Full chapter
Chemical Bonding
Explain particles, molecular shape and material properties through electrostatic attraction.
A-Level 8873, revised syllabus (2026-2027)
One idea behind three strong bonds
Identify the positive and negative charges that attract.
| Bond | Attracting particles | Example |
|---|---|---|
| Ionic | Positive and negative ions throughout a lattice | NaCl; MgO |
| Covalent | A shared electron pair and the positive nuclei of the bonded atoms | H2; Cl2 |
| Metallic | Positive metal ions and delocalised electrons | Copper |
Electron transfer explains ion formation; attraction between the resulting ions is the ionic bond. Sodium transfers one outer electron to chlorine, making Na+ and Cl-. Magnesium transfers two to oxygen, making Mg2+ and O2-. Each compound is neutral overall and extends as a lattice, not a collection of NaCl or MgO molecules.
Sodium chloride
Na (2,8,1) loses one electron. Cl (2,8,7) gains that electron.
Magnesium chloride
Mg (2,8,2) loses two electrons. Two Cl atoms each gain one.
- Electron from chlorine
- Electron from the metal
Each bracket shows the ion's full outer shell. The eight electrons on Na+ and Mg2+ were previously in an inner shell. Dots and crosses track origin; all electrons are the same kind of particle.
These groups show the ion ratio, not separate NaCl or MgCl2 molecules. The ions form an extended lattice.
Magnesium oxide: two electrons transferred
Mg2+ has eight outer-shell crosses. O2- has six original oxygen dots and two transferred crosses. Brackets and ion charges distinguish electron transfer from sharing.
In a covalent bond, electron density between nuclei attracts both nuclei. A shared pair can come one electron from each atom or both from one atom. Dots and crosses record origin; electrons themselves are identical. In a metal, electrons are delocalised through the structure, so attractions are not restricted to one fixed atom pair.
Check your understanding
Which statement best describes the covalent bond in a chlorine molecule, Cl2?
Show the answer
The attraction between a shared pair of electrons and the two Cl nuclei
All bonds are electrostatic. The shared pair lies between the nuclei and attracts both, holding the atoms together.
Make every electron count
Use shared pairs for bonds and keep the remaining electrons as lone pairs.
Count the total outer electrons first. A single bond uses one shared pair, a double bond two pairs, and a triple bond three. Hydrogen has a duet; the other simple molecules below achieve octets. An electron shared in a bond counts towards both bonded atoms, but is counted only once in the total-electron audit.
Each dot or cross is one outer-shell electron. A pair between atoms is shared; a pair beside one atom is a lone pair.
Hydrogen (H2)
- Left H
- Right H
Oxygen (O2)
- Left O
- Right O
Water (H2O)
- Oxygen
- Hydrogens
Methane (CH4)
- Carbon
- Hydrogens
Carbon dioxide (CO2)
- Carbon
- Oxygens
Ammonia (NH3)
- Nitrogen
- Hydrogens
Only outer-shell electrons are shown. Shared electrons count towards both bonded atoms' outer shells. Dots and crosses identify the source atom, not different types of electron.
Nitrogen: three shared pairs
Dots belong to the left atom and crosses to the right. Each mark is one electron. Shared pairs are between atoms; the other pairs are lone pairs.
Chlorine: one shared pair
Dots belong to the left atom and crosses to the right. Each mark is one electron. Shared pairs are between atoms; the other pairs are lone pairs.
Hydrogen chloride: one shared pair
Dots belong to the left atom and crosses to the right. Each mark is one electron. Shared pairs are between atoms; the other pairs are lone pairs.
Ethene: a double bond between the carbons
Each carbon has two C-H shared pairs and two C-C shared pairs. The left carbon supplies four dots and the right carbon four crosses. The two left hydrogens supply crosses and the two right hydrogens dots.
| Molecule | Bonding pairs | Lone pairs | Total outer electrons |
|---|---|---|---|
| H2 | 1 | 0 | 2 |
| O2 | 2 | 2 per O | 12 |
| N2 | 3 | 1 per N | 10 |
| Cl2 | 1 | 3 per Cl | 14 |
| HCl | 1 | 3 on Cl | 8 |
| CO2 | 4 | 2 per O | 16 |
| CH4 | 4 | 0 | 8 |
| C2H4 | 6 | 0 | 12 |
Check your understanding
How many lone pairs are on each oxygen atom in a correct dot-and-cross diagram of CO2?
Show the answer
Two on each oxygen, beside a C=O double bond
CO2 has 16 outer electrons. Two double bonds use 8. The other 8 form two lone pairs on each oxygen, giving every atom an octet.
When one atom supplies the shared pair
A lone-pair donor needs an electron-pair acceptor.
A co-ordinate or dative covalent bond contains a shared pair supplied entirely by one atom. NH3 has a lone pair on N; H+ has an empty orbital and no electron to contribute. Donation forms NH4+. After formation all four N-H bonds are equivalent; the origin of the pair does not create a permanently weaker bond.
Ammonium: nitrogen donates its lone pair
Five dots represent the nitrogen electrons and three crosses come from the original hydrogens. The upper N-H pair contains two dots because its hydrogen arrived as H+. The whole ion is bracketed with a single positive charge.
In gaseous AlCl3 under suitable conditions, an Al atom has only six electrons in its three covalent bonds. Two units join to form Al2Cl6: a Cl from each unit donates a lone pair to the electron-deficient Al in the other unit. Two bridging Cl atoms result; each Al is surrounded by four shared pairs.
Al2Cl6: two chlorine bridges
Two Al atoms each make four bonds. Each has two terminal chlorines and two shared bridging chlorines. Six mixed dot-cross pairs are original bonds; two cross-cross pairs are donated chlorine lone pairs. Terminal chlorines have three lone pairs each; bridging chlorines have two.
Check your understanding
Why does NH4+ have eight outer electrons around nitrogen, not nine?
Show the answer
N and four H atoms supply 9, and the 1+ charge means one electron fewer.
5 + 4 = 9 electrons from the neutral atoms, minus one for the positive charge, gives 8. Or: NH3 supplies 8 and H+ supplies none.
Orbital overlap builds bonds; repulsion sets shape
Separate the type of overlap from the arrangement of electron domains.
A sigma bond forms by head-on overlap along the line joining the nuclei. It can use s-s overlap (H2), s-p overlap (HCl) or p-p overlap. A pi bond forms by sideways overlap of parallel p orbitals, with electron density above and below the internuclear axis. A single bond is sigma; a double bond is one sigma plus one pi; a triple bond is one sigma plus two pi.
Head-on and sideways overlap
The sigma overlap lies on the internuclear axis. Parallel p orbitals overlap sideways in two regions for one pi bond, leaving the axis as a nodal region for that pi orbital.
VSEPR places electron domains as far apart as possible to reduce repulsion. Around a central atom, a single, double or triple bond counts as one bonding domain. Lone pairs also repel, but are omitted from the name of the molecular shape. Lone-pair/bond-pair repulsion is greater than bond-pair/bond-pair repulsion, compressing some angles.
| Species | Bonding domains; lone pairs | Molecular shape | Angles |
|---|---|---|---|
| BF3 | 3; 0 | Trigonal planar | 120 degrees |
| CO2 | 2; 0 | Linear | 180 degrees |
| CH4 | 4; 0 | Tetrahedral | 109.5 degrees |
| NH3 | 3; 1 | Trigonal pyramidal | About 107 degrees |
| H2O | 2; 2 | Bent | About 104.5 degrees |
| SF6 | 6; 0 | Octahedral | 90 and 180 degrees |
The table follows the same repulsion argument throughout: two bonding domains point in opposite directions (linear), three spread in one plane (trigonal planar), and four point towards the corners of a tetrahedron. Six equivalent domains point along three perpendicular axes (octahedral). NH3 and H2O both start from four electron domains. Replace one tetrahedral bonding direction by a lone pair to obtain a pyramidal atom arrangement; replace two to obtain a bent arrangement. Lone pairs occupy more angular space near the central atom, so they compress the remaining bond angles.
Worked exampleTransfer the method to an unfamiliar ion
Predict the shape and bond angle of NH4+.
Try it first. Then show the worked stepsHide the worked steps
- Its four shared pairs are four bonding domains around nitrogen.
- There is no lone pair on nitrogen.
- Four domains adopt a tetrahedral arrangement to minimise repulsion.
NH4+ is tetrahedral, about 109.5 degrees. Its parent NH3 is pyramidal because NH3 has a lone pair.
Worked exampleFind the lone pairs before naming the shape
Deduce the shape of SCl2, taking sulfur as the central atom.
Try it first. Then show the worked stepsHide the worked steps
- Count 6 + 2(7) = 20 outer electrons. Draw two S-Cl single bonds, using four electrons.
- Complete each terminal Cl octet with three lone pairs: twelve more electrons. The four electrons left belong on S as two lone pairs.
- Only count domains around S: two S-Cl bonding domains plus two lone pairs, giving four. The Cl lone pairs do not add to the domain count around S.
- The electron-domain arrangement is tetrahedral. Naming only the atoms gives a bent molecule, analogous to H2O. Lone-pair repulsion compresses the Cl-S-Cl angle below 109.5 degrees.
SCl2 is bent and polar because its equal S-Cl dipoles do not cancel. VSEPR gives an angle smaller than 109.5 degrees; it does not justify copying the exact 104.5-degree water value to every bent molecule.
Sulfur in SF6 has twelve outer electrons, but oxygen in the same group never goes past eight. An element in Period 3 or below has empty d orbitals in its outer shell (3d for Period 3). They are close enough in energy to be used, so the central atom can hold five or six electron domains. A Period 2 atom such as N, O or F has only 2s and 2p orbitals, so it is limited to four domains and eight electrons. That is why PCl5 and SF6 exist, but NCl5 and OF6 do not. This is called expanding the octet.
| Example | Bonding domains; lone pairs | Molecular shape | Angles |
|---|---|---|---|
| PCl5 | 5; 0 | Trigonal bipyramidal | 90, 120 and 180 degrees |
| SF4 | 4; 1 | See-saw | Less than 90 and less than 120 degrees |
| ClF3 | 3; 2 | T-shaped | Slightly less than 90 degrees |
| XeF2 | 2; 3 | Linear | 180 degrees |
| BrF5 | 5; 1 | Square pyramidal | Slightly less than 90 degrees |
| XeF4 | 4; 2 | Square planar | 90 degrees |
Where the lone pairs go decides the shape. With five domains, the three equatorial positions are 120 degrees apart, while the two axial positions are at 90 degrees to three others. A lone pair takes an equatorial position, where it has fewer close neighbours at 90 degrees. So one lone pair gives a see-saw, two give a T-shape and three give a straight line. With six domains every position is the same, so one lone pair gives a square pyramid. A second lone pair goes opposite the first, as far from it as possible, which leaves a square plane.
Check your understanding
Phosphorus forms PCl3 and PCl5. Nitrogen, in the same group, forms NCl3 but not NCl5. Why?
Show the answer
P is in Period 3 and can use empty 3d orbitals; N in Period 2 has no 2d orbitals.
Phosphorus can hold five bonding pairs, ten electrons, by using its 3d orbitals. Nitrogen is limited to four electron pairs in its second shell, so it cannot form five bonds.
Ethene contains five sigma bonds in total (four C-H and one C-C) plus one C-C pi bond. Sideways overlap requires the two p orbitals to remain aligned, so the double bond cannot rotate freely without disrupting that overlap. This explains later cis-trans isomerism.
Exam-style question
Predict the shape of a ClF3 molecule and its F-Cl-F bond angle. Explain your answer. [3]
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A polar bond does not guarantee a polar molecule
Combine bond direction, shape and the possible attractions between particles.
Make a guess. You are not marked.
H2S molecules have more electrons than H2O molecules. Which has the higher boiling point?
Show the answer
H2O, because its molecules form hydrogen bonds
Water molecules form hydrogen bonds, because H is bonded to the small, very electronegative O. These are far stronger than the extra dispersion forces in H2S.
Electronegativity describes an atom's attraction for a bonding electron pair. In H-Cl, Cl attracts the pair more strongly: H is partially positive and Cl partially negative. These are partial charges within a covalent bond, not full H+ and Cl- ions. The molecular dipole is the combined effect of all bond dipoles in three dimensions.
| Molecule | Reason | Overall polarity |
|---|---|---|
| CO2 | Equal C=O bond dipoles point in opposite directions in a linear molecule. | Non-polar |
| BF3 | Three equal B-F dipoles cancel in its trigonal plane. | Non-polar |
| SF6 | Six equal S-F dipoles cancel in an octahedron. | Non-polar |
| H2O | The bent shape prevents the O-H dipoles cancelling. | Polar |
| NH3 | Pyramidal arrangement gives a net dipole. | Polar |
| CHCl3 | The tetrahedral molecule has unlike surrounding atoms, so its dipoles do not cancel. | Polar |
Instantaneous dipole-induced dipole attractions occur in all particles: a momentary uneven electron distribution induces a dipole in a neighbouring particle, and the nearby opposite partial charges attract electrostatically. They are the principal intermolecular attraction in Br2 and the attraction between noble-gas atoms. Larger, more easily polarised electron clouds usually give stronger attractions. CHCl3 also has permanent dipole-permanent dipole attraction between oppositely charged ends of neighbouring molecules.
Hydrogen bonding is an electrostatic attraction between the partially positive H covalently bonded to N, O or F and a lone pair on N, O or F in a neighbouring particle. Water provides O-H donors and oxygen lone pairs; ammonia provides N-H donors and a nitrogen lone pair. Merely containing hydrogen is insufficient: methane cannot hydrogen-bond to itself.
The same possible attractions exist in a gas and its liquid, but gas particles are usually far apart. Cooling reduces kinetic energy and compression brings particles closer, allowing attractions to hold particles in a liquid. A larger attraction generally makes liquefaction easier. Pressure alone does not guarantee liquefaction at every temperature; this is a qualitative explanation, not an ideal-gas calculation.
Hydrogen bonding makes water's boiling point relatively high for such a small molecule because substantial energy is needed to separate molecules. In ordinary ice, each water molecule hydrogen-bonds to four neighbours in an open, approximately tetrahedral network: its two O-H groups donate two hydrogen bonds and its two oxygen lone pairs accept two. On melting, some of that open arrangement collapses, so liquid water is denser than ice and ice floats. The covalent O-H bonds remain intact.
Check your understanding
CH4 and H2O are both small molecules. Why does water boil at a much higher temperature?
Show the answer
Water molecules form hydrogen bonds; methane has only weak instantaneous dipole attractions.
More energy is needed to separate water molecules, held by hydrogen bonds. Boiling does not break the covalent O-H bonds.
Compare the bond you are changing
Bond length, strength and polarity explain different aspects of reactivity.
Bond length is the equilibrium distance between the nuclei of two bonded atoms. Bond energy is the energy required to break one mole of a specified covalent bond in gaseous species; tabulated average bond energies average over molecular environments. Bond breaking is endothermic. For similar bonds, greater length usually means poorer overlap and a weaker bond.
Compare C-Cl, C-Br and C-I: the halogen atom becomes larger down the group, so the bond becomes longer and generally weaker. Breaking C-I is easier than breaking C-Cl, despite C-Cl being more polar. Polarity instead identifies electron-poor and electron-rich sites, which influences how a reagent approaches and interacts with a molecule.
Watch out for this
A student says: "C=C is a stronger bond than C-C, so alkenes are less reactive than alkanes." What is wrong with this?
Show the answer
Bond energy alone does not rank the reactivity of every molecule. A C=C double bond is stronger overall than a C-C single bond, yet its accessible pi electron density enables addition reactions. State which bond or electron-rich region is involved and which reaction is being compared.
Check your understanding
Why does the H-I bond have a lower bond energy than the H-Cl bond?
Show the answer
I is larger, so the H-I bond is longer and the shared pair is held less strongly.
The shared pair is further from the iodine nucleus and the orbital overlap is less effective. The attraction is weaker, so less energy is needed to break one mole of H-I bonds in the gas.
Check your understanding
1-iodobutane reacts with aqueous hydroxide ions faster than 1-chlorobutane does. Which is the best explanation?
Show the answer
C-I is longer with a lower bond energy, so it breaks more easily
The C-X bond must break in the reaction. C-I is the weakest carbon-halogen bond, and bond strength outweighs polarity here.
Turn structures into properties, then reverse the argument
Name the particles, attraction and charge carriers before drawing a conclusion.
| Solid | Particles and structure | Properties explained |
|---|---|---|
| NaCl; MgO | Giant ionic lattice; opposite ions attract. MgO contains 2+ and 2- ions. | High melting points; fixed ions do not conduct in the solid; mobile ions conduct when molten. MgO has stronger attractions than NaCl. |
| Iodine | Discrete I2 molecules in a molecular lattice. | Comparatively low melting point: overcome intermolecular forces, not I-I bonds. No mobile charge carriers. |
| Diamond | Each carbon bonds to four others in a three-dimensional covalent network. | Hard, very high melting temperature; no mobile charged particles. |
| Graphite | Each carbon bonds to three others in a sheet; electrons are delocalised. Weaker attractions act between sheets. | High melting temperature; sheets slide, giving softness; delocalised electrons conduct. |
| Ice | Discrete water molecules joined into an open hydrogen-bonded lattice. | Less dense than liquid water; melting disrupts hydrogen bonding, not O-H bonds. |
| Copper | Positive ions in a lattice with delocalised electrons. | Conducts when solid and molten; layers can slide while metallic attraction remains. |
For melting, identify the attraction that must be overcome. For conductivity, identify mobile charged particles. For solubility, consider whether new solute-solvent attractions can compensate for attractions disrupted in both substances; "ionic" is not a guarantee of water solubility. For malleability, explain whether a displaced layer remains held together or brings like charges into unfavourable proximity.
Worked exampleUse two measurements together
Solid X has a high melting point. It is an electrical insulator as a solid, but its melt conducts. Suggest a structure.
Try it first. Then show the worked stepsHide the worked steps
- A high melting point indicates strong attractions, but is consistent with several structures.
- The molten sample has mobile charged particles.
- An ionic lattice supplies fixed ions in the solid and mobile ions in the melt.
- A metal would usually conduct as a solid; a simple molecular solid would usually melt much lower.
A giant ionic structure best fits both observations. Do not claim a specific compound from this evidence alone.
Worked exampleDistinguish two carbon structures
Two carbon solids both withstand high temperatures. Only one conducts electricity and leaves a slippery mark. Identify them.
Try it first. Then show the worked stepsHide the worked steps
- Conductivity implies mobile delocalised electrons: graphite.
- The slippery mark is consistent with graphite sheets sliding.
- The other can be diamond: a rigid 3-D network with all four outer electrons involved in localised bonds.
Graphite is the conducting, slippery solid; diamond is the non-conducting network. High melting temperature alone does not distinguish them.
Check your understanding
Iodine melts at a low temperature, although the I-I bond is a strong covalent bond. Why?
Show the answer
Melting separates I2 molecules by overcoming weak forces between them; the I-I bonds stay intact.
Iodine is simple molecular. Only the weak instantaneous dipole attractions between molecules are overcome, so little energy is needed.
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Term | Definition to write |
|---|---|
| Ionic bond | The electrostatic attraction between oppositely charged ions. |
| Covalent bond | The electrostatic attraction between a shared pair of electrons and the nuclei of the two bonded atoms. |
| Metallic bond | The electrostatic attraction between a lattice of positive metal ions and a sea of delocalised electrons. |
| Electronegativity | The ability of an atom in a molecule to attract the shared pair of electrons in a covalent bond towards itself. |
| Bond energy | The average energy required to break one mole of a particular covalent bond in gaseous molecules. |
| Bond length | The distance between the nuclei of two covalently bonded atoms. |
| Hydrogen bond | An attraction between an H atom bonded to N, O or F (so carrying a large δ+) and a lone pair on an N, O or F atom of a neighbouring molecule. |
- Particles
Atoms, molecules, ions or ions with delocalised electrons?
- Attraction
Which electrostatic attraction holds the structure?
- Change
Which attraction is overcome, or which charge carrier moves?
- Conclusion
Link that mechanism directly to the measured property.
| Pair of ideas | Difference |
|---|---|
| Bond polarity / molecular polarity | Polar bonds may cancel through symmetry. |
| Sigma / pi | Head-on axial overlap / sideways p overlap. |
| Melting iodine / breaking I2 | Intermolecular separation / covalent bond breaking. |
| NH3 / NH4+ | Three bonds plus lone pair / four bonds, no lone pair. |
| Diamond / graphite | 3-D four-coordinate network / three-coordinate sheets with delocalised electrons. |
Scope
Learning outcomes
2. Chemical Bonding (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
2(a) Explain the electrostatic basis of bonding.
- (i) Ionic: opposite ions.
- (ii) Covalent: shared pair and positive nuclei.
- (iii) Metallic: positive lattice and delocalised electrons.
2(b) Explain bonding using electron diagrams.
- (i) NaCl and MgO.
- (ii) H2, O2, N2, Cl2, HCl, CO2, CH4 and ethene.
- (iii) Dative bonding in NH4+ formation and Al2Cl6.
One idea behind three strong bondsMake every electron countWhen one atom supplies the shared pair
2(c) Connect overlap to sigma and pi bonds.
- s and p orbital overlap only; connection to organic shapes.
2(d) Explain the required shapes with VSEPR.
- BF3 120; CO2 180; CH4 109.5; NH3 about 107; H2O about 104.5; SF6 90/180 degrees.
2(e) Predict analogous shapes and angles.
- Count bonding domains and lone pairs; apply the listed geometries.
2(f) Infer bond polarity.
- Electronegativity; qualitative treatment, not numerical calculations.
2(g) Infer molecular polarity.
- Combine bond dipoles with shapes analogous to 2(d).
2(h) Explain intermolecular attractions.
- (i) Permanent and induced dipoles in CHCl3, Br2 and noble gases, liquid/gas.
- (ii) Hydrogen bonding in NH3 and H2O; N-H and O-H groups.
2(i) Explain gas liquefaction.
- High pressure and/or low temperature make intermolecular attractions important.
2(j) Connect hydrogen bonding to properties.
- Water boiling behaviour and open ice structure/density.
2(k) Explain bond energy and length.
- Covalent bonds; gaseous bond breaking and equilibrium internuclear separation.
2(l) Compare covalent bond reactivity.
- Bond energy, length and polarity; use comparable reaction contexts.
2(m) Describe crystalline lattice types.
- (i) NaCl/MgO ionic.
- (ii) Iodine molecular.
- (iii) Diamond/graphite giant molecular.
- (iv) Ice hydrogen-bonded.
- (v) Copper metallic; no unit cells.
2(n) Predict physical properties from structure.
- Melting, conductivity, mechanical behaviour and qualified solubility reasoning.
2(o) Infer structure from properties.
- Combine evidence; distinguish mobile ions from electrons.