such as bond polarity

Correct. Bond strength is important when that bond changes, but it is only one part of the complete mechanism and activation-energy barrier.

Strong bonds always have zero activation energy

Bond strength and activation energy are different concepts. A strong bond certainly does not imply zero activation energy.

Frequently asked questions

Questions students often ask about bond energies and bond lengths

The basic trends are simple, but applying them correctly requires knowing the limits of each rule and distinguishing bond strength from overall reactivity.

Does a shorter bond always have a higher bond energy?

It is a useful trend within a chemically comparable series. For example, C≡C is shorter and stronger than C=C, which is shorter and stronger than C–C.

However, bond strength also depends on which atoms are bonded, the orbitals involved and the surrounding electronic environment. You should therefore not compare two completely unrelated bonds using bond length alone.

Are bonded atoms completely stationary at the bond length?

No. Bonded atoms vibrate about their equilibrium positions. The bond length represents the characteristic equilibrium separation between the nuclei, rather than a perfectly fixed distance at every instant.

Why must energy be supplied to break a bond?

The bonded atoms are held together by attractive interactions and occupy a lower-potential-energy state than the separated atoms. Energy must therefore be supplied to overcome the bonding interaction and separate them.

Is bond energy the same as activation energy?

No. Bond energy refers to the energy required to break a specified covalent bond in gaseous species.

Activation energy is the energy barrier associated with an entire reaction pathway. It depends on the transition state and all of the interactions occurring as reactants are converted into products.

Why is a double bond shorter than a single bond?

A double bond has a greater bond order and more bonding interaction between the two atoms. This draws the nuclei closer together, giving a shorter equilibrium bond length.

Why is a double bond not exactly twice as strong as a single bond?

A double bond contains one σ bond and one π bond. The π bond is formed by sideways overlap of p orbitals and is weaker than the σ bond formed by head-on overlap. The total bond energy therefore increases, but not by a factor of exactly two.

Can a molecule containing a strong bond still be reactive?

Yes. Reactivity depends on the entire reaction mechanism, not one bond energy in isolation. A reagent may interact strongly with a polar site, another bond may be involved, or a catalyst may provide an alternative lower-energy pathway.

Can a molecule containing a weak bond still be stable?

Yes. A weak bond may be relatively easy to break, but the molecule still needs an accessible reaction pathway and suitable collision partners. Thermodynamic bond strength alone does not determine the observed reaction rate.

Why does bond polarity matter if bond energy tells us how difficult a bond is to break?

Bond polarity produces partial charges that can attract reagents. For example, a nucleophile is attracted towards a δ⁺ atom.

Bond energy answers a different question: how difficult is the bond to separate? A reaction may therefore depend on both how readily the reagent attacks and how readily the bond can subsequently be broken.

Why do iodoalkanes hydrolyse faster than fluoroalkanes even though C–F is more polar?

The C–F bond is more polar, but it is also exceptionally strong. The C–I bond is much longer and has a much lower bond energy, so it is considerably easier to break during hydrolysis.

This is a useful reminder that reactivity should not be predicted from bond polarity alone.

Why are mean bond enthalpy values only approximate?

The strength of a particular type of bond varies slightly depending on the atoms surrounding it. A mean bond enthalpy is an average across several gaseous molecules, so the value may not exactly equal the bond dissociation enthalpy of that bond in one particular molecule.

Do we need s-character to explain bond length and bond energy here?

Not for the central ideas in this chapter. Bond order, atomic size and orbital overlap are sufficient for the main bond-length and bond-energy comparisons required here.

The idea of s-character becomes more useful when hybridisation is discussed in greater depth, particularly when comparing bonds involving sp, sp² and sp³ hybridised carbon atoms.

Bond length and bond energy give us two complementary ways to describe a covalent bond. Bond length tells us the equilibrium separation of the nuclei, while bond energy tells us how difficult the bonded atoms are to separate. Higher bond order and effective orbital overlap generally produce shorter, stronger bonds, while increasing atomic size often produces longer, weaker bonds. When these ideas are combined with bond polarity and reaction mechanism, they become powerful tools for explaining why apparently similar molecules can show very different reactivities.