Eight additional teaching hours, outcomes 2.2.11 to 2.2.16, for higher level only. This part builds on the ten standard level hours in Part A and assumes them.
Guiding question: What determines the covalent nature and properties of a substance?
Structure 2.2 (AHL) · Resonance and benzene
1Resonance and delocalization 2.2.11 HL only Strong
The Lewis model with localized pairs, the octet rule and VSEPR explains a great deal, but some observations do not fit: ozone has two identical O–O bonds although its Lewis formula shows one single and one double bond; benzene does not behave like an alkene; sulfur and phosphorus form SF6 and PCl5 with more than eight valence electrons around the central atom. The higher-level content refines the model in three ways: delocalized electrons (resonance), expanded octets with formal charge, and an orbital picture of bonding (σ and π bonds, hybridization).
For some species, two or more valid Lewis formulas can be drawn that have the same arrangement of atoms but different positions for a double bond (and its associated lone pairs). These are resonance structures. The real species is not one of them, nor does it switch rapidly between them; it is a single resonance hybrid in which the electrons of the multiple bond are spread over several atoms. Electrons shared by more than two nuclei are described as delocalized.
Resonance structures: two or more Lewis formulas for the same species that differ only in the positions of electrons (a double bond can be placed in more than one position). Delocalization: the spreading of bonding electrons (usually π electrons) over three or more atoms. Bond order: the average number of shared pairs between two atoms in the hybrid.
The evidence for delocalization is structural. In ozone both O–O bonds are 128 pm long, between a typical O–O single bond (148 pm) and O=O double bond (121 pm). In the carbonate and nitrate ions all three bonds to oxygen are identical in length and strength. Delocalization lowers the energy of the species, so resonance hybrids are more stable than any single Lewis formula suggests.
| Species | Number of resonance structures | Bond order of equivalent bonds | Charge on each terminal O in the hybrid |
|---|---|---|---|
| O3 | 2 | 1.5 | −½ (central O +1) |
| NO3− | 3 | 1⅓ | −⅔ (central N +1) |
| CO32− | 3 | 1⅓ | −⅔ (central C 0) |
| RCOO− (e.g. HCOO−) | 2 | 1.5 | −½ |
| C6H6 (benzene) | 2 (Kekulé) | 1.5 | not applicable |
| Test | Draw the Lewis formula. If a double bond could equally well be placed between the central atom and a different, equivalent atom without moving any atoms, resonance structures exist. |
| HCOO− | C bonded to H, O and O−: the C=O can be on either oxygen → two resonance structures; the two C–O bonds are identical. |
| HCOOH | The two oxygens are not equivalent (one carries H), so a single Lewis formula is adequate. |
| CH3OH, H2O2, H2S | No multiple bonds at all → no resonance. |
Because the O–O bonds in ozone (bond order 1.5) are weaker than the O=O bond in O2 (bond order 2), less energy is needed to break a bond in O3. Photons of longer wavelength (lower energy) can therefore dissociate ozone, whereas oxygen needs shorter-wavelength ultraviolet radiation (Structure 1.3, Reactivity 3.3).
Exam focus · what the published papers show
The two-headed arrow between resonance structures does not mean the molecule alternates between them. Neither drawing exists. What exists is a single structure in which the electrons are spread out, and the drawings are the best two pictures a Lewis formula can manage. This matters because the examinable consequence is physical: all the equivalent bonds are the same length.
A report describes candidates who gave the two bond lengths 121 pm and 148 pm instead of a single intermediate value. Another spells out what was wanted: that with resonance, all the S–O bonds would be the same length and that this would be intermediate between that of a single and a double bond. If a species has resonance, every equivalent bond in it has one length, and that length lies between the single and double values. Quoting both numbers shows you have described the drawings rather than the molecule.
Question. For which molecule can resonance structures be used to describe the bonding? A parallel item asks which species have resonance structures.
The test is equivalence, not merely the presence of a double bond. Ask whether the multiple bond could sit in more than one position that is indistinguishable from the first. SO₂ passes: either S–O could carry it. NO₃⁻ and CO₃²⁻ pass with three positions each. Benzene passes with two arrangements of its three double bonds.
CO₂ fails, despite having two double bonds — there is no choice about where they go. And H₂CO₃ fails, which is the point: it was the reported distractor, chosen by 34 % of candidates. Its three oxygens are not equivalent, because two of them carry hydrogens, so the double bond has only one place it can be. Reported performance on this outcome ranges from just over 40 % recognising that sulfur dioxide needs resonance to 61.6 % on a similar item.
A scheme for the methanoate ion gives the double/pi/π bond «in the methanoate ion» is delocalized, and notes: Accept drawing showing delocalization of the bond or resonance structures for the mark. So a correct pair of resonance structures earns it, and so does the sentence. A report on that item records about half the cohort succeeding. On a different species, half of the students identified resonance/delocalization as the reason behind the N–O bonds being identical — and identical bonds is exactly the observation resonance explains.
On one question only 20% of the candidates mentioned resonance/delocalization of electrons, and the report gives the reason: many candidates could not solve this part-question because they had drawn a Lewis structure with all nitrogen–oxygen bonds as single bonds or double bonds. If every bond in your structure is the same, there is nothing for resonance to move — the error in the earlier part removes the possibility of answering the later one. This is the clearest case in these notes of one part of a question destroying the next.
2Benzene 2.2.12 HL only Repeated
The formula C6H6 was first explained by Kekulé as a ring with alternating single and double bonds. The modern model is a planar hexagonal ring of six sp2 carbon atoms in which six π electrons are delocalized over the whole ring, usually drawn as a hexagon with a circle inside (Figure 2.17).
| Type | Observation | Kekulé structure predicts | Delocalized model explains |
|---|---|---|---|
| Physical | All six C–C bonds are the same length, 140 pm (X-ray diffraction); the molecule is a planar, regular hexagon with all C–C–C angles 120°. | Two bond lengths, 154 pm and 133 pm; an irregular hexagon. | Every C–C bond has bond order 1.5, intermediate between single and double. |
| The enthalpy change of hydrogenation (and of combustion) is less exothermic than predicted for a structure with three C=C bonds. | Hydrogenation enthalpy ≈ three times that of one C=C. | Delocalization stabilizes benzene (resonance energy), so less energy is released. | |
| Chemical | Benzene does not decolourize bromine water in the dark and undergoes substitution more readily than addition. | Rapid electrophilic addition, like an alkene. | Addition would destroy the stable delocalized system; substitution keeps it intact. |
| Only one isomer of 1,2-disubstituted benzene (e.g. 1,2-dibromobenzene) exists. | Two 1,2-isomers: substituents across a C–C or across a C=C bond. | All ring bonds are identical, so only one 1,2-isomer. |
“Discuss the physical and chemical evidence” questions award one mark for a physical point (equal C–C bond lengths / regular hexagon / bond order 1.5 / all C–C–C angles equal / planar by X-ray) and one for a chemical point (substitution rather than addition / does not decolourize bromine water / one 1,2-disubstituted isomer / more stable than expected / less exothermic hydrogenation). Name the evidence, then say what it shows.
Exam focus · what the published papers show
| Say which electrons | Delocalization / resonance should refer to the pi electrons. "The electrons are delocalized" is not enough; it is the π electrons above and below the ring. |
| Say which evidence | Physical and chemical properties of benzene that are determined experimentally i.e. not bond enthalpy. The evidence must be something measured. |
What that leaves you. Physical: all six carbon–carbon bonds have the same length, intermediate between a single and a double bond; the molecule is planar and regular. Chemical: benzene undergoes substitution rather than the addition an alkene would, and it resists reagents that decolourise bromine water. A report notes that only stronger candidates identified the delocalization of pi electrons and the chemical and physical evidence.
In a substitution mechanism, a published scheme requires indicating delocalized electrons in ring and a curly arrow going from delocalized electrons in benzene to the electrophile. The arrow does not start from a particular double bond, because there is no particular double bond — which is the whole content of this outcome, appearing inside a mechanism question. The mechanism itself belongs to a later topic; the delocalization does not.
For which species can resonance structures be drawn? A. HCOOH B. HCOO− C. CH3OH D. H2CO3
Set on an SL paper under the previous guide; resonance is now additional HL content.
The Lewis (electron dot) structure of the dinitrogen monoxide molecule can be represented as:
(i) State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule. [1]
(ii) State, giving a reason, the shape of the dinitrogen monoxide molecule. [1]
(iii) Deduce the hybridization of the central nitrogen atom in the molecule. [1]
(i) Discuss the bonding in the resonance structures of ozone. [3]
(ii) Deduce one resonance structure of ozone and the corresponding formal charges on each oxygen atom. [2]
The Kekulé structure of benzene suggests it should readily undergo addition reactions.
Discuss two pieces of evidence, one physical and one chemical, which suggest this is not the structure of benzene. [2]
Benzene is an aromatic hydrocarbon.
(a) Discuss the physical evidence for the structure of benzene. [2]
(b) State the typical reactions that benzene and cyclohexene undergo with bromine. [1]
Which statement is correct?
Solutions and mark-scheme guidance · Set G
G1 B
In HCOO− the double bond and the negative charge can be placed on either O, giving two equivalent structures. In HCOOH the two O atoms are not equivalent (one carries H), so no equivalent resonance structures exist.
G2 [3]
(i) The bonding is delocalized / there are delocalized π electrons (resonance accepted) ✓.
(ii) Linear and two electron domains around the central N (two bonds and no lone pair) ✓.
(iii) sp ✓.
G3 [5]
(i) A lone pair in a p orbital on an O atom overlaps / delocalizes with the π electrons of the double bond ✓; both O–O bonds have equal length / bond order 1.5 / intermediate between O–O and O=O ✓; both O–O bonds have equal bond energy ✓.
(ii) O=O+–O−: one correct resonance structure ✓ with formal charges 0 (double-bonded O), +1 (central O), −1 (single-bonded O) ✓. A structure with 1.5 bonds is not accepted for this part.
G4 [2]
Physical ✓ (any one): all C–C bonds have equal length/strength; regular planar hexagon; bond order 1.5, intermediate between single and double (all C–C–C angles equal also accepted).
Chemical ✓ (any one): undergoes substitution more readily than addition; does not decolourise bromine water; only one 1,2-disubstituted isomer exists; more stable than expected for cyclohexa-1,3,5-triene / enthalpy of hydrogenation or combustion less exothermic than predicted.
G5 [3]
(a) Any two ✓✓: planar (X-ray evidence); all C–C bonds equal in length, intermediate between C–C and C=C (equal strength accepted); all C–C–C angles equal.
(b) Benzene undergoes electrophilic substitution and cyclohexene undergoes electrophilic addition ✓.
G6 A
The O–O bonds in O3 (bond order 1.5) are longer and weaker than the O=O bond in O2. Less energy is needed to break them, which corresponds to photons of lower frequency and longer wavelength. B, C and D all state the opposite.
Structure 2.2 (AHL) · Expanded octets and formal charge
3Expanded octets: five and six electron domains 2.2.13 HL only Strong
Atoms of period 3 and beyond (P, S, Cl, As, Se, Br, Xe, I and others) can have more than eight electrons in their valence shell. They have energetically accessible d orbitals in the same shell, so they can accommodate five or six electron domains around the central atom. Period 2 elements (C, N, O, F) cannot, because the second shell has no d sub-level; this is why NCl5 does not exist but PCl5 does.
| Domains | Lone pairs | Electron-domain geometry | Molecular geometry | Bond angles | Examples |
|---|---|---|---|---|---|
| 5 | 0 | trigonal bipyramidal | trigonal bipyramidal | 90°, 120°, 180° | PCl5, PF5 |
| 5 | 1 | see-saw | < 90°, < 120° | SF4 | |
| 5 | 2 | T-shaped | < 90° | ClF3, BrF3 | |
| 5 | 3 | linear | 180° | XeF2, I3− | |
| 6 | 0 | octahedral | octahedral | 90°, 180° | SF6 |
| 6 | 1 | square pyramidal | < 90° | BrF5, XeOF4 | |
| 6 | 2 | square planar | 90° | XeF4, BrF4− |
| SF4 | Valence electrons 6 + 4 × 7 = 34 (17 pairs). Four S–F bonds (4 pairs) + three lone pairs on each F (12 pairs) = 16 pairs; the 17th is a lone pair on S. Five domains, one lone pair (equatorial) → see-saw. |
| XeF4 | 8 + 4 × 7 = 36 (18 pairs). Four bonds + 12 lone pairs on F = 16 pairs; two lone pairs on Xe. Six domains, two lone pairs opposite each other → square planar, 90°. |
| I3− | 3 × 7 + 1 = 22 (11 pairs). Two I–I bonds + 6 lone pairs on the outer I atoms = 8 pairs; three lone pairs on the central I. Five domains, three equatorial lone pairs → linear, 180°. |
| Polarity | SF6 (octahedral) and XeF4 (square planar) are non-polar: identical bond dipoles cancel in these symmetrical arrangements. SF4 (see-saw) and BrF5 (square pyramidal) are polar: the lone pair makes the distribution of bonds unsymmetrical, so the dipoles do not cancel. |
Exam focus · what the published papers show
Period 3 and below can; period 2 cannot. Sulfur, phosphorus, chlorine and xenon can accommodate more than eight electrons. Nitrogen, carbon and oxygen cannot — they are period 2. This is not a detail: a report lists giving nitrogen an expanded octet (using double bonds for all N–O bonds) among the errors on a Lewis-structure question, alongside the opposite error of leaving atoms short. Check the period of the central atom before you draw a fifth bond.
Sulfur trioxide can be drawn two ways — one obeying the octet, one expanding it — and questions ask for a named one. A report records that only 40% of the candidates sketched the Lewis structure of SO₃ that obeys the octet rule, and that drawing structures which both did and did not conform to the octet rule, proved challenging. A related report notes that many candidates believe that the octet rule is followed by all three compounds in a set. Read whether the question wants the octet-obeying structure or the lowest-formal-charge one — they are different drawings of the same formula, and §2.2.14 is how you tell them apart.
A report on the same question: only about half the candidates stated the correct molecular geometry and bond angle based on the Lewis structure they drew in part (i). The geometry is marked as a consequence of your own structure, so a wrong structure can still earn the geometry mark — but only if the geometry matches it. Another item asked candidates to be able to determine the geometry of an extended octet species given its formula, with only just over half succeeding.
4Formal charge 2.2.14 HL only Strong
Often more than one valid Lewis formula can be drawn for a species, for example with or without an expanded octet. Formal charge (FC) is a bookkeeping charge used to decide which Lewis formula is preferred. It assumes that every bonding pair is shared exactly equally between the two atoms, whatever their electronegativities.
FC = V − N − ½B
V = number of valence electrons in the free atom; N = number of non-bonding (lone-pair) electrons on the atom in the structure; B = number of bonding (shared) electrons around the atom.
The sum of the formal charges equals zero for a molecule and the overall charge for an ion. The preferred Lewis formula is the one in which:
- the formal charges are as close to zero as possible (ideally all zero);
- any negative formal charge is on the more electronegative atom.
Both are hypothetical charges, but they rest on opposite assumptions. Formal charge assumes bonding electrons are shared equally. Oxidation state (Structure 3.1, Reactivity 3.2) assumes bonding electrons belong entirely to the more electronegative atom, as if the compound were ionic. In CO2, the carbon atom has formal charge 0 but oxidation state +4.
| Structure A | Octet on S: four S–O single bonds; each O has three lone pairs. FC(S) = 6 − 0 − ½(8) = +2. FC(each O) = 6 − 6 − ½(2) = −1. Sum = +2 + 4(−1) = −2 ✓. |
| Structure B | Expanded octet: two S=O and two S–O. FC(S) = 6 − 0 − ½(12) = 0. FC(O in S=O) = 6 − 4 − ½(4) = 0. FC(O in S–O) = 6 − 6 − ½(2) = −1. Sum = 0 + 0 + 0 − 1 − 1 = −2 ✓. |
| Structure C | Three S=O and one S–O: FC(S) = 6 − 0 − ½(14) = −1. A negative formal charge on S, the less electronegative atom, is unfavourable. |
| Decision | Structure B is preferred by formal charge: the central atom has FC = 0 and the negative formal charges sit on the more electronegative O atoms. Add only as many double bonds as are needed to bring the central atom’s formal charge to zero. |
| Caution | Formal charge is a model. Both A and B are valid Lewis formulas and the real ion is a hybrid with four identical S–O bonds; which description is “better” is still debated. |
Exam focus · what the published papers show
Question. Deduce the formal charge on each of the three oxygen atoms by adding them to the given structure of ozone.
The formula. Formal charge = valence electrons − lone-pair electrons − number of bonds, where a double bond counts as two bonds. Apply it to each atom separately.
| Atom | What it has | Working | FC |
|---|---|---|---|
| central O | 1 lone pair, 3 bonds | 6 − 2 − 3 | +1 |
| singly bonded O | 3 lone pairs, 1 bond | 6 − 6 − 1 | −1 |
| doubly bonded O | 2 lone pairs, 2 bonds | 6 − 4 − 2 | 0 |
Those are the published answers exactly. Check by adding them up: +1 − 1 + 0 = 0, which is the charge on the neutral molecule. For an ion the total must equal the ion's charge — a fast, free check that catches an arithmetic slip. Note that the scheme adds Do not accept delocalized structures: this question wants one specific Lewis formula, not the hybrid.
The clearest report in this topic: Many candidates knew that a lower formal charge indicated the preferred Lewis structure. Still, many tried to assign an overall formal charge instead of identifying the formal charge on each atom in the molecule, and many candidates did not assign formal charges to each atom in the two structures. An overall formal charge is not a finding — it is always just the charge on the species. The information is in how it is distributed. A second report describes candidates only stating the formal charge on the P atom when both P and O were required.
Question. Show which structure is most likely, using the concept of formal charge. [3] The published answer is structure 2 most likely with an explanation based on formal charge — so the choice alone is not the answer; the reasoning carries the marks.
The method, on SO₃. Take the structure with three S=O double bonds: sulfur has 6 − 0 − 6 = 0, and each oxygen 6 − 4 − 2 = 0. Now the octet-obeying structure with one double and two single bonds: sulfur is 6 − 0 − 4 = +2, the doubly bonded oxygen 0, and each singly bonded oxygen −1.
Both sum to zero, as they must — so the total tells you nothing. What separates them is that one structure has every atom at zero while the other has a +2 and two −1 values. Lowest formal charge means closest to zero, atom by atom. A report records just over 60 % succeeding on a question of exactly this shape, and adds that formal charge proved to be challenging especially when resonant structures were involved.
(a) What is the formal charge of the oxygen atom in H3O+? A. −2 B. −1 C. 0 D. +1
(b) What is the molecular geometry of SF4? A. Tetrahedral B. Trigonal bipyramidal C. See-saw D. Square planar
Which combination correctly describes the geometry of BrF4−?
| Electron domain geometry around Br | Molecular geometry around Br | |
|---|---|---|
| A. | Octahedral | Tetrahedral |
| B. | Tetrahedral | Square planar |
| C. | Octahedral | Square planar |
| D. | Tetrahedral | Tetrahedral |
Which elements are capable of forming expanded octets? I. Nitrogen II. Phosphorus III. Arsenic
Bromine can form the bromate(V) ion, BrO3−.
(b)(i) Draw two Lewis (electron dot) structures for BrO3−: Structure I that follows the octet rule, and Structure II that does not follow the octet rule. [2]
(b)(ii) Determine the preferred Lewis structure based on the formal charge on the bromine atom, giving your reasons. [2]
(c) Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O–Br–O bond angles. [3]
SF4Cl2 can form two isomers, one which is polar and another non-polar. Deduce the 3-dimensional representations of both isomers of SF4Cl2. [2]
Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar. [3]
Resonance structures exist when a molecule can be represented by more than one Lewis structure. Carbon dioxide can be represented by at least two resonance structures, I and II.
(i) Calculate the formal charge on each oxygen atom in the two structures. [2]
(ii) Deduce, giving a reason, the more likely structure. [1]
Solutions and mark-scheme guidance · Set H
H1 (a) D (b) C
(a) FC(O) = V − N − ½B = 6 − 2 − ½(6) = +1.
(b) SF4: five electron domains (four bonding, one lone pair); the lone pair occupies an equatorial position, giving a see-saw molecular geometry.
H2 C
Br in BrF4− has 7 + 1 = 8 valence electrons; four are used in Br–F bonds and four form two lone pairs. Six electron domains: octahedral; the two lone pairs sit opposite each other, leaving a square planar arrangement of F atoms.
H3 C
P and As (period 3 and beyond) have accessible d orbitals and are large enough to accommodate more than eight electrons. N, in period 2, has no 2d sub-level and cannot exceed an octet.
H4 [7]
(b)(i) Structure I (octet): Br joined by three single bonds to O, one lone pair on Br, three lone pairs on each O, overall charge −1 ✓. Structure II: at least one Br=O double bond (for example two Br=O and one Br–O−) ✓.
(b)(ii) FC on Br: +2 in structure I; 0 (two Br=O) or +1 (one Br=O) in structure II ✓. Structure II is preferred and its formal charge on Br is closer to zero ✓.
(c) Trigonal pyramidal ✓; three bonding pairs and one lone pair / four electron domains ✓; about 107° (104–109° accepted) ✓.
| Structure | V | N | ½B | FC(Br) |
|---|---|---|---|---|
| I (three Br–O, one lone pair) | 7 | 2 | 3 | +2 |
| II (two Br=O, one Br–O, one lone pair) | 7 | 2 | 5 | 0 |
H5 [2]
Six electron domains, no lone pairs: octahedral. Non-polar isomer: the two Cl atoms opposite each other (trans, Cl–S–Cl 180°) so that dipoles cancel ✓. Polar isomer: the two Cl atoms adjacent (cis, 90°) ✓. Any clear 3-D representation with wedges and dashes is accepted.
H6 [3]
PCl5 is trigonal bipyramidal and symmetrical, so the P–Cl bond dipoles cancel ✓. In PCl4F, the P–F bond has a different polarity from the P–Cl bonds (F is more electronegative) ✓, so the dipoles no longer cancel and the molecule is polar ✓.
H7 [3]
(i) Structure I (O=C=O): O(1) 0, O(2) 0. Structure II (O≡C–O): O(1) = 6 − 2 − 3 = +1; O(2) = 6 − 6 − 1 = −1. One mark for each two correct cells ✓✓.
(ii) Structure I and it has no formal charges ✓.
H8 D
With three S=O double bonds, each O has FC = 6 − 4 − 2 = 0 and S has FC = 6 − 0 − 6 = 0. All other structures give non-zero formal charges (for example structure C: S +1, single-bonded O −1).
Structure 2.2 (AHL) · Orbital overlap
5Sigma (σ) and pi (π) bonds 2.2.15 HL only Strong
The orbital model describes a covalent bond as the overlap of atomic orbitals, each containing one electron, to give a region of increased electron density between the nuclei. The geometry of the overlap gives two types of bond (Figure 2.19).
A sigma (σ) bond forms by the head-on (axial) overlap of atomic orbitals; the electron density is concentrated along the bond axis between the nuclei. It can form from s–s, s–p, p–p (end-on) or hybrid orbitals.
A pi (π) bond forms by the sideways (lateral) overlap of parallel p orbitals; the electron density is concentrated in two regions above and below (on opposite sides of) the bond axis, with none on the axis itself.
- Every single bond is one σ bond. A double bond is one σ + one π. A triple bond is one σ + two π (the two π bonds are at 90° to each other).
- A π bond can only form after a σ bond has brought the atoms close enough together: there is no π bond without a σ bond between the same atoms.
- Sideways overlap is less effective than head-on overlap, so a π bond is weaker than a σ bond. The π electrons are more exposed, which is why alkenes attract electrophiles.
- Rotation about a σ bond does not change the overlap, but rotation about a double bond would destroy the π overlap. This restricted rotation causes cis–trans isomerism (Structure 3.2).
| Rule | Count every bond as one σ; then add one π for each double bond and two π for each triple bond. |
| Ethene, C2H4 | 4 C–H + 1 C=C → σ = 5, π = 1. |
| Propanone, CH3COCH3 | 6 C–H + 2 C–C + 1 C=O → σ = 9, π = 1. |
| Nitrogen, N2 | N≡N → σ = 1, π = 2. |
| Carbon dioxide | O=C=O → σ = 2, π = 2. |
Exam focus · what the published papers show
Question. What are the numbers of sigma and pi bonds in propanenitrile, CH₃CH₂CN?
List the bonds. Three C–H on the first carbon, two C–H on the second, one C–C, one more C–C, and one C≡N. That is eight bonds in total.
Now apply the rule. Eight bonds means eight sigma bonds. The only multiple bond is the triple, which carries two extra shared pairs, so there are two pi bonds. The answer is 8 σ and 2 π. A triple bond is one sigma and two pi — never three pi, and one of the offered wrong answers is exactly that miscount.
A report describes the shape of the loss precisely: candidates who described the hybridisation correctly … often missed the other two marks for the formation of sigma and pi bonds, while the most frequent case was that they described the formation of σ or π bonds accurately gaining 2/3 marks at most and omitted describing hybridisation. Name the hybridization, then the sigma bonds, then the pi bonds. Reported success on the counting itself ranges from just over 50% to about 75%, and one session records that bonding questions were very well answered, both for metallic bonding and sigma/pi bonds.
6Hybridization 2.2.16 HL only Strong
Carbon in its ground state (1s22s22p2) has only two unpaired electrons in p orbitals at 90° to each other, yet methane has four identical C–H bonds at 109.5°. The hybridization model explains this: atomic orbitals of the same atom mix to form a new set of identical hybrid orbitals, which then form σ bonds or hold lone pairs. The number of hybrid orbitals formed equals the number of atomic orbitals mixed, and hybridization is used only to describe the bonding in molecules, never isolated atoms.
| Electron domains on the atom | Hybridization | Orbitals mixed | Electron-domain geometry | Unhybridized p orbitals | Examples (atom in bold) |
|---|---|---|---|---|---|
| 4 | sp3 | one s + three p | tetrahedral, 109.5° | 0 | CH4, NH3, H2O, CH3OH, PCl3 |
| 3 | sp2 | one s + two p | trigonal planar, 120° | 1 (forms π) | C2H4, BF3, H2CO, CO32−, benzene C |
| 2 | sp | one s + one p | linear, 180° | 2 (form two π) | C2H2, HCN, CO2, BeCl2, central N in N2O |
Lewis formula → count electron domains around the atom (lone pairs and bonds; a multiple bond counts once) → electron-domain geometry → hybridization. The route also works in reverse: an sp2 atom has trigonal planar geometry and 120° angles; an sp atom is linear. Lone pairs occupy hybrid orbitals, so N in NH3 and O in H2O are sp3. The oxygen atom of a C=O group has three domains (the double bond and two lone pairs) and is described as sp2.
| Ethanoic acid | CH3 carbon: four σ bonds → sp3. COOH carbon: three domains (C=O counts once) → sp2, trigonal planar. Hydroxyl O: two bonds + two lone pairs → sp3, bent. |
| Inorganic | CO2: carbon has two domains (two C=O) → sp, linear. NH4+: nitrogen has four bonding domains → sp3, tetrahedral. NO3−: nitrogen has three domains → sp2, trigonal planar, with the π electrons delocalized. |
| Check | The count of domains, not the number of atoms attached, fixes the hybridization. |
Exam focus · what the published papers show
Two domains is sp, three is sp², four is sp³. The number after the p is simply how many p-orbitals joined the one s-orbital, and the total always equals the number of domains. So hybridization, electron domain geometry and bond angle are one fact stated three ways — deduce any one of them and you have the other two. That is precisely what the syllabus means by "and vice versa".
Two reported errors, and both come from counting the wrong thing.
Forgetting that a multiple bond is one domain. A report records that many candidates thought carbon dioxide had sp² hybridisation and could not correctly identify the number of sigma and pi bonds. Carbon in CO₂ has two double bonds — but that is two domains, so it is sp. The hybridization error then propagated into the bond count, which is the same question mis-answered twice.
Forgetting that lone pairs are domains. Another report describes candidates who gave the right hybridizations for nitrogen and carbon in a molecule but were incorrectly deciding it was sp for O probably ignoring the lone pairs. An oxygen with two bonds and two lone pairs has four domains: sp³. Count everything around the atom, then name it.
On a geometry-from-hybridization item, about 75% of the candidates correctly remembered the geometry resulting from sp² hybridisation, and the report names what the rest chose: that related to sp³ hybridisation (109.5°) was by far the most effective distractor. Performance across this outcome is wide — almost 90 % on one item asking for geometry and hybridization together, 70 % on identifying sp² in a dimer, 58 % on labelled atoms in a structure — and the difference is whether the domain count was made carefully. Do not recall the hybridization; derive it.
- “A double bond is two σ bonds.” The second bond is a π bond made by sideways overlap of p orbitals, not by hybrid orbitals.
- Drawing a π bond as overlap along the axis, or with the nuclei in the wrong place. Draw two parallel p orbitals with both nuclei on the axis and overlap above and below it.
- Hybridizing isolated atoms, or ignoring lone pairs. Hybridization belongs to atoms in molecules and must count lone pairs as domains.
- Miscounting σ bonds by forgetting C–H bonds in condensed formulas. Draw the displayed formula before counting.
(a) How many sigma (σ) and pi (π) bonds are present in hydrogen cyanide, HCN?
| σ | π | |
|---|---|---|
| A. | 1 | 1 |
| B. | 2 | 2 |
| C. | 2 | 1 |
| D. | 1 | 3 |
(b) What is the hybridization of carbon and oxygen in methanal, H2C=O?
| Hybridization of C | Hybridization of O | |
|---|---|---|
| A. | sp2 | sp2 |
| B. | sp2 | sp |
| C. | sp | sp2 |
| D. | sp3 | sp3 |
(a) What is the number of sigma (σ) and pi (π) bonds in the molecule (NC)2C=C(CN)2?
| σ | π | |
|---|---|---|
| A. | 9 | 9 |
| B. | 5 | 9 |
| C. | 13 | 5 |
| D. | 9 | 5 |
Hybridization of hydrocarbons affects their reactivity.
(i) Distinguish between a sigma and pi bond. [2]
(ii) Identify the hybridization of carbon in ethane, ethene and ethyne. [1]
(i) Sketch the shape of one sigma (σ) and one pi (π) bond. [2]
(ii) Identify the number of sigma and pi bonds in HCN. [1]
(iii) State the hybridization of the carbon atom in HCN. [1]
Compound A is in equilibrium with compound B.
(a) Predict the electron domain and molecular geometries around the oxygen atom of molecule A using VSEPR. [2]
(b) State the type of hybridization shown by the central carbon atom in molecule B. [1]
(c) State the number of sigma (σ) and pi (π) bonds around the central carbon atom in molecule B. [1]
What bond angle is most likely found with an sp2 hybridized carbon as the central atom? A. 90° B. 109.5° C. 120° D. 180°
Solutions and mark-scheme guidance · Set I
I1 (a) B (b) A
(a) H–C≡N: the H–C single bond is one σ; the triple bond is one σ + two π. Total 2 σ, 2 π.
(b) C in methanal has three electron domains (two C–H, one C=O): sp2. O has one bond domain and two lone pairs, and is also treated as sp2 in the mark scheme, with the unhybridized p orbital forming the π bond.
I2 (a) A (b) D
(a) Four C≡N (4 σ + 8 π), four C–C single (4 σ) and one C=C (1 σ + 1 π): 9 σ, 9 π.
(b) The circled CH3 carbon has four single bonds: sp3. The carbonyl O has one double bond and two lone pairs: sp2. The ring N forms a C=N double bond plus one single bond with one lone pair (three domains): sp2.
I3 [3]
(i) σ: overlap along the internuclear axis (head-on / end-to-end), electron density concentrated between the nuclei ✓. π: sideways overlap of p orbitals, electron density above and below the internuclear axis ✓ (a labelled diagram is accepted).
(ii) Ethane sp3, ethene sp2, ethyne sp — all three are required ✓.
I4 [4]
(i) σ sketch: two orbitals (s–s, s–p or p–p) overlapping head-on along the axis ✓; π sketch: two parallel p orbitals overlapping sideways, lobes above and below the axis ✓.
(ii) 2 σ and 2 π ✓. (iii) sp ✓ (two electron domains on C).
I5 [4]
(a) The O in A (an –OH group) has two bonding pairs and two lone pairs: electron domain geometry tetrahedral ✓; molecular geometry bent ✓.
(b) The central C of B (C=O, bonded to two C) has three electron domains: sp2 ✓.
(c) The C=O double bond is 1 σ and 1 π; together with the two C–C bonds the central carbon forms 3 σ and 1 π ✓.
I6 C
An sp2 carbon has three hybrid orbitals directed to the corners of a triangle: 120°.
Review · Structure 2.2, higher level
7The examiner’s view, and the higher level layer in one page
Evidence base: principal examiner reports for Cambridge International AS & A Level Chemistry (2016–2024) on the same chemistry.
Identification of π bonds is more accurate than counting σ bonds; common wrong σ totals arise from omitting C–H bonds. The term hybridization is frequently misunderstood: answers to “state the hybridization” were sometimes split almost equally between sp, sp2 and sp3, and candidates often described how σ and π bonds form instead of how hybrid orbitals form. Diagrams of orbital overlap were often inadequate or unlabelled; a π bond requires adjacent parallel p orbitals overlapping sideways, and many candidates showed a poor grasp of where the nuclei lie relative to the π bond. Good answers state that hybridization describes atoms (for example, the carbon atoms are sp2), that σ bonds form by head-on overlap of hybrid orbitals, and that π bonds form by sideways overlap of unhybridized p orbitals. For benzene, many answers stated the planar ring, 120° angles and sp2 carbon but did not explain how the σ and π bonds form or that the π electrons are delocalized; some described the Kekulé structure instead. With expanded octets, many candidates did not recognise that sulfur can accommodate ten electrons when forming two double bonds with oxygen, and placed lone pairs on sulfur instead.
| More than one Lewis formula? | If the multiple bond has more than one equivalent position, the species has resonance. The consequence to state is that all those bonds have one length, intermediate between single and double. |
| Central atom in period 3 or below? | Then five or six domains are allowed. Period 2 atoms — C, N, O — are limited to four. |
| Two structures to choose between? | Formal charge, per atom: valence − lone-pair electrons − bonds. The preferred structure is the one whose atoms sit closest to zero. |
| Asked for sigma and pi? | Count the bonds — that is the sigma total. Add one pi for every extra shared pair. |
| Asked for hybridization? | Count the domains around that atom, lone pairs included. Two is sp, three sp², four sp³ — and each carries its geometry and angle with it. |
| Asked about benzene? | Delocalized pi electrons, and evidence that was measured: equal bond lengths, and substitution rather than addition. |
Every outcome in this document runs off the Lewis formula you drew and the domains you counted. Resonance is whether an alternative formula exists; formal charge is which formula wins; hybridization and the sigma/pi count are read off the finished structure. So the habit from Part A is the habit here too, with one addition: after drawing the structure, count the domains around every atom the question names — lone pairs included — and write those numbers down before answering anything. The reports show that almost every loss in this half of the topic traces back to a domain miscounted or a lone pair not drawn.
8Quick check
- Which of these has resonance structures: CO₂, SO₂, CH₄, H₂CO₃? Give the test you applied.
- A single N–O bond is 148 pm and a double N–O bond is 121 pm. What can you say about the N–O bonds in the nitrate ion?
- Give two pieces of evidence for the structure of benzene, one physical and one chemical.
- Why can sulfur have six electron domains but nitrogen cannot?
- State the molecular geometry of a species with five domains and one lone pair, and of one with six domains and two lone pairs.
- Calculate the formal charge on the nitrogen of the ammonium ion, NH₄⁺, and on each hydrogen. Check your answer.
- Two Lewis formulas of SO₃ are proposed: one with three double bonds, one with a double bond and two single bonds. Which is preferred, and why?
- How many sigma and how many pi bonds are in ethene, C₂H₄, and in ethyne, C₂H₂?
- What is the hybridization of the carbon in CO₂, and of the oxygen in water? What is the trap in each?
- An atom is described as sp². What is its geometry and its bond angle?
| 1 | SO₂ only. The test is whether the multiple bond has more than one equivalent position. SO₂ has two. CO₂ has double bonds but no choice about where they go; CH₄ has no multiple bond; H₂CO₃'s three oxygens are not equivalent because two carry hydrogens — which is why it was the popular wrong answer. |
| 2 | All three N–O bonds are identical, with a single length between 121 and 148 pm. Giving two separate lengths describes the drawings rather than the ion. |
| 3 | Physical: all six carbon–carbon bonds have the same length, intermediate between a single and a double bond. Chemical: benzene undergoes substitution rather than addition. Both must be things that were measured — a report rules out bond enthalpy — and the delocalization must be described as being of the pi electrons. |
| 4 | Sulfur is in period 3 and can accommodate more than eight electrons; nitrogen is period 2 and cannot. "Giving nitrogen an expanded octet" is a reported error. |
| 5 | Five domains with one lone pair: see-saw. Six domains with two lone pairs: square planar. The electron domain geometries remain trigonal bipyramidal and octahedral. |
| 6 | Nitrogen: 5 − 0 − 4 = +1. Each hydrogen: 1 − 0 − 1 = 0. Check: +1 + 4(0) = +1, which is the charge on the ion — as it must be. |
| 7 | The three-double-bond structure. There, sulfur is 6 − 0 − 6 = 0 and every oxygen 6 − 4 − 2 = 0. In the other, sulfur is 6 − 0 − 4 = +2 and two oxygens are −1. Both sum to zero, so the total decides nothing; what decides it is that every atom in the first is already at zero. |
| 8 | Ethene: 4 C–H plus 1 C=C is 5 bonds, so 5 σ, and the double bond adds 1 π. Ethyne: 2 C–H plus 1 C≡C is 3 bonds, so 3 σ, and the triple bond adds 2 π. |
| 9 | Carbon in CO₂ is sp — two domains, because each double bond is one domain; the trap is answering sp². Oxygen in water is sp³ — two bonds and two lone pairs make four domains; the trap is ignoring the lone pairs and answering sp. |
| 10 | Trigonal planar, 120°. Three domains. The reported distractor is 109.5°, which belongs to sp³. |
9Summary and knowledge organiser
Additional HL knowledge
- Resonance structures describe delocalized electrons; the real species is a single hybrid with intermediate, equal bond lengths (benzene, O3, CO32−, carboxylate ions).
- Formal charge (FC = V − N − ½B) selects the preferred Lewis structure: formal charges closest to zero, with any negative charge on the more electronegative atom.
- Expanded octets are possible for period 3 and later elements; five and six electron domains give trigonal bipyramidal and octahedral electron domain geometries.
- σ bonds form by head-on overlap along the internuclear axis; π bonds by sideways overlap of p orbitals above and below the axis. A double bond is 1 σ + 1 π; a triple bond is 1 σ + 2 π.
- Hybridization: 2, 3 and 4 electron domains correspond to sp, sp2 and sp3 (180°, 120°, 109.5°).
Essential definitions and relationships
| Term or relationship | Meaning and use |
|---|---|
| Resonance | Two or more valid Lewis structures differing only in the position of electrons; the real structure is the hybrid. |
| Hybridization | Mixing of atomic orbitals to form new, equivalent hybrid orbitals used in σ bonding. |
| FC = V − N − ½B | V = valence electrons of the free atom; N = non-bonding electrons on the atom; B = bonding electrons around the atom. The sum of formal charges equals the charge on the species. |
Essential shapes · five and six electron domains
| Domains | Lone pairs | Molecular geometry | Angle | Example |
|---|---|---|---|---|
| 5 | 0 / 1 / 2 / 3 | Trigonal bipyramidal / see-saw / T-shaped / linear | 90°, 120°, 180° | PCl5 / SF4 / ClF3 / XeF2 |
| 6 | 0 / 1 / 2 | Octahedral / square pyramidal / square planar | 90° | SF6 / BrF5 / XeF4 |
Common misconception
- Resonance structures flip between forms. The real structure is one hybrid with delocalized electrons.
Examination checklist
- Show formal-charge arithmetic; count σ and π bonds separately; state hybridization for the named atom.
Knowledge organiser · the covalent model, HL
| Model | Particles and attraction | Key facts and trends | Must-remember distinctions and common errors |
|---|---|---|---|
| HL covalent 2.2.11–2.2.16 | Delocalized π electrons; hybrid orbitals. | FC = V − N − ½B. 5 domains: TBP, see-saw, T-shaped, linear. 6: octahedral, square pyramidal, square planar. sp 180°, sp2 120°, sp3 109.5°. | Double = 1σ + 1π; triple = 1σ + 2π. Benzene: equal C–C lengths, substitution not addition. |