IB DP Chemistry Structure 2.2: The covalent model: formal charge, sigma and pi bonds, and hybridization

This study guide teaches the higher level end of Structure 2.2, The covalent model, as listed in the IB Chemistry guide for Diploma Programme candidates examined from 2025 (issued February 2023). It works through understandings Structure 2.2.11 to 2.2.16. All six belong to the guide’s additional higher level (AHL) material, so everything on this page is HL only. It applies to the course as first assessed in 2025.

It builds on the SL skills of drawing Lewis formulas and using VSEPR for up to four electron domains. Afterwards, test yourself with the revision notes and the practice questions.

What this unit covers

Understanding	Skills to show	Level
Structure 2.2.11	Deduce resonance structures of molecules and ions; use the term delocalization	HL only
Structure 2.2.12	Use physical and chemical evidence to discuss benzene’s structure	HL only
Structure 2.2.13	Draw Lewis formulas with five and six electron domains on the central atom; deduce electron domain and molecular geometry	HL only
Structure 2.2.14	Calculate formal charges and use them to pick the preferred Lewis formula	HL only
Structure 2.2.15	Describe sigma (σ) and pi (π) bond formation; count σ and π bonds in molecules and ions	HL only
Structure 2.2.16	Link Lewis formula, electron domains, geometry and sp, sp² or sp³ hybridization	HL only

Resonance and delocalization (2.2.11)

Sometimes one Lewis formula cannot describe a species, because a double bond could sit in more than one place. Each acceptable drawing is a resonance structure. You show them side by side with a double-headed arrow (↔) between them.

The real species does not flip between the drawings. It is one resonance hybrid, with π electrons spread over three or more atoms instead of held between two. This spreading is delocalization. It has three consequences:

• The bonds that take part are all the same length.
• That length lies between a single and a double bond of the same atoms.
• The species is more stable (lower in energy) than any single resonance structure suggests.

Worked example 1: ozone, O₃

1. Count valence electrons: 3 × 6 = 18, so nine pairs.
2. Draw one O=O and one O–O, then complete octets: one lone pair on the central O, two on the double-bonded O, three on the single-bonded O.
3. The double bond could be on either side, so there are two resonance structures: O=O–O ↔ O–O=O. The two real bonds are identical.

This links to Structure 1.3. Each bond in O₃ is only partly double, so it is weaker than the O=O bond in O₂. Ozone is therefore split by lower-energy, longer-wavelength ultraviolet light than oxygen.

Worked example 2: the methanoate ion, HCOO⁻

1. Valence electrons: 1 + 4 + 6 + 6 + 1 (charge) = 18.
2. Carbon is central, bonded to H and both O atoms. One C–O is double and one is single; the single-bonded O carries three lone pairs and the negative charge.
3. Swap which O holds the double bond and you get the second resonance structure.
4. Three shared pairs are spread over two C–O positions, so each C–O bond is worth 3/2 = 1.5 bonds. Both bonds have the same length, shorter than the C–O in methanol and longer than C=O in methanal.

Resonance structures move electrons only, never atoms.

Benzene (2.2.12)

Benzene, C₆H₆, is a ring of six carbon atoms, each bonded to one hydrogen. Kekulé drew it with alternating single and double bonds. Those two resonance structures combine into a hybrid, drawn as a hexagon with a circle inside.

In the delocalized model each carbon uses three of its four valence electrons in σ bonds (two to ring carbons, one to H). Each carbon’s fourth electron occupies a p orbital perpendicular to the plane of the ring. These p orbitals overlap side by side around the ring, giving a ring of delocalized π electron density above and below the plane.

The evidence

Physical evidence

• Bond lengths. X-ray diffraction shows all six C–C bonds are the same length, between a single and a double bond. Alternating bonds would give two different lengths.
• Enthalpy of hydrogenation. Adding hydrogen to benzene releases much less energy than expected for three separate C=C bonds (see the example below).

Chemical evidence

• Only one 1,2-disubstituted isomer. Fixed double bonds would give two forms of 1,2-dichlorobenzene (Cl atoms across a single or a double bond). Only one exists.
• Reactions. Benzene does not readily decolourize bromine water, unlike alkenes. It tends to react by substitution, which keeps the delocalized ring intact, rather than addition.

Worked example 3: resonance energy from hydrogenation data

Use these values. Cyclohexene + H₂ → cyclohexane, ΔH = −116 kJ mol⁻¹. Benzene + 3H₂ → cyclohexane, ΔH = −204 kJ mol⁻¹.

1. A ring with three isolated C=C bonds should give 3 × (−116) = −348 kJ mol⁻¹.
2. Benzene actually gives −204 kJ mol⁻¹.
3. Difference = −204 − (−348) = +144 kJ mol⁻¹.
4. Both routes end at cyclohexane, so benzene is 144 kJ mol⁻¹ lower in energy than the Kekulé structure. This resonance energy explains its relative unreactivity and preference for substitution.

Expanded octets: five and six electron domains (2.2.13)

Atoms from period 3 downwards are larger and can hold more than eight valence electrons, so they can carry five or six electron domains. Period 2 atoms such as C, N and O cannot.

VSEPR works as before: each bond (single or multiple) and each lone pair is one domain. Choose the electron domain geometry, then name the shape from the atoms only.

Domains	Electron domain geometry	Bonding + lone	Molecular geometry
5	trigonal bipyramidal	5 + 0	trigonal bipyramidal (90°, 120°)
5	trigonal bipyramidal	4 + 1	seesaw
5	trigonal bipyramidal	3 + 2	T-shaped
5	trigonal bipyramidal	2 + 3	linear (180°)
6	octahedral	6 + 0	octahedral (90°)
6	octahedral	5 + 1	square pyramidal
6	octahedral	4 + 2	square planar (90°)

In a trigonal bipyramid, lone pairs go in the equatorial positions, where they have more room. In an octahedron, two lone pairs sit opposite each other.

Worked example 4: SeF₄

1. Se is in group 16, so it has 6 valence electrons. Four go into Se–F bonds, leaving 2: one lone pair.
2. Domains: 4 bonding + 1 lone pair = 5. Electron domain geometry: trigonal bipyramidal.
3. The lone pair takes an equatorial position, so the four F atoms make a seesaw shape. Its stronger repulsion closes the bond angles to below 120° and 90°.

Worked example 5: ICl₄⁻

1. I has 7 valence electrons, plus 1 for the charge: 8. Four are used in I–Cl bonds, leaving 4: two lone pairs.
2. Domains: 4 + 2 = 6. Electron domain geometry: octahedral.
3. The lone pairs sit opposite each other, so the four Cl atoms form a square planar ion with Cl–I–Cl angles of 90°. Put square brackets and the charge round the Lewis formula.

Formal charge (2.2.14)

Formal charge (FC) treats every bond as perfectly shared and compares the electrons each atom “owns” with the free atom:

FC = V − (L + ½B)

where V is the number of valence electrons in the free atom, L is the number of electrons in lone pairs on that atom, and B is the number of electrons in bonds to that atom. The formal charges in a species always add up to its overall charge.

To choose between possible Lewis formulas, the preferred one has:

1. formal charges as close to zero as possible on every atom;
2. any negative formal charge on the most electronegative atom.

Worked example 6: the cyanate ion, OCN⁻

Valence electrons: 6 + 4 + 5 + 1 = 16. Carbon is central. Three Lewis formulas give every atom an octet:

Structure	FC on O	FC on C	FC on N
[O=C=N]⁻	0	0	−1
[O≡C–N]⁻	+1	0	−2
[O–C≡N]⁻	−1	0	0

The second structure has the largest charges, so reject it. The first and third both have only one −1 charge. Oxygen is more electronegative than nitrogen, so [O–C≡N]⁻ is preferred.

Sulfate: why formal charge favours double bonds

The guide links formal charge to the structure of sulfate, SO₄²⁻. With four S–O single bonds, S has FC = 6 − (0 + 4) = +2 and each O has −1. With two S=O and two S–O bonds (an expanded octet), S has FC = 6 − (0 + 6) = 0, the two double-bonded O atoms have 0, and the two single-bonded O atoms have −1. The total is still −2. Formal charge prefers this second structure.

Formal charge versus oxidation state

Formal charge assumes bonding electrons are shared equally. Oxidation state assumes they all go to the more electronegative atom. So S in the preferred sulfate structure has a formal charge of 0 but an oxidation state of +6.

Sigma and pi bonds (2.2.15)

• A sigma (σ) bond forms by head-on overlap of two orbitals (s, p or hybrid). Its electron density lies mainly on the bond axis.
• A pi (π) bond forms by sideways (lateral) overlap of two parallel p orbitals. Electron density lies on opposite sides of the bond axis.

Counting rule:

Bond	σ	π
single	1	0
double	1	1
triple	1	2

Worked example 7: counting bonds

Propenenitrile, CH₂=CH–C≡N (organic). Bonds: three C–H, one C=C, one C–C and one C≡N. σ bonds = 3 + 1 + 1 + 1 = 6. π bonds = 1 (in C=C) + 2 (in C≡N) = 3.

Carbonyl fluoride, COF₂ (inorganic). One C=O and two C–F bonds: 3 σ and 1 π.

Hybridization (2.2.16)

Hybridization is the mixing of atomic orbitals on one atom to make new, equivalent hybrid orbitals for bonding. A carbon atom has one 2s and three 2p orbitals in its valence shell:

• sp³: s + three p mix to give four sp³ orbitals, pointing to the corners of a tetrahedron (109.5°). No p orbitals left over.
• sp²: s + two p give three sp² orbitals in a plane at 120°. One unhybridized p orbital is left at right angles to the plane.
• sp: s + one p give two sp orbitals at 180°. Two unhybridized p orbitals are left.

Hybrid orbitals form σ bonds or hold lone pairs; unhybridized p orbitals form π bonds. Only sp, sp² and sp³ are needed.

The link you must know

Electron domains on the atom	Hybridization	Electron domain geometry
4	sp³	tetrahedral
3	sp²	trigonal planar
2	sp	linear

The chain runs Lewis formula → domains → hybridization → geometry, and backwards too: an sp² atom has three domains.

Worked example 8: but-1-en-3-yne, CH₂=CH–C≡CH

1. C1 (CH₂=): three domains (two C–H, one C=C) → sp², 120°.
2. C2 (=CH–): three domains → sp², 120°.
3. C3 (–C≡): two domains → sp, 180°.
4. C4 (≡CH): two domains → sp, 180°.
5. σ bonds: four C–H + C=C + C–C + C≡C = 7. π bonds: 1 + 2 = 3, all formed from unhybridized p orbitals.

Inorganic examples. O in H₃O⁺ (three bonds, one lone pair) is sp³ and trigonal pyramidal. B in BCl₃ is sp², trigonal planar. C in CO₂ is sp, linear.

Common errors

• Counting a double bond as two electron domains in VSEPR or hybridization. It is one domain.
• Placing lone pairs in axial positions of a trigonal bipyramid.
• Getting B wrong in the formal charge formula: use half the bonding electrons, not all of them.
• Writing that a triple bond is three σ bonds.
• Assigning hybridization from the number of atoms bonded instead of the number of electron domains.

Next steps

• Revision notes for a fast recap
• Practice questions with worked answers
• The subject hub and a checklist of every understanding
• Short free diagnostics that show which units need work

Official syllabus

Source: the IB Chemistry guide for the Diploma Programme (International Baccalaureate Organization), issued in February 2023 for first assessment in 2025. This page covers Structure 2.2, The covalent model, understandings 2.2.11 to 2.2.16, all of them additional higher level.