Chemical Bonding

Learn and apply Chemical Bonding in the published Chemistry course sequence.

GCE A-Level H1 Chemistry 8873-2027

Learning goals

show understanding that all chemical bonds are electrostatic in nature and describe: — ionic bond as the electrostatic attraction between oppositely charged ions
show understanding that all chemical bonds are electrostatic in nature and describe: — covalent bond as the electrostatic attraction between a shared pair of electrons and positively charged nuclei
show understanding that all chemical bonds are electrostatic in nature and describe: — metallic bond as the electrostatic attraction between a lattice of positive ions and delocalised electrons
describe, including the use of ‘dot-and-cross’ diagrams, — ionic bonding as in sodium chloride and magnesium oxide
describe, including the use of ‘dot-and-cross’ diagrams, — covalent bonding as in hydrogen; oxygen; nitrogen; chlorine; hydrogen chloride; carbon dioxide; methane; ethene
describe, including the use of ‘dot-and-cross’ diagrams, — co-ordinate (dative covalent) bonding, as in formation of the ammonium ion and in the Al2Cl6 molecule
describe covalent bonding in terms of orbital overlap (limited to s and p orbitals only), giving σ and π bonds (see also Section 9)
explain the shapes of, and bond angles in, molecules such as BF3 (trigonal planar); CO2 (linear); CH4 (tetrahedral); NH3 (trigonal pyramidal); H2O (bent); SF6 (octahedral) by using the Valence Shell Electron Pair Repulsion theory
predict the shapes of, and bond angles in, molecules analogous to those specified in (d)
explain and deduce bond polarity using the concept of electronegativity [quantitative treatment of electronegativity is not required]
deduce the polarity of a molecule using bond polarity and its molecular shape (analogous to those specified in (d))
describe the following forces of attraction (electrostatic in nature): — intermolecular forces, based on permanent and induced dipoles, as in liquid and gaseous CHCl3, Br2 and the noble gases
describe the following forces of attraction (electrostatic in nature): — hydrogen bonding, using ammonia and water as examples of molecules containing –NH and –OH groups
outline the importance of intermolecular forces to the liquefaction of gases when subjected to high pressure and/or low temperature
outline the importance of hydrogen bonding to the physical properties of substances, including ice and water
explain the terms bond energy and bond length for covalent bonds
compare the reactivities of covalent bonds in terms of bond energy, bond length and bond polarity
describe, in simple terms, the lattice structure of a crystalline solid which is: — ionic, as in sodium chloride and magnesium oxide
describe, in simple terms, the lattice structure of a crystalline solid which is: — simple molecular, as in iodine
describe, in simple terms, the lattice structure of a crystalline solid which is: — giant molecular, as in graphite and diamond
describe, in simple terms, the lattice structure of a crystalline solid which is: — hydrogen-bonded, as in ice
describe, in simple terms, the lattice structure of a crystalline solid which is: — metallic, as in copper [the concept of the ‘unit cell’ is not required]
describe, interpret and/or predict the effect of different types of structure and bonding on the physical properties of substances
suggest the type of structure and bonding present in a substance from given information

H1 Chemical Bonding learning outcomes

Chemical bonding connects electron behaviour to shape, intermolecular attraction and bulk properties. Build every explanation through particles and structure → electrostatic attraction → observed property.

Follow the seven lessons in order for first learning. For revision, decide whether you need more help with a definition, a shape or polarity decision, or a structure–property explanation. H1 8873 also requires gas-liquefaction reasoning in the intermolecular-forces lesson.

What You’ll Learn

Describe ionic, covalent, dative covalent and metallic bonding as electrostatic attractions, using dot-and-cross diagrams where required.
Relate orbital overlap to σ and π bonds, and connect bond energy, bond length and electronegativity to bond polarity.
Predict molecular shape and bond angle from electron-pair repulsion, including the stronger repulsion of lone pairs.
Combine bond dipoles with molecular shape to decide whether a molecule is polar.
Compare London forces, permanent dipole–dipole attractions and hydrogen bonding, then apply them to boiling and gas liquefaction.
Relate simple molecular, giant ionic, giant covalent and metallic structures to melting point, solubility and electrical conductivity.

Lessons (Recommended Order)

The sequence moves from bonding models to molecular geometry, intermolecular behaviour and finally whole-solid properties.

Ionic and Covalent Bonding Models
Define ionic, covalent and metallic attraction precisely and construct the required dot-and-cross representations.
Open lesson
Dative Bonding and Common Examples
Track electron-pair donation in ammonium ions and aluminium chloride without treating the formed bond as a different species.
Open lesson
Sigma and Pi Bonds: Orbital Overlap
Relate head-on and sideways orbital overlap to sigma and pi bonding.
Open lesson
Bond Energy, Bond Length and Bond Polarity
Connect bond energy, length and electronegativity to bond and molecular polarity.
Open lesson
Molecular Shapes and Bond Angles
Predict required shapes and angles using electron-region repulsion, including lone-pair effects.
Open lesson
Intermolecular Forces and Properties
Compare induced and permanent dipoles, hydrogen bonding, water and ice; H1 also applies these ideas to gas liquefaction.
Open lesson
Solids: Structure and Physical Properties
Explain lattice properties and deduce bonding and structure from combined physical evidence.
Open lesson

Quick Reference

Idea	Exam-ready reminder
Ionic bond	electrostatic attraction between oppositely charged ions in a giant lattice
Covalent bond	electrostatic attraction between a shared pair of electrons and the nuclei of the bonded atoms
Dative covalent bond	a shared pair of electrons supplied by one atom; after formation, treat it like any other covalent bond
Double bond	one σ bond and one π bond
Repulsion order	lone pair–lone pair > lone pair–bond pair > bond pair–bond pair
Molecular polarity	combine bond polarity with molecular shape; equal bond dipoles can cancel
Boiling	overcomes intermolecular attractions, not the covalent bonds within each molecule
Structure–property answer	state structure and particles → identify attraction or mobile carrier → explain the observed property
Liquefaction	cooling lowers kinetic energy; pressure reduces particle separation

Name the particles and structure, identify the attraction or mobile charge carrier, then state why that feature produces the observed property.

Common Exam Traps

Calling intermolecular forces covalent bonds, or saying that boiling breaks covalent bonds within simple molecules.
Treating a dative covalent bond as a different type of bond after it has formed.
Counting a multiple bond as more than one electron region when applying VSEPR.
Giving the electron-pair geometry but forgetting that a molecular shape names only atom positions.
Using bond polarity alone to decide molecular polarity without considering whether the dipoles cancel.
Naming hydrogen bonding whenever hydrogen is present instead of checking for an H–N, H–O or H–F bond and a lone pair.
Explaining conductivity without identifying mobile ions or delocalised electrons in the stated physical state.
Saying pressure makes an individual intermolecular force stronger; pressure instead brings molecules closer so attractions become effective more often.

Check especially that boiling overcomes intermolecular forces rather than covalent bonds, and that molecular polarity depends on both bond dipoles and shape.

Practice and Check Your Understanding

Chemical Bonding Topic Practice
Course-specific 8873 questions across bonding, shape, polarity, forces and structures.
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Chemical Bonding: Check Your Understanding
Check each link in your explanation, read the feedback, then try a fresh question independently.
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Calcium (Group 2) reacts with fluorine (Group 17) to form calcium fluoride. Which statement about its formation and bonding is correct?

Each Ca atom transfers one electron to each of two F atoms, forming Ca²⁺ (2,8,8) and two F⁻ (2,8); the ions are held by electrostatic attraction in a giant lattice.
One F atom gains both of calcium's outer electrons, forming Ca²⁺ and F²⁻.
Ca and each F share a pair of electrons, forming two Ca–F covalent bonds.
Ca²⁺ and F⁻ form, and the ionic bond is the process of electron transfer, which ends once the ions have formed.

A dot-and-cross diagram of carbon dioxide, O=C=O, shows outer electrons only. How many shared pairs and lone pairs does it show in total?

4 shared pairs and 4 lone pairs
2 shared pairs and 6 lone pairs
4 shared pairs and 2 lone pairs
4 shared pairs and 0 lone pairs

Ammonia reacts with a hydrogen ion: NH₃ + H⁺ → NH₄⁺. Which statement about the bond formed is correct?

N donates its lone pair into the empty orbital of H⁺, forming a co-ordinate bond; once formed, it is identical to the other three N–H bonds.
H⁺ donates an electron pair to N.
N transfers one electron to H⁺, forming an ionic bond.
A co-ordinate bond forms, and it remains longer and weaker than the other three N–H bonds.

Which statement describes the N≡N bond in N₂ in terms of orbital overlap?

One σ bond from head-on overlap of p orbitals along the internuclear axis and two π bonds from sideways overlap of the remaining p orbitals, in perpendicular planes.
Three σ bonds from head-on overlap of three pairs of p orbitals.
One π bond from sideways overlap and two σ bonds from head-on overlap.
One σ bond and two π bonds, with both π bonds lying in the same plane.

Bond data: H–Cl 0.127 nm, 431 kJ mol⁻¹; H–Br 0.141 nm, 366 kJ mol⁻¹; H–I 0.161 nm, 299 kJ mol⁻¹. A hot wire is placed in each gas. Which prediction is correct?

HCl has the most polar bond, but HI decomposes into its elements most readily because H–I is the longest and weakest bond.
HI has the most polar bond and decomposes most readily.
HCl has the most polar bond and therefore decomposes most readily.
HCl has the most polar bond and decomposes most readily because the shortest bond needs the least energy to break.

Oxygen (Group 16) forms OF₂, in which O has two bonding pairs. Using VSEPR theory, which shape and F–O–F bond angle are expected?

Bent, with an angle less than 109.5°
Linear, with an angle of 180°
Bent, with an angle of 120°
Tetrahedral, with an angle of 109.5°

CCl₄ and CHCl₃ are both tetrahedral about carbon. C–Cl bonds are polar and C–H bonds are almost non-polar. Which statement is correct?

CCl₄ is non-polar because its four identical bond dipoles cancel; CHCl₃ is polar because replacing one Cl with H leaves the dipoles unbalanced.
Both are polar because both contain polar C–Cl bonds.
Both are non-polar because both are tetrahedral.
CHCl₃ is non-polar because it has fewer polar bonds than CCl₄.

The boiling points of neon, argon and krypton are −246 °C, −186 °C and −153 °C. Which explanation of the trend is correct?

Down the group the atoms have more electrons, so their electron clouds are more easily distorted and instantaneous dipole–induced dipole attractions become stronger.
Heavier atoms form covalent bonds between atoms, which must be broken on boiling.
Larger atoms have larger permanent dipoles.
The outer electrons are held more tightly down the group, so atoms attract each other more strongly.

Ammonia, NH₃ (10 electrons), boils at −33 °C, whereas phosphine, PH₃ (18 electrons), boils at −88 °C. Which explanation is correct?

N is highly electronegative, so a δ+ H on one NH₃ is attracted to the lone pair on N of another NH₃, forming hydrogen bonds; PH₃ cannot form hydrogen bonds.
PH₃ has more electrons, so ammonia's higher boiling point must come from stronger N–H covalent bonds breaking on boiling.
NH₃ has no lone pair, so its molecules pack more closely.
Hydrogen bonds form between the N atoms of neighbouring NH₃ molecules.

Which statement correctly compares solid magnesium oxide with solid magnesium metal?

MgO is a giant lattice of Mg²⁺ and O²⁻ ions held by attraction between oppositely charged ions; magnesium is a lattice of Mg²⁺ ions held by attraction to delocalised electrons.
MgO is a giant ionic lattice; in magnesium each delocalised electron is attracted only to the atom it came from.
MgO contains delocalised electrons as well as ions; magnesium is a lattice of positive ions in delocalised electrons.
MgO consists of discrete MgO molecules held by weak attractions; magnesium is a lattice of positive ions in delocalised electrons.

Which description of the crystalline solids graphite, iodine and ice is correct?

Graphite: layers of hexagonally arranged C atoms, each covalently bonded to three others, with weak attractions between layers. Iodine: I₂ molecules held by weak attractions. Ice: H₂O molecules hydrogen-bonded in an open lattice.
Graphite: each C covalently bonded to four others in a three-dimensional network. Iodine: I₂ molecules held by weak attractions. Ice: open hydrogen-bonded lattice.
Graphite: layers of C atoms with weak attractions between layers. Iodine: I⁻ and I⁺ ions in a giant ionic lattice. Ice: open hydrogen-bonded lattice.
Graphite: layers of C atoms with weak attractions between layers. Iodine: I₂ molecules held by weak attractions. Ice: H₂O molecules held only by instantaneous dipole–induced dipole attractions.

Substance Y melts at 801 °C, does not conduct when solid but conducts when molten or dissolved in water, and is brittle. Which structure and explanation fit these data?

A giant ionic lattice: strong attraction between ions gives a high melting point, and ions are fixed in the solid but mobile when molten or dissolved.
A metallic lattice: delocalised electrons carry the current.
A giant molecular lattice: many strong covalent bonds give the high melting point.
A giant ionic lattice: electrons become delocalised when the lattice melts, so the melt conducts.

Nitrogen stays a gas in a cylinder at 20 °C even at very high pressure, but it liquefies at 1 atm when cooled to −196 °C. Which explanation is correct?

N₂ has only weak instantaneous dipole–induced dipole attractions; cooling lowers the molecules' kinetic energy until these weak attractions can hold them together, which compression at 20 °C cannot achieve.
Cooling breaks the N≡N triple bond so that nitrogen atoms can pack together.
Cooling makes N₂ molecules polar, so they attract each other.
High pressure pushes N₂ molecules further apart, which stops them liquefying.

Practise after feedback

After a check, practise the skills it showed you need to work on.

About 10 minutes

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Calcium (Group 2) reacts with fluorine (Group 17) to form calcium fluoride. Which statement about its formation and bonding is correct?

Each Ca atom transfers one electron to each of two F atoms, forming Ca²⁺ (2,8,8) and two F⁻ (2,8); the ions are held by electrostatic attraction in a giant lattice.
One F atom gains both of calcium's outer electrons, forming Ca²⁺ and F²⁻.
Ca and each F share a pair of electrons, forming two Ca–F covalent bonds.
Ca²⁺ and F⁻ form, and the ionic bond is the process of electron transfer, which ends once the ions have formed.

A dot-and-cross diagram of carbon dioxide, O=C=O, shows outer electrons only. How many shared pairs and lone pairs does it show in total?

4 shared pairs and 4 lone pairs
2 shared pairs and 6 lone pairs
4 shared pairs and 2 lone pairs
4 shared pairs and 0 lone pairs

Ammonia reacts with a hydrogen ion: NH₃ + H⁺ → NH₄⁺. Which statement about the bond formed is correct?

N donates its lone pair into the empty orbital of H⁺, forming a co-ordinate bond; once formed, it is identical to the other three N–H bonds.
H⁺ donates an electron pair to N.
N transfers one electron to H⁺, forming an ionic bond.
A co-ordinate bond forms, and it remains longer and weaker than the other three N–H bonds.

Which statement describes the N≡N bond in N₂ in terms of orbital overlap?

One σ bond from head-on overlap of p orbitals along the internuclear axis and two π bonds from sideways overlap of the remaining p orbitals, in perpendicular planes.
Three σ bonds from head-on overlap of three pairs of p orbitals.
One π bond from sideways overlap and two σ bonds from head-on overlap.
One σ bond and two π bonds, with both π bonds lying in the same plane.

Bond data: H–Cl 0.127 nm, 431 kJ mol⁻¹; H–Br 0.141 nm, 366 kJ mol⁻¹; H–I 0.161 nm, 299 kJ mol⁻¹. A hot wire is placed in each gas. Which prediction is correct?

HCl has the most polar bond, but HI decomposes into its elements most readily because H–I is the longest and weakest bond.
HI has the most polar bond and decomposes most readily.
HCl has the most polar bond and therefore decomposes most readily.
HCl has the most polar bond and decomposes most readily because the shortest bond needs the least energy to break.

Oxygen (Group 16) forms OF₂, in which O has two bonding pairs. Using VSEPR theory, which shape and F–O–F bond angle are expected?

Bent, with an angle less than 109.5°
Linear, with an angle of 180°
Bent, with an angle of 120°
Tetrahedral, with an angle of 109.5°

CCl₄ and CHCl₃ are both tetrahedral about carbon. C–Cl bonds are polar and C–H bonds are almost non-polar. Which statement is correct?

CCl₄ is non-polar because its four identical bond dipoles cancel; CHCl₃ is polar because replacing one Cl with H leaves the dipoles unbalanced.
Both are polar because both contain polar C–Cl bonds.
Both are non-polar because both are tetrahedral.
CHCl₃ is non-polar because it has fewer polar bonds than CCl₄.

The boiling points of neon, argon and krypton are −246 °C, −186 °C and −153 °C. Which explanation of the trend is correct?

Down the group the atoms have more electrons, so their electron clouds are more easily distorted and instantaneous dipole–induced dipole attractions become stronger.
Heavier atoms form covalent bonds between atoms, which must be broken on boiling.
Larger atoms have larger permanent dipoles.
The outer electrons are held more tightly down the group, so atoms attract each other more strongly.

Ammonia, NH₃ (10 electrons), boils at −33 °C, whereas phosphine, PH₃ (18 electrons), boils at −88 °C. Which explanation is correct?

N is highly electronegative, so a δ+ H on one NH₃ is attracted to the lone pair on N of another NH₃, forming hydrogen bonds; PH₃ cannot form hydrogen bonds.
PH₃ has more electrons, so ammonia's higher boiling point must come from stronger N–H covalent bonds breaking on boiling.
NH₃ has no lone pair, so its molecules pack more closely.
Hydrogen bonds form between the N atoms of neighbouring NH₃ molecules.

Which statement correctly compares solid magnesium oxide with solid magnesium metal?

MgO is a giant lattice of Mg²⁺ and O²⁻ ions held by attraction between oppositely charged ions; magnesium is a lattice of Mg²⁺ ions held by attraction to delocalised electrons.
MgO is a giant ionic lattice; in magnesium each delocalised electron is attracted only to the atom it came from.
MgO contains delocalised electrons as well as ions; magnesium is a lattice of positive ions in delocalised electrons.
MgO consists of discrete MgO molecules held by weak attractions; magnesium is a lattice of positive ions in delocalised electrons.

Which description of the crystalline solids graphite, iodine and ice is correct?

Graphite: layers of hexagonally arranged C atoms, each covalently bonded to three others, with weak attractions between layers. Iodine: I₂ molecules held by weak attractions. Ice: H₂O molecules hydrogen-bonded in an open lattice.
Graphite: each C covalently bonded to four others in a three-dimensional network. Iodine: I₂ molecules held by weak attractions. Ice: open hydrogen-bonded lattice.
Graphite: layers of C atoms with weak attractions between layers. Iodine: I⁻ and I⁺ ions in a giant ionic lattice. Ice: open hydrogen-bonded lattice.
Graphite: layers of C atoms with weak attractions between layers. Iodine: I₂ molecules held by weak attractions. Ice: H₂O molecules held only by instantaneous dipole–induced dipole attractions.

Substance Y melts at 801 °C, does not conduct when solid but conducts when molten or dissolved in water, and is brittle. Which structure and explanation fit these data?

A giant ionic lattice: strong attraction between ions gives a high melting point, and ions are fixed in the solid but mobile when molten or dissolved.
A metallic lattice: delocalised electrons carry the current.
A giant molecular lattice: many strong covalent bonds give the high melting point.
A giant ionic lattice: electrons become delocalised when the lattice melts, so the melt conducts.

Nitrogen stays a gas in a cylinder at 20 °C even at very high pressure, but it liquefies at 1 atm when cooled to −196 °C. Which explanation is correct?

N₂ has only weak instantaneous dipole–induced dipole attractions; cooling lowers the molecules' kinetic energy until these weak attractions can hold them together, which compression at 20 °C cannot achieve.
Cooling breaks the N≡N triple bond so that nitrogen atoms can pack together.
Cooling makes N₂ molecules polar, so they attract each other.
High pressure pushes N₂ molecules further apart, which stops them liquefying.

Check what I know

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Magnesium (Group 2) reacts with oxygen to form magnesium oxide, and nitrogen exists as N₂ molecules. Which statement describes the bonding in both substances correctly?

MgO: each Mg transfers two electrons to an O, giving Mg²⁺ and O²⁻ held by electrostatic attraction between oppositely charged ions. N₂: the atoms share three electron pairs, each N keeps one lone pair, and each shared pair is attracted to both nuclei.
MgO: Mg and O share two electron pairs in a Mg=O double bond. N₂: three shared pairs attracted to both nuclei, with one lone pair on each N.
MgO: one electron is transferred, giving Mg⁺ and O⁻ held by attraction between the ions. N₂: three shared pairs attracted to both nuclei, with one lone pair on each N.
MgO: Mg²⁺ and O²⁻ held by attraction between oppositely charged ions. N₂: one shared pair, with two lone pairs on each N.
MgO: Mg²⁺ and O²⁻ held by attraction between oppositely charged ions. N₂: three shared pairs and one lone pair on each N, with the atoms held by attraction between the two nuclei.

Two AlCl₃ units join to form the Al₂Cl₆ molecule. Which statement correctly describes the co-ordinate bonds in Al₂Cl₆?

A Cl atom bonded to one Al donates a lone pair into an empty orbital of the other Al; there are two such bonds, giving each Al eight outer electrons.
An Al atom donates a lone pair to a Cl atom of the other AlCl₃ unit.
The two AlCl₃ units are joined by a single Al–Al co-ordinate bond.
There is one co-ordinate bond, so one Al has eight outer electrons and the other has six.

Which statement about the covalent bonds in propene, CH₂=CH–CH₃, is correct?

There are 8 σ bonds and 1 π bond; the π bond forms by sideways overlap of p orbitals on the two doubly bonded carbon atoms.
There are 7 σ bonds and 2 π bonds, because the C=C double bond consists of two π bonds.
There are 9 σ bonds and no π bonds, because both bonds of C=C form by head-on overlap.
There are 8 σ bonds and 1 π bond; the π bond forms by head-on overlap of p orbitals along the C=C axis.

Bond data: C–Cl 0.177 nm, 340 kJ mol⁻¹; C–I 0.214 nm, 240 kJ mol⁻¹. Iodoalkanes are hydrolysed faster than chloroalkanes, a reaction in which the carbon–halogen bond breaks. Which explanation is correct?

Cl is more electronegative than I, so C–Cl is more polar; but C–I is longer and has the lower bond energy, so less energy is needed to break it, and bond strength decides the rate.
I is more electronegative than Cl, so C–I is the more polar bond and attracts the attacking species more strongly.
C–I is shorter than C–Cl, so its bonding electrons are held less tightly and the bond breaks more easily.
C–Cl releases more energy (340 kJ mol⁻¹) when it breaks, so breaking it is less favourable than breaking C–I.

Phosphorus (Group 15) forms PCl₃, in which P has three bonding pairs and each P–Cl bond is polar. Using VSEPR theory, which description of its shape, Cl–P–Cl bond angle and polarity is correct?

Trigonal pyramidal; less than 109.5°; a polar molecule
Trigonal planar; 120°; a non-polar molecule
Tetrahedral; 109.5°; a non-polar molecule
Trigonal pyramidal; less than 109.5°; a non-polar molecule because the three P–Cl bonds are identical
Trigonal pyramidal; greater than 109.5°; a polar molecule

Br₂ (boiling point 59 °C) and ICl (boiling point 97 °C) both have 70 electrons per molecule. ICl is a polar molecule. Why does ICl have the higher boiling point?

With equal numbers of electrons their instantaneous dipole–induced dipole attractions are similar, but ICl molecules also attract through permanent dipole–permanent dipole attractions.
The I–Cl covalent bond is stronger than the Br–Br bond, and boiling breaks these bonds.
ICl molecules form hydrogen bonds with each other.
Br₂ is non-polar, so there are no attractions between Br₂ molecules.

Water, H₂O, boils at 100 °C but hydrogen sulfide, H₂S, which has more electrons, boils at −60 °C. Which explanation is correct?

O is highly electronegative and small, so a δ+ H on one H₂O molecule is strongly attracted to a lone pair on O of another, forming hydrogen bonds; S–H bonds cannot do this, and more energy is needed to overcome hydrogen bonds.
The O–H covalent bonds are stronger than S–H bonds, and they must be broken when water boils.
H₂S has more electrons, so its molecules should attract more strongly; water's high boiling point shows it is an ionic compound.
Hydrogen bonds form between the δ+ hydrogen atoms of neighbouring water molecules.

Which description of the crystalline solids sodium chloride, copper, iodine, diamond and ice is correct?

NaCl: alternating Na⁺ and Cl⁻ ions, each with six oppositely charged neighbours. Copper: positive ions attracted to a sea of delocalised electrons. Iodine: I₂ molecules held by weak intermolecular attractions. Diamond: each C covalently bonded to four others tetrahedrally. Ice: H₂O molecules hydrogen-bonded in an open lattice.
NaCl: separate Na–Cl molecules packed together. Copper: positive ions in delocalised electrons. Iodine: I₂ molecules held by weak attractions. Diamond: tetrahedral covalent network. Ice: open hydrogen-bonded lattice.
NaCl: alternating ions with six neighbours. Copper: Cu atoms joined by localised shared electron pairs. Iodine: I₂ molecules held by weak attractions. Diamond: tetrahedral covalent network. Ice: open hydrogen-bonded lattice.
NaCl: alternating ions with six neighbours. Copper: positive ions in delocalised electrons. Iodine: a giant lattice of I atoms each bonded to four others. Diamond: tetrahedral covalent network. Ice: open hydrogen-bonded lattice.
NaCl: alternating ions with six neighbours. Copper: positive ions in delocalised electrons. Iodine: I₂ molecules held by weak attractions. Diamond: hexagonal layers held together by weak attractions. Ice: open hydrogen-bonded lattice.

Substance X melts at 1710 °C, does not conduct electricity when solid or molten, and is insoluble in water. Which structure and explanation fit these data?

A giant molecular lattice: melting breaks many strong covalent bonds, and there are no mobile ions or delocalised electrons.
A giant ionic lattice: strong electrostatic attraction between ions gives the high melting point.
A simple molecular lattice: strong covalent bonds inside the molecules give the high melting point.
A metallic lattice: strong metallic bonding gives the high melting point.

Butane, C₄H₁₀, is stored as a liquid under pressure in lighters at room temperature, but methane, CH₄, cannot be liquefied by compression alone at room temperature. Which explanation is correct?

Butane has more electrons, so its instantaneous dipole–induced dipole attractions are stronger; when compression brings the molecules close, these attractions can hold them together as a liquid at room temperature.
High pressure breaks C–H bonds in butane, so the fragments form a liquid.
Methane molecules are polar and repel each other, so they resist compression.
Any gas can be liquefied by compression alone; methane is simply kept at lower pressure.

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Chemical Bonding

H1 Chemical Bonding learning outcomes

What You’ll Learn

Lessons (Recommended Order)

Ionic and Covalent Bonding Models

Dative Bonding and Common Examples

Sigma and Pi Bonds: Orbital Overlap

Bond Energy, Bond Length and Bond Polarity

Molecular Shapes and Bond Angles

Intermolecular Forces and Properties

Solids: Structure and Physical Properties

Quick Reference

Practice and Check Your Understanding