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9476 / 2027
Chemical Bonding

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Chemical Bonding

Use electrons, molecular shape and electrostatic attraction to explain structures, properties and reactivity.

A-Level 9476 (2026-2027)

01

All chemical bonds involve attraction between charges

Identify the positive and negative particles before naming the bond.

The electrostatic explanation
BondAttracting particlesWhat not to confuse it with
IonicOppositely charged ions in an extended structure.Electron transfer creates the ions; transfer itself is not the attractive force.
CovalentA shared electron pair and the positively charged nuclei of both bonded atoms.It is not an attraction between two negative electrons.
MetallicA 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.

Dots come from the non-metal; crosses come from the metal. Metal cations expose a previously inner shell after losing electrons. The nuclei do not change.

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.

Check your understanding

Why does MgO have a much higher melting point than NaCl, although both are giant ionic lattices?

  1. MgO ions are larger than NaCl ions, so more of each ion's surface touches its neighbours.
  2. Its ions carry larger charges (2+ and 2-) and are smaller, so the attractions are stronger.
  3. MgO is made of separate Mg-O molecules joined by strong covalent bonds inside each one.
Show the answer

Its ions carry larger charges (2+ and 2-) and are smaller, so the attractions are stronger.

Lattice attraction grows with the size of the ionic charges and falls with the distance between ions. Mg2+ and O2- have double the charges of Na+ and Cl- and are closer together.

02

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)

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.

Required examples H2, O2, CO2 and CH4, with H2O and NH3 to support the shape discussion. Count lone pairs as well as bonding pairs.

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.

N2: a triple bond and one lone pair on each nitrogen.

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.

Cl2 has 14 outer electrons; HCl has 8. Hydrogen does not need an octet.

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.

Dots: left C and right-hand H atoms. Crosses: right C and left-hand H atoms. The C=C bond contains two shared pairs; each C-H bond contains one.

Check your understanding

How many outer (valence) electrons should a correct dot-and-cross diagram of N2 show in total?

  1. 10: three shared pairs and one lone pair on each N
  2. 6: just the three shared pairs of the triple bond
  3. 16: an octet of eight around each nitrogen atom
Show the answer

10: three shared pairs and one lone pair on each N

Each N brings 5 outer electrons, so 10 in total. Three shared pairs (6) plus a lone pair on each atom (4) gives each N an octet.

03

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.

NH3 + H+ → NH4+. Five dots from nitrogen and three crosses from the original hydrogens give eight electrons. The added proton contributes no electron.

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.

Dots are chlorine electrons; crosses are aluminium electrons. Each two-dot bond pair was donated by a bridging Cl. This is a flat connectivity diagram, not the actual three-dimensional shape.

Check your understanding

Does NH4+ keep one weaker N-H bond because that bond was formed by donation?

  1. No. Once formed, all four N-H bonds are identical; donation only says where the pair came from.
  2. No. The dative bond is the strongest of the four, since both its electrons come from N.
  3. Yes. The dative bond has only one electron from each atom, so it is weaker than the others.
Show the answer

No. Once formed, all four N-H bonds are identical; donation only says where the pair came from.

A dative bond is a shared pair, like any covalent bond. In NH4+ the four bonds are equivalent, and the ion is tetrahedral.

04

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.

Keep the p orbitals parallel. Rotation would remove sideways overlap, explaining restricted rotation about C=C.
Count bonds by overlap
ConnectionSigma bondsPi bonds
Single bond10
Double bond11
Triple bond12 in mutually perpendicular planes
Whole ethene molecule5: four C-H and one C-C1: 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.

Check your understanding

How many sigma and pi bonds are in one molecule of propene, CH2=CHCH3?

  1. 9 sigma and 0 pi
  2. 8 sigma and 1 pi
  3. 7 sigma and 2 pi
Show the answer

8 sigma and 1 pi

Count each bond once: six C-H and two C-C bonds each contain one sigma bond (8). The C=C also contains one pi bond.

05

Count electron regions, then describe the atoms

Multiple bonds count as one region. Lone pairs affect shape even though they are not atoms.

Make a guess. You are not marked.

Methane, CH4, has a bond angle of 109.5 degrees. Water also has four electron pairs around its central atom. What is the H-O-H angle?

  1. Smaller, about 104.5 degrees, as lone pairs repel more
  2. 180 degrees, because there are just two bonds
  3. The same, 109.5 degrees, as there are four pairs
Show the answer

Smaller, about 104.5 degrees, as lone pairs repel more

Two of the four pairs on oxygen are lone pairs. Lone pairs repel more strongly than bonding pairs, so they push the O-H bonds closer together.

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.

Required reference shapes
MoleculeBonding regions / lone pairsMolecular shapeTypical bond angle
BF33 / 0Trigonal planar120 degrees
CO22 / 0Linear180 degrees
CH44 / 0Tetrahedral109.5 degrees
NH33 / 1Trigonal pyramidalAbout 107 degrees
H2O2 / 2BentAbout 104.5 degrees
SF66 / 0Octahedral90 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 exampleDeduce an unfamiliar ion

Predict the shape of H3O+.

Try it first. Then show the worked stepsHide the worked steps
  1. Count valence electrons: oxygen contributes six; three hydrogens add three; the positive charge subtracts one. Total = eight.
  2. Three O-H shared pairs use six electrons, leaving one lone pair on O.
  3. Four electron regions give a tetrahedral arrangement; three bonded atoms and one lone pair give a trigonal pyramidal molecular shape.
Answer

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 exampleA double bond is one region

SO2 has two S-O bonding regions and one lone pair on S. What shape follows?

Try it first. Then show the worked stepsHide the worked steps
  1. There are three electron regions, so their reference arrangement is trigonal planar.
  2. Only two of those regions contain bonded atoms.
  3. The molecule is bent, with an O-S-O angle slightly below the 120-degree reference.
Answer

Bent, not linear. Counting each double bond as two directions would give the wrong model.

Why can sulfur have more than eight outer electrons when oxygen, in the same group, cannot? 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 regions. A Period 2 atom such as N, O or F has only 2s and 2p orbitals, so it is limited to four regions 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 regions
ExampleBonding regions / lone pairsMolecular shapeBond angles
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
SF66 / 0Octahedral90 degrees
BrF55 / 1Square pyramidalSlightly less than 90 degrees
XeF44 / 2Square planar90 degrees

Where the lone pairs go decides the shape. With five regions, 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 regions 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.

Worked exampleSix regions with two lone pairs

Predict the shape and bond angle of the ICl4- ion.

Try it first. Then show the worked stepsHide the worked steps
  1. Count electrons around iodine: seven valence electrons plus one for the negative charge = eight.
  2. Four I-Cl bonds use four of these electrons. The other four make two lone pairs.
  3. Four bonding pairs and two lone pairs give six regions, so the electron arrangement is octahedral.
  4. The two lone pairs sit opposite each other to keep as far apart as possible. The four Cl atoms lie in one plane.
Answer

Square planar, with Cl-I-Cl bond angles of 90 degrees. Iodine is in Period 5, so it can hold more than eight outer electrons.

Check your understanding

Phosphorus forms PCl3 and PCl5. Nitrogen, in the same group, forms NCl3 but not NCl5. Why?

  1. P is in Period 3 and can use empty 3d orbitals; N in Period 2 has no 2d orbitals.
  2. N atoms are larger than P atoms, so five Cl atoms cannot fit around a nitrogen atom.
  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 regions in its second shell, so it cannot form five bonds.

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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06

A polar bond need not make a polar molecule

Add the bond-dipole directions using the actual three-dimensional shape.

Make a guess. You are not marked.

Each C=O bond in carbon dioxide is polar, because oxygen is more electronegative than carbon. Is the CO2 molecule polar overall?

  1. Yes, because each of its bonds is polar
  2. No; it is linear, so the bond dipoles cancel
  3. No; carbon and oxygen have equal electronegativity
Show the answer

No; it is linear, so the bond dipoles cancel

CO2 is linear, so its two equal bond dipoles point in opposite directions and cancel.

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.

Use bond polarity and shape together
MoleculeBond dipolesOverall result
CO2Two C-O dipoles point in opposite directions along a straight line.Cancel: non-polar molecule.
BF3Three identical B-F dipoles arranged symmetrically at 120 degrees.Cancel: non-polar.
CH4 / CCl4Four equivalent bonds point to the corners of a tetrahedron.Cancel: non-polar.
SF6Opposite S-F dipoles pair in an octahedron.Cancel: non-polar.
H2OO-H dipoles meet at a bent angle.Do not cancel: polar.
NH3Three N-H bonds form a pyramid, not a flat triangle.Do not cancel: polar.
CHCl3A tetrahedron with one H and three Cl substituents is not dipole-symmetric.Polar.

Worked exampleSame general shape, different polarity

Both CO2 and OCS are linear. Why can OCS be polar?

Try it first. Then show the worked stepsHide the worked steps
  1. Linearity can make equal and opposite dipoles cancel.
  2. In CO2 the two C-O bonds are equivalent.
  3. OCS has different bonds at the two ends, so equal magnitudes cannot be assumed. Their vector sum need not be zero.
Answer

Shape alone does not decide polarity. Cancellation requires the appropriate symmetry and equivalent opposing bond dipoles.

07

Attractions between molecules have an electrostatic origin

All particles can form instantaneous dipoles; only suitable molecules form hydrogen-bond networks.

Make a guess. You are not marked.

HCl molecules have more electrons than HF molecules. Which has the higher boiling point?

  1. HCl, because it has more electrons
  2. HCl, because its molecules are heavier
  3. HF, because it forms hydrogen bonds
Show the answer

HF, because it forms hydrogen bonds

HF molecules form hydrogen bonds, because H is bonded to the small, very electronegative F. These are much stronger than the extra dispersion forces in HCl.

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.

Identify the force before explaining a boiling point
  1. 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.

  2. Check molecular polarity

    Polar molecules also have permanent dipole attractions.

  3. Compare polarizability and shape

    Electron-cloud size and contact between molecules affect instantaneous-dipole attractions.

  4. State what separates on boiling

    Molecules separate; ordinary covalent bonds inside them remain intact.

Check your understanding

Neither Br2 nor argon has a permanent dipole. Why can both be liquefied?

  1. Cooling makes their molecules polar, so permanent dipole attractions then hold them together.
  2. Their electron clouds fluctuate and induce dipoles in neighbours, which attract each other.
  3. Their atoms form covalent bonds with each other when they are cooled enough to liquefy.
Show the answer

Their electron clouds fluctuate and induce dipoles in neighbours, which attract each other.

Instantaneous dipole-induced dipole attractions act between all particles. At low enough temperature they hold the particles together as a liquid. Argon is made of single atoms.

08

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.

Structure explains the observation
ObservationMolecular 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.

Check your understanding

Why does ice float on liquid water?

  1. Hydrogen bonds hold ice in an open network, so it has more space and a lower density.
  2. Ice has trapped air bubbles from freezing, which lower its density below that of the liquid.
  3. Melting breaks every hydrogen bond, so liquid water molecules pack with no gaps between them.
Show the answer

Hydrogen bonds hold ice in an open network, so it has more space and a lower density.

In ice each molecule hydrogen-bonds to four others in a roughly tetrahedral network with gaps. On melting, part of the network breaks and molecules pack more closely.

09

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 exampleWhy polarity alone fails for C-halogen reactivity

C-F is very polar. Does that make a fluoroalkane the easiest halogenoalkane to hydrolyse?

Try it first. Then show the worked stepsHide the worked steps
  1. Polarity creates an electron-deficient carbon that can attract a nucleophile.
  2. However, the C-F bond is particularly short and strong, so breaking it is difficult.
  3. Down the common Cl, Br, I series, C-halogen bonds generally get longer and weaker; comparable iodoalkanes often hydrolyse faster.
Answer

No. Bond polarity identifies a possible attack site; bond strength and the reaction pathway also control the rate.

Check your understanding

C=C has a larger total bond energy than C-C. Why is ethene still more reactive than ethane?

  1. A larger bond energy means the bond is longer, so it is easier for other molecules to reach.
  2. The pi bond is weaker than the sigma bond and its electrons are exposed, so it reacts first.
  3. Ethene has more hydrogen atoms than ethane, so it has more sites for an attacking species.
Show the answer

The pi bond is weaker than the sigma bond and its electrons are exposed, so it reacts first.

Total bond energy is the sigma and pi together. An electrophile attacks the exposed pi electrons, and only the pi bond breaks; the sigma framework stays.

10

Different solids contain different moving charges and attractions

Name the particles, arrangement and attraction before predicting a property.

The required crystalline structures
Example and structureParticles and attractionProperties to explain
NaCl and MgO: ionicAn 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 molecularDiscrete 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 covalentEach 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 layersEach 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 molecularDiscrete H2O molecules in an open hydrogen-bond network.Lower density than liquid water; melting changes intermolecular organisation.
Copper: metallicPositive metal ions and delocalised electrons throughout the metal.Electrical/thermal conduction; layers can shift while metallic attraction persists.

Diamond: four bonds per carbon

Diamond network fragment with four bonds at the central carbonThe central carbon bonds to four surrounding carbon atoms. Two plain bonds, a wedge towards the viewer and a dashed bond away from the viewer indicate a three-dimensional arrangement. Each outer carbon has three additional bond stubs continuing into the network.CCCCC
Strong covalent bonds extend through a three-dimensional network. There are no small diamond molecules.

Graphite: three bonds per carbon

One hexagon within a graphite sheetSix carbon atoms form a hexagon. Each has two bonds around the hexagon and a third bond continuing outwards into the sheet. These are covalent bonds. Each carbon also contributes one electron to delocalisation within the sheet; those electrons are not individually drawn here.CCCCCCGraphite sheets stacked with weaker attractions between themThree broad parallelograms represent whole sheets, not individual bonds. Dashed amber lines between sheets represent weaker attractions. The sheets can slide relative to each other.
Covalent bonds within sheets; weaker attractions between sheets. Each carbon supplies one delocalised electron within its sheet.

Silicon dioxide: a repeating network

Silicon dioxide coordination fragmentThe central silicon bonds to four oxygen atoms. Each of these oxygens bridges to a second silicon. Additional bonds from the four outer silicons continue beyond this fragment. The flat arrangement shows connectivity only; the real network is three-dimensional and the bonds are not at these drawn angles.SiSiSiSiSiOOOO
Each Si bonds to four O atoms; each O joins two Si atoms. Shared oxygen atoms give the ratio Si:O = 1:2.

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.

Diamond and graphite are required H2 examples. The silica comparison reinforces the distinction between an extended network and discrete molecules.

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.

Check your understanding

Graphite conducts electricity but diamond does not. Why?

  1. Graphite contains ions between its layers, which move when a voltage is applied.
  2. Diamond's covalent bonds are too strong to break, while graphite's bonds break to free electrons.
  3. Each C in graphite bonds to three others, leaving one delocalised electron per atom.
Show the answer

Each C in graphite bonds to three others, leaving one delocalised electron per atom.

In graphite the fourth outer electron of each carbon is delocalised within the layers and can move. In diamond all four electrons are in localised covalent bonds.

11

Use several observations to identify a structure

A property is evidence; it is rarely a unique label by itself.

Worked exampleAn unknown solid

A solid melts at a high temperature, does not conduct when solid, but conducts when molten. What is the most plausible structure?

Try it first. Then show the worked stepsHide the worked steps
  1. High melting temperature suggests strong attractions, but could fit several giant structures.
  2. Failure to conduct as a solid argues against an ordinary metal and conducting graphite.
  3. Conduction after melting is consistent with charged ions becoming mobile.
Answer

A giant ionic lattice is the best explanation from these observations. Say which result distinguishes it; high melting point alone is insufficient.

Worked exampleA soft conductor

Another solid conducts electricity, is soft and has very high thermal stability. Why is graphite plausible?

Try it first. Then show the worked stepsHide the worked steps
  1. Delocalised electrons account for conduction.
  2. Weak interlayer attractions permit sliding and account for softness.
  3. Strong covalent bonds within layers explain the high thermal stability.
Answer

Different parts of the structure explain different properties. Calling the whole structure "weakly bonded" contradicts its thermal behaviour.

A practical evidence sequence
  1. Observe before identifying

    Record appearance, melting behaviour, solubility and conductivity under stated conditions.

  2. Control the conductivity comparison

    Use comparable temperature, electrode arrangement and solution concentration; avoid confusing a wet sample with the dry solid.

  3. Match carriers to state

    Electron conduction can occur in a solid metal; ionic conduction requires mobile ions.

  4. Keep conclusions proportional

    Suggest a structure consistent with the evidence and identify any extra test needed to distinguish alternatives.

Check your understanding

A solid has a very high melting point and does not conduct as a solid or when molten. Which structure fits best?

  1. Simple molecular, such as iodine or sulfur
  2. Giant covalent, such as diamond or silicon dioxide
  3. Giant ionic, such as sodium chloride or MgO
Show the answer

Giant covalent, such as diamond or silicon dioxide

A very high melting point needs strong bonds throughout. No conduction even when molten rules out ions and delocalised electrons.

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.
Dative (coordinate) bondA covalent bond in which both shared electrons come from the same atom.
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.
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.
The shortest useful explanation
  1. Name the particles and arrangement

    Ions, atoms or molecules; discrete units, sheets or an extended network.

  2. Name the relevant attraction

    Ionic, covalent, metallic or a specific intermolecular attraction.

  3. Connect to the observation

    Energy to overcome attraction explains melting; mobile charges explain conductivity.

Keep these distinctions
Do not confuseReason
Bond polarity / molecular polarityDipoles may cancel by symmetry.
Covalent bond / intermolecular forceMelting a molecular solid does not break its molecules.
Sigma / piDouble = 1 sigma + 1 pi; triple = 1 sigma + 2 pi.
Electron arrangement / molecular shapeLone pairs count as regions but are omitted from the atom-based shape name.
Dative formation / finished bondThe donor supplies the pair; the finished bond is covalent.
Bond energy / reaction rateThe pathway and activation barrier also matter.

Scope

Learning outcomes

2. Chemical Bonding. Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map
  1. 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

    All chemical bonds involve attraction between charges

  2. 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

  3. 2(c) Explain covalent bonds using orbital overlap.

    • s and p overlap only
    • Sigma and pi bonds
    • Connection to organic structures

    End-on overlap gives sigma; sideways overlap gives pi

  4. 2(d) Explain required molecular shapes and angles by VSEPR.

    • BF3 trigonal planar
    • CO2 linear
    • CH4 tetrahedral
    • NH3 trigonal pyramidal
    • H2O bent
    • SF6 octahedral

    Count electron regions, then describe the atoms

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

    • Electron-region count
    • Bond pairs and lone pairs
    • Analogous unfamiliar molecules and ions

    Count electron regions, then describe the atoms

  6. 2(f) Explain bond polarity using electronegativity.

    • Qualitative electronegativity
    • Partial charges; no numerical electronegativity calculation

    A polar bond need not make a polar molecule

  7. 2(g) Deduce molecular polarity.

    • Bond dipoles
    • Three-dimensional molecular shape
    • Symmetry and cancellation in analogous shapes

    A polar bond need not make a polar molecule

  8. 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

    Attractions between molecules have an electrostatic origin

  9. 2(i) Link hydrogen bonding to physical properties.

    • Ice and liquid water
    • Density and melting
    • Thermal properties and suitable solubility examples

    Hydrogen bonding makes ice an unusually open solid

  10. 2(j) Define covalent bond energy and length.

    • Gas-phase bond breaking per mole
    • Mean energy versus a specific bond
    • Equilibrium internuclear distance

    A strong bond is not automatically a non-polar bond

  11. 2(k) Compare bond reactivity using structure.

    • Bond energy
    • Bond length
    • Bond polarity and limits of one-factor predictions

    A strong bond is not automatically a non-polar bond

  12. 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

  13. 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

  14. 2(n) Infer structure and bonding from supplied evidence.

    • Use multiple observations
    • Distinguish evidence from a unique identification
    • State-dependent conductivity

    Use several observations to identify a structure