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

These are recall notes for the HL part of Structure 2.2, The covalent model, as listed in the IB Chemistry guide for Diploma Programme candidates examined from 2025 (issued February 2023). They cover understandings Structure 2.2.11 to 2.2.16: resonance, benzene, expanded octets, formal charge, σ and π bonds, and hybridization. All six are additional higher level, so this whole page is HL only. The reasoning and longer examples behind each point are in the study guide.

Then try the full practice questions. Course links: the subject hub page, a syllabus tick-list, and our free diagnostics for spotting weak areas.

Key definitions

Term	Meaning
Resonance structures	Two or more valid Lewis formulas for one species that differ only in where the double bond (π electrons) sits
Resonance hybrid	The real species: one structure, an average of the resonance structures
Delocalization	π electrons spread over more than two atoms
Resonance energy	Extra stability of the real species compared with one resonance structure
Expanded octet	More than eight electrons round a central atom (period 3 and below)
Formal charge	V − (L + ½B), assuming bonding electrons are shared equally
σ bond	Head-on overlap of orbitals; electron density along the bond axis
π bond	Sideways overlap of parallel p orbitals; electron density on opposite sides of the axis
Hybridization	Mixing atomic orbitals on one atom to make equivalent hybrid orbitals for bonding

2.2.11 Resonance in four lines

• Draw the Lewis formula, then ask: could a double bond go to another equivalent atom?
• Draw each option, joined by ↔. Move electrons, never atoms.
• Delocalized bonds are equal in length, between single and double.
• Bonds per position = (total shared pairs over those positions) ÷ (number of positions). Two positions sharing three pairs gives 1.5.

2.2.12 Benzene: evidence checklist

A good “discuss” answer gives both kinds of evidence and links each to the delocalized model.

Evidence	Type	What it shows
All six C–C bonds equal, between single and double in length	Physical	No alternating single and double bonds
Planar hexagon, 120° angles	Physical	Each C is sp²; its leftover p orbital stands perpendicular to the ring plane
Enthalpy of hydrogenation much less negative than 3 × that of cyclohexene	Physical (thermochemical)	Extra stability: resonance energy
Only one 1,2-disubstituted isomer	Chemical	Every ring bond is the same
Does not readily decolourize bromine water; reacts by substitution	Chemical	Addition would destroy the delocalized ring

Method: resonance energy from data

1. Expected ΔH = 3 × ΔH(hydrogenation of cyclohexene).
2. Resonance energy = measured ΔH(benzene) − expected ΔH. Give it as a positive stabilization.
3. State that both routes form cyclohexane, so benzene starts lower in energy.

2.2.13 Five and six electron domains

Method in steps

1. Valence electrons of the central atom, adding an electron for each negative charge and removing one for each positive charge.
2. Subtract one electron per bond to a monovalent atom such as F or Cl. Halve the rest for lone pairs.
3. Domains = bonding domains + lone pairs.
4. Electron domain geometry from the domain count; molecular geometry from the atoms only.

Domains	Lone pairs	Molecular geometry	Example
5	0	trigonal bipyramidal	SbCl₅
5	1	seesaw	TeCl₄
5	2	T-shaped	ClF₃
5	3	linear	XeF₂
6	0	octahedral	TeF₆
6	1	square pyramidal	IF₅
6	2	square planar	XeF₄

Rules: lone pairs go equatorial in a trigonal bipyramid; two lone pairs go opposite each other in an octahedron. Lone pairs shrink neighbouring bond angles slightly.

2.2.14 Formal charge

FC = V − (L + ½B)

Symbol	Meaning
V	valence electrons in the free atom
L	electrons in lone pairs on the atom
B	electrons in all bonds to the atom

Shortcut: FC = V − (number of lone-pair electrons) − (number of bonds).

Choosing the preferred Lewis formula

1. Check every candidate has the right electron total.
2. Work out FC on every atom. Check they add up to the overall charge.
3. Prefer formal charges closest to zero.
4. If two are tied, prefer the negative FC on the more electronegative atom.

Formal charge vs oxidation state

	Formal charge	Oxidation state
Bonding electrons	shared equally	all given to the more electronegative atom
S in preferred SO₄²⁻	0	+6

2.2.15 σ and π bonds

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

• σ: s–s, s–p, p–p end-on, or hybrid orbitals. Density on the axis.
• π: p–p side-on only. Density above and below (or either side of) the axis.
• Count σ = number of bonds of any kind; count π = extra bonds in double and triple bonds.

2.2.16 Hybridization

Domains on atom	Hybridization	Unhybridized p left	Electron domain geometry	Ideal angle
4	sp³	0	tetrahedral	109.5°
3	sp²	1	trigonal planar	120°
2	sp	2	linear	180°

• Hybrid orbitals make σ bonds and hold lone pairs. Leftover p orbitals make π bonds.
• Count domains, not atoms: N in NH₃ has three atoms but four domains, so it is sp³.
• Work backwards too: “sp” means two domains and a linear arrangement around that atom.
• Only sp, sp² and sp³ are in the guide. Do not assign hybridization to five- or six-domain atoms.
• Organic and inorganic examples are both expected: CH₃ carbon sp³, C=O carbon sp², nitrile C and N sp; O in H₂O sp³, B in BF₃ sp², Be in BeCl₂ sp.

Must-know distinctions

• Resonance structure vs resonance hybrid: drawings on paper vs the one real species.
• Electron domain geometry vs molecular geometry: arrangement of all domains vs arrangement of atoms only.
• σ vs π: head-on vs sideways; on the axis vs either side of it.
• Hybrid vs unhybridized orbitals: σ bonds and lone pairs vs π bonds.
• Formal charge vs oxidation state: equal sharing vs electrons to the more electronegative atom.

Where marks are usually lost

• Drawing resonance structures with a moved H atom, which makes a different compound.
• Writing that the double bond “switches back and forth” instead of describing one delocalized hybrid.
• Giving only physical evidence for benzene when the question asks for physical and chemical evidence.
• Forgetting the charge when counting electrons on ions such as ICl₂⁻ or PF₆⁻.
• Putting a lone pair in an axial position of a trigonal bipyramid, which gives the wrong shape.
• Naming the electron domain geometry when the molecular geometry was asked for.
• Using all the bonding electrons instead of half of them in the formal charge formula.
• Picking a structure with a negative formal charge on the less electronegative atom.
• Counting a double bond as two σ bonds.
• Giving sp² to an atom with three bonded atoms plus a lone pair (that is four domains: sp³).

Quick self-test

1. Draw the resonance structures for the nitric acid molecule, HNO₃, and state how many there are.
2. Calculate the formal charge on N in NH₄⁺.
3. Predict the molecular geometry of XeF₂.
4. Predict the electron domain geometry and molecular geometry of IF₅.
5. State the number of σ and π bonds in CS₂.
6. State the number of σ and π bonds in propyne, CH₃C≡CH.
7. Calculate the formal charges on each N atom in the azide ion drawn as [N=N=N]⁻.
8. State the hybridization of the carbon atom in CO₂ and of the nitrogen atom in CH₃CN.
9. Rank the C–O bond length in CH₃OH, CO₂ and the ethanoate ion, CH₃COO⁻, from shortest to longest.
10. State the formal charge and the oxidation state of S in the preferred Lewis formula of SO₄²⁻.
11. Explain why the C=C bond in an alkene contains one π bond, not two σ bonds.
12. State the electron domain geometry of SbCl₅ and the Cl–Sb–Cl angles present.

Answers

1. Two. H–O–N(=O)–O ↔ H–O–N(–O)=O, with the double bond moving between the two O atoms not bonded to H. The N carries +1 and the single-bonded O carries −1 in each.
2. FC = 5 − (0 + 4) = +1.
3. Five domains (two bonds, three lone pairs, all lone pairs equatorial): linear, 180°.
4. Six domains: octahedral electron domain geometry; one lone pair gives a square pyramidal molecule.
5. 2 σ, 2 π (two C=S double bonds).
6. 6 σ, 2 π (four C–H, one C–C, one σ in C≡C; two π in C≡C).
7. Each terminal N: 5 − (4 + 2) = −1. Central N: 5 − (0 + 4) = +1. Total −1, matching the charge.
8. C in CO₂: two domains, sp. N in CH₃CN: one triple bond and one lone pair, two domains, sp.
9. CO₂ < CH₃COO⁻ < CH₃OH. Double bonds are shortest, the delocalized 1.5 bonds in ethanoate are intermediate, the single bond is longest.
10. Formal charge 0; oxidation state +6.
11. Only one head-on overlap fits between two nuclei along the axis. The second bond comes from sideways overlap of parallel unhybridized p orbitals, which is a π bond.
12. Trigonal bipyramidal; 90° (axial to equatorial), 120° (between equatorial) and 180° (axial to axial).

Next steps

• Work through the practice questions with full worked answers.
• Go back over shaky topics in the study guide.
• Mark this unit as done on your course checklist.
• Run a free diagnostic to choose your next unit.

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.