Chemical Bonding and Molecular Structure · Lesson 4 of 12
Measuring Bonds and Understanding Resonance
“Relate bond length, energy, and order to experimental evidence, then use resonance to explain equivalent bonds.”
• Define bond length, bond angle, bond enthalpy, and bond order. • Distinguish covalent radius from van der Waals radius and average bond enthalpy from successive dissociation values. • Use bond multiplicity to compare related bond lengths and strengths. • Explain canonical structures, resonance hybrids, and bond equivalence without treating resonance as an equilibrium.
Measurements test our drawings
Lewis structures are useful pictures, but a bond line is not a measurement. Experiments can reveal how far apart nuclei are, how much energy is needed to separate bonded atoms, and how the bonds are arranged. Comparing those measurements with the drawings tells us both what the model explains and where it needs improvement.
Bond length, atomic radius, and bond angle
The nuclei in a bonded pair have an equilibrium separation at the minimum-energy arrangement. This separation is called the bond length. Spectroscopy and diffraction methods can determine such distances. A bond length is a nuclear separation, rather than the distance from the outer edge of one electron cloud to the other.
The equilibrium distance between the nuclei of two bonded atoms.
For two identical covalently bonded atoms, half their nuclear separation is their covalent radius in that bonding situation. For unlike atoms, a bond length can often be estimated by adding the covalent radii. These radii are useful approximations, and their values depend on the bonding context.
A van der Waals radius describes an atom in a nonbonded contact. For two identical nonbonded atoms in neighbouring molecules, half the relevant contact distance gives this radius. It is generally larger than the covalent radius because bonded atoms are held closer together.
The angle between two bond directions meeting at an atom, such as the H–O–H angle at oxygen in water.
Bond angles describe spatial arrangement. A water molecule has an H–O–H angle of about 104.5°, so its atoms form a bent arrangement. Bond length and angle answer different questions: one is a distance, the other a direction difference.
Problem
A homonuclear single bond has length 200 pm. Estimate the covalent radius of either atom.
- 1.The two bonded atoms are identical, so they contribute equally to the nuclear separation.
- 2.Divide the bond length by two: .
- 3.This is a covalent radius for that bonded situation, not automatically the atom’s nonbonded radius.
| Bond or molecule | Representative length from the chapter (pm) |
|---|---|
| O–H | 96 |
| C–H | 107 |
| C–C | 154 |
| C=C | 133 |
| C≡C | 120 |
| C–N | 143 |
| C=N | 128 |
| C≡N | 116 |
| C–O | 143 |
| C=O | 121 |
| N–O | 136 |
| N=O | 122 |
| H–H | 74 |
| F–F | 144 |
| Cl–Cl | 199 |
| Br–Br | 228 |
| I–I | 267 |
| N≡N | 109 |
| O=O | 121 |
| H–F / H–Cl / H–Br / H–I | 92 / 127 / 141 / 160 |
The chapter’s covalent-radius reference table supports the same way of reading data: specify the element, its bonding situation, and the unit. Comparing H–F with H–I shows how a larger bonded atom can produce a longer bond. Comparing C–C, C=C, and C≡C shows the effect of multiplicity while keeping the element pair unchanged. These are different comparisons and should not be mixed.
| Element | Covalent radius in pm; parenthesis gives bond order |
|---|---|
| H | 37 |
| C | 77 (1), 67 (2), 60 (3) |
| N | 74 (1), 65 (2), 55 (3) |
| O | 66 (1), 57 (2) |
| F | 64 |
| P | 110 |
| S | 104 (1), 95 (2) |
| Cl | 99 |
| Br | 114 |
| I | 133 |
| As | 121 |
| Sb | 141 |
| Se | 104 |
| Te | 137 |
Use single-bond radii unless a different order is specified. Carbon’s decreasing listed radius from single through triple bonding reflects the closer nuclear approach in higher-order bonds. The table also shows why “atomic radius” is not one unchanging number independent of bonding.
Bond enthalpy measures the energy cost of separation
Bond enthalpy is the energy needed to break one mole of a specified type of bond between gaseous species. Its unit is kilojoules per mole. Breaking a bond requires energy; the corresponding bond-forming process releases energy. A larger dissociation enthalpy indicates a stronger bond for the stated process.
The chapter gives about for O₂ and for N₂ dissociation. These high values help us recognise strong multiple bonds. In a polyatomic species, apparently similar bonds need not have identical successive dissociation energies because removing one atom changes the chemical environment of the remaining fragment.
Problem
The successive O–H dissociation enthalpies of gaseous water are approximately 502 and 427 kJ per mole. Find their average.
- 1.The first step is . The second is . The second bond is broken in a different species.
- 2.The total energy for removing both H atoms is of water.
- 3.Divide by two O–H bonds: . This is an average, not the claim that each successive step actually requires 464.5 kJ per mole.
Bond order links multiplicity with properties
In a simple Lewis drawing, bond order is the number of shared pairs between a particular pair of atoms. Thus a single bond has order one, a double bond order two, and a triple bond order three. For comparable bonds between the same element pair, increasing order generally shortens and strengthens the bond.
This trend works well for the C–C series in the table. It is not a rule that every triple bond in any species must be shorter than every double bond between unrelated elements. Atomic size, electron distribution, and molecular environment also matter. Molecular orbital theory later provides an electron-occupancy definition that can produce fractional orders. Particular isoelectronic species provide useful checks: F₂ and peroxide each have bond order one in the simple MO treatment, while N₂, CO, and NO⁺ each have order three. Isoelectronic means having the same total electron count. It is not a guarantee that all their measured properties are identical.
Problem
Arrange C–C, C=C, and C≡C in increasing length and explain the trend.
- 1.Their Lewis bond orders are 1, 2, and 3 respectively.
- 2.Higher order generally holds the same atom pair more closely. The chapter’s values are 154, 133, and 120 pm.
- 3.The order of increasing length is C≡C, C=C, C–C. For this related series, the expected strength trend is the reverse.
Resonance explains equivalent bonds
Ozone can be drawn with a double bond on either side of its central oxygen. Either drawing by itself predicts one long single bond and one short double bond. Experiment instead finds two equivalent O–O bonds of intermediate length. The molecule therefore needs a description that is not limited to one fixed Lewis drawing.
A description using two or more valid electron arrangements with the same atomic skeleton when a single Lewis structure is insufficient to represent the species.
The alternative Lewis drawings are canonical structures. The real species is the resonance hybrid, with an electron distribution spread over the relevant bonded region. For ozone, the equivalent bonds are commonly described by an average order of 1.5. The hybrid has lower energy than either isolated canonical description would represent, so resonance contributes to stabilisation. The source compares a reference O–O single-bond length near 148 pm with an O=O length near 121 pm; ozone’s two equal bonds are about 128 pm. The measured intermediate value supports the shared, delocalised description.
Problem
Why are the three C–O bonds in carbonate equivalent?
- 1.A usual octet Lewis drawing contains one C=O bond and two C–O bonds, but any of the three oxygen atoms can be shown double-bonded.
- 2.The three equivalent canonical structures retain the same nuclei and differ only in electron placement. The real ion is their resonance hybrid.
- 3.For these equivalent contributions, the average order is . All three C–O bonds therefore have the same intermediate character; there is no permanently selected C=O bond.
Carbon dioxide also has contributing canonical forms in addition to its dominant neutral O=C=O drawing. Charge-separated forms can be drawn with a triple bond on one side and a single bond on the other. Their relative contributions are not necessarily equal: formal charge and energy help us judge the importance of a contributor. The physical molecule has equivalent C–O bonds. Comparisons of its measured distances with reference double- and triple-bond distances motivate these additional contributions. Reference lengths vary with chemical context, so they should not be treated as universal constants for every C–O bond.
Canonical structures are drawings, not separate molecules in equilibrium. The real molecule does not spend one moment in one drawing and the next in another. Moving nuclei or changing which atom a hydrogen is attached to produces a different structure, not an ordinary resonance contributor.
Check your understanding
Use the measurement or model appropriate to the question. Distinguish a measured distance from a radius, an energy per mole from an electron count, and a resonance hybrid from its individual drawings.
Quiz
Covalent radius for two identical bonded atoms is approximately:
Why do the successive O–H dissociation energies in water differ?
For the same carbon pair, which normally has the shortest bond?
Which statement correctly describes ozone resonance?
The average C–O bond order in the equivalent carbonate resonance model is:
Practice Problems
- Define bond length and bond angle using a water molecule to illustrate the difference.
- Use covalent radii 77 pm for C and 99 pm for Cl to estimate a C–Cl single-bond length. State why the answer is approximate.
- Explain why an atom’s van der Waals radius is generally larger than its covalent radius.
- Two successive bond dissociation enthalpies are 480 and 440 kJ per mole. Calculate their mean and explain why the mean need not equal either measured step.
- Draw the resonance contributors of ozone and carbonate, and explain the experimental bond-equivalence argument for each.
- Explain why moving a hydrogen nucleus from one oxygen to another is not the same operation as drawing another resonance contributor.
- Draw resonance contributors for nitrate and explain why all three N–O bonds are equivalent in the simple resonance model.
Key Takeaways
• Bond length, bond angle, bond enthalpy, and bond order describe different aspects of bonding. • Covalent radii describe bonded contacts; van der Waals radii describe nonbonded contacts. • Polyatomic molecules often require average bond enthalpies because successive dissociations have different environments. • Higher order generally means shorter, stronger bonds within a comparable series. • A resonance hybrid is one real structure represented by several canonical drawings with unchanged nuclei.