Bonding

:::info Board Coverage AQA Paper 1 | Edexcel Paper 1 | OCR A Gateway C2 | WJEC C2 :::

1. Ionic Bonding

1.1 Formation of Ions

Ionic bonding occurs between metals and non-metals. Metal atoms lose electrons to form positive Ions (cations), and non-metal atoms gain electrons to form negative ions (anions). The oppositely Charged ions are attracted to each other by strong electrostatic forces.

The driving force is the achievement of a full outer shell. A metal atom in Group 1 loses one Electron to attain the electron configuration of the previous noble gas. A non-metal atom in Group 7 Gains one electron to attain the configuration of the next noble gas. Both species achieve a lower Energy state, and the energy released when the ions come together (the lattice energy) is what holds The ionic compound together.

Example: Sodium chloride (NaCl)

$\mathrm{Na \to \mathrm{Na^+ + e^-$ $\mathrm{Cl + e^- \to \mathrm{Cl^-$ $\mathrm{Na^+ + \mathrm{Cl^- \to \mathrm{NaCl$

Sodium (Group 1) loses one electron to achieve a full outer shell (like neon). Chlorine (Group 7) Gains one electron to achieve a full outer shell (like argon).

1.2 Ionic Lattice Structure

Ionic compounds form giant ionic lattices in which positive and negative ions are arranged in a Regular, repeating pattern. The electrostatic attraction between all the ions acts in all Directions, giving ionic compounds high melting and boiling points.

In sodium chloride, each Na $^+$ ion is surrounded by six Cl $^-$ ions, and each Cl $^-$ ion is Surrounded by six Na $^+$ ions. This is a 6:6 coordination. The lattice extends in three dimensions — it is not a collection of discrete NaCl molecules.

Properties of ionic compounds:

Property	Explanation
High melting/boiling point	Strong electrostatic forces between ions require a lot of energy to overcome
Conduct electricity when molten or dissolved	Ions are free to move and carry charge
Do NOT conduct when solid	Ions are fixed in position in the lattice
Brittle	A force displaces layers of ions, bringing like charges together, causing repulsion
soluble in water	Water molecules attract and separate the ions

The brittleness of ionic compounds has a clear structural explanation. When a force is applied, one Layer of ions shifts relative to another. Ions of the same charge are brought into proximity, and The resulting electrostatic repulsion causes the crystal to shatter. This is fundamentally different From the response of metals, where layers can slide past each other without breaking bonds.

1.3 Dot and Cross Diagrams

Dot and cross diagrams show the transfer of electrons in ionic bonding. The original electrons of Each atom are shown as dots or crosses, and transferred electrons are shown in the colour/symbol of The receiving atom.

Example: Magnesium oxide (MgO)

Magnesium ( $2, 8, 2$ ) loses 2 electrons to form Mg $^{2+}$ . Oxygen ( $2, 6$ ) gains 2 electrons to form O $^{2-}$ .

$\mathrm{Mg + \mathrm{O \to \mathrm{Mg^{2+} + \mathrm{O^{2-}$

Example: Calcium chloride (CaCl $_2$ )

Calcium ( $2, 8, 8, 2$ ) loses 2 electrons. Each chlorine ( $2, 8, 7$ ) gains 1 electron. Two chlorine Atoms are needed.

$\mathrm{Ca + 2\mathrm{Cl \to \mathrm{Ca^{2+} + 2\mathrm{Cl^-$

Example: Sodium oxide (Na $_2$ O)

Each sodium ( $2, 8, 1$ ) loses 1 electron. Oxygen ( $2, 6$ ) gains 2 electrons. Two sodium atoms are Needed.

$2\mathrm{Na + \mathrm{O \to 2\mathrm{Na^+ + \mathrm{O^{2-}$

Example: Aluminium oxide (Al $_2$ O $_3$ )

Each aluminium ( $2, 8, 3$ ) loses 3 electrons to form Al $^{3+}$ . Each oxygen gains 2 electrons to Form O $^{2-}$ . The lowest common multiple of 3 and 2 is 6, so we need two aluminium atoms and three Oxygen atoms.

$2\mathrm{Al + 3\mathrm{O \to 2\mathrm{Al^{3+} + 3\mathrm{O^{2-}$

Worked Example. Draw the dot and cross diagram for calcium fluoride (CaF $_2$ ).

Calcium ( $2, 8, 8, 2$ ) loses 2 electrons to form Ca $^{2+}$ . Each fluorine ( $2, 7$ ) gains 1 electron To form F $^-$ . Two fluorine atoms are needed to accept the 2 electrons from calcium. The Ca $^{2+}$ Ion achieves the configuration of argon; each F $^-$ achieves the configuration of neon.

Worked Example. Draw the dot and cross diagram for lithium oxide (Li $_2$ O).

Each lithium ( $2, 1$ ) loses 1 electron to form Li $^+$ . Oxygen ( $2, 6$ ) gains 2 electrons to form O $^{2-}$ . Two lithium atoms are needed.

Worked Example. Draw the dot and cross diagram for potassium sulfide (K $_2$ S).

Each potassium ( $2, 8, 8, 1$ ) loses 1 electron to form K $^+$ . Sulfur ( $2, 8, 6$ ) gains 2 electrons To form S $^{2-}$ . Two potassium atoms are needed. Each K $^+$ achieves the configuration of argon; S $^{2-}$ achieves the configuration of argon.

1.4 Formulae of Ionic Compounds

The formula of an ionic compound is determined by balancing the charges:

Cation	Anion	Formula
Na $^+$	Cl $^-$	NaCl
Ca $^{2+}$	Cl $^-$	CaCl $_2$
Al $^{3+}$	O $^{2-}$	Al $_2$ O $_3$
Mg $^{2+}$	SO $_4^{2-}$	MgSO $_4$
Na $^+$	CO $_3^{2-}$	Na $_2$ CO $_3$

The method is: write the ions with their charges, then find the smallest whole number ratio that Makes the total charge zero. For Al $_2$ O $_3$ : $2 \times (+3) + 3 \times (-2) = 0$ .

Worked Example. Determine the formula of the compound formed between magnesium ions and oxide Ions.

Mg $^{2+}$ and O $^{2-}$ . The charges are equal and opposite, so the ratio is 1:1. Formula: MgO.

Worked Example. Determine the formula of the compound formed between aluminium ions and sulfate Ions.

Al $^{3+}$ and SO $_4^{2-}$ . The lowest common multiple of 3 and 2 is 6. We need 2 Al $^{3+}$ (total $+6$ ) and 3 SO $_4^{2-}$ (total $-6$ ). Formula: Al $_2$ (SO $_4$ ) $_3$ .

Worked Example. Determine the formula of the compound formed between calcium ions and phosphate Ions.

Ca $^{2+}$ and PO $_4^{3-}$ . The lowest common multiple of 2 and 3 is 6. We need 3 Ca $^{2+}$ (total $+6$ ) and 2 PO $_4^{3-}$ (total $-6$ ). Formula: Ca $_3$ (PO $_4$ ) $_2$ .

1.5 Polyatomic Ions

Some ions consist of more than one atom bonded covalently but carrying an overall charge:

Ion	Formula	Charge
Hydroxide	OH $^-$	$1-$
Nitrate	NO $_3^-$	$1-$
Carbonate	CO $_3^{2-}$	$2-$
Sulfate	SO $_4^{2-}$	$2-$
Ammonium	NH $_4^+$	$1+$
Hydrogencarbonate	HCO $_3^-$	$1-$

When writing formulae with polyatomic ions, use brackets if more than one of the polyatomic ion is Needed: Ca(NO $_3$ ) $_2$ Not CaNO $\_3$ 2.

1.6 Derivation: Why Ionic Compounds Have High Melting Points

The melting point of an ionic compound depends on the strength of the electrostatic forces between Ions, given by Coulomb’s law:

$F \propto \frac{q_1 \cdot q_2}{r^2}$

Where $q_1$ and $q_2$ are the ion charges and $r$ is the distance between ion centres. For compounds With higher charges (e.g. MgO with $2+$ and $2-$ ) and smaller ions (shorter $r$ ), the forces are Much stronger and the melting point is much higher. This explains why MgO melts at $2852^{\circ}\mathrm{C$ while NaCl melts at only $801^{\circ}\mathrm{C$ .

1.7 Comparison of Ionic Compound Properties

Compound	Ion Charges	Melting Point ( $^{\circ}\mathrm{C$ )	Solubility in Water
NaCl	$+1, -1$	801	Soluble
MgO	$+2, -2$	2852	Slightly soluble
CaCl $_2$	$+2, -1$	772	Soluble
Al $_2$ O $_3$	$+3, -2$	2072	Insoluble

The trend is clear: higher ionic charges produce stronger electrostatic attraction and higher Melting points. This is a direct consequence of Coulomb’s law.

2. Covalent Bonding

2.1 Formation of Covalent Bonds

Covalent bonding occurs between non-metals. Atoms share pairs of electrons so that each atom Achieves a full outer shell. Each shared pair of electrons is called a single covalent bond.

A covalent bond is a localised region of high electron density between two nuclei. The shared Electrons are attracted to both nuclei simultaneously, which lowers the energy and holds the atoms Together.

2.2 Simple Covalent Molecules

Examples of dot and cross diagrams:

Hydrogen (H $_2$ ): Each hydrogen atom has 1 electron. They share one pair.

$\mathrm{H \cdot + \cdot \mathrm{H \to \mathrm{H - \mathrm{H$

Water (H $_2$ O): Oxygen has 6 outer electrons and needs 2 more. Each hydrogen has 1 electron.

Oxygen forms two single covalent bonds, one with each hydrogen. The two lone pairs on oxygen are not Shared.

Carbon dioxide (CO $_2$ ): Carbon has 4 outer electrons and needs 4 more. Each oxygen has 6 outer Electrons and needs 2 more. Each oxygen forms a double bond with carbon.

Carbon forms two double bonds, one with each oxygen. Each double bond consists of two shared pairs Of electrons.

Methane (CH $_4$ ): Carbon forms four single bonds with four hydrogen atoms. Carbon achieves a Full outer shell of 8 electrons, and each hydrogen achieves a full outer shell of 2 electrons.

Ammonia (NH $_3$ ): Nitrogen has 5 outer electrons and needs 3 more. Each hydrogen has 1 electron. Nitrogen forms three single bonds and retains one lone pair.

Hydrogen chloride (HCl): Chlorine has 7 outer electrons and needs 1 more. Hydrogen has 1 Electron. They share one pair.

Worked Example. Draw the dot and cross diagram for hydrogen sulfide (H $_2$ S).

Sulfur is in group 6, with 6 outer electrons. It needs 2 more to complete its octet. Each hydrogen Contributes 1 electron. Sulfur forms two single covalent bonds, one with each hydrogen, and retains Two lone pairs. The structure is analogous to water.

Worked Example. Draw the dot and cross diagram for nitrogen (N $_2$ ).

Nitrogen is in group 7 with 5 outer electrons. Each nitrogen atom needs 3 more to complete its Octet. They share three pairs of electrons, forming a triple bond. Each nitrogen retains one lone Pair.

Worked Example. Draw the dot and cross diagram for phosphorus trichloride (PCl $_3$ ).

Phosphorus is in group 5 with 5 outer electrons. It needs 3 more to complete its octet. Each Chlorine has 7 outer electrons and needs 1 more. Phosphorus forms three single covalent bonds, one With each chlorine, and retains one lone pair. The structure is analogous to ammonia.

2.3 Properties of Simple Covalent Molecules

Property	Explanation
Low melting/boiling point	Weak intermolecular forces between molecules (not the covalent bonds)
Do NOT conduct electricity	No free ions or electrons
gases or liquids at room temperature	Weak forces between molecules

:::caution It is the intermolecular forces that are overcome when a simple covalent substance Melts or boils, NOT the covalent bonds within the molecules. Covalent bonds are strong, but there Are only weak forces between molecules. Breaking covalent bonds would decompose the molecule into Atoms — this does not happen during melting or boiling. :::

The distinction between intramolecular bonds (within molecules) and intermolecular forces (between Molecules) is one of the most important ideas in chemistry. The covalent bonds inside a water Molecule are strong (about 464 kJ/mol for O-H), but the hydrogen bonds between water molecules are Much weaker (about 20 kJ/mol). It is these weaker forces that are overcome when water boils at $100^{\circ}\mathrm{C$ .

2.4 Bonding in Giant Covalent Structures

Diamond: Each carbon atom forms four covalent bonds with four other carbon atoms in a Tetrahedral arrangement. This creates a very rigid, three-dimensional lattice.

Very hard (hardest natural substance)
Very high melting point ( $\gt 3500^{\circ}\mathrm{C$ )
Does not conduct electricity (no free electrons)
Insoluble in all solvents

Every carbon in diamond is sp $^3$ hybridised. The tetrahedral arrangement leaves no gaps and no free Electrons, explaining both the hardness and the lack of electrical conductivity.

Graphite: Each carbon atom forms three covalent bonds with three other carbon atoms, arranged in Layers of hexagonal rings. The fourth electron of each carbon is delocalised (free to move).

Soft and slippery (layers can slide over each other)
High melting point
Conducts electricity (delocalised electrons)
Used as a lubricant and in pencils

Graphite and diamond are both composed entirely of carbon atoms, yet their properties are Dramatically different. This illustrates a fundamental principle: the properties of a substance Depend not only on what atoms it contains but on how those atoms are bonded and arranged.

Silicon dioxide (SiO $_2$ ): Each silicon atom bonds to four oxygen atoms, and each oxygen bonds To two silicon atoms, forming a giant covalent structure similar to diamond.

Very high melting point
Hard
Insulator

Graphene: A single layer of graphite, one atom thick. It is the thinnest known material and has Exceptional strength and electrical conductivity.

Fullerenes: Molecules of carbon arranged in closed cages (e.g. C $_{60}$ Buckminsterfullerene). They have applications in drug delivery and materials science.

Nanotubes: Cylindrical fullerenes with very high tensile strength and useful electrical Properties. They are used in composite materials and electronics.

2.5 Comparison of Diamond and Graphite

Property	Diamond	Graphite
Bonding	4 covalent bonds per carbon	3 covalent bonds + 1 delocalised
Structure	Tetrahedral, 3D giant lattice	Layers of hexagonal rings
Hardness	Extremely hard	Soft and slippery
Electrical conductivity	Does not conduct	Conducts (delocalised electrons)
Melting point	Very high ( $\gt 3500^{\circ}\mathrm{C$ )	Very high
Uses	Cutting tools, jewellery	Pencils, lubricants, electrodes

2.6 Bond Polarity

When two different non-metals form a covalent bond, the shared electrons may be pulled more towards One atom than the other (because of different electronegativities).

If the electrons are shared equally, the bond is non-polar (e.g. H-H, Cl-Cl)
If the electrons are pulled towards one atom, the bond is polar (e.g. H-Cl)

In H-Cl, the chlorine is more electronegative, so it has a slight negative charge ( $\delta^-$ ) and Hydrogen has a slight positive charge ( $\delta^+$ ). The bond has a dipole.

2.7 Bond Energy

The bond energy is the energy required to break one mole of a particular bond in the gaseous State.

Bond	Bond energy (kJ/mol)
H-H	436
C-C	348
C=C	612
O-H	463
C=O	805
Cl-Cl	243

Multiple bonds are stronger than single bonds because there are more shared electrons holding the Atoms together: C=C (612 kJ/mol) is stronger than C-C (348 kJ/mol), and C $\equiv$ C (839 kJ/mol) is Stronger still.

2.8 Derivation: Why Multiple Bonds Are Stronger

A double bond consists of one sigma bond and one pi bond. A triple bond consists of one sigma bond And two pi bonds. Each additional shared pair adds more electron density between the nuclei, Increasing the net attractive force and pulling the nuclei closer together. The shorter bond length Also contributes to the higher bond energy because the electrons are closer to both nuclei.

The relationship between bond order, bond length, and bond energy:

Bond Order	Bond Length	Bond Energy
1 (single)	Longest	Weakest
2 (double)	Intermediate	Intermediate
3 (triple)	Shortest	Strongest

3. Metallic Bonding

3.1 The Model

In metals, the outer electrons of each atom are delocalised — they are free to move throughout The metallic structure. This creates a “sea” of delocalised electrons surrounding positive metal Ions.

The metallic bond is the strong electrostatic attraction between the positive metal ions and the Delocalised electrons. This model accounts for the key properties of metals and distinguishes them From both ionic and covalent substances.

3.2 Properties of Metals

Property	Explanation
High melting/boiling point	Strong metallic bonds require a lot of energy to overcome
Good conductors of electricity	Delocalised electrons are free to carry charge
Good conductors of heat	Delocalised electrons transfer kinetic energy
Malleable and ductile	Layers of ions can slide over each other without breaking the metallic bonds
Shiny	Delocalised electrons absorb and re-emit light
high density	Atoms are closely packed in the lattice

The malleability and ductility of metals are consequences of the non-directional nature of the Metallic bond. When a force is applied, layers of metal ions can slide past each other. The Delocalised electrons adjust their positions to maintain the electrostatic attraction, so the metal Does not shatter. This is in stark contrast to ionic solids, where displacement of layers brings Like charges together and causes fracture.

3.3 Alloys

An alloy is a mixture of a metal with one or more other elements ( another metal or Carbon). The different-sized atoms distort the regular lattice, preventing layers from sliding . This makes alloys harder than pure metals.

Alloy	Composition	Use
Steel	Iron + carbon	Construction, tools
Brass	Copper + zinc	Musical instruments, fittings
Bronze	Copper + tin	Statues, medals
Stainless steel	Iron + chromium + nickel	Cutlery, medical instruments
Solder	Tin + lead	Joining wires
Amalgam	Mercury + other metals	Dental fillings

The mechanism of hardening in alloys is straightforward. In a pure metal, all atoms are the same Size, so layers can slide over each other . In an alloy, the atoms of the added element are a Different size and disrupt the regular arrangement. This creates “roughness” in the lattice that Impedes the sliding of layers, making the material harder and less ductile.

Worked Example. Explain why gold is often alloyed with other metals to make jewellery.

Pure gold (24 carat) is too soft for everyday wear. Alloying gold with copper or silver distorts the Lattice, making the metal harder and more durable. 18 carat gold is 75% gold, 25% other metals.

Worked Example. Explain why steel is used for construction but pure iron is not.

Pure iron is relatively soft and malleable because the layers of iron ions can slide past each other . Adding a small amount of carbon (to make steel) introduces carbon atoms of a different size Into the iron lattice, distorting it and preventing the layers from sliding. This makes steel much Harder and stronger.

3.4 Superconductors

Some materials, when cooled below a critical temperature, lose all electrical resistance and become superconductors. Metals and metallic alloys can be superconductors. Superconductors have Applications in MRI scanners, maglev trains, and power transmission (because no energy is lost as Heat).

4. Types of Bonding and Properties

4.1 Summary Table

Type of Bonding	Between	Structure	Melting Point	Conductivity
Ionic	Metal + non-metal	Giant ionic lattice	High	Yes (molten/dissolved)
Covalent (simple)	Non-metal + non-metal	Simple molecules	Low	No
Covalent (giant)	Non-metal + non-metal	Giant covalent lattice	Very high	Graphite: yes; others: no
Metallic	Metal + metal	Giant metallic lattice	High	Yes (solid)

4.2 Predicting Bonding Type

To predict the type of bonding in a substance:

Check if it contains a metal: if yes, likely metallic or ionic bonding
If it is a metal + non-metal compound: ionic bonding
If it contains only non-metals: covalent bonding
Check the structure: simple molecules vs. Giant structures

Worked Example. Predict the type of bonding and state in each case: (a) Cu, (b) NaCl, (c) CO $_2$ (d) SiO $_2$ .

(a) Cu: metallic bonding (metal + metal, giant metallic lattice). (b) NaCl: ionic bonding (metal + Non-metal, giant ionic lattice). (c) CO $_2$ : covalent bonding (non-metal + non-metal, simple Molecule). (d) SiO $_2$ : covalent bonding (non-metal + non-metal, giant covalent lattice).

Worked Example. Predict the bonding type and state of: (a) Si, (b) Na $_2$ O, (c) NH $_3$ (d) Al.

(a) Si: covalent bonding, giant covalent lattice (silicon is a non-metal with a diamond-like Structure). (b) Na $_2$ O: ionic bonding, giant ionic lattice. (c) NH $_3$ : covalent bonding, simple Molecule (gas at room temperature). (d) Al: metallic bonding, giant metallic lattice.

4.3 Predicting Properties from Bonding Type

Given the type of bonding, you should be able to predict:

Melting and boiling point (high for ionic, metallic, and giant covalent; low for simple covalent)
Electrical conductivity (yes for metallic and molten/dissolved ionic; no for covalent)
State at room temperature (solid for ionic, metallic, and giant covalent; often gas or liquid for simple covalent)
Solubility (ionic compounds tend to be soluble in water; giant covalent structures tend to be insoluble)

4.4 Comparison: Ionic vs. Metallic Bonding

Property	Ionic Bonding	Metallic Bonding
Type of attraction	Between positive and negative ions	Between positive ions and delocalised electrons
Conductivity	Only when molten/dissolved	Always conducts (solid, liquid)
Malleability	Brittle (like charges repel)	Malleable (layers slide)
Melting point	High	High
Solubility in water	soluble	insoluble
Example	NaCl, MgO	Iron, copper

4.5 Comparison: Ionic vs. Covalent vs. Metallic

Feature	Ionic	Simple Covalent	Giant Covalent	Metallic
Constituents	Positive and negative ions	Discrete molecules	Atoms bonded in a lattice	Positive ions + delocalised e
Melting point	High	Low	Very high	High
Electrical cond.	Molten/dissolved only	None	Graphite only	Yes (always)
Solubility	soluble in water	Variable	Insoluble	Insoluble
Example	NaCl	CO $_2$ H $_2$ O	Diamond, SiO $_2$	Cu, Fe

5. States of Matter and Changes of State

5.1 The Three States

Property	Solid	Liquid	Gas
Arrangement	Regular, fixed	Random, close	Random, far apart
Movement	Vibrate in fixed positions	Move freely, sliding past each other	Move rapidly in all directions
Forces	Strong	Weaker than solids	Very weak
Shape	Fixed	Takes container shape	Fills container
Volume	Fixed	Fixed	Fills available space

5.2 Changes of State

Change	Name	Energy Change
Solid to liquid	Melting	Energy absorbed
Liquid to solid	Freezing	Energy released
Liquid to gas	Evaporation/Boiling	Energy absorbed
Gas to liquid	Condensation	Energy released
Solid to gas	Sublimation	Energy absorbed
Gas to solid	Deposition	Energy released

:::info Changes of state are physical changes, not chemical changes. No new substances are Formed, and the process is reversible. The identity of the substance does not change; only the Arrangement and energy of its particles. :::

5.3 Heating and Cooling Curves

A heating curve for a pure substance shows:

Temperature increasing as the substance is heated
A plateau at the melting point (energy goes into breaking bonds, not raising temperature)
Temperature increasing again
A plateau at the boiling point (energy goes into breaking intermolecular forces)

The plateau occurs because the energy supplied is used to overcome the forces between particles (rather than increase their kinetic energy). The temperature remains constant until all the Substance has changed state.

Worked Example. Explain why the temperature stays constant during the boiling of water.

During boiling, the energy supplied is used to break the hydrogen bonds and other intermolecular Forces between water molecules, converting liquid water to water vapour. This energy does not Increase the kinetic energy of the molecules, so the temperature remains constant at $100^{\circ}\mathrm{C$ until all the water has boiled.

Worked Example. Explain the shape of a cooling curve for a pure substance.

As the substance cools, its temperature decreases steadily. When the freezing point is reached, a Plateau appears because energy is released as particles form bonds and arrange into a solid lattice. The temperature remains constant until all the substance has solidified. The temperature then Decreases again as the solid continues to cool.

5.4 The Kinetic Theory and States of Matter

In a solid, particles vibrate about fixed positions. As temperature increases, the vibrations become Larger. At the melting point, the particles have enough energy to break free from their fixed Positions and slide past each other.

In a liquid, particles can move freely but remain in close contact. As temperature increases Further, the particles gain more kinetic energy. At the boiling point, they have enough energy to Escape the liquid entirely and become a gas.

5.5 Derivation: Kinetic Energy and Temperature

The average kinetic energy of gas particles is directly proportional to the absolute temperature:

$E_k = \frac{3}{2}k_BT$

Where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin. This Relationship explains why increasing temperature causes particles to move faster, collide more Frequently, and exert greater pressure.

6. Intermolecular Forces

6.1 Types of Intermolecular Force

Type	Strength	Between
London (van der Waals)	Weak	All molecules
Permanent dipole-dipole	Moderate	Polar molecules
Hydrogen bonding	Strong	Molecules with H bonded to N, O, or F

London dispersion forces arise from temporary, instantaneous dipoles caused by the uneven Distribution of electrons within a molecule at any given instant. These temporary dipoles induce Dipoles in neighbouring molecules, creating a weak attractive force. The strength of London forces Increases with the number of electrons (and therefore the molecular size).

Permanent dipole-dipole forces occur between molecules that have a permanent separation of Charge (a permanent dipole). These are stronger than London forces but weaker than hydrogen bonds.

Hydrogen bonds are the strongest type of intermolecular force. They occur when hydrogen is Covalently bonded to a highly electronegative atom (N, O, or F), creating a large dipole. The Partially positive hydrogen can then interact with a lone pair on a neighbouring N, O, or F atom.

6.2 Predicting Boiling Points

Substances with stronger intermolecular forces have higher boiling points.

Example: Water has a much higher boiling point than hydrogen sulfide, despite H $_2$ S having a Larger relative molecular mass. This is because water forms hydrogen bonds, which are much stronger Than the dipole-dipole forces in H $_2$ S.

Example: Explain why the boiling point increases from F $_2$ to Cl $_2$ to Br $_2$ to I $_2$ .

All four are simple covalent molecules with only London dispersion forces between them. As the Number of electrons increases down the group, the electron cloud becomes more polarizable, London Forces become stronger, and more energy is needed to separate the molecules. Hence the boiling point Increases.

Example: Explain why NH $_3$ has a higher boiling point than CH $_4$ despite having a similar Molecular mass.

NH $_3$ can form hydrogen bonds (H bonded to N), whereas CH $_4$ can only form London forces. Hydrogen Bonding is much stronger than London forces, so more energy is required to boil NH $_3$ .

Worked Example. Arrange the following in order of increasing boiling point and explain your Reasoning: F $_2$ Cl $_2$ Br $_2$ .

F $_2$ < Cl $_2$ < Br $_2$ . All three have only London dispersion forces. As the number of Electrons increases (F $_2$ : 18, Cl $_2$ : 34, Br $_2$ : 70), London forces become stronger, requiring More energy to overcome. Hence the boiling point increases.

Worked Example. Explain why ethanol (C $_2$ H $_5$ OH, boiling point $78^{\circ}\mathrm{C$ ) has a Much higher boiling point than dimethyl ether (CH $_3$ OCH $_3$ Boiling point $-24^{\circ}\mathrm{C$ ) Despite both having the molecular formula C $_2$ H $_6$ O.

Both molecules have the same number of electrons, so London forces are similar. However, ethanol can Form hydrogen bonds (it has an O-H group), while dimethyl ether cannot (it has no O-H group; the Oxygen is bonded to two carbons). The hydrogen bonds in ethanol require much more energy to Overcome, giving it a significantly higher boiling point.

6.3 Nanoparticles

Nanoparticles are particles with at least one dimension between 1 nm and 100 nm. They have a Very large surface area to volume ratio compared to bulk materials, which gives them different Properties.

Applications of nanoparticles include catalysts (large surface area increases reaction rate), Cosmetics, sunscreens, and medicine (drug delivery).

Worked Example. Calculate the surface area to volume ratio for a cube of side 1 cm and a cube of Side 1 nm.

For 1 cm cube: surface area = $6 \times 1^2 = 6$ cm $^2$ Volume = $1^3 = 1$ cm $^3$ . Ratio = 6 Cm $^{-1}$ .

For 1 nm cube: surface area = $6 \times (10^{-7})^2 = 6 \times 10^{-14}$ cm $^2$ Volume = $(10^{-7})^3 = 10^{-21}$ cm $^3$ . Ratio = $6 \times 10^7$ cm $^{-1}$ .

The nanoparticle has a surface area to volume ratio about $10^7$ times larger.

6.4 Derivation: Surface Area to Volume Ratio Scaling

For a cube of side length $s$ :

$\mathrm{Surface area = 6s^2, \quad \mathrm{Volume = s^3$

$\frac{\mathrm{Surface area}{\mathrm{Volume} = \frac{6s^2}{s^3} = \frac{6}{s}$

The ratio is inversely proportional to the side length. As $s$ decreases, the ratio increases. This Is why nanoparticles have such large surface area to volume ratios: their small size means a large Proportion of atoms are on the surface, available for catalysis or other reactions.

6.5 Summary: Intermolecular Forces and Boiling Points

Substance	Molar Mass (g/mol)	IMF Type	Boiling Point ( $^{\circ}\mathrm{C$ )
CH $_4$	16	London only	-161
NH $_3$	17	Hydrogen bonding	-33
H $_2$ O	18	Hydrogen bonding	100
H $_2$ S	34	Dipole-dipole + London	-60
HF	20	Hydrogen bonding	20
F $_2$	38	London only	-188
Cl $_2$	71	London only	-34
Br $_2$	160	London only	59
I $_2$	254	London only	184

This table demonstrates two key principles: hydrogen bonding produces anomalously high boiling Points compared to London forces, and London forces increase with molar mass.

Common Pitfalls

Confusing intermolecular forces with covalent bonds. It is the intermolecular forces that break when simple covalent molecules melt or boil, not the covalent bonds inside the molecules.
Drawing dot and cross diagrams for ionic compounds incorrectly. Show the electrons being transferred, not shared. Ensure that the correct number of electrons are transferred.
Thinking that metals do not have high melting points. All metals (except mercury) have high melting points due to strong metallic bonding.
Confusing ionic and metallic conductivity. Ionic compounds conduct only when molten or dissolved (ions must be free to move); metals conduct as solids (delocalised electrons are always free).
Forgetting that graphite conducts electricity. Although it is a giant covalent structure, the delocalised electrons between the layers allow conduction.
Stating that alloys are softer than pure metals. Alloys are HARDER because the different-sized atoms prevent the layers from sliding.
Assuming all covalent substances have low melting points. Giant covalent structures (diamond, graphite, SiO $_2$ ) have very high melting points because covalent bonds must be broken throughout the entire structure.
Using the wrong type of bonding for a given substance. Always check whether the substance contains a metal and whether it forms a giant or simple structure.
Forgetting brackets in formulae with polyatomic ions. Ca(NO $_3$ ) $_2$ requires brackets because there are two nitrate ions.
Confusing the terms “molecular” and “ionic” for giant structures. Ionic lattices are not molecular; they do not contain discrete molecules.
Confusing diamond and graphite properties. Diamond is hard and insulating; graphite is soft and conducting. Both are forms of carbon.
Stating that hydrogen bonding occurs between any molecule containing hydrogen. Hydrogen bonding requires H bonded to N, O, or F specifically.

Practice Questions

Draw a dot and cross diagram to show the bonding in magnesium fluoride (MgF $_2$ ).
Explain why sodium chloride has a high melting point, whereas carbon dioxide is a gas at room temperature.
Explain, in terms of structure and bonding, why graphite can conduct electricity but diamond cannot.
Describe the bonding in aluminium and explain why aluminium is a good conductor of electricity.
Explain why steel is harder than pure iron.
State the type of bonding in each of the following: (a) NaBr, (b) H $_2$ O, (c) SiO $_2$ (d) Cu.
Explain why ice melts at $0^{\circ}\mathrm{C$ but water does not boil until $100^{\circ}\mathrm{C$ in terms of the energy required to overcome forces.
Draw a dot and cross diagram for ammonia (NH $_3$ ).
Explain why ionic compounds are brittle but metals are malleable.
Predict and explain which has the higher boiling point: Br $_2$ or I $_2$ .
Explain why H $_2$ O is a liquid at room temperature but H $_2$ S is a gas.
State and explain three differences in properties between diamond and graphite.
Explain, with reference to bonding, why copper is used for electrical wiring but sodium chloride is not.
A student claims that “when NaCl melts, the covalent bonds between sodium and chlorine break.” Identify the error in this statement and correct it.
Describe the bonding in an alloy and explain why alloys are generally harder than pure metals.
Draw a dot and cross diagram for nitrogen (N $_2$ ), showing all outer electrons .
Explain why the boiling point of ethanol ( $78^{\circ}\mathrm{C$ ) is much higher than that of dimethyl ether ( $-24^{\circ}\mathrm{C$ ) despite both having the same molecular formula (C $_2$ H $_6$ O).
Compare and contrast the structure and bonding in diamond and silicon dioxide.
Explain why nanoparticles are more effective catalysts than the same material in bulk form.
Draw a dot and cross diagram for carbon dioxide (CO $_2$ ), showing all outer shell electrons .
Potassium oxide has the formula K $_2$ O. Draw a dot and cross diagram showing the transfer of electrons and explain why two potassium atoms are needed for each oxygen atom.
Explain, using ideas about bonding, why a piece of magnesium ribbon can be bent but a crystal of sodium chloride shatters when a force is applied.
A student says that silicon dioxide has a low melting point because it contains covalent bonds. Explain why this statement is incorrect.
Compare and contrast the bonding in sodium chloride, diamond, and copper. Your answer should include the types of particles present, the forces between them, and how these forces account for the electrical conductivity of each substance.

Practice Problems

Question 1: Ionic vs covalent bonding

Explain the difference between ionic and covalent bonding. Use sodium chloride and water as examples.

Answer

Ionic bonding involves the transfer of electrons from a metal to a non-metal, forming positive and negative ions held together by electrostatic attraction. In NaCl, sodium transfers one electron to chlorine, forming $\mathrm{Na^+$ and $\mathrm{Cl^-$ .

Covalent bonding involves the sharing of electron pairs between non-metal atoms. In water ( $\mathrm{H_2\mathrm{O$ ), oxygen shares electrons with two hydrogen atoms, forming two covalent bonds.

Question 2: Dot-cross diagrams

Draw dot-cross diagrams for (a) magnesium oxide ( $\mathrm{MgO$ ) and (b) carbon dioxide ( $\mathrm{CO_2$ ).

Answer

(a) $\mathrm{MgO$ : $\mathrm{Mg$ donates 2 electrons (shown as x) to $\mathrm{O$ Forming $\mathrm{Mg^{2+}$ and $\mathrm{O^{2-}$ . The diagram shows the oxide ion with 8 electrons (6 of its own + 2 from Mg) and the magnesium ion with none in its outer shell.

(b) $\mathrm{CO_2$ : $\mathrm{O=\mathrm{C=\mathrm{O$ . Each oxygen shares 2 electron pairs with carbon (double bonds). Carbon shares 4 electrons total (2 with each oxygen), achieving a full outer shell of 8.

Question 3: Properties of ionic compounds

Sodium chloride has a high melting point, conducts electricity when molten but not when solid, and dissolves in water. Explain these properties in terms of its structure and bonding.

Answer

High melting point: the ionic bonds between $\mathrm{Na^+$ and $\mathrm{Cl^-$ in the giant ionic lattice are strong and require a lot of energy to break.

Conducts when molten but not solid: in the solid state, ions are fixed in position and cannot move. When molten, the lattice breaks down and ions are free to move, carrying charge.

Dissolves in water: water molecules are polar. The $\delta^+$ hydrogen atoms are attracted to $\mathrm{Cl^-$ ions and the $\delta^-$ oxygen atoms to $\mathrm{Na^+$ ions, pulling ions away from the lattice.

Question 4: Simple covalent molecules vs giant covalent structures

Compare the structure and properties of methane ( $\mathrm{CH_4$ ) and diamond. Explain why diamond is much harder and has a much higher melting point.

Answer

Methane is a simple covalent molecule with weak London forces between molecules. It has a low melting point and is a gas at room temperature.

Diamond has a giant covalent (macromolecular) structure where each carbon atom is covalently bonded to four others in a tetrahedral arrangement. The entire structure is one giant molecule held together by strong covalent bonds throughout. Breaking these bonds requires enormous energy, so diamond has a very high melting point and is extremely hard.

Question 5: Metallic bonding

Explain why metals are good conductors of electricity and are malleable, using the concept of metallic bonding.

Answer

In metallic bonding, metal atoms lose their outer electrons to form positive ions surrounded by a “sea” of delocalised electrons. These free electrons can move throughout the structure, carrying electrical charge, making metals good conductors.

Metals are malleable because the layers of positive ions can slide over each other when a force is applied. The delocalised electrons adjust their positions to maintain the metallic bonds between the layers, so the metal does not shatter. This is different from ionic compounds, where layers of ions of the same charge repel when forced to slide.

Worked Examples

Example 1:

A typical exam question on Bonding requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Bonding often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

Summary

This topic covers the essential chemistry of bonding, including key reactions, underlying theories, and practical applications.

Key concepts include:

ionic, covalent, and metallic bonding
electronegativity and bond polarity
intermolecular forces
giant and simple molecular structures
VSEPR theory

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.

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