Chemical Bonding and Structure

Introduction

Why Atoms Bond

Atoms interact to achieve lower potential energy. This is a stability argument. An isolated atom is a high-energy state; when atoms rearrange their electrons to form bonds, the resulting configuration sits in an energy well. The depth of that well is the bond enthalpy.

There are three broad categories of chemical bonding:

Bond Type	Mechanism	Typical Participants	Directionality
Ionic	Electron transfer	Metal + non-metal	Non-directional
Covalent	Electron sharing	Non-metal + non-metal	Directional
Metallic	Delocalised electron pool	Metal atoms	Non-directional

Beyond intramolecular bonds, intermolecular forces govern how molecules interact with each other. These are weaker by one to two orders of magnitude but are critical for determining physical properties such as melting point, boiling point, and solubility.

Definition. The bond enthalpy is the average enthalpy change when one mole of a specified type of bond is broken in the gaseous phase, measured in kJ/mol.

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Ionic Bonding

Mechanism

Ionic bonding results from the electrostatic attraction between cations and anions formed by complete electron transfer from a metal atom to a non-metal atom.

The driving force is the attainment of noble gas electron configurations:

$$\mathrm{Na}(s) \to \mathrm{Na}^+(g) + e^- \quad \Delta H_{\mathrm{at}}^\circ = +108\mathrm{ kJ/mol}$$ $$\frac{1}{2}\mathrm{Cl}_2(g) \to \mathrm{Cl}(g) \quad \Delta H_{\mathrm{at}}^\circ = +122\mathrm{ kJ/mol}$$ $$\mathrm{Cl}(g) + e^- \to \mathrm{Cl}^-(g) \quad \Delta H_{\mathrm{EA}} = -349\mathrm{ kJ/mol}$$ $$\mathrm{Na}^+(g) + \mathrm{Cl}^-(g) \to \mathrm{NaCl}(s) \quad \Delta H_{\mathrm{LE}} = -787\mathrm{ kJ/mol}$$

Definition. Lattice energy ($\Delta H_{\mathrm{LE}}$) is the enthalpy change when one mole of an ionic solid is formed from its gaseous ions. It is always exothermic. A more negative lattice energy indicates a stronger ionic bond.

Factors Affecting Lattice Energy

The Born-Lande equation captures the key variables:

$$\Delta H_{\mathrm{LE}} \propto -\frac{|z^+| \cdot |z^-|}{r_+ + r_-}$$

Factor	Effect on Lattice Energy	Example
Higher ion charge	More negative (stronger)	$\mathrm{MgO} \gt \mathrm{NaCl}$
Smaller ion radii	More negative (stronger)	$\mathrm{LiF} \gt \mathrm{NaF}$

Compound	z⁺	z⁻	r⁺ + r⁻ (pm)	Lattice Energy (kJ/mol)
NaCl	+1	-1	276	-787
MgO	+2	-2	210	-3795
LiF	+1	-1	201	-1036
CaO	+2	-2	241	-3414

The Born-Haber Cycle

The Born-Haber cycle is an application of Hess's law that links lattice energy to thermodynamic data you can measure experimentally.

Definition. The Born-Haber cycle is a thermochemical cycle that decomposes the formation of an ionic solid into a series of sequential steps, allowing calculation of lattice energy from measurable quantities.

For NaCl:

$$\Delta H_f^\circ = \Delta H_{\mathrm{at}}^\circ(\mathrm{Na}) + \frac{1}{2}\Delta H_{\mathrm{at}}^\circ(\mathrm{Cl}_2) + \mathrm{IE}_1(\mathrm{Na}) + \mathrm{EA}_1(\mathrm{Cl}) + \Delta H_{\mathrm{LE}}$$

Rearranging for lattice energy:

$$\Delta H_{\mathrm{LE}} = \Delta H_f^\circ - \Delta H_{\mathrm{at}}^\circ(\mathrm{Na}) - \frac{1}{2}\Delta H_{\mathrm{at}}^\circ(\mathrm{Cl}_2) - \mathrm{IE}_1(\mathrm{Na}) - \mathrm{EA}_1(\mathrm{Cl})$$

Substituting values:

$$\Delta H_{\mathrm{LE}} = -411 - 108 - 122 - 496 - (-349) = -788\mathrm{ kJ/mol}$$

IB Exam Tip

When constructing a Born-Haber cycle diagram, every arrow must be labelled with the correct enthalpy term. The most common error is confusing $\Delta H_{\mathrm{at}}^\circ$ (atomisation of the solid element) with $\Delta H_{\mathrm{sub}}$ (sublimation) -- for metals they are the same quantity, but the terminology matters.

Physical Properties of Ionic Compounds

Property	Explanation
High melting/boiling point	Strong electrostatic forces in the lattice require large energy input to break
Brittle	Shifting one layer of ions places like charges adjacent, causing repulsion and fracture
Conduct electricity when molten or aqueous	Ions are free to move and carry charge; in the solid state ions are fixed in the lattice
Soluble in polar solvents	Polar water molecules can surround and stabilise individual ions (solvation/hydration)

Solubility Rules

Ion Group	Solubility Pattern
Group 1 cations, NH$_4^+$	Always soluble
Nitrates (NO$_3^-$)	Always soluble
Acetates (CH$_3$COO$^-$)	Always soluble
Chlorides, bromides, iodides	Soluble except with Ag$^+$, Pb$^{2+}$, Hg$_2^{2+}$
Sulfates (SO$_4^{2-}$)	Soluble except with Ba$^{2+}$, Pb$^{2+}$, Ca$^{2+}$ (slightly)
Hydroxides (OH$^-$)	Insoluble except with Group 1, Ba$^{2+}$, Ca$^{2+}$ (slightly)
Carbonates (CO$_3^{2-}$)	Insoluble except with Group 1, NH$_4^+$
Phosphates (PO$_4^{3-}$)	Insoluble except with Group 1, NH$_4^+$

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Covalent Bonding

Lewis Structures

Lewis structures represent valence electrons as dots and show bonding pairs as lines (each line = 2 shared electrons).

Rules for drawing Lewis structures:

1. Count the total number of valence electrons from all atoms.
2. Identify the central atom (lowest electronegativity, excluding H which is always terminal).
3. Connect all atoms with single bonds (use 2 electrons per bond).
4. Complete the octets of terminal atoms first.
5. Place any remaining electrons on the central atom.
6. If the central atom lacks an octet, form double or triple bonds by converting lone pairs on terminal atoms into bonding pairs.

Common Mistake

Hydrogen only needs 2 electrons (duet rule). Beryllium can be stable with 4 electrons, and boron with 6. Do not force an octet on these atoms.

Exceptions to the Octet Rule

Element	Reason	Example
Be	Only 4 valence electrons available	BeCl$_2$
B	Only 6 valence electrons in stable compounds	BF$_3$
P, S, Cl, Xe	Can expand octet using d-orbitals (period 3+)	PCl$_5$, SF$_6$, XeF$_4$

Sigma and Pi Bonds

Definition. A sigma bond ($\sigma$) is formed by end-to-end (head-on) overlap of atomic orbitals. Electron density is concentrated along the internuclear axis.

Definition. A pi bond ($\pi$) is formed by side-to-side (lateral) overlap of parallel p-orbitals. Electron density is concentrated above and below the internuclear axis.

Feature	Sigma Bond	Pi Bond
Orbital overlap	Head-on	Side-to-side
Electron density	Along internuclear axis	Above and below the axis
Rotation	Free rotation about the bond	No rotation (locks the bond)
Strength	Stronger	Weaker
First bond	Always sigma	Never pi
Additional	Can exist alone	Only with an existing sigma

A single bond = 1 sigma. A double bond = 1 sigma + 1 pi. A triple bond = 1 sigma + 2 pi.

Bond Polarity and Electronegativity

Definition. Electronegativity is the ability of an atom to attract the bonding electron pair towards itself within a covalent bond.

The Pauling scale assigns fluorine (the most electronegative element) a value of 4.0.

Element	Electronegativity (Pauling)
F	4.0
O	3.5
N, Cl	3.0
C	2.5
S	2.6
P	2.2
H	2.2
Na, K	0.9, 0.8

Electronegativity Trends

• Across a period (left to right): Increases. Nuclear charge increases with constant shielding, pulling electron density in.
• Down a group: Decreases. Shielding increases and atomic radius increases, reducing the nucleus-electron attraction.

Classifying Bond Type by Electronegativity Difference

$\Delta\mathrm{EN}$	Bond Classification
0.0	Non-polar covalent
0.1 -- 1.7	Polar covalent
$\gt 1.7$	Ionic (predominantly)

IB Exam Tip

The threshold of 1.7 is a guideline, not an absolute boundary. For example, H-Cl has $\Delta\mathrm{EN} = 0.9$ (polar covalent), but Al-Cl has $\Delta\mathrm{EN} = 1.55$ (still considered covalent in AlCl$_3$, a molecular compound). Always consider the compound's actual properties.

Dipole Moments

A bond dipole is represented by an arrow pointing towards the more electronegative atom, with a cross at the less electronegative end.

The molecular dipole moment ($\mu$) is the vector sum of all individual bond dipoles. A molecule can have polar bonds but be non-polar overall if the bond dipoles cancel by symmetry.

$$\vec{\mu}_{\mathrm{net}} = \sum \vec{\mu}_i$$

Molecule	Bond Dipoles	Molecular Dipole	Reason
CO$_2$	Present	Zero	Linear geometry, dipoles cancel
H$_2$O	Present	Present	Bent geometry, dipoles do not cancel
CCl$_4$	Present	Zero	Tetrahedral symmetry, cancellation
CHCl$_3$	Present	Present	Asymmetric substitution

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Metallic Bonding

The Sea of Electrons Model

In a metallic lattice, metal atoms release their valence electrons into a delocalised "sea" or "cloud" of electrons. The resulting cations are held in a regular lattice by electrostatic attraction to this delocalised electron pool.

This model explains the key properties of metals:

Property	Explanation
High melting points	Strong metallic bonding throughout the lattice
Electrical conductivity	Delocalised electrons are free to move under an applied potential
Malleability and ductility	Layers of cations can slide past each other without breaking metallic bonds
Thermal conductivity	Delocalised electrons transfer kinetic energy efficiently
Lustrous appearance	Delocalised electrons absorb and re-emit photons across the visible spectrum
Alloy formation	Atoms of different sizes distort the lattice, preventing layer sliding

Factors Affecting Metallic Bond Strength

Factor	Effect	Example
Number of valence electrons	More delocalised electrons = stronger bond	Al $\gt$ Na
Nuclear charge	Higher charge = stronger attraction	Ca $\gt$ K
Ionic radius	Smaller radius = stronger bond	Mg $\gt$ Ca

Metal	Melting Point ($\degree$C)	Reason
Na	98	1 valence electron, large radius
Mg	650	2 valence electrons
Al	660	3 valence electrons
W	3422	Many valence electrons, small radius

Alloys

Definition. An alloy is a homogeneous mixture of two or more elements, at least one of which is a metal.

Alloy Type	Description	Effect on Properties
Substitutional	Atoms of similar size replace host atoms in the lattice	Distorts lattice, increases hardness, reduces malleability
Interstitial	Small atoms (C, N) fit into gaps in the lattice	Blocks dislocation movement, increases hardness

Steel is an interstitial alloy of iron with carbon. Brass is a substitutional alloy of copper and zinc.

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Intermolecular Forces

Intermolecular forces (IMFs) are the attractions between molecules. They are much weaker than intramolecular bonds (typically 2--50 kJ/mol vs 150--1000 kJ/mol for covalent bonds).

Types of Intermolecular Forces

IMF Type	Strength (kJ/mol)	Mechanism	Present In
London dispersion	0.05 -- 40	Temporary dipole from electron cloud fluctuation	All molecules
Dipole-dipole	5 -- 20	Permanent dipole-dipole attraction	Polar molecules
Hydrogen bonding	10 -- 40	H bonded to N, O, or F attracted to lone pair	Molecules with N-H, O-H, or F-H
Ion-dipole	10 -- 50	Ion interacts with molecular dipole	Ionic compounds in polar solvents

London Dispersion Forces

Definition. London dispersion forces (also called induced dipole-induced dipole forces or van der Waals forces) arise from temporary, instantaneous dipoles created by the uneven distribution of electrons at any given moment.

Factors affecting London dispersion force strength:

1. Number of electrons: More electrons = larger electron cloud = stronger temporary dipoles.
2. Molecular shape (surface area): Larger contact area between molecules = stronger forces.

Molecule	Electrons	Boiling Point ($\degree$C)	Reason
CH$_4$	10	-161	Few electrons, small surface
C$_2$H$_6$	18	-89	More electrons
C$_4$H_\\{10\\}	50	-1	Many more electrons

Dipole-Dipole Forces

Polar molecules have a permanent separation of charge. The positive end of one molecule is attracted to the negative end of another.

Definition. Dipole-dipole forces are the electrostatic attractions between the positive end of one polar molecule and the negative end of another.

Hydrogen Bonding

Definition. Hydrogen bonding is a particularly strong dipole-dipole interaction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (N, O, or F) and is simultaneously attracted to a lone pair on another N, O, or F atom.

Requirements:

• A hydrogen atom bonded to N, O, or F.
• A lone pair on an N, O, or F atom on a neighbouring molecule.

Substance	Boiling Point ($\degree$C)	Why so high?
H$_2$O	100	Extensive hydrogen bonding network
HF	20	Strong H-bonds (1 per molecule)
NH$_3$	-33	Fewer H-bonds per molecule
H$_2$S	-60	No hydrogen bonding (S not EN enough)
CH$_4$	-161	Only London dispersion forces

IB Exam Tip

Water has an anomalously high boiling point compared to H$_2$S, H$_2$Se, and H$_2$Te. The expected trend (boiling point increases down the group due to increasing electrons) is overridden by hydrogen bonding in water. This is a classic IB exam question.

Ion-Dipole Forces

When an ionic compound dissolves in a polar solvent like water, the ions interact with the molecular dipoles. This is the force responsible for the solvation of ions.

$$\mathrm{Na}^+ \cdots \delta^-\mathrm{O}(\mathrm{H}_2\mathrm{O}) \qquad \mathrm{Cl}^- \cdots \delta^+\mathrm{H}(\mathrm{H}_2\mathrm{O})$$

Trends in Boiling Points

For comparing boiling points of similar molecules:

1. Check for hydrogen bonding first (dominant IMF).
2. Among non-H-bonding molecules, compare dipole-dipole vs London dispersion.
3. For non-polar molecules, boiling point increases with molar mass (more electrons = stronger London forces).
4. For isomers, the more branched isomer has a lower boiling point (smaller surface area).

Effect of IMF on Physical Properties

Property	Strong IMF	Weak IMF
Melting point	High	Low
Boiling point	High	Low
Vapour pressure	Low	High
Viscosity	High	Low
Surface tension	High	Low
Volatility	Low	High

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Molecular Geometry

VSEPR Theory

Definition. Valence Shell Electron Pair Repulsion (VSEPR) theory states that electron pairs (bonding and non-bonding) around a central atom will arrange themselves to minimise repulsion, adopting geometries that maximise the angles between them.

The repulsion order is:

$$\mathrm{lone pair--lone pair} \gt \mathrm{lone pair--bond pair} \gt \mathrm{bond pair--bond pair}$$

This is because lone pairs are held by only one nucleus and occupy more space, while bonding pairs are constrained between two nuclei.

AXnEm Notation

• A = central atom
• X = bonded atom (bonding pair)
• n = number of bonding pairs
• E = lone pair on the central atom
• m = number of lone pairs

Electron Domain Geometries

The base geometries depend on the total number of electron domains ($n + m$):

Total Domains	Base Geometry	Bond Angles
2	Linear	180$\degree$
3	Trigonal planar	120$\degree$
4	Tetrahedral	109.5$\degree$
5	Trigonal bipyramidal	90$\degree$, 120$\degree$
6	Octahedral	90$\degree$

Molecular Shapes and Examples

2 Electron Domains

Notation	Shape	Bond Angle	Example
AX$_2$	Linear	180$\degree$	CO$_2$, BeCl$_2$

3 Electron Domains

Notation	Shape	Bond Angle	Example
AX$_3$	Trigonal planar	120$\degree$	BF$_3$, AlCl$_3$
AX$_2$E	Bent/V-shaped	$\lt 120\degree$	SO$_2$, O$_3$

Common Mistake

Students often forget that lone pairs repel more strongly, so AX$_2$E has a bond angle less than 120$\degree$, not exactly 120$\degree$. SO$_2$ has a bond angle of approximately 119.5$\degree$.

4 Electron Domains

Notation	Shape	Bond Angle	Example
AX$_4$	Tetrahedral	109.5$\degree$	CH$_4$, CCl$_4$
AX$_3$E	Trigonal pyramidal	$\lt 109.5\degree$	NH$_3$, PCl$_3$
AX$_2$E$_2$	Bent	$\lt 109.5\degree$	H$_2$O, H$_2$S

Molecule	Measured Angle	Deviation from 109.5$\degree$
CH$_4$	109.5$\degree$	0$\degree$
NH$_3$	107.0$\degree$	-2.5$\degree$
H$_2$O	104.5$\degree$	-5.0$\degree$

The increasing deviation reflects the increasing number of lone pairs compressing the bonding pairs.

5 Electron Domains

Notation	Shape	Bond Angles	Example
AX$_5$	Trigonal bipyramidal	90$\degree$, 120$\degree$	PCl$_5$
AX$_4$E	Seesaw	$\lt 90\degree$, $\lt 120\degree$	SF$_4$
AX$_3$E$_2$	T-shaped	$\lt 90\degree$	ClF$_3$
AX$_2$E$_3$	Linear	180$\degree$	XeF$_2$

In a trigonal bipyramidal arrangement, lone pairs always occupy equatorial positions because this minimises repulsion (equatorial positions have two 90$\degree$ interactions vs three 90$\degree$ interactions for axial positions).

6 Electron Domains

Notation	Shape	Bond Angles	Example
AX$_6$	Octahedral	90$\degree$	SF$_6$
AX$_5$E	Square pyramidal	$\lt 90\degree$	BrF$_5$
AX$_4$E$_2$	Square planar	90$\degree$	XeF$_4$

In octahedral geometry, all positions are equivalent. Lone pairs occupy positions 180$\degree$ apart to maximise separation.

Polarity Prediction from Geometry

To determine if a molecule is polar:

1. Draw the Lewis structure and identify polar bonds.
2. Determine the molecular geometry.
3. Check whether bond dipoles cancel by symmetry.

Geometry	Polar Bonds Cancel?	Result
Linear (AX$_2$)	Yes (opposing)	Non-polar
Trigonal planar (AX$_3$)	Yes (symmetric)	Non-polar
Tetrahedral (AX$_4$)	Yes (symmetric)	Non-polar
Trigonal pyramidal (AX$_3$E)	No (asymmetric)	Polar
Bent (AX$_2$E or AX$_2$E$_2$)	No (asymmetric)	Polar
Octahedral (AX$_6$)	Yes (symmetric)	Non-polar
Square planar (AX$_4$E$_2$)	Yes (symmetric)	Non-polar

IB Exam Tip

A common exam question asks whether a molecule like CHCl$_3$ or CH$_2$Cl$_2$ is polar. Even though C-H and C-Cl bonds have different polarities, the key is whether the vector sum of all bond dipoles equals zero. CHCl$_3$ is polar (no symmetry), but CCl$_4$ is non-polar (perfect tetrahedral symmetry). CH$_2$Cl$_2$ is polar because the two C-Cl dipoles and two C-H dipoles do not cancel.

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Hybridization

SL Content: sp, sp$^2$, sp$^3$

Definition. Hybridization is the mathematical mixing of atomic orbitals on a central atom to form a new set of equivalent hybrid orbitals that match the observed geometry.

Hybridization	Atomic Orbitals Mixed	Number of Hybrid Orbitals	Geometry	Bond Angle
sp	1s + 1p	2	Linear	180$\degree$
sp$^2$	1s + 2p	3	Trigonal planar	120$\degree$
sp$^3$	1s + 3p	4	Tetrahedral	109.5$\degree$

How to Determine Hybridization

Count the number of electron domains (bonding pairs + lone pairs) around the central atom:

• 2 domains = sp
• 3 domains = sp$^2$
• 4 domains = sp$^3$
• 5 domains = sp$^3$d
• 6 domains = sp$^3$d$^2$

Examples

Molecule	Central Atom	Domains	Hybridization	Geometry
BeCl$_2$	Be	2	sp	Linear
BF$_3$	B	3	sp$^2$	Trigonal planar
CH$_4$	C	4	sp$^3$	Tetrahedral
NH$_3$	N	4	sp$^3$	Trigonal pyramidal
H$_2$O	O	4	sp$^3$	Bent

Hybridization and Multiple Bonds

In a double bond, one bond is sigma (hybrid orbital overlap) and one is pi (unhybridized p-orbital overlap). The hybridization of the central atom is determined by the total number of domains (not bonds).

Molecule	Domains on C	Hybridization	Sigma Bonds	Pi Bonds
C$_2$H$_4$ (ethene)	3	sp$^2$	5	1
C$_2$H$_2$ (ethyne)	2	sp	3	2
CO$_2$	2	sp	2	2
HCN	2	sp	2	2

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Resonance

Delocalization

Definition. Resonance occurs when a molecule or ion can be represented by two or more valid Lewis structures that differ only in the positions of electrons (not atoms). The actual structure is a hybrid -- an average of all resonance forms.

Resonance stabilises a molecule. The more resonance structures, the greater the delocalisation energy (lower energy, more stable).

Ozone (O$_3$)

Ozone has two equivalent resonance structures:

$$\chemfig{O=O^{-}-O^{+}} \longleftrightarrow \chemfig{^{-}O-O^{+}=O}$$

The actual O-O bond order is 1.5, and the bond length is intermediate between a single and double bond (127.8 pm vs 121 pm for O=O and 148 pm for O-O).

Carbonate Ion (CO$_3^{2-}$)

Three equivalent resonance structures, each with one C=O double bond and two C-O single bonds. The actual bond order is 1.33 for each C-O bond.

$$\mathrm{Bond length measured: } 136\mathrm{ pm (between } 123\mathrm{ pm for C=O and } 143\mathrm{ pm for C-O)}$$

Benzene (C$_6$H$_6$)

Benzene has two Kekule structures with alternating single and double bonds. The actual structure has:

• Six equivalent C-C bonds with bond order 1.5
• All bond lengths identical: 140 pm (between 134 pm for C=C and 154 pm for C-C)
• A delocalised pi electron system above and below the ring
• Planar geometry (sp$^2$ hybridised carbons)

IB Exam Tip

The enthalpy of hydrogenation of benzene (-208 kJ/mol, for 3 moles of H$_2$) is less exothermic than expected from three isolated C=C bonds (-360 kJ/mol). The difference (152 kJ/mol) is the resonance energy (or delocalisation energy), which is a direct measure of the extra stability gained from electron delocalisation.

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HL-Only Extensions

Formal Charge

Definition. Formal charge is the charge assigned to an atom in a Lewis structure, calculated by comparing the number of valence electrons in the free atom with the number assigned to it in the structure.

$$\mathrm{Formal charge} = V - N_B - \frac{N_L}{2}$$

where:

• $V$ = number of valence electrons in the free atom
• $N_B$ = number of bonding electrons (shared) assigned to the atom
• $N_L$ = number of lone pair (non-bonding) electrons on the atom

Equivalently:

$$\mathrm{Formal charge} = V - (\mathrm{number of bonds}) - (\mathrm{number of lone pair electrons})$$

Rules for choosing the best Lewis structure:

1. The best structure minimises formal charges.
2. Negative formal charges should reside on the most electronegative atoms.
3. Like charges should not be adjacent.
4. Formal charges closest to zero are preferred.

Example: SO$_4^{2-}$

Sulfur has 6 valence electrons. With four single bonds to oxygen and no lone pairs:

$$\mathrm{FC}(\mathrm{S}) = 6 - 4 - 0 = +2$$

Each singly-bonded oxygen: $\mathrm{FC} = 6 - 1 - 6 = -1$

Total charge: $+2 + 4(-1) = -2$. This is valid but has large formal charges. Adding double bonds reduces the formal charges.

With two S=O double bonds:

$$\mathrm{FC}(\mathrm{S}) = 6 - 6 - 0 = 0$$

The two double-bonded oxygens: $\mathrm{FC} = 6 - 2 - 4 = 0$

The two single-bonded oxygens: $\mathrm{FC} = 6 - 1 - 6 = -1$

Total charge: $0 + 0 + 2(-1) = -2$. This is the preferred structure.

sp$^3$d and sp$^3$d$^2$ Hybridization (HL)

These hybridizations involve d-orbitals and are used for expanded octet species:

Hybridization	Orbitals Mixed	Domains	Geometry	Bond Angles
sp$^3$d	1s + 3p + 1d	5	Trigonal bipyramidal	90$\degree$, 120$\degree$
sp$^3$d$^2$	1s + 3p + 2d	6	Octahedral	90$\degree$

Molecule	Central Atom	Domains	Hybridization
PCl$_5$	P	5	sp$^3$d
SF$_4$	S	5	sp$^3$d
ClF$_3$	Cl	5	sp$^3$d
SF$_6$	S	6	sp$^3$d$^2$
BrF$_5$	Br	6	sp$^3$d$^2$
XeF$_4$	Xe	6	sp$^3$d$^2$

Molecular Orbital Theory (HL)

Definition. Molecular orbital (MO) theory describes bonding in terms of the combination of atomic orbitals to form molecular orbitals that belong to the entire molecule.

Key principles:

1. Atomic orbitals combine to form molecular orbitals.
2. The number of molecular orbitals equals the number of atomic orbitals combined.
3. Bonding orbitals are lower in energy than the parent atomic orbitals.
4. Antibonding orbitals are higher in energy than the parent atomic orbitals.

MO Diagrams for Homonuclear Diatomic Molecules

For elements in period 2:

• Li$_2$ through N$_2$: $\sigma_{2s} \lt \sigma^*_{2s} \lt \pi_{2p_x} = \pi_{2p_y} \lt \sigma_{2p_z} \lt \pi^*_{2p_x} = \pi^*_{2p_y} \lt \sigma^*_{2p_z}$
• O$_2$ through Ne$_2$: $\sigma_{2s} \lt \sigma^*_{2s} \lt \sigma_{2p_z} \lt \pi_{2p_x} = \pi_{2p_y} \lt \pi^*_{2p_x} = \pi^*_{2p_y} \lt \sigma^*_{2p_z}$

The s-p mixing in Li$_2$ through N$_2$ pushes the $\sigma_{2p_z}$ above the $\pi_{2p}$ orbitals. For O$_2$ and F$_2$, the energy gap is large enough that s-p mixing is negligible.

Bond Order from MO Theory

$$\mathrm{Bond order} = \frac{1}{2}(N_{\mathrm{bonding}} - N_{\mathrm{antibonding}})$$

Molecule	Bonding Electrons	Antibonding Electrons	Bond Order	Stability
H$_2$	2	0	1	Stable
He$_2$	2	2	0	Not stable
Li$_2$	2	0	1	Stable
Be$_2$	2	2	0	Not stable
B$_2$	4	2	1	Stable
C$_2$	6	2	2	Stable
N$_2$	8	2	3	Very stable
O$_2$	8	4	2	Stable
F$_2$	8	6	1	Stable
Ne$_2$	8	8	0	Not stable

IB Exam Tip

MO theory explains why O$_2$ is paramagnetic (has unpaired electrons in the $\pi^*$ orbitals). Lewis structures cannot predict this. This is a classic HL exam question.

Paramagnetism vs Diamagnetism

• Paramagnetic: Contains unpaired electrons. Attracted to a magnetic field. Examples: O$_2$, B$_2$.
• Diamagnetic: All electrons are paired. Repelled by a magnetic field. Examples: N$_2$, F$_2$, C$_2$.

Band Theory of Metals and Semiconductors (HL)

When many metal atoms come together in a lattice, their atomic orbitals combine to form bands -- a huge number of closely spaced molecular orbitals.

Definition. The valence band is the highest energy band that is occupied by electrons at 0 K. The conduction band is the next higher band, which is empty at 0 K.

Classification by Band Gap

Material Type	Band Gap	Conductivity at 0 K	Example
Metal	Overlapping	Conducts	Cu, Na
Semiconductor	Small (0.1 -- 3 eV)	Does not conduct (at 0 K)	Si, Ge
Insulator	Large ($\gt 3$ eV)	Does not conduct	Diamond

In metals, the valence and conduction bands overlap, so electrons are always available for conduction. In semiconductors, thermal energy can promote electrons across the band gap, creating charge carriers.

Intrinsic vs Extrinsic Semiconductors

Intrinsic semiconductors are pure materials (e.g., pure Si) where conductivity increases with temperature as more electrons are promoted across the band gap.

Extrinsic semiconductors have been doped with impurities:

Doping Type	Dopant	Effect	Carrier Type
n-type	Group 15 (P)	Extra electron enters conduction band	Electron
p-type	Group 13 (B)	Electron vacancy (hole) in valence band	Hole (positive)

Common Mistake

The "n" in n-type stands for "negative" (electron carriers), not the element nitrogen. The "p" in p-type stands for "positive" (hole carriers). Doping does not make the material charged -- the overall crystal remains electrically neutral.

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Exam Practice

Question 1: Ionic Bonding (SL, 4 marks)

Explain why magnesium oxide has a much higher melting point than sodium chloride.

Markscheme:

• Mg$^{2+}$ and O$^{2-}$ have higher charges than Na$^+$ and Cl$^-$ (1 mark).
• Higher ionic charge leads to stronger electrostatic attraction (1 mark).
• More energy is required to overcome the stronger lattice (1 mark).
• Mg$^{2+}$ has a smaller ionic radius than Na$^+$, which further increases lattice energy (1 mark).

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Question 2: Lewis Structures and VSEPR (SL, 5 marks)

(a) Draw the Lewis structure of the phosphate ion, PO$_4^{3-}$. (2 marks)

Markscheme:

• P is the central atom with 5 valence electrons. Each O contributes 6 valence electrons. Add 3 for the -3 charge.
• Total valence electrons: $5 + 4(6) + 3 = 32$.
• P forms single bonds to all four O atoms (8 electrons used, 24 remaining).
• Each O gets 3 lone pairs to complete its octet (24 electrons used).
• P has a formal charge of $+1$; each singly-bonded O has a formal charge of $-1$.
• One P=O double bond is added to reduce formal charges. The structure can have resonance with 4 equivalent forms.

(b) Determine the shape and bond angle of PO$_4^{3-}$. (3 marks)

Markscheme:

• Four bonding pairs, zero lone pairs around P (1 mark).
• Shape is tetrahedral (1 mark).
• Bond angle is approximately 109.5$\degree$ (1 mark).

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Question 3: Intermolecular Forces (SL, 4 marks)

Explain why propan-1-ol (CH$_3$CH$_2$CH$_2$OH) has a higher boiling point than propane (CH$_3$CH$_2$CH$_3$), despite having a similar molar mass.

Markscheme:

• Propan-1-ol can form hydrogen bonds between molecules due to the O-H group (1 mark).
• Propane can only form London dispersion forces (1 mark).
• Hydrogen bonding is a much stronger intermolecular force than London dispersion (1 mark).
• More energy is therefore required to separate propan-1-ol molecules (1 mark).

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Question 4: Polarity (SL, 3 marks)

Determine whether sulfur tetrafluoride (SF$_4$) is a polar or non-polar molecule, explaining your reasoning.

Markscheme:

• SF$_4$ has a seesaw geometry (AX$_4$E) (1 mark).
• The bond dipoles do not cancel due to the asymmetric shape and presence of a lone pair (1 mark).
• Therefore, SF$_4$ is a polar molecule (1 mark).

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Question 5: Resonance (SL/HL, 3 marks)

The carbonate ion, CO$_3^{2-}$, has a measured C-O bond length of 136 pm. Explain this value.

Markscheme:

• CO$_3^{2-}$ has three equivalent resonance structures (1 mark).
• Each C-O bond is intermediate between a single and a double bond (bond order = 1.33) (1 mark).
• The bond length of 136 pm is between a typical C-O single bond (143 pm) and a C=O double bond (123 pm) (1 mark).

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Question 6: Born-Haber Cycle (HL, 6 marks)

Calculate the lattice energy of calcium fluoride, CaF$_2$, using the following data:

Quantity	Value (kJ/mol)
$\Delta H_f^\circ$(CaF$_2$)	-1220
$\Delta H_{\mathrm{at}}^\circ$(Ca)	+178
$\Delta H_{\mathrm{at}}^\circ$(F$_2$)	+159
IE$_1$(Ca)	+590
IE$_2$(Ca)	+1145
EA$_1$(F)	-328

Markscheme:

$$\Delta H_f^\circ = \Delta H_{\mathrm{at}}^\circ(\mathrm{Ca}) + \Delta H_{\mathrm{at}}^\circ(\mathrm{F}_2) + \mathrm{IE}_1 + \mathrm{IE}_2 + 2 \times \mathrm{EA}_1(\mathrm{F}) + \Delta H_{\mathrm{LE}}$$ $$-1220 = 178 + 159 + 590 + 1145 + 2(-328) + \Delta H_{\mathrm{LE}}$$ $$-1220 = 1416 + \Delta H_{\mathrm{LE}}$$ $$\Delta H_{\mathrm{LE}} = -1220 - 1416 = -2636\mathrm{ kJ/mol}$$

(6 marks for correct cycle setup, correct substitution of all values, and correct arithmetic.)

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Question 7: MO Theory (HL, 5 marks)

(a) Draw the molecular orbital energy level diagram for O$_2$. Indicate the electron configuration and label all orbitals. (3 marks)

Markscheme:

Using the O$_2$/F$_2$ ordering (no s-p mixing):

$$\sigma_{2s}^2\; \sigma^{*2}_{2s}\; \sigma^2_{2p_z}\; \pi^2_{2p_x} = \pi^2_{2p_y}\; \pi^{*1}_{2p_x} = \pi^{*1}_{2p_y}$$

• Correct orbital energy ordering (1 mark).
• 12 valence electrons correctly placed (1 mark).
• Two unpaired electrons in $\pi^*$ orbitals shown (1 mark).

(b) Explain why O$_2$ is paramagnetic, referring to your diagram. (2 marks)

Markscheme:

• O$_2$ has two unpaired electrons in the degenerate $\pi^*_{2p}$ antibonding orbitals (1 mark).
• Species with unpaired electrons are paramagnetic (attracted to a magnetic field) (1 mark).

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Question 8: Formal Charge (HL, 4 marks)

Three possible Lewis structures can be drawn for sulfur dioxide, SO$_2$:

1. O=S=O (no formal charges)
2. O=S-O with formal charges of 0 on S, -1 on single-bonded O, +1 on double-bonded O
3. O-S=O with formal charges of 0 on S, -1 on single-bonded O, +1 on double-bonded O

Identify the most stable resonance structure and explain your reasoning.

Markscheme:

• Structure 1 (O=S=O) is the most significant contributor to the resonance hybrid (1 mark).
• It has zero formal charge on all atoms (1 mark).
• Structures with formal charges closest to zero are more stable (1 mark).
• The actual structure of SO$_2$ is a resonance hybrid of all three forms (1 mark).

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Question 9: Hybridization and Bonding (HL, 4 marks)

Describe the bonding in ethyne, C$_2$H$_2$, including hybridization and the types of bonds formed.

Markscheme:

• Each carbon is sp hybridised (2 electron domains: one C-C bond, one C-H bond) (1 mark).
• The C-C bond consists of one sigma bond (sp-sp overlap) and two pi bonds (p-p overlap) (1 mark).
• Each C-H bond is a sigma bond (sp-s overlap) (1 mark).
• The molecule is linear with a bond angle of 180$\degree$ (1 mark).

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Question 10: Comprehensive -- Boiling Points (SL, 5 marks)

The following substances have the boiling points shown:

Substance	Boiling Point ($\degree$C)
CH$_4$	-161
SiH$_4$	-112
NH$_3$	-33
PH$_3$	-88
H$_2$O	100
H$_2$S	-60

(a) Explain the trend in boiling points from CH$_4$ to SiH$_4$. (2 marks)

Markscheme:

• Both are non-polar molecules with only London dispersion forces (1 mark).
• SiH$_4$ has more electrons than CH$_4$, so London dispersion forces are stronger (1 mark).

(b) Explain why H$_2$O has a much higher boiling point than H$_2$S, but NH$_3$ and PH$_3$ show the expected trend. (3 marks)

Markscheme:

• H$_2$O can form extensive hydrogen bonding due to two O-H bonds and two lone pairs on oxygen (1 mark).
• H$_2$S cannot form hydrogen bonding because S is not electronegative enough (1 mark).
• NH$_3$ can form hydrogen bonding but only has one N-H bond per molecule, limiting the extent; the trend from NH$_3$ to PH$_3$ is dominated by increasing London dispersion forces (1 mark).

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Summary: Quick Reference Tables

Bond Type Comparison

Feature	Ionic	Covalent (Molecular)	Covalent (Network)	Metallic
Bond type	Electrostatic attraction	Shared electron pairs	Continuous covalent	Metallic bonding
Constituents	Cations and anions	Discrete molecules	Giant lattice	Metal cations + sea
Melting point	High	Low	Very high	Variable (often high)
Electrical	Conducts when molten/aqueous	Generally insulators	Insulators (except graphite)	Conducts (solid and liquid)
Solubility	Polar solvents	Non-polar (molecular), varies	Insoluble	Generally insoluble
Hardness	Hard but brittle	Soft (molecular), hard (network)	Very hard	Malleable, ductile
Example	NaCl, MgO	H$_2$O, CO$_2$	Diamond, SiO$_2$	Cu, Fe, Al

IMF Strength Ranking

$$\mathrm{Ion-dipole} \gt \mathrm{Hydrogen bonding} \gt \mathrm{Dipole-dipole} \gt \mathrm{London dispersion}$$

VSEPR Quick Reference

Domains	Lone Pairs	Shape	Angle	Example
2	0	Linear	180$\degree$	CO$_2$
3	0	Trigonal planar	120$\degree$	BF$_3$
3	1	Bent	$\lt 120\degree$	SO$_2$
4	0	Tetrahedral	109.5$\degree$	CH$_4$
4	1	Trigonal pyramidal	$\lt 109.5\degree$	NH$_3$
4	2	Bent	$\lt 109.5\degree$	H$_2$O
5	0	Trigonal bipyramidal	90$\degree$, 120$\degree$	PCl$_5$
5	1	Seesaw	$\lt 90\degree$, $\lt 120\degree$	SF$_4$
5	2	T-shaped	$\lt 90\degree$	ClF$_3$
5	3	Linear	180$\degree$	XeF$_2$
6	0	Octahedral	90$\degree$	SF$_6$
6	1	Square pyramidal	$\lt 90\degree$	BrF$_5$
6	2	Square planar	90$\degree$	XeF$_4$

Hybridization Quick Reference

Domains	Hybridization	Geometry
2	sp	Linear
3	sp$^2$	Trigonal planar
4	sp$^3$	Tetrahedral
5	sp$^3$d	Trigonal bipyramidal
6	sp$^3$d$^2$	Octahedral

Practice Problems

Question 1: Lattice Energy Comparison

Explain why $\mathrm{MgO}$ has a much higher lattice energy ($-3795\mathrm{ kJ/mol}$) than $\mathrm{NaCl}$ ($-787\mathrm{ kJ/mol}$), even though the ionic radii of $\mathrm{Mg}^{2+}$ and $\mathrm{Na}^+$ are similar.

Answer

Lattice energy depends on the product of ionic charges and inversely on the sum of ionic radii:

$\Delta H_{\mathrm{LE}} \propto -\frac{|z^+| \cdot |z^-|}{r_+ + r_-}$

In $\mathrm{MgO}$, both ions are doubly charged ($\mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$), so $|z^+| \cdot |z^-| = 2 \times 2 = 4$. In $\mathrm{NaCl}$, both ions are singly charged ($\mathrm{Na}^+$ and $\mathrm{Cl}^-$), so $|z^+| \cdot |z^-| = 1 \times 1 = 1$. The electrostatic attraction is approximately four times stronger for $\mathrm{MgO}$. Additionally, $\mathrm{O}^{2-}$ is smaller than $\mathrm{Cl}^-$, further increasing the lattice energy.

Question 2: VSEPR and Molecular Polarity

Determine the molecular geometry and polarity of $\mathrm{BrF}_3$ and $\mathrm{XeF}_4$. Justify your answers using VSEPR theory.

Answer

$\mathrm{BrF}_3$: $\mathrm{Br}$ has 7 valence electrons, each $\mathrm{F}$ contributes 1 bonding pair. Total domains = 3 bonding pairs + 2 lone pairs = 5 domains. This is $\mathrm{AX}_3\mathrm{E}_2$ (T-shaped). The bond dipoles do not cancel due to the asymmetric shape and lone pairs, so $\mathrm{BrF}_3$ is polar.

$\mathrm{XeF}_4$: $\mathrm{Xe}$ has 8 valence electrons, each $\mathrm{F}$ contributes 1 bonding pair. Total domains = 4 bonding pairs + 2 lone pairs = 6 domains. The lone pairs occupy positions 180$^\circ$ apart (axial). This is $\mathrm{AX}_4\mathrm{E}_2$ (square planar). The four bond dipoles cancel by symmetry, so $\mathrm{XeF}_4$ is non-polar.

Question 3: Boiling Point Trends

Explain why $\mathrm{H}_2\mathrm{O}$ ($100\degree\mathrm{C}$) has a much higher boiling point than $\mathrm{H}_2\mathrm{S}$ ($-60\degree\mathrm{C}$), despite $\mathrm{H}_2\mathrm{S}$ having a higher molar mass.

Answer

$\mathrm{H}_2\mathrm{O}$ can form extensive hydrogen bonding because oxygen is highly electronegative and has two lone pairs. Each water molecule can form up to four hydrogen bonds, creating a strong three-dimensional network. $\mathrm{H}_2\mathrm{S}$ cannot form hydrogen bonds because sulfur is not electronegative enough (EN = 2.6 vs O = 3.5). $\mathrm{H}_2\mathrm{S}$ molecules are held together only by weaker dipole-dipole interactions and London dispersion forces. The hydrogen bonding in water requires significantly more energy to overcome, resulting in a much higher boiling point.

Question 4: MO Theory and Bond Order

Use molecular orbital theory to determine the bond order of $\mathrm{O}_2$, $\mathrm{O}_2^+$, and $\mathrm{O}_2^{2-}$. Arrange them in order of increasing bond length.

Answer

For $\mathrm{O}_2$ (12 valence electrons, O$_2$/F$_2$ ordering):

$\sigma_{2s}^2\; \sigma^{*2}_{2s}\; \sigma^2_{2p_z}\; \pi^2_{2p_x} = \pi^2_{2p_y}\; \pi^{*1}_{2p_x} = \pi^{*1}_{2p_y}$

Bonding electrons = 8, Antibonding electrons = 4:

$\mathrm{Bond order} = \frac{8 - 4}{2} = 2$

$\mathrm{O}_2^+$ (11 valence electrons): Bonding = 8, Antibonding = 3:

$\mathrm{Bond order} = \frac{8 - 3}{2} = 2.5$

$\mathrm{O}_2^{2-}$ (14 valence electrons): Bonding = 8, Antibonding = 6:

$\mathrm{Bond order} = \frac{8 - 6}{2} = 1$

Higher bond order means shorter bond length:

$\mathrm{O}_2^{2-} \lt \mathrm{O}_2 \lt \mathrm{O}_2^+$

(increasing bond length order)

Question 5: Hybridization and Bonding in Ethene

Describe the bonding in ethene ($\mathrm{C}_2\mathrm{H}_4$), including the hybridization of each carbon atom, the types of bonds formed, and the molecular geometry.

Answer

Each carbon in ethene has 3 electron domains (2 C-H bonds + 1 C=C bond), so each carbon is $sp^2$ hybridised. The three $sp^2$ hybrid orbitals form sigma bonds: two C-H sigma bonds and one C-C sigma bond. The remaining unhybridized $p$ orbital on each carbon overlaps side-to-side to form a pi ($\pi$) bond. The molecule is trigonal planar around each carbon with bond angles of approximately $120^\circ$, and the entire molecule is planar. The C=C double bond consists of one sigma bond and one pi bond. The pi bond restricts rotation about the C=C bond.

For the A-Level treatment of this topic, see Bonding & Structure.