Chemical Bonding — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: Ionic, covalent and metallic bonding, Lewis dot structures with dative bonds, VSEPR molecular geometry, electronegativity and bond polarity, and intermolecular forces including hydrogen bonds, dipole-dipole and London dispersion forces.

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 A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Chemical Bonding?

Chemical bonding is the strong electrostatic attraction between subatomic or atomic particles (ions, atomic nuclei, shared electrons, delocalised electrons) that holds chemical substances together in stable, low-energy configurations. All bonds form to allow atoms to achieve a more stable electronic state, usually a full outer valence shell of electrons. This is a core A-Level Chemistry topic tested across Papers 1, 2, and 4, accounting for 20-25% of AS level marks and often integrated with other topics like periodicity, organic chemistry, and energetics.

2. Ionic, covalent, metallic bonding

All three primary bond types are based on electrostatic attraction, but differ in the particles involved:

1. Ionic bonding: Electrostatic attraction between oppositely charged cations (positive, formed by electron loss from metals) and anions (negative, formed by electron gain from non-metals). Ionic compounds form giant 3D lattices, have high melting/boiling points, conduct electricity only when molten or aqueous (when ions are free to move), and are often soluble in polar solvents.
2. Covalent bonding: Electrostatic attraction between a shared pair of valence electrons and the positively charged nuclei of the two bonded atoms, formed between non-metal atoms. Covalent substances can be simple molecular (low melting points, no electrical conductivity) or giant covalent lattices (very high melting points, e.g. diamond, SiO₂).
3. Metallic bonding: Electrostatic attraction between a lattice of positively charged metal ions and a "sea" of delocalised valence electrons. Metals have high melting points, are malleable/ductile (layers of ions slide over each other without breaking bonds), and conduct electricity in all states due to mobile delocalised electrons.

Worked example: State the bonding type in aluminium and explain why it has a higher melting point than sodium. Solution: Aluminium has metallic bonding. Each Al atom loses 3 valence electrons to form Al³⁺ ions, while Na only loses 1 electron to form Na⁺. The higher charge on Al ions and higher number of delocalised electrons create stronger electrostatic attraction, so more energy is required to break the metallic lattice, leading to a higher melting point.

3. Lewis dot structures and dative bonds

Lewis dot structures are 2D representations of valence electrons in molecules, using dots/crosses to represent electrons, lines for shared bonding pairs, and pairs of dots for non-bonding lone pairs. Follow these steps to draw them:

1. Calculate total valence electrons for all atoms, adjusting for ionic charge (add electrons for negative charge, subtract for positive charge).
2. Place the least electronegative atom (except H) as the central atom, surrounded by outer atoms.
3. Form single covalent bonds between the central and outer atoms, using 2 electrons per bond.
4. Fill the octet (8 valence electrons, except H which only needs 2) of all outer atoms first, then fill the octet of the central atom.
5. If the central atom has fewer than 8 electrons, form double or triple bonds by moving lone pairs from outer atoms to form shared pairs. Period 3 elements and below can have expanded octets (more than 8 electrons) as they have available d-orbitals.

A dative (coordinate) covalent bond is a covalent bond where both electrons in the shared pair come from one donor atom, while the acceptor atom contributes no electrons. It is represented by an arrow pointing from the donor to the acceptor. Once formed, dative bonds are identical in strength and length to regular covalent bonds. A common example is the formation of ammonium: the N atom in NH₃ donates its lone pair to an H⁺ ion (which has no electrons) to form NH₄⁺.

Worked example: Draw the Lewis structure for the hydroxonium ion H₃O⁺, including any dative bonds. Solution: Total valence electrons: O has 6, 3 H have 1 each, subtract 1 for +1 charge: total = 6 + 3 -1 = 8. Central O forms 2 regular covalent bonds to 2 H atoms, 1 dative bond from O to H⁺, with 1 remaining lone pair on O. The structure is drawn with O at the centre, 3 single bonds to H, one lone pair on O, and a +1 charge on the overall ion.

4. Shapes of molecules — VSEPR theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the 3D shape of molecules by stating that electron domains (bonding pairs or lone pairs) around a central atom repel each other, so arrange themselves as far apart as possible to minimise repulsion. Repulsion strength follows the order: $$\text{Lone pair-lone pair}>\text{Lone pair-bonding pair}>\text{Bonding pair-bonding pair}$$ To determine molecular shape:

1. Calculate the steric number of the central atom: $\text{Steric number}=\frac{\text{Valence electrons of central atom + number of monovalent bonded atoms ± ionic charge}}{2}$
2. Assign electron geometry based on steric number: 2 = linear, 3 = trigonal planar, 4 = tetrahedral, 5 = trigonal bipyramidal, 6 = octahedral.
3. Subtract the number of lone pairs from the steric number to assign molecular geometry, adjusting bond angles for the higher repulsion of lone pairs.

Worked example: State and explain the shape of the H₂S molecule, including its approximate bond angle. Solution: Central S has 6 valence electrons, 2 monovalent H atoms, no charge. Steric number = (6 + 2)/2 = 4, so electron geometry is tetrahedral. Number of lone pairs = 4 - 2 bonding pairs = 2. The two lone pairs repel the bonding pairs more strongly than bonding pairs repel each other, so the bond angle is reduced from the tetrahedral 109.5° to ~104.5°, giving a bent (V-shaped) molecular geometry.

5. Electronegativity and bond polarity

Electronegativity is the ability of an atom in a covalent bond to attract the shared pair of electrons to itself, measured on the Pauling scale (0.7 for Cs, the least electronegative, 4.0 for F, the most electronegative). The difference in electronegativity ($\Delta EN$) between two bonded atoms determines bond type:

• $\Delta EN<0.5$: Non-polar covalent bond, electrons shared almost equally
• $0.5<\Delta EN<1.7$: Polar covalent bond, electrons shared unequally, creating partial charges $\delta +$ on the less electronegative atom and $\delta -$ on the more electronegative atom, forming a bond dipole
• $\Delta EN>1.7$: Ionic bond, electrons transferred fully

A molecule is polar if it has a net overall dipole, which depends on both bond polarity and molecular symmetry. Even if a molecule has polar bonds, it will be non-polar if the bond dipoles cancel out due to symmetric geometry (e.g. CO₂, CCl₄).

Worked example: Explain why CCl₄ is non-polar even though it contains polar C-Cl bonds. Solution: The C-Cl bond has $\Delta EN=3.2-2.5=0.7$, so it is polar, with a dipole pointing from C to Cl. CCl₄ has a tetrahedral geometry, so the four C-Cl dipoles are arranged symmetrically, pointing in opposite directions relative to each other. The dipoles cancel out, giving no net dipole, so the molecule is non-polar.

6. Intermolecular forces — H-bonds, dipole-dipole, London

Intermolecular forces are weak electrostatic attractions between molecules, much weaker than intramolecular covalent/ionic/metallic bonds. They determine physical properties like boiling point, solubility, and viscosity:

1. London dispersion forces (instantaneous dipole-induced dipole): Present in all atoms and molecules, caused by temporary uneven distribution of electrons creating a transient dipole, which induces a dipole in adjacent particles. Strength increases with increasing number of electrons (molar mass) and increasing molecular surface area.
2. Dipole-dipole forces: Attractions between permanent dipoles of polar molecules, stronger than London forces for molecules of similar molar mass.
3. Hydrogen bonds: A special, very strong type of dipole-dipole interaction, formed when a hydrogen atom covalently bonded to N, O, or F (highly electronegative, small atoms) is attracted to a lone pair of electrons on a N, O, or F atom of an adjacent molecule. Hydrogen bonds are ~10x stronger than other intermolecular forces, causing significantly higher boiling points for compounds that exhibit them (e.g. H₂O, NH₃, ethanol).

Worked example: Arrange the following in order of increasing boiling point: Cl₂, Br₂, ICl, H₂O. Justify your answer. Solution: Order: Cl₂ < Br₂ < ICl < H₂O. Cl₂ and Br₂ are non-polar, so only have London forces. Br₂ has more electrons than Cl₂, so stronger London forces, higher boiling point. ICl is polar, so has dipole-dipole forces plus London forces, stronger than the London forces in Br₂ (similar molar mass), so higher boiling point. H₂O has hydrogen bonds, the strongest intermolecular force, so highest boiling point.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Counting double or triple bonds as multiple electron domains for VSEPR. Why students do it: Confusing number of electron pairs with number of domains. Correct move: All bonding pairs (single, double, triple) count as one electron domain each; lone pairs count as one domain each.
• Wrong move: Stating that dative covalent bonds are stronger than regular covalent bonds. Why students do it: Assuming the source of electrons affects bond strength. Correct move: Once formed, dative bonds are identical in all properties (strength, length) to regular covalent bonds.
• Wrong move: Claiming that covalent bonds are broken when a simple covalent substance boils. Why students do it: Confusing intermolecular and intramolecular forces. Correct move: Boiling only breaks weak intermolecular forces, not strong intramolecular covalent bonds.
• Wrong move: Identifying any polar molecule with a H atom as having hydrogen bonds. Why students do it: Forgetting the strict requirements for H bonding. Correct move: H bonds only form if H is covalently bonded to N, O or F, and is attracted to a lone pair on N, O or F of another molecule.
• Wrong move: Assuming all molecules with polar bonds are polar. Why students do it: Forgetting to check molecular symmetry. Correct move: Always check if bond dipoles cancel due to symmetric geometry before assigning molecular polarity.

8. Practice Questions (A-Level Chemistry Style)

Question 1 (3 marks)

Magnesium fluoride (MgF₂) has a melting point of 1263°C, while sodium fluoride (NaF) has a melting point of 993°C. State the type of bonding in both compounds, and explain the difference in melting points.

Solution: Both compounds have ionic bonding (1 mark). Mg²⁺ ions have a higher charge than Na⁺ ions, so the electrostatic attraction between Mg²⁺ and F⁻ ions is stronger than between Na⁺ and F⁻ (1 mark). More energy is required to overcome the stronger ionic bonds in MgF₂, so it has a higher melting point (1 mark).

Question 2 (4 marks)

(a) Draw the Lewis structure for the nitrate ion NO₃⁻. (2 marks) (b) State the shape and bond angle of the nitrate ion. (2 marks)

Solution: (a) Central N atom, one double bond to one O, single bonds to the other two O atoms, with a negative charge on each single-bonded O, and overall -1 charge. All O atoms have full octets, N has 8 electrons, total valence electrons = 5 + 3*6 +1 = 24 (2 marks, 1 mark for correct bonding, 1 for correct charge/lone pairs). (b) Steric number of N is 3, no lone pairs, so trigonal planar shape with 120° bond angle (2 marks, 1 for shape, 1 for angle).

Question 3 (3 marks)

Explain why ethanoic acid (CH₃COOH) has a higher boiling point than ethanol (C₂H₅OH), even though both have similar molar masses and both form hydrogen bonds.

Solution: Both molecules form hydrogen bonds via their -OH groups (1 mark). Ethanoic acid can form hydrogen-bonded dimers, where two ethanoic acid molecules form two hydrogen bonds between their carboxylic acid groups, effectively doubling the molecular mass of the species present (1 mark). Stronger overall intermolecular forces require more energy to overcome, so ethanoic acid has a higher boiling point (1 mark).

9. Quick Reference Cheatsheet

10. What's Next

Chemical bonding is a foundational topic that connects to almost every other part of the A-Level Chemistry syllabus. You will use your understanding of ionic lattices to explain enthalpy changes of solution and solubility in physical chemistry, covalent bonding to predict the reactivity of organic functional groups, and intermolecular forces to explain trends in periodic properties and the boiling points of organic homologous series. Mastering this topic is non-negotiable for scoring well in both AS and A Level papers, as questions on bonding are often integrated with other topics to test higher-order thinking skills.

If you are struggling with any part of this guide, from drawing Lewis structures to identifying intermolecular forces, you can ask Ollie, our AI tutor, for personalised explanations, extra practice questions, or feedback on your answers. You can also find more topic guides and full mock exams aligned to the A-Level Chemistry syllabus on the homepage to help you prepare for your exams.

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