Resonance and Delocalisation
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You should be comfortable drawing Lewis structures, calculating formal charges using \(FC=V-L-\tfrac{B}{2}\), and identifying the most stable arrangement of electrons for main-group elements.
Why this matters
Many molecules cannot be described by a single static picture of electron pairs.
In the previous lecture, we learned to select the "best" Lewis structure by minimising formal charge. However, for molecules like the carbonate ion or benzene, no single structure satisfies all bonding requirements simultaneously. The true electronic structure is a hybrid of multiple valid Lewis structures, a phenomenon known as resonance. This delocalisation of electrons leads to increased stability and unique physical properties that cannot be predicted from any single contributing structure alone.
The Limitation of Static Structures
Consider the carbonate ion, \(\mathrm{CO_3^{2-}}\). Carbon has four valence electrons, and each oxygen has six. Including the two extra electrons from the charge, we have a total of \(4 + 3(6) + 2 = 24\) valence electrons. If we draw a structure with one double bond and two single bonds to oxygen, we satisfy the octet rule for all atoms. However, this implies that one carbon–oxygen bond is shorter and stronger (a double bond) while the other two are longer and weaker (single bonds). Experimental data, however, shows that all three carbon–oxygen bonds in the carbonate ion are identical in length and strength. This discrepancy arises because the electrons are not fixed in one position; they are delocalised over the entire ion. The single Lewis structure is merely a snapshot, not the complete reality.
Resonance Structures and the Hybrid
Resonance structures are valid Lewis structures that differ only in the placement of electrons, not in the positions of atoms. For \(\mathrm{CO_3^{2-}}\), we can draw three equivalent structures where the double bond is placed on a different oxygen atom in each case. These are called resonance contributors. The actual molecule is a resonance hybrid, which is a weighted average of these contributors. We represent this by drawing the sigma bond framework and using a dashed line or a circle to indicate delocalised pi electrons. It is crucial to understand that the molecule does not oscillate between these structures; rather, it exists in a single, stable state that is distinct from any individual contributor.
Delocalisation Energy and Stability
The stability gained from electron delocalisation is quantified as resonance energy. This is the difference in energy between the actual delocalised molecule and the most stable single Lewis structure. Delocalisation lowers the potential energy of the system because the electrons are spread out over a larger volume, reducing electron–electron repulsion and allowing for better electrostatic stabilisation by the nuclei. This is why benzene is significantly more stable than a hypothetical cyclohexatriene with three isolated double bonds. The resonance energy of benzene is approximately \(150\ \mathrm{kJ\,mol^{-1}}\), which explains its reluctance to undergo addition reactions that would destroy the aromatic pi system.
Rules for Drawing Valid Resonance Structures
Not all Lewis structures are valid resonance contributors. To be valid, structures must have the same number of unpaired electrons, the same net charge, and the same atomic positions. Only electrons (lone pairs or pi bonds) may move. You cannot move atoms or change the sigma bond framework. Additionally, the octet rule should be respected wherever possible. For example, in the nitrate ion \(\mathrm{NO_3^-}\), valid resonance structures involve moving pi electrons between nitrogen and oxygen atoms. Invalid structures would involve breaking a sigma bond or creating a structure where nitrogen has more than eight electrons (unless it is an exception to the octet rule, which is covered in the next lecture).
The result
Lewis structures and the octet rule
A molecule's total valence electrons can be distributed as bonding pairs (shared between two atoms) and lone pairs (localised on one atom) so that, wherever possible, every main-group atom is surrounded by eight electrons (two, for hydrogen). Where more than one such distribution is possible, the formal charge \(FC=V-L-\tfrac{B}{2}\) (valence electrons \(V\), lone-pair electrons \(L\), bonding electrons \(B\)) on each atom selects the best structure: minimise the magnitude of formal charges, and place any necessary negative formal charge on the more electronegative atom.
A common misconception is that resonance structures represent different isomers or rapidly interconverting species. In reality, resonance is a mathematical description of a single quantum mechanical state. The molecule does not "flip" between structures; the electrons are permanently delocalised. Treating resonance as dynamic equilibrium leads to incorrect predictions about reaction mechanisms and intermediate stability. C
Worked example
Problem. Determine the bond order of the carbon–oxygen bonds in the carbonate ion \(\mathrm{CO_3^{2-}}\) and calculate the average bond length if a C=O double bond is \(1.23\ \mathrm{\AA}\) and a C–O single bond is \(1.43\ \mathrm{\AA}\).
- Identify the number of resonance structures. For \(\mathrm{CO_3^{2-}}\), there are three equivalent resonance structures.
- Count the total number of bonds between carbon and oxygen across all resonance structures. In each structure, there is one double bond (counted as 2) and two single bonds (counted as 1 each). Total bond count per structure = \(2 + 1 + 1 = 4\).
- Calculate the average bond order. Since there are 3 C–O connections in the ion, and the total "bond order sum" is 4 (from the delocalised pi system plus sigma framework), the average bond order is \(4/3\).
- Estimate the bond length. Bond length is inversely proportional to bond order. We can interpolate linearly between single and double bond lengths. The bond order of a single bond is 1, and a double bond is 2. The carbonate bond order is \(1.33\). The difference in bond order from single is \(0.33\). The total difference in length between single and double is \(1.43 - 1.23 = 0.20\ \mathrm{\AA}\). The reduction in length is \(0.33 \times 0.20 = 0.066\ \mathrm{\AA}\).
- Subtract this reduction from the single bond length: \(1.43 - 0.066 = 1.364\ \mathrm{\AA}\). Rounding to two decimal places gives \(1.36\ \mathrm{\AA}\).
Units check. Bond order is dimensionless, and bond length is in angstroms (\(\mathrm{\AA}\)), which is consistent with standard chemical bond lengths.
Reality check: the measured C–O length in carbonate is about 1.29 Å. Bond length is not linear in bond order, so this interpolation only shows the trend (between the single- and double-bond lengths), not the exact value.
Three quick tests
- How many resonance structures does the nitrate ion \(\mathrm{NO_3^-}\) have?
Answer
Three equivalent structures, each with one N=O double bond and two N–O single bonds. - Can you move a hydrogen atom to a different position to create a resonance structure?
Answer
No. Resonance involves only the movement of electrons. Moving atoms creates a different isomer, not a resonance structure. - Why is the bond length in benzene intermediate between a single and double bond?
Answer
Because the pi electrons are delocalised over all six carbon atoms, giving each C–C bond a bond order of 1.5, which results in a bond length shorter than a single bond but longer than a double bond.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.