Bonding and Structure Problem Solving

This lesson brings together everything you have learned about bonding and structure into a problem-solving framework. A-Level Chemistry exams frequently ask you to draw structures, predict shapes, explain properties, and compare substances. The key to success is a systematic approach that connects structure to properties through clear reasoning.

Drawing Dot-and-Cross Diagrams for Complex Molecules

Dot-and-cross diagrams (Lewis structures) are the starting point for understanding bonding. Here is a reliable method for drawing them correctly:

Step-by-Step Method

Count the total number of valence electrons. Sum the outer shell electrons of all atoms. For ions, add electrons for negative charges and subtract for positive charges.
Identify the central atom. This is usually the least electronegative atom (excluding hydrogen, which is always terminal).
Place single bonds between the central atom and each surrounding atom. Each bond uses 2 electrons.
Distribute remaining electrons as lone pairs. Start with the outer atoms (give each atom an octet, or 2 for hydrogen). Then place any leftover electrons on the central atom.
Check octets. If the central atom has fewer than 8 electrons, convert lone pairs on outer atoms into double or triple bonds to the central atom.
Check for expanded octets. If the central atom is in Period 3 or below and has more than 8 electrons, this is acceptable.

Worked Example: SO₃ (Sulfur Trioxide)

Valence electrons: S = 6, each O = 6, total = 6 + 3(6) = 24
Central atom: S (less electronegative than O)
Three S–O single bonds: uses 6 electrons, leaving 18
Give each O three lone pairs (completing octets): uses 18 electrons. S has only 6 electrons — needs more.
Convert one lone pair from an O into a double bond: S now has 8 electrons. Final structure has two S–O single bonds and one S=O double bond.

Note: In reality, SO₃ has resonance structures where the double bond can be in any of the three positions, but for A-Level one valid Lewis structure is sufficient.

Worked Example: NO₃⁻ (Nitrate Ion)

Valence electrons: N = 5, each O = 6, plus 1 for the charge = 5 + 3(6) + 1 = 24
Central atom: N
Three N–O single bonds: uses 6, leaving 18
Give each O three lone pairs: uses 18. N has only 6 electrons.
Convert one O lone pair to a N=O double bond: N now has 8 electrons. Structure has one N=O double bond and two N–O single bonds, with an overall charge of 1−.

Worked Example: XeF₄

Valence electrons: Xe = 8, each F = 7, total = 8 + 4(7) = 36
Central atom: Xe (least electronegative)
Four Xe–F single bonds: uses 8 electrons, leaving 28
Give each F three lone pairs (completing octets): uses 24 electrons. Remaining: 4 electrons = 2 lone pairs on Xe.
Xe has 12 electrons around it — this is an expanded octet, acceptable for Period 5 element.
Shape: 6 electron domains (4 bp + 2 lp) → square planar.

Predicting Shapes and Bond Angles

Once you have the dot-and-cross diagram, predicting shape follows directly:

Count electron domains around the central atom (lone pairs + bonding regions; double/triple bonds count as one domain)
Determine the base geometry (linear, trigonal planar, tetrahedral, etc.)
Account for lone pairs (which alter the visible shape)
Adjust bond angles for lone pair effects

Common Exam Molecules

Molecule	Electron domains	Bonding pairs	Lone pairs	Shape	Bond angle
CO₂	2	2	0	Linear	180°
BF₃	3	3	0	Trigonal planar	120°
CH₄	4	4	0	Tetrahedral	109.5°
NH₃	4	3	1	Pyramidal	107°
H₂O	4	2	2	Bent	104.5°
PCl₅	5	5	0	Trigonal bipyramidal	90°, 120°
SF₆	6	6	0	Octahedral	90°
XeF₂	5	2	3	Linear	180°
ClF₃	5	3	2	T-shaped	~87.5°
XeF₄	6	4	2	Square planar	90°

Predicting Polarity: The Complete Method

Predicting whether a molecule is polar requires both shape and electronegativity information. Here is the systematic approach:

Determine if the bonds are polar. Check the electronegativity difference. If ΔEN = 0, the bond is non-polar.
Determine the molecular shape using VSEPR.
Check for symmetry. If all bond dipoles are identical AND the shape is symmetric, the dipoles cancel → non-polar molecule. If the shape is asymmetric (due to lone pairs or mixed atoms), dipoles do NOT cancel → polar molecule.

Quick Polarity Reference

Molecule	Shape	Identical bonds?	Symmetric?	Polar?
CO₂	Linear	Yes (2 C=O)	Yes	No
H₂O	Bent	Yes (2 O–H)	No (lone pairs)	Yes
CCl₄	Tetrahedral	Yes (4 C–Cl)	Yes	No
CHCl₃	Tetrahedral	No (3 C–Cl, 1 C–H)	No	Yes
BF₃	Trigonal planar	Yes (3 B–F)	Yes	No
NF₃	Pyramidal	Yes (3 N–F)	No (lone pair)	Yes
SF₆	Octahedral	Yes (6 S–F)	Yes	No
SF₄	Seesaw	Yes (4 S–F)	No (lone pair)	Yes
PCl₅	Trigonal bipyramidal	Yes (5 P–Cl)	Yes	No
PCl₃	Pyramidal	Yes (3 P–Cl)	No (lone pair)	Yes

Explaining Physical Properties from Structure

This is where the marks are in exams. You need to make a clear chain of reasoning:

Structure → Type of forces → Strength of forces → Physical property

Template for Explaining Boiling Points

"Substance X has a [low/high] boiling point because it has a [structure type] structure. The [type of intermolecular force/bonding] between [molecules/ions/atoms] are [weak/strong]. [More/less] energy is required to overcome these forces, so the boiling point is [low/high]."

Example: Why does NaCl have a higher melting point than I₂?

"NaCl has a giant ionic structure with strong electrostatic attraction between Na⁺ and Cl⁻ ions throughout the lattice. I₂ is a simple molecular substance with only weak London dispersion forces between I₂ molecules. Much more energy is required to overcome the strong ionic bonds in NaCl than the weak London forces in I₂, so NaCl has a much higher melting point."

Notice the structure: identify both structures, name the specific forces, compare the strengths, and link to the property.

Comparing Boiling Points

Boiling point comparisons are among the most common exam questions. Here is a systematic approach:

Substances with Different Structure Types

Giant ionic/covalent/metallic > Simple molecular (almost always). The forces within giant structures are much stronger than intermolecular forces.

Simple Molecular Substances with Different Intermolecular Forces

Hydrogen bonding > Permanent dipole–dipole > London forces (for molecules of similar size). Always check whether hydrogen bonding is possible first.

Simple Molecular Substances with the Same Intermolecular Forces

Larger molecules (more electrons, more surface area) have stronger London forces and higher boiling points. Compare molecular mass and shape (branching reduces surface area).

Comprehensive Worked Example: Five Substances

Consider the following five substances: NaCl, SiO₂, I₂, H₂O, Cu

Arrange in order of increasing melting point and explain each:

H₂O (0°C) — Simple molecular. Strong hydrogen bonds between molecules, but still much weaker than giant structure bonds. The relatively high mp for a molecule of Mr = 18 is due to extensive H-bonding.
I₂ (114°C) — Simple molecular. Only London forces, but I₂ has 106 electrons creating a very polarisable electron cloud. The cumulative London forces are strong enough to make it a solid, with a higher mp than water despite lacking H-bonding.
NaCl (801°C) — Giant ionic. Strong electrostatic attraction between Na⁺ and Cl⁻ ions throughout the lattice requires substantial energy to overcome.
Cu (1085°C) — Metallic. Cu²⁺ ions with 2 delocalised electrons per atom and relatively small ionic radius give strong metallic bonding. Transition metals also contribute d electrons.
SiO₂ (1713°C) — Giant covalent. Very strong Si–O covalent bonds extend throughout the entire 3D network. Must break many covalent bonds to melt.

The correct order: H₂O (0°C) < I₂ (114°C) < NaCl (801°C) < Cu (1085°C) < SiO₂ (1713°C)

Common Exam Mistakes to Avoid

"Strong bonds between molecules" — Intermolecular forces are not bonds. Use "forces between molecules" or "intermolecular forces."
"Breaking covalent bonds when boiling" — Boiling a simple molecular substance does not break covalent bonds. Only intermolecular forces are overcome.
"NaCl has a high melting point because of strong covalent bonds" — NaCl is ionic, not covalent. Use "strong electrostatic attraction between ions."
"Diamond has strong van der Waals forces" — Diamond is a giant covalent structure. There are no intermolecular forces because there are no separate molecules. The strong C–C covalent bonds throughout the structure explain its high melting point.
Forgetting to specify the type of force — Do not just say "strong forces." Specify: London dispersion forces, permanent dipole–dipole, hydrogen bonding, ionic bonding, covalent bonding, or metallic bonding.
Confusing conductivity explanations — Metals conduct due to delocalised electrons. Ionic compounds conduct when molten due to mobile ions. Graphite conducts due to delocalised electrons in layers. Never mix these up.
"Hydrogen bonding is the strongest force so it always gives the highest boiling point" — This ignores molecular size. Large non-polar molecules (I₂) can have stronger total intermolecular forces than small molecules with H-bonding (NH₃).
"CO₂ is non-polar because carbon is not very electronegative" — CO₂ has very polar C=O bonds (ΔEN = 1.0). It is non-polar because the linear shape causes the bond dipoles to cancel.

A Structured Approach for Any Problem

When faced with a bonding and structure question, follow this checklist:

Identify the structure type (simple molecular, giant ionic, giant covalent, metallic)
Identify the specific forces or bonds (London, dipole–dipole, hydrogen bonding, ionic, covalent, metallic)
Assess the strength (considering charge, size, number of electrons, number of hydrogen bonds, etc.)
Link to the property being asked about (melting point, conductivity, solubility, hardness)
Use comparative language when comparing substances (stronger than, weaker than, more energy required, less energy required)

Final Exam Checklist: Common Mark-Scheme Points

Property to explain	Key phrases examiners look for
High mp (ionic)	"Strong electrostatic attraction between oppositely charged ions throughout the lattice"
High mp (giant covalent)	"Many strong covalent bonds throughout the structure that must be broken"
High mp (metallic)	"Strong electrostatic attraction between positive ions and delocalised electrons"
Low mp (molecular)	"Weak intermolecular forces (name the type) between molecules"
Conducts (metal)	"Delocalised electrons free to move and carry charge"
Conducts (molten ionic)	"Ions free to move and carry charge"
Does not conduct (molecular)	"No free ions or delocalised electrons; no mobile charge carriers"
Soluble in water (ionic)	"Ion-dipole interactions between ions and polar water molecules overcome the lattice energy"
Insoluble in water (non-polar)	"Cannot form strong enough interactions with water to compensate for breaking water–water hydrogen bonds"

A-Level Deep Dive: Bonding and Structure Problem Solving

Spec mapping

This synoptic capstone draws on Edexcel 9CH0 Topic 2 in its entirety — Bonding and Structure (sub-topics 2.1 through 2.7): ionic bonding and giant ionic lattices; covalent bonding (including dative); molecular shapes via electron-pair repulsion; electronegativity and bond polarity; the four bulk structure types (giant ionic, giant covalent, simple molecular, giant metallic); intermolecular forces (London/dispersion, permanent dipole–dipole, hydrogen bonding); and the structure–property bridge that lets you predict melting point, conductivity, solubility and hardness from a structural model. Cross-paper, the same toolkit is examined throughout Paper 1 (period 3 chlorides and oxides in Topic 4 inorganic), throughout Paper 2 (every organic compound in Topics 6, 9, 16–17 is simple molecular and lives or dies by intermolecular forces), and explicitly in Paper 3 (synoptic, where unknown-substance identification from physical-property data is a recurring style). Topics 8 (energetics, lattice and hydration enthalpies), 12 (acid–base equilibria, hydroxide basicity), and 18–19 (analytical spectroscopy) all import bonding-and-structure reasoning wholesale (refer to the official Pearson Edexcel specification document for exact wording).

Worked example with full mark scheme

Question (12 marks): The chlorides of the period 3 elements include NaCl, MgCl $_2$ , AlCl $_3$ (also written as the dimer Al $_2$ Cl $_6$ ), SiCl $_4$ , PCl $_5$ , S $_2$ Cl $_2$ and Cl $_2$ .

(a) Classify each chloride as ionic, polar covalent or essentially non-polar covalent. (3)

(b) Predict and explain the trend in melting point across the period from NaCl to Cl $_2$ . (3)

(c) Predict the approximate pH of the aqueous solution that results when each of NaCl, MgCl $_2$ , AlCl $_3$ , SiCl $_4$ and PCl $_5$ is added to water, justifying each prediction in terms of bonding. (4)

(d) Draw a dot-and-cross diagram or shape diagram for one of the covalent chlorides above and justify the shape and bond angle. (2)

Solution with full mark scheme:

(a) NaCl is ionic (Na has low electronegativity $\chi \approx 0.9$ ; $\Delta\chi > 2$ with Cl). MgCl $_2$ is predominantly ionic but with measurable covalent character ( $\Delta\chi \approx 1.9$ , the small Mg $^{2+}$ polarises Cl $^-$ ). AlCl $_3$ is polar covalent despite involving a metal — Al $^{3+}$ is so polarising that the Al–Cl bond becomes covalent, and the substance forms covalent dimers Al $_2$ Cl $_6$ in the vapour. SiCl $_4$ , PCl $_5$ , S $_2$ Cl $_2$ are all polar covalent molecular compounds (each Si/P/S–Cl bond is polar but molecular geometry determines overall dipole). Cl $_2$ is non-polar covalent ( $\Delta\chi = 0$ ).

B1 for correctly placing NaCl, MgCl $_2$ as ionic (or NaCl ionic, MgCl $_2$ ionic-with-character).
B1 for correctly identifying AlCl $_3$ as covalent / polar covalent, citing high charge density of Al $^{3+}$ .
B1 for correctly classifying SiCl $_4$ , PCl $_5$ , S $_2$ Cl $_2$ as polar covalent molecular and Cl $_2$ as non-polar covalent.

(b) The trend is high (NaCl ~801 °C, MgCl $_2$ ~714 °C) — falls sharply at AlCl $_3$ (sublimes ~180 °C) — low and decreasing for SiCl $_4$ (b.p. 57 °C), PCl $_5$ (sublimes 167 °C), S $_2$ Cl $_2$ (b.p. 137 °C) and Cl $_2$ (b.p. −34 °C).

M1 — recognising the structural transition from giant ionic lattice (NaCl, MgCl $_2$ ) to molecular (AlCl $_3$ onwards).
M1 — explaining that breaking a giant ionic lattice requires overcoming many strong electrostatic attractions throughout the structure, whereas breaking a molecular lattice only requires overcoming weak intermolecular forces.
A1 — within the molecular series, m.p./b.p. tracks molecular size and shape (number of electrons → strength of London forces), with PCl $_5$ exceptional because it sublimes via an ionic [PCl $_4$ ] $^+$ [PCl $_6$ ] $^-$ phase.

(c) NaCl in water gives a neutral solution (pH ≈ 7) — the Na $^+$ and Cl $^-$ ions are neither significantly acidic nor basic. MgCl $_2$ gives a slightly acidic solution (pH ≈ 5–6) because the small, highly charged Mg $^{2+}$ aqua ion [Mg(H $_2$ O) $_6$ ] $^{2+}$ polarises coordinated water and releases H $^+$ . AlCl $_3$ hydrolyses violently in water giving a strongly acidic solution (pH ≈ 2–3) — Al $^{3+}$ has very high charge density and the [Al(H $_2$ O) $_6$ ] $^{3+}$ ion is a strong acid; HCl gas is also evolved. SiCl $_4$ and PCl $_5$ undergo complete covalent hydrolysis: SiCl $_4$ + 2H $_2$ O → SiO $_2$ + 4HCl (pH ≈ 1–2); PCl $_5$ + 4H $_2$ O → H $_3$ PO $_4$ + 5HCl (strongly acidic).

B1 — NaCl neutral; MgCl $_2$ slightly acidic via aqua-ion polarisation.
B1 — AlCl $_3$ strongly acidic with explicit reference to charge density of Al $^{3+}$ .
M1 — SiCl $_4$ and PCl $_5$ hydrolyse to give acidic solutions.
A1 — at least one balanced hydrolysis equation written out (SiCl $_4$ + 2H $_2$ O → SiO $_2$ + 4HCl, or equivalent for PCl $_5$ ).

(d) PCl $_5$ in the gas phase is trigonal bipyramidal (5 bonding pairs, 0 lone pairs on P; bond angles 120° equatorial and 90° axial–equatorial). Justification: the central P atom has 5 bonding pairs, which arrange themselves to minimise mutual electron-pair repulsion; trigonal bipyramidal geometry maximises bond-pair separations. (Equally acceptable: SiCl $_4$ tetrahedral with 109.5° bond angles, 4 bonding pairs and 0 lone pairs on Si.)

M1 — correct shape with bond angle(s) labelled.
A1 — justification using electron-pair repulsion; "lone-pair repulsion > bond-pair repulsion" cited if a lone pair is present in the chosen example.

Total: 12 marks (3 + 3 + 4 + 2). Synoptic across sub-topics 2.1, 2.3, 2.5 and links forward to Topic 4 (period-3 trends) and Topic 12 (acid–base behaviour).

Specimen question modelled on the Edexcel 9CH0 Paper 3 format

Question (9 marks): A student is given three unknown white solids labelled X, Y and Z, with the following data:

Test	X	Y	Z
Melting point	801 °C	1610 °C	113 °C
Conductivity (solid)	none	none	none
Conductivity (molten)	high	none	none
Solubility in water	high	insoluble	insoluble
Solubility in hexane	insoluble	insoluble	high
Hardness	brittle crystalline	very hard	soft, waxy
IR spectrum	featureless	featureless	strong absorption near 1700 cm $^{-1}$

Identify the bonding/structure type of each substance and suggest a possible identity. Justify each identification by explicit reference to the data. (9)

Mark scheme decomposition by AO: