Structure 2. Models of bonding and structure

Use the electron configuration to see how many electrons are needed to reach the relevant stable arrangement. The number lost or gained determines the magnitude of the ion charge.

Loss of electrons leaves more protons than electrons and gives a positive charge. Gain of electrons gives more electrons than protons and gives a negative charge.

Read the outer-shell electrons before predicting charge: Al loses three electrons to form Al³⁺, while O gains two to form O²⁻. Check the sign by recounting protons and electrons after transfer; do not assume that every metal, especially a transition metal, has only one possible charge.

Ionic bonding is the electrostatic attraction between oppositely charged cations and anions. Electron transfer can create the ions, but it is not by itself the definition of the bond.

For a metal ion and a polyatomic ion, balance the charges rather than copying the numerical charge into a subscript without checking the whole formula.

Balance total charge, not ion numbers. Al³⁺ and O²⁻ require 2(+3) + 3(−2) = 0, giving Al₂O₃; Ca²⁺ and NO₃⁻ give Ca(NO₃)₂. These formulae state the simplest ion ratio in a lattice, not the composition of one molecule.

An ionic lattice is a three-dimensional, repeating arrangement of cations and anions. Its empirical formula gives the simplest whole-number ion ratio, not a molecule.

Lattice dissociation enthalpy is the positive enthalpy change for separating one mole of a solid lattice into gaseous ions. It becomes larger when ionic charges are higher or ionic radii are smaller, because the electrostatic attraction is stronger.

Use charge density to compare lattice strength: MgO has stronger attractions than NaCl because both ions carry ±2 rather than ±1, so its melting point is higher. For conductivity, the presence of charged particles is not enough—solid NaCl does not conduct until its ions can move. Water often hydrates ions, but solubility still depends on the energy balance rather than on polarity alone.

A covalent bond is the electrostatic attraction between a shared electron pair and the nuclei of the bonded atoms. Lewis formulas show valence electrons and shared pairs.

Use the octet tendency to place shared and lone pairs, while remembering that the assessed Lewis scope allows up to four electron pairs around an atom.

Build a Lewis formula by counting all valence electrons, choosing a skeleton, completing outer octets and placing any remainder on the central atom. Then recount electrons and formal charge; a shared pair contributes to the attraction between both nuclei rather than belonging exclusively to either atom.

Charged-species check: NH₄⁺ has 5 + 4(1) − 1 = 8 valence electrons, so draw four N–H shared pairs, no lone pair on N, enclose the ion in brackets and write the overall + charge. A final audit must match the available electron total, complete the required outer-shell arrangements and reproduce the species charge; lone pairs, shared pairs and formal charges answer different parts of that audit.

More shared electron pairs increase electron density between nuclei, so the bond becomes shorter and stronger.

For the same pair of atoms, increasing bond order generally shortens and strengthens the bond because more electron density lies between the nuclei. Use that comparison locally: bond enthalpy also depends on the atoms and molecular environment, so any triple bond is not automatically stronger than every unrelated single bond.

In a coordination bond, both electrons in the shared pair come from the same atom. The bond can be represented by an arrow from the electron-pair donor to the acceptor.

Identify the donor atom or ligand and the acceptor, including coordination bonds in transition-element complexes at HL.

In NH₃→BF₃ the nitrogen lone pair supplies both bonding electrons, so the arrow starts at N and ends at B. The arrow records how the pair originated; after formation it is a shared covalent pair, and in a complex the same donor logic identifies each ligand–metal bond.

VSEPR predicts molecular shape by arranging electron domains around a central atom to minimize repulsion. Lone pairs repel more strongly than bonding pairs and therefore alter common bond angles.

NH₃ has four electron domains around N, so its electron-domain geometry is tetrahedral but its molecular shape is trigonal pyramidal; the lone pair compresses the H–N–H angle below 109.5°. Count a double bond as one domain and distinguish electron geometry from the shape named using atoms only.

A bond is polar when a difference in electronegativity gives unequal sharing of the bonding electrons. The more electronegative atom carries partial negative character.

Compare electronegativities, assign partial charges, and draw the bond-dipole arrow toward the more electronegative atom.

For H–Cl, chlorine is more electronegative, so label Hδ⁺–Clδ⁻ and point the dipole arrow toward Cl. Electronegativity difference predicts unequal sharing within that bond; it does not by itself decide the polarity of the whole molecule.

Molecular polarity depends on both the polarity of individual bonds and the three-dimensional geometry of the molecule or ion. Bond dipoles can cancel or produce a net dipole moment.

Draw or infer the geometry, place each bond dipole, and check whether the vector sum is zero. Do not decide molecular polarity from a single bond alone.

CO₂ contains polar C=O bonds, but their equal opposite dipoles cancel in a linear molecule. In bent H₂O they do not cancel, so the molecule has a net dipole. Always establish the three-dimensional geometry before adding dipoles as vectors.

Explain a network material property by connecting the structure and bonding arrangement to the relevant mobility, strength, or dimensional feature.

Diamond is hard because each carbon is held in a three-dimensional covalent network, while graphite conducts along layers through delocalized electrons and its layers can slide. Silicon dioxide is also an extended network: describe network atoms, not discrete SiO₂ molecules, when explaining its high melting point.

Start with molecular size and polarity, then check for the structural requirement for hydrogen bonding. More than one IMF can be present.

All molecules have London dispersion forces. Add permanent dipole–dipole attraction when a net molecular dipole exists, and add hydrogen bonding only when the required H–N, H–O or H–F environment and an acceptor lone pair are present. Name every relevant force before deciding which dominates.

For the relative comparison in this topic, London dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonding. Molecular size also affects dispersion strength.

Stronger intermolecular attractions generally reduce volatility. Explain conductivity and solubility by considering whether charged particles are available and whether solute–solvent attractions are favourable.

Compare like evidence: pentane has stronger dispersion forces and a higher boiling point than butane because its electron cloud is larger. The simple London < dipole–dipole < hydrogen-bond ordering is a guide for comparable molecules, not a rule that ignores molecular size and the number of interaction sites.

Chromatography separates components because they have different attractions to the stationary and mobile phases. A component that is more strongly attracted to the mobile phase travels farther.

Rf is a dimensionless ratio of distances measured from the same origin under the same conditions. Because a component cannot pass the solvent front, a valid result lies from 0 to 1; a value above 1 signals a distance or origin error. Operational details beyond the separation principle are not assessed here.

If a spot moves 3.2 cm while the solvent front moves 8.0 cm, Rf = 0.40. A larger Rf means greater relative affinity for the mobile phase under those conditions, but values from different solvents, stationary phases or temperatures are not directly interchangeable.

Resonance structures represent alternative valid positions for multiple bonds while preserving the atom framework. The actual bonding is described using delocalized electrons, not a molecule switching between drawings.

Keep atom connectivity and total electrons consistent, draw each valid arrangement, and use delocalization to describe the shared bonding picture.

For CO₃²⁻, place the C=O bond in each of three valid positions while keeping atom positions and total charge fixed. The observed C–O bonds are equivalent because the π electrons are delocalized; the ion does not alternate among three localized structures.

Benzene is an important resonance example: alternative ring structures place the double bonds in different positions, while the actual π electrons are delocalized across the ring.

Use the resonance drawings as representations of one delocalized bonding system, not as separate rapidly changing molecular forms.

Benzene's six equal C–C bonds and greater stability than a localized triene support a delocalized π system above and below the ring. Use the two Kekulé drawings or a circle as representations of the same molecule, never as an equilibrium between two species.

Some HL species have five or six electron domains around a central atom. Draw the Lewis formula, count the domains, and apply VSEPR to assign the corresponding geometry.

For five domains, start from trigonal-bipyramidal electron geometry; lone pairs prefer equatorial positions because that reduces 90° interactions, producing common shapes such as seesaw and T-shaped. For six domains, start from octahedral geometry; removing one or two lone-pair positions gives common square-pyramidal or square-planar molecular shapes. Name molecular shape from atom positions after the electron-domain arrangement is fixed.

Five bonding domains give trigonal-bipyramidal PCl₅ and six give octahedral SF₆. If lone pairs are present, keep the electron-domain arrangement but remove lone-pair positions when naming molecular shape, choosing positions that minimize the strongest repulsions.

Calculate formal charges for each valid Lewis structure, then compare the charge distribution when choosing a preferred representation. Preserve the total charge of the species.

After calculating every atom, verify that formal charges sum to the overall species charge. Prefer a valid structure that minimizes charge magnitude and separation, placing negative formal charge on the more electronegative atom when other evidence is comparable; formal charge is bookkeeping, not measured partial charge.

A multiple bond contains one sigma bond plus one or more pi bonds. Use the overlap and density location to distinguish the two.

Ethene contains five σ bonds—four C–H and one C–C—and one π bond; ethyne contains three σ and two π bonds. The σ framework fixes the bond axis, while sideways p-orbital overlap creates π density and restricts rotation about a double bond.

Hybridization analysis follows a chain: Lewis formula → electron domains → geometry → hybrid orbital description. It describes mixing atomic orbitals to form hybrid orbitals used in bonding.

Count the electron domains around the relevant atom, identify the geometry, then assign the corresponding hybridization description for the molecule or ion.

Treat each single, double or triple bond as one electron domain. Four domains map to sp³, three to sp² and two to sp; thus methane carbon is sp³, each ethene carbon sp² and each ethyne carbon sp. Hybridization follows the Lewis/VSEPR model rather than replacing it.

Metallic bonding is the electrostatic attraction between a lattice of positive metal ions and delocalized electrons.

Delocalized electrons can move through the structure and carry charge and thermal energy. Non-directional attraction allows layers of cations to slide while the bonding remains.

When a potential difference is applied, delocalized electrons drift through the fixed cation lattice and carry charge; the positive ions do not travel through the metal. When layers shift under force, non-directional attraction to the electron sea persists, explaining malleability rather than brittle fracture.

Property-to-use link: copper is used for electrical wiring because its delocalized electrons carry charge through the solid; aluminium can be rolled into foil because ion layers can shift while non-directional metallic attraction is maintained. A use must be justified by the relevant property, not merely by stating that the substance is a metal.

Metallic-bond strength depends on the attraction between metal ions and delocalized electrons. Ion charge, ion radius, and the number of delocalized electrons affect charge density and attraction.

A larger ion radius generally lowers attraction; greater charge or more delocalized electrons can strengthen metallic bonding. Use the stated comparison rather than a memorized trend alone.

Compare Na and Mg using the model: Mg supplies more delocalized electrons and forms smaller, more highly charged ions, giving stronger attraction and a higher melting point. State all relevant factors before predicting; across broader sets, lattice structure can prevent a perfectly smooth trend.

Transition elements have delocalized d electrons as well as s electrons in their metallic structure. These mobile electrons are attracted to a lattice of positive metal ions.

More delocalized electrons can strengthen the electrostatic attraction, so substantial energy is often needed to disrupt the lattice, helping to explain high melting points. The same mobile electrons carry charge through the solid, explaining electrical conductivity.

Use this as a causal model, not a universal ranking: electron contribution, ion radius and crystal structure vary across the transition series, so melting points need not form a perfectly smooth trend. The chemical reactions of transition elements belong to Reactivity 3.4, not this card.

Ionic, covalent, and metallic bonding are models whose contributions can vary across materials. The bonding continuum represents mixed character rather than completely separate categories.

Use the relative contributions of the three bonding types to explain a material's position and its likely properties.

Treat the three bonding models as coordinates rather than sealed boxes. A material may combine electron sharing with partial charge separation, so its properties can fall between idealized categories. Explain which model contribution accounts for each observed property instead of assigning one label and stopping.

The bonding triangle is a model for mixed ionic, covalent and metallic character. Its horizontal coordinate uses average electronegativity and its vertical coordinate uses electronegativity difference.

Calculate the two coordinates from the supplied electronegativities, place the substance, then check whether the predicted properties fit the indicated bonding contribution. The triangle is a qualitative model, not a requirement to memorise percentage boundaries.

For NaCl, using χ(Na)=0.9 and χ(Cl)=3.2 gives average χ = (0.9+3.2)/2 = 2.05 and Δχ = 3.2−0.9 = 2.3. The large difference places it near the ionic apex; that placement is consistent with a high-melting lattice and conduction only when ions can move.

Use the coordinates to compare materials rather than treating labels as absolute. A small Δχ can still sit at different average electronegativities, so metallic versus covalent character depends on both axes. Test the interpretation against conductivity, melting behaviour and mechanical response.

An alloy is a mixture containing a metal and one or more other metals or non-metals. Different-sized atoms disrupt the regular lattice, making layer sliding more difficult while non-directional metallic bonding remains.

Explain an alloy property by referring to composition, lattice disruption, and the restricted movement of layers; do not call the alloy a compound.

In brass, differently sized Cu and Zn atoms disturb regular layer alignment, so dislocations move less easily and the alloy can be harder than pure copper. Material choice still involves trade-offs—an alloy may gain strength while losing ductility or conductivity—and its variable composition confirms that it is a mixture.

A polymer is a macromolecule built from repeating monomer-derived units. Chain structure and cross-links influence plastic properties.

Compare chain mobility: weakly interacting, unlinked chains can soften and be reshaped, whereas extensive cross-linking restricts movement and gives thermoset behaviour. A useful structure–property explanation names the repeat-chain feature, the permitted molecular motion and the resulting macroscopic response.

Cellulose is a natural polymer, whereas polyethene is synthetic. Both are macromolecules with repeating units, but origin alone does not determine a plastic's properties or biodegradability: chain structure, functional groups, intermolecular attractions, branching and cross-linking control packing and molecular motion.

Addition polymerization forms a chain by opening the monomer C=C bond. The substituents remain attached to the backbone carbons and continuation bonds show the repeating unit extends.

Remove the double bond in the monomer, preserve every substituent, and draw bonds out of the repeating unit at both ends.

Propene forms the repeat unit [–CH₂–CH(CH₃)–]ₙ: open the C=C, keep CH₃ on the same backbone carbon and draw continuation bonds through the brackets. No small molecule is eliminated, so atom accounting should match the monomer exactly.

Condensation polymerization joins difunctional monomers and releases a small molecule. Polyamides and polyesters form through the corresponding functional-group linkages.

Use both functional groups to connect the monomers, show continuation bonds, and account for the eliminated small molecule. A diol plus dicarboxylic acid forms ester links; a diamine plus dicarboxylic acid forms amide links, –CO–NH–, with water eliminated. Check that the repeat unit contains residues from both monomers, that no unreacted end group is trapped inside the repeat, and that the linkage plus by-product conserves every atom.

For a diol and a dicarboxylic acid, join –OH and –COOH groups to make ester links and release water at each new link. The repeat unit must contain residues from both monomers with continuation bonds through the functional links; do not leave unreacted end groups inside the repeat.