Skip to notes
8873 / 2027
Chemical Bonding

Full chapter

Chemical Bonding

Explain particles, molecular shape and material properties through electrostatic attraction.

A-Level 8873, revised syllabus (2026-2027)

01

One idea behind three strong bonds

Identify the positive and negative charges that attract.

All three bond types are electrostatic
BondAttracting particlesExample
IonicPositive and negative ions throughout a latticeNaCl; MgO
CovalentA shared electron pair and the positive nuclei of the bonded atomsH2; Cl2
MetallicPositive metal ions and delocalised electronsCopper

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.

Sodium chloride: ions after electron transferA sodium ion with charge plus one and a chloride ion with charge minus one. Each has a full outer shell of eight electrons. The chloride has seven dots from chlorine and one cross transferred from sodium. Sodium's remaining outer shell has eight crosses, which were previously in an inner shell.Na+Cl-
One Na+ for each Cl-. The total charge is zero.

Magnesium chloride

Mg (2,8,2) loses two electrons. Two Cl atoms each gain one.

Magnesium chloride: ions after electron transferOne magnesium ion with charge plus two and two chloride ions, each with charge minus one. All three ions have full outer octets. Each chloride shows seven dots from chlorine and one cross transferred from magnesium. Magnesium's eight remaining outer-shell electrons were previously in an inner shell.Mg2+Cl-Cl-
One Mg2+ for every two Cl- ions: +2 - 1 - 1 = 0.
  • 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.

NaCl shows the one-electron transfer. The additional MgCl2 comparison shows why a 2+ ion needs two singly charged negative ions.

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.

The Mg2+ outer shell was previously an inner shell. The 1:1 ion ratio balances +2 and -2.

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?

  1. The attraction between Cl+ and Cl- ions formed by electron transfer
  2. The attraction between a shared pair of electrons and the two Cl nuclei
  3. The attraction between one Cl2 molecule and the Cl2 molecules next to it
  4. The attraction between a lattice of positive ions and delocalised electrons
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.

02

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)

Hydrogen dot-and-cross diagramTwo hydrogen atoms share one pair of electrons: one dot from the left hydrogen and one cross from the right. Each hydrogen has two electrons in its shared first shell.HH
  • Left H
  • Right H
One shared pair. Each hydrogen has a full first shell of two electrons.

Oxygen (O2)

Oxygen dot-and-cross diagramTwo oxygen atoms share two electron pairs, a double bond. The left oxygen contributes six dots and the right six crosses. Each oxygen has two lone pairs as well as the two shared pairs, giving eight electrons around each atom.OO
  • Left O
  • Right O
Two shared pairs form a double bond. Each oxygen also has two lone pairs.

Water (H2O)

Water dot-and-cross diagramOxygen contributes six dots: one in each of two bonds and four in two lone pairs. Each hydrogen contributes one cross. Oxygen has an octet and each hydrogen a duet. The flat layout shows electron accounting; a water molecule is bent.HOH
  • Oxygen
  • Hydrogens
Two shared pairs and two lone pairs around oxygen. Water is bent; this layout only counts electrons.

Methane (CH4)

Methane dot-and-cross diagramCarbon contributes four dots, one to each of four shared pairs. Each of four hydrogens contributes one cross. Carbon has an octet and each hydrogen a duet. Methane is tetrahedral; the flat cross layout does not show its shape.CHHHH
  • Carbon
  • Hydrogens
Four shared pairs, with no lone pair on carbon. Methane is tetrahedral, not flat.

Carbon dioxide (CO2)

Carbon dioxide dot-and-cross diagramCarbon shares two pairs with each oxygen, making two double bonds. Carbon contributes four dots in total. Each oxygen contributes six crosses: two in the shared pairs and four in two lone pairs. All three atoms have octets.OCO
  • Carbon
  • Oxygens
Two double bonds. Each oxygen has two lone pairs; carbon has none.

Ammonia (NH3)

Ammonia dot-and-cross diagramNitrogen contributes five dots: three in shared pairs and two in one lone pair. Each hydrogen contributes one cross. Nitrogen has an octet and every hydrogen a duet. Ammonia is pyramidal, not flat.NHHH
  • Nitrogen
  • Hydrogens
Deduce three shared pairs and one lone pair from nitrogen's five outer electrons. Ammonia is pyramidal.

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.

The H2, O2, CO2 and CH4 panels cover the required patterns. Water and ammonia additionally prepare the shape discussion.

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.

Outer electrons only. Shapes and spacing are not molecular geometry.

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.

Outer electrons only. Shapes and spacing are not molecular geometry.

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.

Outer electrons only. Shapes and spacing are not molecular geometry.

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.

Dots: left carbon and right hydrogens. Crosses: right carbon and left hydrogens. Ethene has 12 outer electrons, six shared pairs, and no lone pairs on carbon.
Check the electron ledger
MoleculeBonding pairsLone pairsTotal outer electrons
H2102
O222 per O12
N231 per N10
Cl213 per Cl14
HCl13 on Cl8
CO242 per O16
CH4408
C2H46012

Check your understanding

How many lone pairs are on each oxygen atom in a correct dot-and-cross diagram of CO2?

  1. Three on each oxygen, beside a C-O single bond
  2. One on each oxygen, beside a C=O triple bond
  3. Two on each oxygen, beside a C=O double bond
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.

03

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.

NH3 + H+ → NH4+. Eight outer electrons form four shared pairs; the ion has no lone pair on nitrogen.

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.

Dots: Al electrons. Crosses: Cl electrons. Count 48 outer electrons: 16 in eight bonds and 32 in lone pairs. The flat arrangement shows connectivity, not bond angles.

Check your understanding

Why does NH4+ have eight outer electrons around nitrogen, not nine?

  1. N and four H atoms supply 9, and the 1+ charge means one electron fewer.
  2. Nitrogen loses one of its own 5 electrons before it forms any bonds.
  3. One H atom does not share its electron, so that H is left out of the count.
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.

04

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.

Schematic overlap regions, not measured electron density. Only s and p overlap is required.

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.

Required shapes and angles
SpeciesBonding domains; lone pairsMolecular shapeAngles
BF33; 0Trigonal planar120 degrees
CO22; 0Linear180 degrees
CH44; 0Tetrahedral109.5 degrees
NH33; 1Trigonal pyramidalAbout 107 degrees
H2O2; 2BentAbout 104.5 degrees
SF66; 0Octahedral90 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
  1. Its four shared pairs are four bonding domains around nitrogen.
  2. There is no lone pair on nitrogen.
  3. Four domains adopt a tetrahedral arrangement to minimise repulsion.
Answer

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
  1. Count 6 + 2(7) = 20 outer electrons. Draw two S-Cl single bonds, using four electrons.
  2. Complete each terminal Cl octet with three lone pairs: twelve more electrons. The four electrons left belong on S as two lone pairs.
  3. 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.
  4. 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.
Answer

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.

Five and six electron domains
ExampleBonding domains; lone pairsMolecular shapeAngles
PCl55; 0Trigonal bipyramidal90, 120 and 180 degrees
SF44; 1See-sawLess than 90 and less than 120 degrees
ClF33; 2T-shapedSlightly less than 90 degrees
XeF22; 3Linear180 degrees
BrF55; 1Square pyramidalSlightly less than 90 degrees
XeF44; 2Square planar90 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?

  1. N atoms are larger than P atoms, so five Cl atoms cannot fit around a nitrogen atom.
  2. P is in Period 3 and can use empty 3d orbitals; N in Period 2 has no 2d orbitals.
  3. N is more electronegative than P, so it holds its electrons too tightly to share five.
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]

Saved on this browser only. Anyone using this browser can see it.

05

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?

  1. H2O, because its molecules form hydrogen bonds
  2. H2S, because its S-H bonds are stronger than O-H
  3. H2S, because it has more electrons per molecule
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.

Add the dipoles, not just the labels
MoleculeReasonOverall polarity
CO2Equal C=O bond dipoles point in opposite directions in a linear molecule.Non-polar
BF3Three equal B-F dipoles cancel in its trigonal plane.Non-polar
SF6Six equal S-F dipoles cancel in an octahedron.Non-polar
H2OThe bent shape prevents the O-H dipoles cancelling.Polar
NH3Pyramidal arrangement gives a net dipole.Polar
CHCl3The 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?

  1. Water molecules form hydrogen bonds; methane has only weak instantaneous dipole attractions.
  2. Water has more electrons than methane, so its dispersion forces are much stronger.
  3. Boiling water breaks its strong covalent O-H bonds, which takes a lot of energy.
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.

06

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?

  1. I is larger, so the H-I bond is longer and the shared pair is held less strongly.
  2. I is more electronegative than Cl, so it pulls the shared pair away from H.
  3. HI has stronger forces between molecules, which weaken the H-I bond inside.
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?

  1. I is more electronegative than Cl, so it leaves more readily
  2. C-I is longer with a lower bond energy, so it breaks more easily
  3. C-I is more polar than C-Cl, so OH- attacks it more strongly
  4. C-Cl is weaker than C-I, so the chloro compound reacts last
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.

07

Turn structures into properties, then reverse the argument

Name the particles, attraction and charge carriers before drawing a conclusion.

Five crystalline patterns
SolidParticles and structureProperties explained
NaCl; MgOGiant 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.
IodineDiscrete I2 molecules in a molecular lattice.Comparatively low melting point: overcome intermolecular forces, not I-I bonds. No mobile charge carriers.
DiamondEach carbon bonds to four others in a three-dimensional covalent network.Hard, very high melting temperature; no mobile charged particles.
GraphiteEach 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.
IceDiscrete water molecules joined into an open hydrogen-bonded lattice.Less dense than liquid water; melting disrupts hydrogen bonding, not O-H bonds.
CopperPositive 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
  1. A high melting point indicates strong attractions, but is consistent with several structures.
  2. The molten sample has mobile charged particles.
  3. An ionic lattice supplies fixed ions in the solid and mobile ions in the melt.
  4. A metal would usually conduct as a solid; a simple molecular solid would usually melt much lower.
Answer

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
  1. Conductivity implies mobile delocalised electrons: graphite.
  2. The slippery mark is consistent with graphite sheets sliding.
  3. The other can be diamond: a rigid 3-D network with all four outer electrons involved in localised bonds.
Answer

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?

  1. The I-I bond is weaker than most covalent bonds, so melting breaks the molecules into atoms.
  2. Melting separates I2 molecules by overcoming weak forces between them; the I-I bonds stay intact.
  3. Iodine is a giant covalent lattice in which few of its bonds need to break to melt it.
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.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Ionic bondThe electrostatic attraction between oppositely charged ions.
Covalent bondThe electrostatic attraction between a shared pair of electrons and the nuclei of the two bonded atoms.
Metallic bondThe electrostatic attraction between a lattice of positive metal ions and a sea of delocalised electrons.
ElectronegativityThe ability of an atom in a molecule to attract the shared pair of electrons in a covalent bond towards itself.
Bond energyThe average energy required to break one mole of a particular covalent bond in gaseous molecules.
Bond lengthThe distance between the nuclei of two covalently bonded atoms.
Hydrogen bondAn 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.
A structure-property explanation
  1. Particles

    Atoms, molecules, ions or ions with delocalised electrons?

  2. Attraction

    Which electrostatic attraction holds the structure?

  3. Change

    Which attraction is overcome, or which charge carrier moves?

  4. Conclusion

    Link that mechanism directly to the measured property.

Fast distinctions
Pair of ideasDifference
Bond polarity / molecular polarityPolar bonds may cancel through symmetry.
Sigma / piHead-on axial overlap / sideways p overlap.
Melting iodine / breaking I2Intermolecular separation / covalent bond breaking.
NH3 / NH4+Three bonds plus lone pair / four bonds, no lone pair.
Diamond / graphite3-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
  1. 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.

    One idea behind three strong bonds

  2. 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

  3. 2(c) Connect overlap to sigma and pi bonds.

    • s and p orbital overlap only; connection to organic shapes.

    Orbital overlap builds bonds; repulsion sets shape

  4. 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.

    Orbital overlap builds bonds; repulsion sets shape

  5. 2(e) Predict analogous shapes and angles.

    • Count bonding domains and lone pairs; apply the listed geometries.

    Orbital overlap builds bonds; repulsion sets shape

  6. 2(f) Infer bond polarity.

    • Electronegativity; qualitative treatment, not numerical calculations.

    A polar bond does not guarantee a polar molecule

  7. 2(g) Infer molecular polarity.

    • Combine bond dipoles with shapes analogous to 2(d).

    A polar bond does not guarantee a polar molecule

  8. 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.

    A polar bond does not guarantee a polar molecule

  9. 2(i) Explain gas liquefaction.

    • High pressure and/or low temperature make intermolecular attractions important.

    A polar bond does not guarantee a polar molecule

  10. 2(j) Connect hydrogen bonding to properties.

    • Water boiling behaviour and open ice structure/density.

    A polar bond does not guarantee a polar molecule

  11. 2(k) Explain bond energy and length.

    • Covalent bonds; gaseous bond breaking and equilibrium internuclear separation.

    Compare the bond you are changing

  12. 2(l) Compare covalent bond reactivity.

    • Bond energy, length and polarity; use comparable reaction contexts.

    Compare the bond you are changing

  13. 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.

    Turn structures into properties, then reverse the argument

  14. 2(n) Predict physical properties from structure.

    • Melting, conductivity, mechanical behaviour and qualified solubility reasoning.

    Turn structures into properties, then reverse the argument

  15. 2(o) Infer structure from properties.

    • Combine evidence; distinguish mobile ions from electrons.

    Turn structures into properties, then reverse the argument