Chemical Bonding and Structure — IB Chemistry HL HL Study Guide

For: IB Chemistry HL candidates sitting IB Chemistry HL.

Covers: Ionic, covalent and metallic bonding, Lewis dot structures, VSEPR theory and molecular geometry, bond and molecular polarity, intermolecular forces, and higher-level hybridisation concepts for IB Chemistry HL assessments.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Chemistry HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Chemical Bonding and Structure?

Chemical bonding describes the electrostatic forces of attraction that hold atoms, ions, or molecules together, while molecular structure refers to the three-dimensional arrangement of these particles in space. This core HL topic makes up ~12% of your final exam grade and links directly to later syllabus content including energetics, organic reaction mechanisms, and materials chemistry. All bonding interactions arise from attractive forces between positively charged species (nuclei, cations) and negatively charged species (electrons, anions).

2. Ionic, covalent, metallic bonding

The three primary bonding types are distinguished by the behaviour of valence electrons between participating atoms:

Ionic bonding

Formed by electrostatic attraction between oppositely charged ions, produced when a metal atom transfers valence electrons to a more electronegative nonmetal atom. The IB rule of thumb is that ionic bonding occurs when the electronegativity difference between two atoms is >1.7. Ionic compounds form giant ionic lattices, with properties including high melting/boiling points, solubility in polar solvents, and electrical conductivity only when molten or dissolved (free moving ions). Worked example: Sodium chloride (NaCl) forms when a Na atom loses its single valence electron to form $N{a}^{+}$, and a Cl atom gains this electron to form $C{l}^{-}$. The lattice energy of NaCl is $-787{\text{kJ mol}}^{-1}$, meaning 787 kJ of energy is released when 1 mole of solid NaCl forms from gaseous $N{a}^{+}$ and $C{l}^{-}$ ions.

Covalent bonding

Formed by the sharing of valence electron pairs between two nonmetal atoms, with electronegativity difference <1.7. Covalent substances exist as either simple molecular structures (e.g. $H_{2}O$, $C{O}_2$) held together by weak intermolecular forces, or giant covalent lattices (e.g. diamond, $Si{O}_2$) held together by a continuous network of strong covalent bonds. Simple covalent substances have low melting/boiling points, while giant covalent substances have extremely high melting/boiling points and are almost entirely insoluble.

Metallic bonding

Formed by electrostatic attraction between positively charged metal cations and a delocalised "sea" of mobile valence electrons. Metallic structures are giant metallic lattices, with properties including high melting/boiling points, malleability, ductility, and high thermal and electrical conductivity. Exam tip: Examiners frequently ask you to explain melting point differences between metals: Magnesium has a higher melting point than sodium because $M{g}^{2+}$ has a higher charge and smaller ionic radius than $N{a}^{+}$, leading to stronger electrostatic attraction between cations and delocalised electrons.

3. Lewis dot structures

Lewis dot structures are 2D representations of the valence electron arrangement in molecules or polyatomic ions, with dots for lone electrons, lines for covalent bonds, and lone pairs written as pairs of dots. Follow these IB-endorsed steps to draw them correctly:

1. Calculate total valence electrons: Sum the valence electrons of all constituent atoms, add 1 electron for each negative charge, and subtract 1 electron for each positive charge.
2. Arrange atoms: The least electronegative atom is the central atom (H is always terminal, as it can only form 1 bond).
3. Draw single bonds between the central and terminal atoms, subtracting 2 electrons per bond from the total.
4. Distribute remaining electrons as lone pairs to terminal atoms first to satisfy the octet rule (H only needs 2 valence electrons), then add remaining electrons to the central atom.
5. If the central atom does not have a full octet, form double or triple bonds by moving lone pairs from terminal atoms into bonding positions.
6. For ions, add square brackets around the structure and write the charge outside the brackets.

Exceptions to the octet rule: Period 3 and higher elements can have expanded octets (up to 12 valence electrons) as they have access to d orbitals; small atoms like B can have incomplete octets (6 valence electrons); odd-electron species (e.g. $NO$) have unpaired electrons. Worked example: Lewis structure of $N{O}_3^{-}$:

1. Total valence electrons: $5+(3\times 6)+1=24$
2. N is the central atom, with 3 terminal O atoms.
3. 3 single N-O bonds use 6 electrons, 18 remaining.
4. Each O gets 6 lone electrons, using all remaining electrons.
5. N only has 6 valence electrons, so move one lone pair from an O to form a N=O double bond.
6. Add square brackets and ${}^{-}$ charge outside. Resonance structures exist, so all N-O bonds are equal length, intermediate between single and double bonds.

4. VSEPR and molecular shapes

Valence Shell Electron Pair Repulsion (VSEPR) theory states that electron domains (bonding pairs + lone pairs on the central atom) repel each other, so they arrange themselves as far apart as possible to minimise repulsion. Repulsion strength follows the order: $$\text{Lone pair-lone pair}>\text{Lone pair-bonding pair}>\text{Bonding pair-bonding pair}$$ Note: Single, double, and triple bonds all count as 1 electron domain each. Steps to determine molecular shape:

1. Draw the Lewis structure and identify the central atom.
2. Count total electron domains on the central atom to find the electron domain geometry:

   Electron domains	Electron domain geometry	Bond angle
   2	Linear	$180^{\circ }$
   3	Trigonal planar	$120^{\circ }$
   4	Tetrahedral	$109.5^{\circ }$
   5	Trigonal bipyramidal	$90^{\circ },120^{\circ },180^{\circ }$
   6	Octahedral	$90^{\circ },180^{\circ }$

3. Adjust for lone pairs to find molecular geometry: Lone pairs compress bond angles, as they exert stronger repulsion than bonding pairs.

Worked example: $N{H}_3$: Central N has 3 bonding pairs and 1 lone pair, so 4 total electron domains. Electron domain geometry is tetrahedral, but the lone pair compresses bond angles to ~$107^{\circ }$, giving a trigonal pyramidal molecular geometry. Exam tip: Examiners require you to state both electron domain geometry and molecular geometry when asked; you will lose half marks if you only provide one.

5. Polarity and intermolecular forces

Bond and molecular polarity

A covalent bond is polar if the electronegativity difference between the two atoms is between 0.4 and 1.7: electrons are shared unequally, creating a dipole moment with a partial positive charge (${\delta }^{+}$) on the less electronegative atom and partial negative charge (${\delta }^{-}$) on the more electronegative atom. A molecule is polar if it has a net dipole moment, meaning polar bonds are not symmetrically arranged so their dipoles cancel out. For example, $C{O}_2$ has polar C=O bonds but is linear, so dipoles cancel, making it nonpolar; $H_{2}O$ has polar O-H bonds and is bent, so dipoles do not cancel, making it polar.

Intermolecular forces (IMFs)

IMFs are weak attractive forces between molecules, responsible for the melting and boiling points of simple covalent substances. They are significantly weaker than covalent, ionic, or metallic bonds, and follow this strength order:

1. Hydrogen bonding (strongest): Occurs only when H is covalently bonded to N, O, or F (small, highly electronegative atoms). The ${\delta }^{+}$ H is attracted to a lone pair of electrons on a N/O/F atom of an adjacent molecule.
2. Dipole-dipole forces: Attractive forces between the ${\delta }^{+}$ end of one polar molecule and the ${\delta }^{-}$ end of another polar molecule.
3. London dispersion forces (weakest): Present in all molecules, caused by temporary dipoles from random electron movement. Strength increases with number of electrons (molar mass) and molecular surface area.

Worked example: Explain why $H_{2}O$ has a boiling point of $100^{\circ }C$ while $H_{2}S$ has a boiling point of $-60^{\circ }C$: Both are polar, so they have dipole-dipole and London dispersion forces, but $H_{2}O$ has hydrogen bonding between molecules, which is far stronger than the dipole-dipole forces in $H_{2}S$. More energy is required to overcome the stronger IMFs in $H_{2}O$, leading to a higher boiling point.

6. Hybridisation (HL)

Hybridisation is the mixing of atomic orbitals to form new, equal-energy hybrid orbitals suitable for pairing electrons to form covalent bonds. It is exclusive to HL content, and you can directly determine hybridisation from the number of electron domains on the central atom:

Unhybridised p orbitals overlap side-on to form pi bonds, while hybrid orbitals overlap end-on to form sigma bonds. A single bond is 1 sigma bond, a double bond is 1 sigma + 1 pi bond, and a triple bond is 1 sigma + 2 pi bonds. Worked example: Hybridisation of C in ethene ($C_2{H}_4$): Each C has 3 electron domains (2 single C-H bonds, 1 double C=C bond), so it is $s{p}^2$ hybridised. The double bond consists of 1 sigma bond from overlap of $s{p}^2$ orbitals, and 1 pi bond from overlap of unhybridised p orbitals perpendicular to the plane of the hybrid orbitals.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Stating ionic compounds conduct electricity in solid state. Why: Students remember ionic compounds conduct, but forget ions are fixed in a lattice in solid form. Correct move: Only state ionic compounds conduct when molten or dissolved in water, where ions are free to move.
• Wrong move: Counting double/triple bonds as multiple electron domains for VSEPR. Why: Confusing number of bonding pairs with number of electron domains. Correct move: All bonding pairs (single, double, triple) count as 1 electron domain each, regardless of bond order.
• Wrong move: Assuming all molecules with polar bonds are polar. Why: Forgetting symmetric arrangement cancels dipole moments. Correct move: Always check molecular geometry first; if polar bonds are symmetrically arranged, the molecule is nonpolar.
• Wrong move: Identifying hydrogen bonding in molecules where H is bonded to C, Cl, or other atoms not N/O/F. Why: Overgeneralising the hydrogen bond rule. Correct move: Only state hydrogen bonding exists if H is covalently bonded to N, O, or F, and there is a lone pair on N/O/F in an adjacent molecule.
• Wrong move: Drawing Lewis structures for period 2 elements with more than 8 valence electrons. Why: Confusing period 3 expanded octet rules with period 2, which has no d orbitals to expand its octet. Correct move: Period 2 elements (Li to Ne) can never have more than 8 valence electrons in Lewis structures.

8. Practice Questions (IB Chemistry HL Style)

Question 1 (Paper 1, 1 mark)

Which of the following lists the intermolecular forces between $C{H}_{3}C{H}_{2}OH$, $C{H}_{3}C{H}_{2}C{H}_3$ and $C{H}_{3}CHO$ in order of increasing strength? A. $C{H}_{3}C{H}_{2}C{H}_3<C{H}_{3}CHO<C{H}_{3}C{H}_{2}OH$ B. $C{H}_{3}CHO<C{H}_{3}C{H}_{2}C{H}_3<C{H}_{3}C{H}_{2}OH$ C. $C{H}_{3}C{H}_{2}OH<C{H}_{3}CHO<C{H}_{3}C{H}_{2}C{H}_3$ D. $C{H}_{3}C{H}_{2}C{H}_3<C{H}_{3}C{H}_{2}OH<C{H}_{3}CHO$

Worked solution: First identify IMFs for each compound:

1. $C{H}_{3}C{H}_{2}C{H}_3$ (propane) is nonpolar, so only has London dispersion forces.
2. $C{H}_{3}CHO$ (ethanal) is polar, so has dipole-dipole + London dispersion forces.
3. $C{H}_{3}C{H}_{2}OH$ (ethanol) has an O-H bond, so has hydrogen bonding + dipole-dipole + London dispersion forces. IMF strength order: LDF < dipole-dipole < H-bond, so the correct order is A. (1 mark awarded for A)

Question 2 (Paper 2, 3 marks)

(a) Draw the Lewis structure of the sulfite ion, $S{O}_3^{2-}$. (2 marks) (b) State the hybridisation of the central S atom. (1 mark)

Worked solution: (a) Step 1: Total valence electrons = $6+(3\times 6)+2=26$. Step 2: S is central, 3 O terminal. Step 3: 3 single S-O bonds use 6 electrons, 20 remaining. Step 4: Each O gets 6 lone electrons (18 total used), remaining 2 electrons form a lone pair on S. Step 5: Add square brackets and ${}^{2-}$ charge outside. (2 marks awarded for correct electron count, lone pair on S, and charge notation) (b) S has 4 electron domains (3 bonding pairs, 1 lone pair), so hybridisation is $s{p}^3$. (1 mark awarded for $s{p}^3$)

Question 3 (Paper 2, 4 marks)

Explain why the boiling point of magnesium ($650^{\circ }C$) is significantly higher than that of sodium ($98^{\circ }C$), and both are higher than that of phosphorus ($P_4$, $280^{\circ }C$).

Worked solution:

1. Mg and Na are metallic solids: $M{g}^{2+}$ has a higher charge (+2 vs +1) and smaller ionic radius than $N{a}^{+}$, so electrostatic attraction between Mg cations and delocalised electrons is stronger than in Na, requiring more energy to overcome, leading to a higher boiling point. (2 marks)
2. $P_4$ is a simple molecular solid, held together by weak London dispersion forces between $P_4$ molecules. These forces are much weaker than the strong metallic bonds in Na and Mg, so less energy is required to overcome them, leading to a lower boiling point. (2 marks) Total: 4 marks.

9. Quick Reference Cheatsheet

10. What's Next

This topic is foundational to almost all subsequent IB Chemistry HL content: you will use bonding principles to explain organic reaction mechanisms, the acidity and basicity of organic and inorganic compounds, the properties of polymers and nanomaterials in Option B, and enthalpy changes of reaction in the Energetics topic. A solid grasp of hybridisation is also required to understand delocalised pi systems in aromatic compounds and conjugated polymers, which are frequently tested in Paper 3.

If you struggle with any of the concepts in this guide, or want to practise more exam-style questions tailored to your weak areas, you can ask Ollie, our AI tutor, for personalised help at any time. You can also find more study guides and past paper walkthroughs for IB Chemistry HL on the homepage.