Full chapter
Chemical Bonding
Use electrons, molecular shape and electrostatic attraction to explain structures, properties and reactivity.
A-Level 9476 (2026-2027)
All chemical bonds involve attraction between charges
Identify the positive and negative particles before naming the bond.
| Bond | Attracting particles | What not to confuse it with |
|---|---|---|
| Ionic | Oppositely charged ions in an extended structure. | Electron transfer creates the ions; transfer itself is not the attractive force. |
| Covalent | A shared electron pair and the positively charged nuclei of both bonded atoms. | It is not an attraction between two negative electrons. |
| Metallic | A lattice of positive ions and delocalised electrons. | The electrons are not confined to individual two-atom bonds. |
In NaCl, sodium loses one electron and chlorine gains one, forming Na+ and Cl-. In MgO, magnesium loses two and oxygen gains two, forming Mg2+ and O2-. Both products have full outer shells and zero overall charge. The formula gives an ion ratio; neither solid consists of isolated two-ion molecules.
NaCl and MgO: ions after transfer
The upper row shows Na+ and Cl-; the lower row Mg2+ and O2-. Each bracketed ion has eight outer-shell electrons. Chloride shows seven dots and one transferred cross; oxide six dots and two transferred crosses. Metal ions show eight remaining electrons, all crosses, from the formerly inner shell.
An ionic lattice is held by many attractions in all directions. Larger charge magnitudes and smaller interionic distances strengthen electrostatic attraction. MgO therefore has much stronger lattice attraction than NaCl: its ions have charges of magnitude two rather than one. Compare the whole structure rather than imagining a single Na-Cl covalent bond.
Count shared pairs and every lone pair
A single, double or triple bond contains one, two or three shared pairs.
Dots and crosses track which atom supplied each electron; they do not identify different kinds of electron. A shared pair contributes to the outer-shell count of both atoms. Hydrogen needs a duet in its first shell; the C, N, O and Cl examples below reach octets. The diagram counts electrons and does not necessarily show molecular shape.
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
The two nitrogen atoms share three electron pairs. Each has one lone pair. Five dots come from the left nitrogen and five crosses from the right nitrogen, giving ten outer electrons in total and an octet around each atom.
Chlorine and hydrogen chloride
The first row shows Cl2 with one shared pair and three lone pairs on each chlorine. The second shows HCl with one shared pair and three lone pairs on chlorine. Hydrogen has only two electrons around it.
Ethene: account for twelve outer electrons
Two carbons share two electron pairs. Each carbon also shares one pair with each of two hydrogens. All carbons have octets and all hydrogens duets. Dots come from the left carbon and the two right-hand hydrogens. Crosses come from the right carbon and the two left-hand hydrogens.
A dative bond uses two electrons from one donor
Formation differs; after formation it is a covalent shared pair.
In a co-ordinate (dative covalent) bond, one atom supplies both electrons of the shared pair. The donor must have a lone pair, and the acceptor must have an available orbital. An arrow points from donor to acceptor when showing bond formation. It does not mean that the finished bond contains a different kind of electron.
NH3 donates its lone pair to H+
The ammonium ion has four N-H shared pairs and no nitrogen lone pair. The upper N-H pair contains two dots donated by nitrogen. The other three pairs contain one nitrogen dot and one hydrogen cross. The whole structure is bracketed with charge plus one.
AlCl3 is electron deficient in its simple monomer description: Al has three shared pairs, only six electrons around it. Two units can associate to make Al2Cl6. A chlorine lone pair from each unit donates to the other aluminium. Two chlorine atoms bridge the Al centres, and each Al then has four shared pairs.
Al2Cl6: two chlorine bridges
Two aluminium atoms and six chlorine atoms form a dimer. Four chlorines are terminal with three lone pairs each. Two bridging chlorines each have two lone pairs and two shared pairs. Each aluminium supplies three crosses to ordinary shared pairs and receives a two-dot donated pair. The complete drawing contains 48 outer electrons.
Check your understandingDoes NH4+ keep one unusually weak N-H bond because it was formed by donation?Think it through, then reveal the answer
End-on overlap gives sigma; sideways overlap gives pi
A double bond contains one sigma and one pi bond.
A sigma (σ) bond has electron density along the line joining the nuclei. It forms by end-on overlap: s-s in H2, s-p in a simple H-Cl description, or p-p. A pi (π) bond forms by sideways overlap of parallel p orbitals, producing electron density above and below that line. The two pi regions together constitute one bond.
The extra bond in ethene comes from sideways overlap
Two nuclei lie on the horizontal internuclear axis. A sigma bond occupies the region directly between them. Parallel vertical p orbitals on the two carbons overlap above and below that axis to form one pi bond.
| Connection | Sigma bonds | Pi bonds |
|---|---|---|
| Single bond | 1 | 0 |
| Double bond | 1 | 1 |
| Triple bond | 1 | 2 in mutually perpendicular planes |
| Whole ethene molecule | 5: four C-H and one C-C | 1: between the carbon atoms |
A carbon-carbon single bond can usually rotate without destroying its sigma overlap. Rotation about a double bond requires loss of the pi overlap and is restricted. Bond rotation is therefore not the same as rotating the entire molecule in space. Detailed d-orbital overlap is outside the overlap requirement here.
Count electron regions, then describe the atoms
Multiple bonds count as one region. Lone pairs affect shape even though they are not atoms.
VSEPR places regions of electron density around a central atom as far apart as possible. A single, double or triple bond counts as one region for this purpose. A lone pair is another region. The electron-region arrangement includes lone pairs; the molecular shape names the positions of the atoms only.
| Molecule | Bonding regions / lone pairs | Molecular shape | Typical bond angle |
|---|---|---|---|
| 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 degrees between adjacent bonds; 180 between opposite bonds |
A lone pair is held by only one nucleus and occupies more space near the central atom than a bonding pair. The usual repulsion order is lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. The tetrahedral electron arrangement therefore gives progressively smaller bond angles from CH4 to NH3 to H2O. Exact angles also depend on the atoms and bonds: do not assign water's angle to every bent molecule.
Worked example
Deduce an unfamiliar ion
Predict the shape of H3O+.
- Count valence electrons: oxygen contributes six; three hydrogens add three; the positive charge subtracts one. Total = eight.
- Three O-H shared pairs use six electrons, leaving one lone pair on O.
- Four electron regions give a tetrahedral arrangement; three bonded atoms and one lone pair give a trigonal pyramidal molecular shape.
Trigonal pyramidal. A basic VSEPR comparison predicts an angle near the tetrahedral value, modified by its lone pair; it does not justify copying the exact NH3 angle.
Worked example
A double bond is one region
SO2 has two S-O bonding regions and one lone pair on S. What shape follows?
- There are three electron regions, so their reference arrangement is trigonal planar.
- Only two of those regions contain bonded atoms.
- The molecule is bent, with an O-S-O angle slightly below the 120-degree reference.
Bent, not linear. Counting each double bond as two directions would give the wrong model.
A polar bond need not make a polar molecule
Add the bond-dipole directions using the actual three-dimensional shape.
Electronegativity is an atom's ability to attract the shared electrons of a covalent bond. Unequal attraction creates partial charges, δ+ and δ-. In H-Cl, chlorine attracts the pair more strongly, so H is δ+ and Cl is δ-. These are partial charges within a covalent bond, not a declaration that HCl is an ionic lattice.
| Molecule | Bond dipoles | Overall result |
|---|---|---|
| CO2 | Two C-O dipoles point in opposite directions along a straight line. | Cancel: non-polar molecule. |
| BF3 | Three identical B-F dipoles arranged symmetrically at 120 degrees. | Cancel: non-polar. |
| CH4 / CCl4 | Four equivalent bonds point to the corners of a tetrahedron. | Cancel: non-polar. |
| SF6 | Opposite S-F dipoles pair in an octahedron. | Cancel: non-polar. |
| H2O | O-H dipoles meet at a bent angle. | Do not cancel: polar. |
| NH3 | Three N-H bonds form a pyramid, not a flat triangle. | Do not cancel: polar. |
| CHCl3 | A tetrahedron with one H and three Cl substituents is not dipole-symmetric. | Polar. |
Worked example
Same general shape, different polarity
Both CO2 and OCS are linear. Why can OCS be polar?
- Linearity can make equal and opposite dipoles cancel.
- In CO2 the two C-O bonds are equivalent.
- OCS has different bonds at the two ends, so equal magnitudes cannot be assumed. Their vector sum need not be zero.
Shape alone does not decide polarity. Cancellation requires the appropriate symmetry and equivalent opposing bond dipoles.
Attractions between molecules have an electrostatic origin
All particles can form instantaneous dipoles; only suitable molecules form hydrogen-bond networks.
At any instant an electron cloud may be uneven, creating an instantaneous dipole. It induces a dipole in a neighbour, and opposite partial charges attract. These instantaneous dipole-induced dipole attractions occur in all atoms and molecules, including non-polar Br2 and the monatomic liquid noble gases. A larger, more easily distorted electron cloud is generally more polarizable and gives stronger attractions.
Permanent dipole-dipole attractions also occur between polar molecules, such as CHCl3. Favourable orientations bring a δ+ region near a δ- region. CHCl3 still has instantaneous-dipole attractions as well; do not choose exactly one force for every substance.
Hydrogen bonding is a particularly strong, directional interaction involving H covalently bonded to N, O or F and a suitable lone pair on N, O or F of another molecule (or another part of the same molecule). In water, O-H groups are donors and oxygen lone pairs are acceptors. In ammonia, N-H groups donate and the nitrogen lone pair accepts.
- Find possible hydrogen-bond donors and acceptors
An O atom alone is not enough to make a pure substance donate hydrogen bonds; it needs an appropriate O-H, N-H or F-H group.
- Check molecular polarity
Polar molecules also have permanent dipole attractions.
- Compare polarizability and shape
Electron-cloud size and contact between molecules affect instantaneous-dipole attractions.
- State what separates on boiling
Molecules separate; ordinary covalent bonds inside them remain intact.
Check your understandingWhy can Br2 and argon liquefy although neither has a permanent molecular dipole?Think it through, then reveal the answer
Hydrogen bonding makes ice an unusually open solid
An open network explains floating ice; persistent attractions help explain water's thermal properties.
In ordinary ice, water molecules form an open hydrogen-bonded network. Each water molecule can donate two hydrogen bonds through its O-H groups and accept two through oxygen lone pairs. The roughly tetrahedral arrangement leaves more empty space than in liquid water.
On melting, enough of the network is disrupted for molecules to pack more closely. Ice is therefore less dense than liquid water and floats. Liquid water still contains hydrogen bonds: melting does not remove every attraction, and it does not break water into H and O atoms.
| Observation | Molecular explanation |
|---|---|
| Water boils unusually high for a small molecule. | Substantial energy is required to overcome its extensive intermolecular hydrogen bonding. |
| Ice floats. | The open solid network has a larger volume per molecule and lower density. |
| Water has a high heat capacity and enthalpy of vaporisation. | Energy changes involve an extensive network of intermolecular attractions as well as molecular motion. |
| Small alcohols mix readily with water. | Their O-H groups can hydrogen-bond with water; larger non-polar portions can reduce solubility. |
Ammonia also hydrogen-bonds, but its available lone-pair sites and the strength and extent of the network differ from water. Do not explain every boiling-point comparison by molar mass alone, or claim that any molecule containing hydrogen can hydrogen-bond.
A strong bond is not automatically a non-polar bond
Energy, length and polarity answer different questions about a reaction.
Bond length is the equilibrium distance between two bonded nuclei. Bond energy is the energy required to break one mole of specified covalent bonds in gaseous species. A mean bond energy averages the same bond type across compounds; its use in a calculation is approximate, not an exact measurement for every molecule.
For bonds between the same elements, higher bond order generally means a shorter, stronger bond: C≡C is shorter and has greater total bond energy than C=C, which is shorter and stronger than C-C. But the pi part of a multiple bond can react while the sigma framework remains. A large total double-bond energy does not make an alkene generally less reactive than an alkane.
Worked example
Why polarity alone fails for C-halogen reactivity
C-F is very polar. Does that make a fluoroalkane the easiest halogenoalkane to hydrolyse?
- Polarity creates an electron-deficient carbon that can attract a nucleophile.
- However, the C-F bond is particularly short and strong, so breaking it is difficult.
- Down the common Cl, Br, I series, C-halogen bonds generally get longer and weaker; comparable iodoalkanes often hydrolyse faster.
No. Bond polarity identifies a possible attack site; bond strength and the reaction pathway also control the rate.
Different solids contain different moving charges and attractions
Name the particles, arrangement and attraction before predicting a property.
| Example and structure | Particles and attraction | Properties to explain |
|---|---|---|
| NaCl and MgO: ionic | An extended array of oppositely charged ions; electrostatic lattice attraction. | High melting points. Fixed ions prevent solid conduction; mobile ions conduct when molten or suitably dissolved. MgO has stronger attractions than NaCl. |
| I2: simple molecular | Discrete I2 molecules; strong covalent bonds within each, weaker attractions between them. | Relatively low melting point; no mobile charged particles in the pure solid or liquid. |
| Diamond: giant covalent | Each C covalently bonds to four C atoms in a three-dimensional network. | Very hard; high thermal stability; no mobile electrons for electrical conduction. |
| Graphite: giant covalent layers | Each C bonds to three in a sheet; delocalised electrons within sheets and weaker attractions between sheets. | Conducts along sheets; layers slide, so it is soft despite strong in-plane bonding. |
| Ice: hydrogen-bonded molecular | Discrete H2O molecules in an open hydrogen-bond network. | Lower density than liquid water; melting changes intermolecular organisation. |
| Copper: metallic | Positive metal ions and delocalised electrons throughout the metal. | Electrical/thermal conduction; layers can shift while metallic attraction persists. |
Diamond: four bonds per carbon
Graphite: three bonds per carbon
Silicon dioxide: a repeating network
Original network fragments, not to scale. Bond stubs continue beyond each drawing. Diamond and silica extend in three dimensions; graphite forms extended sheets. The drawings show connections, not measured bond angles.
Crystalline means a regular, repeating arrangement; it does not mean that every crystalline material is ionic. The unit-cell construction is not required. Electrical conductivity needs mobile charged particles, so identify whether the carrier is an electron or ion and whether it can move in the stated physical state.
Solubility depends on the balance of attractions broken and formed, as well as disorder, not a universal "ionic means soluble" rule. Some ionic compounds dissolve very little. Molecular substances capable of strong interactions with water can dissolve; non-polar substances often dissolve better in non-polar solvents.
Use several observations to identify a structure
A property is evidence; it is rarely a unique label by itself.
Worked example
An unknown solid
A solid melts at a high temperature, does not conduct when solid, but conducts when molten. What is the most plausible structure?
- High melting temperature suggests strong attractions, but could fit several giant structures.
- Failure to conduct as a solid argues against an ordinary metal and conducting graphite.
- Conduction after melting is consistent with charged ions becoming mobile.
A giant ionic lattice is the best explanation from these observations. Say which result distinguishes it; high melting point alone is insufficient.
Worked example
A soft conductor
Another solid conducts electricity, is soft and has very high thermal stability. Why is graphite plausible?
- Delocalised electrons account for conduction.
- Weak interlayer attractions permit sliding and account for softness.
- Strong covalent bonds within layers explain the high thermal stability.
Different parts of the structure explain different properties. Calling the whole structure "weakly bonded" contradicts its thermal behaviour.
- Observe before identifying
Record appearance, melting behaviour, solubility and conductivity under stated conditions.
- Control the conductivity comparison
Use comparable temperature, electrode arrangement and solution concentration; avoid confusing a wet sample with the dry solid.
- Match carriers to state
Electron conduction can occur in a solid metal; ionic conduction requires mobile ions.
- Keep conclusions proportional
Suggest a structure consistent with the evidence and identify any extra test needed to distinguish alternatives.
Quick revision
Revisit the essentials, then return to an explanation when you need it.
- Name the particles and arrangement
Ions, atoms or molecules; discrete units, sheets or an extended network.
- Name the relevant attraction
Ionic, covalent, metallic or a specific intermolecular attraction.
- Connect to the observation
Energy to overcome attraction explains melting; mobile charges explain conductivity.
| Do not confuse | Reason |
|---|---|
| Bond polarity / molecular polarity | Dipoles may cancel by symmetry. |
| Covalent bond / intermolecular force | Melting a molecular solid does not break its molecules. |
| Sigma / pi | Double = 1 sigma + 1 pi; triple = 1 sigma + 2 pi. |
| Electron arrangement / molecular shape | Lone pairs count as regions but are omitted from the atom-based shape name. |
| Dative formation / finished bond | The donor supplies the pair; the finished bond is covalent. |
| Bond energy / reaction rate | The pathway and activation barrier also matter. |
Scope and references
Learning outcomes and sources
2. Chemical Bonding. Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
2(a) Explain the electrostatic nature of chemical bonds.
- (i) Oppositely charged ions
- (ii) Shared pair and positive nuclei
- (iii) Positive-ion lattice and delocalised electrons
2(b) Describe bonding with complete electron diagrams.
- (i) NaCl and MgO
- (ii) H2, O2, N2, Cl2, HCl, CO2, CH4 and ethene
- (iii) NH4+ formation and Al2Cl6 dative bonds
All chemical bonds involve attraction between chargesCount shared pairs and every lone pairA dative bond uses two electrons from one donor
2(c) Explain covalent bonds using orbital overlap.
- s and p overlap only
- Sigma and pi bonds
- Connection to organic structures
2(d) Explain required molecular shapes and angles by VSEPR.
- BF3 trigonal planar
- CO2 linear
- CH4 tetrahedral
- NH3 trigonal pyramidal
- H2O bent
- SF6 octahedral
2(e) Predict analogous shapes and angles.
- Electron-region count
- Bond pairs and lone pairs
- Analogous unfamiliar molecules and ions
2(f) Explain bond polarity using electronegativity.
- Qualitative electronegativity
- Partial charges; no numerical electronegativity calculation
2(g) Deduce molecular polarity.
- Bond dipoles
- Three-dimensional molecular shape
- Symmetry and cancellation in analogous shapes
2(h) Explain electrostatic attractions between particles.
- (i) Permanent and induced dipoles: CHCl3(l), Br2(l), liquid noble gases
- (ii) Hydrogen bonding in NH3 and H2O; N-H and O-H groups
2(i) Link hydrogen bonding to physical properties.
- Ice and liquid water
- Density and melting
- Thermal properties and suitable solubility examples
2(j) Define covalent bond energy and length.
- Gas-phase bond breaking per mole
- Mean energy versus a specific bond
- Equilibrium internuclear distance
2(k) Compare bond reactivity using structure.
- Bond energy
- Bond length
- Bond polarity and limits of one-factor predictions
2(l) Describe the required crystal structures.
- (i) Ionic NaCl/MgO
- (ii) Simple molecular iodine
- (iii) Giant molecular diamond/graphite
- (iv) Hydrogen-bonded ice
- (v) Metallic copper
- Unit-cell concepts not required
Different solids contain different moving charges and attractions
2(m) Predict properties from structure and bonding.
- Strength of attractions
- Mobile charge carriers
- State, melting, conductivity, mechanical behaviour and solubility
Different solids contain different moving charges and attractionsUse several observations to identify a structure
2(n) Infer structure and bonding from supplied evidence.
- Use multiple observations
- Distinguish evidence from a unique identification
- State-dependent conductivity
- SEAB H2 Chemistry 9476, examination 2026
Topic 2, printed pages 14-15. All 14 outcomes, named compounds and exclusions inspected. Electron diagrams and examples are original.