IB Chemistry SL — Bonding & Structure

Complete Study Guide

Topics Covered

1. Ionic Bonding & Ionic Compounds (S2.1)
2. Covalent Bonding (S2.2)
3. Metallic Bonding (S2.3)
4. VSEPR Theory & Molecular Geometry (S2.2)
5. Electronegativity & Polarity (S2.2)
6. Intermolecular Forces (S2.2)
7. The Bonding Triangle (S2.1–S2.3)

Videos on this page: Ionic Bonding — Ion Formation & Structure · Covalent Bonding, Lewis Structures, VSEPR & Polar Bonds

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Prerequisites: This guide assumes familiarity with electron configurations, sublevels ($s$, $p$, $d$), and periodic trends (electronegativity, atomic radius). If you need a refresher, start with Atomic Structure & Periodicity (Structure 1) before continuing here.

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1. Ionic Bonding & Ionic Compounds (S2.1)

Ionic bonding occurs when one or more electrons are transferred from a metal atom to a non-metal atom. The metal loses electrons to form a positive ion (cation) and the non-metal gains electrons to form a negative ion (anion). The oppositely charged ions attract each other through strong electrostatic forces, forming an ionic bond. Ionic bonding typically occurs between elements with a large difference in electronegativity — usually a metal from Groups 1 or 2 and a non-metal from Groups 16 or 17.

Electron Transfer and Ion Formation

The driving force for ionic bonding is that both atoms achieve a more stable electron configuration (usually a full outer shell, like a noble gas). For example, in sodium chloride:

• Sodium ($\text{Na}$, Group 1) has 1 valence electron. It loses this electron to form $\text{Na}^+$ (with the electron configuration of neon).
• Chlorine ($\text{Cl}$, Group 17) has 7 valence electrons. It gains 1 electron to form $\text{Cl}^-$ (with the electron configuration of argon).

$$\text{Na} \rightarrow \text{Na}^+ + e^-$$ $$\text{Cl} + e^- \rightarrow \text{Cl}^-$$ $$\text{Na} + \text{Cl} \rightarrow \text{Na}^+\text{Cl}^-$$

For magnesium oxide, magnesium (Group 2) loses 2 electrons and oxygen (Group 16) gains 2:

$$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$$ $$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$

Key Ion Charges to Know

Element	Group	Ion Formed	Example
Sodium	1	$\text{Na}^+$	$\text{NaCl}$
Potassium	1	$\text{K}^+$	$\text{KBr}$
Magnesium	2	$\text{Mg}^{2+}$	$\text{MgO}$
Calcium	2	$\text{Ca}^{2+}$	$\text{CaCl}_2$
Fluoride	17	$\text{F}^-$	$\text{NaF}$
Chloride	17	$\text{Cl}^-$	$\text{NaCl}$
Oxide	16	$\text{O}^{2-}$	$\text{MgO}$
Sulfide	16	$\text{S}^{2-}$	$\text{Na}_2\text{S}$

The Ionic Lattice

Ions do not exist as simple pairs in a solid ionic compound. Instead, they form a giant ionic lattice — a regular three-dimensional arrangement in which each positive ion is surrounded by negative ions and vice versa. In sodium chloride, each $\text{Na}^+$ is surrounded by six $\text{Cl}^-$ ions, and each $\text{Cl}^-$ is surrounded by six $\text{Na}^+$ ions.

The strong electrostatic attractions between all the ions in the lattice hold it together. This is why we write $\text{NaCl}$ as an empirical formula (the simplest ratio of ions) rather than a molecular formula — there are no discrete molecules in an ionic solid.

Properties of Ionic Compounds

Property	Explanation
High melting and boiling points	Strong electrostatic forces between many ions require a large amount of energy to overcome
Brittle	Layers of ions can shift so that like charges align; repulsion then shatters the crystal
Do not conduct electricity as solids	Ions are fixed in the lattice and cannot move
Conduct electricity when molten or dissolved in water	Ions are free to move and carry charge
Often soluble in water	Polar water molecules attract and surround ions, pulling them away from the lattice

Quick Recall — Section 1

Try to answer without scrolling up:

1. What is an ionic bond?
2. Why do ionic compounds have high melting points?
3. When do ionic compounds conduct electricity?

Reveal answers

1. The electrostatic attraction between a positive ion (cation) and a negative ion (anion), formed by the transfer of electrons from a metal to a non-metal.
2. Strong electrostatic forces between many oppositely charged ions in the lattice require a large amount of energy to overcome.
3. Only when molten or dissolved in water (ions must be free to move).

Watch: Ionic Bonding — Ion Formation & Structure

MSJChem · 5 min · How cations and anions form — electron transfer, charge, and notation

MSJChem · 6 min · Ionic bonding, giant ionic lattices, and properties of ionic compounds

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2. Covalent Bonding (S2.2)

Covalent bonding occurs when two non-metal atoms share pairs of electrons. Neither atom has enough tendency to completely take electrons from the other (both have high electronegativity), so instead they share, and both atoms gain a complete outer shell through the shared electrons. The shared pair of electrons is attracted simultaneously to both nuclei, holding the atoms together.

Single, Double, and Triple Bonds

Atoms can share more than one pair of electrons:

Bond Type	Shared Pairs	Bond Strength	Bond Length	Example
Single bond	1 pair (2 electrons)	Weakest	Longest	$\text{H}_2$, $\text{F}_2$, $\text{CH}_4$
Double bond	2 pairs (4 electrons)	Intermediate	Intermediate	$\text{O}_2$, $\text{CO}_2$, $\text{C}_2\text{H}_4$
Triple bond	3 pairs (6 electrons)	Strongest	Shortest	$\text{N}_2$, $\text{C}_2\text{H}_2$

Key relationship: As the number of shared pairs increases, bond energy increases and bond length decreases. More electron density between the nuclei = stronger attraction = shorter, stronger bond.

Lewis (Dot-and-Cross) Structures

A Lewis structure shows all valence electrons in a molecule — both the bonding pairs (shared) and the lone pairs (non-bonding). Drawing Lewis structures correctly is one of the most tested skills in IB Chemistry SL.

Rules for drawing Lewis structures:

1. Count the total number of valence electrons from all atoms.
2. Arrange atoms — the least electronegative atom is usually the central atom (hydrogen is always terminal).
3. Place single bonds between connected atoms first (each bond uses 2 electrons from your total).
4. Complete octets on outer atoms first, then the central atom.
5. If the central atom still lacks an octet, convert lone pairs on outer atoms into double or triple bonds.
6. Check: total electrons used = total valence electrons counted in Step 1.

Valence electron counts by group:

Group	Valence electrons	Common atoms
1	1	H, Na (but Na rarely in Lewis structures)
14	4	C
15	5	N, P
16	6	O, S
17	7	F, Cl, Br, I

Dative (Coordinate) Covalent Bonds

In a dative bond (also called a coordinate bond), both electrons in the shared pair come from the same atom. Once formed, a dative bond is identical to a regular covalent bond — the difference is only in how it formed, not in its nature. In Lewis structures, dative bonds are sometimes shown with an arrow pointing from the donor atom to the acceptor atom.

Example: The ammonium ion $\text{NH}_4^+$ forms when $\text{NH}_3$ donates its lone pair to a proton $\text{H}^+$. The N atom donates both electrons to form a fourth N–H bond. All four N–H bonds in $\text{NH}_4^+$ are equivalent.

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3. Metallic Bonding (S2.3)

In metals, the valence (outer) electrons are not fixed to individual atoms. Instead, they become delocalised — they leave their atoms and move freely throughout the entire metallic structure. The result is a lattice of positive metal ions (also called cations or metal cores) surrounded by a “sea” of freely moving electrons. The electrostatic attraction between the positive metal ions and the delocalised electrons holds the metal together. This is the metallic bond.

The Sea of Electrons Model

• Positive ion lattice: The metal atoms lose their valence electrons and become positive ions arranged in a regular, closely packed pattern.
• Delocalised electron cloud: The released valence electrons move freely through the entire lattice, not belonging to any single atom.
• Metallic bond: The overall electrostatic attraction between the positive ions and the electron cloud.

The strength of the metallic bond depends on:

1. Charge on the metal ion — higher charge = stronger attraction (e.g. $\text{Mg}^{2+}$ forms stronger bonds than $\text{Na}^+$)
2. Number of delocalised electrons per atom — more electrons = stronger bond
3. Size of the metal ion — smaller ion = closer approach of electrons = stronger attraction

Properties of Metals Explained by the Model

Property	Explanation using sea of electrons model
High melting and boiling points	Strong electrostatic attraction between positive ion lattice and delocalised electrons requires much energy to break
Good electrical conductivity	Delocalised electrons can move freely and carry charge throughout the lattice; this happens in both solid and liquid state
Good thermal conductivity	Delocalised electrons can rapidly transfer kinetic energy through the lattice
Malleability and ductility	Layers of positive ions can slide past each other without breaking bonds; the electron sea re-adjusts around the new positions
Lustre (shiny appearance)	Delocalised electrons absorb and re-emit light across all visible wavelengths

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4. VSEPR Theory & Molecular Geometry (S2.2)

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the shape of a molecule based on the idea that electron pairs (both bonding pairs and lone pairs) around a central atom repel each other and arrange themselves as far apart as possible to minimise repulsion.

Electron Domains

An electron domain (or electron group) is any region of electron density around the central atom. Each of the following counts as ONE electron domain:

• A single bond
• A double bond (counts as 1 domain, even though it contains 2 pairs)
• A triple bond (counts as 1 domain)
• A lone pair

The number of electron domains determines the electron geometry (the arrangement of all electron pairs). The molecular shape (or molecular geometry) only counts the positions of atoms, not lone pairs — so lone pairs affect bond angles but do not appear in the shape name.

Lone Pair vs Bonding Pair Repulsion

Not all electron pairs repel equally:

Lone pair–lone pair repulsion > Lone pair–bonding pair repulsion > Bonding pair–bonding pair repulsion

This is because lone pairs are closer to the nucleus and occupy more space. The presence of lone pairs compresses the bond angles below the ideal geometry.

• In $\text{CH}_4$ (no lone pairs): bond angle = 109.5° (perfect tetrahedral)
• In $\text{NH}_3$ (1 lone pair): bond angle ≈ 107° (compressed from 109.5°)
• In $\text{H}_2\text{O}$ (2 lone pairs): bond angle ≈ 104.5° (compressed further)

VSEPR Shapes — Complete Reference Table

Quick Recall — Sections 2-4

Try to answer without scrolling up:

1. What molecular shape does a molecule with 4 bonding pairs and 0 lone pairs have?
2. Why is water bent and not linear?
3. What is a dative (coordinate) covalent bond?

Reveal answers

1. Tetrahedral (bond angle approximately 109.5 degrees).
2. Water has 2 bonding pairs and 2 lone pairs around oxygen. The lone pairs repel more strongly, pushing the bonds closer together into a bent shape (~104.5 degrees).
3. A covalent bond where both shared electrons come from the same atom.

Watch: Covalent Bonding, Lewis Structures, VSEPR & Polar Bonds

MSJChem · 6 min · Drawing Lewis (electron dot) structures for molecules and polyatomic ions

MSJChem · 6 min · VSEPR theory — predicting molecular and electron domain geometries from Lewis structures

MSJChem · 5 min · Polar vs. non-polar covalent bonds — electronegativity difference and bond dipoles

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5. Electronegativity & Polarity (S2.2)

Electronegativity is the ability of an atom to attract the shared pair of electrons in a covalent bond towards itself. It is a property of an atom within a molecule (not a property of an isolated atom). The most widely used electronegativity scale is the Pauling scale, where fluorine ($\text{F}$) is the most electronegative element at 4.0.

Electronegativity Trends

Trend	Direction	Reason
Across a period (left → right)	Increases	More protons → greater nuclear charge → stronger pull on bonding electrons
Down a group (top → bottom)	Decreases	More electron shells → greater shielding → bonding electrons further from nucleus

Most electronegative elements: F (4.0) > O (3.5) > N (3.0) ≈ Cl (3.0)

Polar Covalent Bonds

When two atoms with different electronegativities share electrons, the electrons are not shared equally. The more electronegative atom pulls the shared pair towards itself, developing a partial negative charge ($\delta-$), while the less electronegative atom develops a partial positive charge ($\delta+$). This is a polar covalent bond.

• If the electronegativity difference is small (roughly < 0.4): the bond is considered non-polar (e.g. C–H).
• If the electronegativity difference is intermediate (roughly 0.4–1.7): the bond is polar covalent (e.g. O–H, N–H, C–Cl).
• If the electronegativity difference is large (roughly > 1.7): the bond is considered ionic.

Example: In $\text{HCl}$, chlorine is more electronegative than hydrogen:

$$\delta+\text{H} - \text{Cl}\delta-$$

Polar vs Non-Polar Molecules

A molecule is polar if it has a net dipole moment — an overall separation of charge. A molecule is non-polar if the individual bond dipoles cancel out due to symmetrical molecular geometry.

A molecule is non-polar if:

• All bonds are non-polar (e.g. $\text{Cl}_2$), OR
• All bonds are polar BUT the molecule is symmetrical so dipoles cancel (e.g. $\text{CO}_2$, $\text{BF}_3$, $\text{CCl}_4$)

A molecule is polar if:

• It has polar bonds AND an asymmetrical shape so dipoles do NOT cancel (e.g. $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{HCl}$, $\text{CHCl}_3$)

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6. Intermolecular Forces (S2.2)

Intermolecular forces (IMFs) are the attractive forces between molecules. They are much weaker than the covalent bonds within a molecule (intramolecular bonds). IMFs are responsible for the physical properties of molecular substances — particularly boiling points, melting points, and solubility.

Three Types of Intermolecular Forces

London (Dispersion) Forces

London dispersion forces (a type of van der Waals force, also called induced-dipole–induced-dipole forces) are present in all molecules — polar and non-polar alike. They arise from temporary dipoles:

• At any instant, the electrons in a molecule may be unevenly distributed, creating a temporary (instantaneous) dipole.
• This temporary dipole induces a dipole in a neighbouring molecule.
• The two temporary dipoles then attract each other.

Factors affecting strength of London forces:

• More electrons (higher molar mass / larger molecule) → stronger London forces — more electrons means larger, more easily polarised electron clouds and bigger temporary dipoles.
• Shape — more elongated/flat molecules have greater surface area contact → stronger London forces than compact/spherical molecules of the same molar mass.

Effect on boiling point: As London forces increase (larger molecules, more electrons), more energy is needed to separate molecules → higher boiling point. This explains why boiling points increase down a group for non-polar molecules (e.g. noble gases: He < Ne < Ar < Kr < Xe).

Dipole–Dipole Forces

Dipole–dipole forces act between polar molecules. Because polar molecules have a permanent $\delta+$ end and a $\delta-$ end, neighbouring molecules align so that the positive end of one molecule is attracted to the negative end of another.

Dipole–dipole forces are stronger than London forces for molecules of similar size.

Example: $\text{HCl}$ molecules attract each other via dipole–dipole forces (in addition to London forces which are present too).

Hydrogen Bonding

Hydrogen bonding is a special, particularly strong type of dipole–dipole interaction. It forms when:

1. A hydrogen atom is directly bonded to a highly electronegative atom: N, O, or F.
2. The $\delta+$ H is then attracted to a lone pair on an N, O, or F atom of a neighbouring molecule.

Hydrogen bonds are approximately 5–10 times stronger than typical dipole–dipole forces and significantly stronger than London forces (for small molecules).

Examples of molecules that form hydrogen bonds: $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{HF}$, alcohols ($\text{-OH}$), amines ($\text{-NH}_2$)

Effect of IMFs on Boiling Points

The stronger the intermolecular forces, the higher the energy needed to separate molecules into a gas, and therefore the higher the boiling point.

Ranking IMF strength (weakest → strongest):

London forces < Dipole–dipole forces < Hydrogen bonds

However, London forces can overcome H-bonding if the molecule is very large (many electrons). For instance, large alkanes with hundreds of electrons have higher boiling points than small hydrogen-bonded molecules like water.

IMF Comparison Table

Water’s Anomalous Properties

Water’s unusually high boiling point (100°C) for such a small molecule ($M_r = 18$) is entirely due to its extensive hydrogen bonding. Each water molecule can form up to 4 hydrogen bonds (2 from the two O–H hydrogen atoms, 2 using the 2 lone pairs on oxygen as acceptors). This network of hydrogen bonds gives water its high boiling point, high surface tension, and ability to dissolve many ionic and polar substances.

Properties Comparison: Ionic, Covalent, Metallic

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7. The Bonding Triangle (S2.1–S2.3)

The three types of bonding — ionic, covalent, and metallic — are not completely distinct categories. They form a continuum, and real substances often sit somewhere between pure ionic, pure covalent, or pure metallic character. The bonding triangle (also called Van Arkel–Ketelaar triangle) represents this continuum graphically, with each corner representing one idealised bond type.

Position on the Triangle

The position of a compound or element on the bonding triangle depends on two factors:

1. Electronegativity difference between the atoms — determines how ionic or covalent a bond is.
2. Average electronegativity of the atoms — determines how metallic or covalent the substance is.

High electronegativity difference + high average electronegativity	→ Ionic (e.g. $\text{NaF}$, $\text{MgO}$, $\text{CsF}$)
Low electronegativity difference + high average electronegativity	→ Covalent (e.g. $\text{F}_2$, $\text{Cl}_2$, $\text{CO}_2$, $\text{H}_2\text{O}$)
Low average electronegativity (metals)	→ Metallic (e.g. Na, Fe, Cu, Mg)

Key Points About the Bonding Triangle

• Pure ionic bonds do not truly exist — all “ionic” bonds have some covalent character because electrons are never 100% transferred; there is always some sharing.
• Pure covalent bonds between identical atoms (e.g. $\text{Cl}_2$, $\text{H}_2$) come closest to pure covalent character, since there is no electronegativity difference.
• Ionic vs covalent character is a spectrum based on electronegativity difference. Greater electronegativity difference → more ionic character.
• Polar covalent bonds sit between pure covalent and ionic on the triangle.

Examples of Where Substances Sit

Substance	Position on triangle	Reason
$\text{CsF}$	Most ionic corner	Cs (low EN) + F (highest EN) = largest possible EN difference
$\text{NaCl}$	Ionic (EN diff ≈ 2.1)	Large EN difference between Na and Cl
$\text{HCl}$	Polar covalent	Intermediate EN difference; H and Cl are both non-metals
$\text{Cl}_2$	Non-polar covalent	Identical atoms; EN difference = 0
$\text{Na}$	Metallic corner	Metal with low EN; electrons delocalised
$\text{Fe}$	Metallic corner	Transition metal; delocalised electrons
$\text{MgO}$	Ionic (closer to ionic than NaCl)	$\text{Mg}^{2+}$ and $\text{O}^{2-}$ — large charge; high EN difference

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Written Question Answers

W1 — Ionic lattice conductivity: Solid $\text{NaCl}$ does not conduct electricity because the ions are fixed in the lattice and cannot move. When melted or dissolved in water, the ions become free to move and can carry charge, so the melt/solution conducts electricity.

W2 — Bond energy and length: Bond energy increases and bond length decreases as bond order increases (single < double < triple). More shared electron pairs create greater electron density between the nuclei, resulting in stronger, shorter bonds.

W3 — Polarity of $\text{CHCl}_3$ vs $\text{CCl}_4$: $\text{CCl}_4$ is non-polar because its tetrahedral symmetry causes all four equal C–Cl dipoles to cancel. $\text{CHCl}_3$ is polar because substituting one Cl with H (which has a different electronegativity) breaks the symmetry, so the bond dipoles no longer cancel and a net dipole moment exists.

W4 — Hydrogen bonding in water: Water’s anomalously high boiling point is due to extensive hydrogen bonding. Each water molecule has two O–H bonds (H-bond donors) and two lone pairs on oxygen (H-bond acceptors), allowing each molecule to form up to four hydrogen bonds. Overcoming this hydrogen bond network requires significantly more energy than overcoming the weaker London/dipole–dipole forces in molecules of similar molar mass.

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Mixed Practice — Exam Style

1. [Bonding Types] Which of the following compounds contains both ionic and covalent bonds?

   A. $\text{NaCl}$

   B. $\text{NH}_3$

   C. $\text{Na}_2\text{SO}_4$ (sodium sulfate)

   D. $\text{HCl}$

2. [VSEPR — Molecular Shape] The molecule $\text{SF}_4$ has 4 bonding pairs and 1 lone pair around the central sulfur atom. Its molecular geometry is:

   A. Tetrahedral

   B. Trigonal bipyramidal

   C. See-saw (sawhorse)

   D. Square planar

3. [Electronegativity and Polarity] $\text{CCl}_4$ has four polar C–Cl bonds but is a non-polar molecule overall. The reason is:

   A. The electronegativity difference between C and Cl is too small to create dipoles

   B. The four C–Cl bond dipoles are arranged symmetrically (tetrahedral), and they cancel exactly, giving a zero net dipole moment

   C. Carbon’s electronegativity equals chlorine’s, so no dipoles exist

   D. The molecule is non-polar because it contains no lone pairs on the central atom

4. [Intermolecular Forces] Arrange the following in order of increasing boiling point: $\text{CH}_4$, $\text{HF}$, $\text{CH}_3\text{CH}_2\text{OH}$.

   A. $\text{HF} < \text{CH}_4 < \text{CH}_3\text{CH}_2\text{OH}$

   B. $\text{CH}_4 < \text{HF} < \text{CH}_3\text{CH}_2\text{OH}$

   C. $\text{CH}_4 < \text{CH}_3\text{CH}_2\text{OH} < \text{HF}$

   D. $\text{CH}_3\text{CH}_2\text{OH} < \text{HF} < \text{CH}_4$

5. [VSEPR — Molecular Geometry] Using VSEPR theory, predict the molecular geometry and bond angle of $\text{CO}_2$.

   A. Bent, ~104.5°

   B. Linear, 180°

   C. Trigonal planar, 120°

   D. Trigonal pyramidal, ~107°

6. [Bonding Triangle — Distractor] A student places $\text{NaCl}$ at the “100% ionic” corner of the Van Arkel–Ketelaar bonding triangle, arguing that the electronegativity difference (2.1) is very large and therefore the bonding is purely ionic. Evaluate this claim:

   A. Correct — any electronegativity difference above 1.7 defines purely ionic bonding

   B. Incorrect — the bonding triangle represents a continuum. $\text{NaCl}$ has predominantly ionic character but still has some covalent character. No real compound occupies the pure ionic corner.

   C. Correct — $\text{NaCl}$ is always placed at the ionic apex because its ions are fully charge-separated

   D. Incorrect — $\text{NaCl}$ should be placed at the metallic corner because sodium is a metal

7. [VSEPR — Bond Angles] The H–N–H bond angle in $\text{NH}_3$ is approximately 107°, less than the ideal tetrahedral angle of 109.5°. The reason is:

   A. Nitrogen is more electronegative than hydrogen, which compresses the bonding pairs

   B. The lone pair on nitrogen occupies more space than a bonding pair (greater electron-electron repulsion), pushing the bonding pairs closer together

   C. Three hydrogen atoms are larger than nitrogen, forcing the bond angles to decrease

   D. The molecule is trigonal planar, so the angle must be less than 120°

8. [Intermolecular Forces — Hydrogen Bonding] Which of the following can act as a hydrogen bond donor?

   A. $\text{CH}_4$

   B. $\text{F}_2$

   C. $\text{NH}_3$

   D. $\text{CCl}_4$

9. [Structure and Properties] Diamond and graphite are both allotropes of carbon. Diamond is hard and does not conduct electricity, while graphite is soft and conducts electricity. The difference in properties arises because:

   A. Diamond is ionic, graphite is metallic

   B. In diamond, each carbon forms 4 covalent bonds in a rigid 3D network; in graphite, each carbon forms 3 covalent bonds in flat layers, leaving one delocalised electron per carbon that can conduct electricity; the layers are held by weak London dispersion forces, allowing them to slide

   C. Graphite has a higher molar mass than diamond, making it denser and more conductive

   D. Diamond has a molecular (simple covalent) structure; graphite has a giant ionic lattice

10. [Electronegativity — Bond Character] Place the following bonds in order from most ionic to most covalent character: C–H, Na–Cl, O–H, F–F.

    A. Na–Cl > O–H > C–H > F–F

    B. F–F > C–H > O–H > Na–Cl

    C. O–H > Na–Cl > F–F > C–H

    D. Na–Cl > C–H > O–H > F–F

Show Answers

1. C — $\text{Na}_2\text{SO}_4$ contains ionic bonds between Na$^+$ and SO$_4^{2-}$, and covalent bonds within the sulfate ion ($\text{SO}_4^{2-}$). A (NaCl) has only ionic bonds. B (NH$_3$) has only covalent bonds. D (HCl) has a polar covalent bond.

2. C — See-saw geometry. With 5 electron domains (4 bonding + 1 lone pair), the electron geometry is trigonal bipyramidal. The lone pair occupies an equatorial position, distorting the shape to a see-saw. A (tetrahedral) would require 4 bonding pairs and 0 lone pairs.

3. B — Symmetry cancels the bond dipoles. In $\text{CCl}_4$, the four C–Cl bonds point to the corners of a tetrahedron. The vector sum of the four equal dipoles is zero, giving no net dipole moment. A is wrong — the C–Cl electronegativity difference (~0.5) is sufficient to create individual bond dipoles.

4. B — $\text{CH}_4 < \text{HF} < \text{CH}_3\text{CH}_2\text{OH}$. CH$_4$ has only weak London dispersion forces (very low boiling point, −161°C). HF has strong hydrogen bonds but only one H-bond donor per molecule (b.p. 19°C). Ethanol has both hydrogen bonding and larger London forces from its ethyl group (b.p. 78°C). C is wrong — ethanol has a higher boiling point than HF.

5. B — Linear, 180°. In VSEPR, each double bond counts as one electron domain regardless of the number of bonding pairs within it. CO$_2$ has 2 bonding domains (one per C=O bond) and 0 lone pairs on the central carbon. Two domains with no lone pairs give linear electron geometry and linear molecular geometry, with a bond angle of exactly 180°. A (bent, ~104.5°) describes water, which has 2 bonding domains and 2 lone pairs. C (trigonal planar, 120°) would require 3 domains. D (trigonal pyramidal, ~107°) describes NH$_3$, which has 3 bonding domains and 1 lone pair.

6. B — The bonding triangle represents a continuum, and no real compound is purely ionic. NaCl is predominantly ionic but retains some covalent character. The IB first assessment of 2025 explicitly tests this concept. A is incorrect — there is no strict electronegativity cutoff for “pure” ionic bonding.

7. B — The lone pair on nitrogen has greater electron density than a bonding pair (which is shared and pulled away from N by H). Greater repulsion from the lone pair compresses the H–N–H angle below the ideal tetrahedral value. A is incorrect — electronegativity affects bond polarity, not angle directly. D is wrong — NH$_3$ is pyramidal, not trigonal planar.

8. C — $\text{NH}_3$. A hydrogen bond donor must have a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F). NH$_3$ has N–H bonds, making it a donor. CH$_4$ (A) and CCl$_4$ (D) have no N–H, O–H, or F–H bonds. F$_2$ (B) has no hydrogen.

9. B — Diamond: 4 covalent bonds per carbon, 3D giant covalent network — hard, non-conductor. Graphite: 3 covalent bonds per carbon in layers + 1 delocalised electron per carbon — good conductor, soft (layers slide due to weak London forces between them). A is wrong — neither allotrope is ionic or metallic.

10. A — Na–Cl > O–H > C–H > F–F. Na–Cl has the largest electronegativity difference (~2.1) — most ionic. O–H (~1.4 difference) is polar covalent. C–H (~0.4) is weakly polar covalent. F–F (0 difference, same element) — purely covalent and non-polar. D is wrong because it places C–H above O–H.

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Structure 2: Models of Bonding and Structure — IB Chemistry SL

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May 2026 Prediction Questions

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