3.1.3 Bonding

Describe the attraction as electrostatic attraction between oppositely charged ions acting throughout a giant ionic lattice. Avoid describing an ionic bond as a single shared pair or as a bond between just one ion pair.

Do not picture an “NaCl molecule” inside the crystal. Each ion is surrounded by several oppositely charged ions, so the electrostatic attraction extends in all directions through the lattice.

Know the formula and charge together: sulfate SO₄²⁻, hydroxide OH⁻, nitrate NO₃⁻, carbonate CO₃²⁻ and ammonium NH₄⁺. Keep each polyatomic ion intact when building formulae.

The formula becomes useful when you combine ions: Ca²⁺ needs two nitrate ions, so write Ca(NO₃)₂. The brackets keep NO₃⁻ intact and the outside subscript tells you how many nitrate ions are present.

For simple main-group ions, use the electron configuration needed to reach a noble-gas arrangement. Metals on the left tend to lose electrons; non-metals on the right tend to gain them.

The periodic-table shortcut works well for the simple main-group ions you meet here: Group 1 → +1, Group 2 → +2, Group 6 → −2 and Group 7 → −1. For example, Mg forms Mg²⁺ and Cl forms Cl⁻.

Make the total charge zero. For Al³⁺ and O²⁻, for example, 2 × (+3) and 3 × (−2) cancel, giving Al₂O₃. Change the number of ions, not the formula inside a polyatomic ion.

With polyatomic ions, keep the ion itself intact and use brackets when more than one is needed: Ca²⁺ with NO₃⁻ gives Ca(NO₃)₂. The subscript changes the number of nitrate ions; it does not alter NO₃⁻.

A single covalent bond is a shared pair of electrons. The bond is the electrostatic attraction between both nuclei and that shared pair.

Take H₂ as the simplest example: each H contributes one electron to the pair between the nuclei. In a displayed formula that shared pair is shown by one line, H–H.

When you see a double or triple bond, count shared electron pairs: a double bond has two and a triple bond has three. In displayed formulae, that means two lines for a double bond and three for a triple bond.

O₂ and N₂ are useful checks: O=O contains two shared pairs and N≡N contains three. If the question asks for shared electron pairs, count the bonding pairs, not the number of bonded atoms.

A dative covalent bond is still a shared pair of electrons, but both electrons in that pair were donated by the same atom. Once formed, it behaves as an ordinary covalent bond.

NH₄⁺ formation is the standard simple example: the lone pair on NH₃ supplies both electrons to bond to H⁺. The distinction tells you where the bonding pair came from.

If the question asks you to show an ordinary covalent bond with a line, one line means one shared electron pair. Keep structural, displayed and skeletal conventions separate so the diagram says exactly what the question asks for.

A displayed formula should show every bond; a structural formula may group atoms. Follow the representation requested. A chemically correct molecule drawn in the wrong convention can still miss what the question asks for.

Draw the arrow from the atom donating the lone pair towards the atom accepting it. The arrow direction tracks where the electron pair came from.

The arrow points from the lone-pair donor to the electron-pair acceptor. Reversing it changes the meaning, so decide which species supplies both bonding electrons before drawing.

I’d explain metallic bonding as electrostatic attraction between positive metal ions and delocalised electrons throughout a giant lattice. Those mobile delocalised electrons also explain why metals conduct electricity.

For comparisons, use the attraction in the metallic lattice. A higher ionic charge and/or smaller positive ion can increase attraction to the delocalised electrons; that is the route to explaining stronger metallic bonding.

Classify the structure first, then identify the particles and attractions present. That gives you a reliable route to melting point, conductivity and other physical properties.

Ask two questions: what particles are present, and what attractions hold them together? That separates ions in an ionic lattice, atoms joined covalently in a network, metal ions plus delocalised electrons, and discrete molecules with intermolecular forces.

Use the standard examples as anchors: diamond and graphite are giant covalent, ice and iodine are simple molecular, magnesium is metallic and sodium chloride is giant ionic.

Two useful traps sit in that list. Ice is still a molecular solid despite its extended hydrogen-bond network, while diamond and graphite are both giant covalent even though their structures and properties differ markedly.

For a physical property, name the particles and the interaction that must be overcome or the charged particles that can move. “Strong bonds” on its own is rarely enough.

For example, molten NaCl conducts because mobile ions carry charge; solid NaCl does not because the ions are fixed in the lattice. Graphite conducts because delocalised electrons can move through the structure.

Keep the molecules themselves chemically intact when you explain a change of state in a simple molecular substance. What changes is their arrangement and separation: energy is absorbed when intermolecular attractions are overcome and released when those attractions form.

So boiling water overcomes intermolecular attractions between H₂O molecules. It does not break O–H covalent bonds inside the molecules.

Show the correct repeating arrangement and exactly the number of particles requested. Include charges for ions and avoid drawing isolated “molecules” for giant structures.

For an ionic lattice show alternating oppositely charged ions; for a molecular solid show discrete molecules; for metals show positive ions with delocalised electrons; for giant covalent structures show the continuous bonded network.

Treat each bonding pair, lone pair and multiple bond direction as a region of electron density around the central atom. These regions repel because they contain negative charge.

Lone pairs usually repel more strongly than bonding pairs because their electron density is concentrated closer to the central atom. That is why replacing a bonding pair with a lone pair can reduce a bond angle even when the total number of electron regions is unchanged.

Arrange electron-density regions as far apart as possible. Start from the total number of regions, then account for lone pairs when naming the molecular shape.

Once you have the electron-region arrangement, convert it into the named molecular shape by looking at atom positions. Lone pairs still occupy regions and usually compress bond angles, but they are not included as vertices in the shape name.

Use LP–LP > LP–BP > BP–BP. Lone pairs are held closer to the central atom and occupy more space, so they repel neighbouring pairs more strongly.

This ordering explains why lone pairs compress bond angles. Replacing a bond pair by a lone pair therefore usually reduces the angle between the remaining bonds.

When you explain a bond-angle change, say what is doing the extra repelling: a lone pair repels bonding pairs more strongly than a bonding pair does, so it compresses the bond angle.

For NH₄⁺, NH₃ and H₂O the electron-pair geometry is tetrahedral, but increasing lone-pair count strengthens repulsion and compresses H–N–H/H–O–H angles.

Count electron-density regions around the central atom, choose the electron-pair geometry, then remove lone-pair positions from the shape name. Learn the associated ideal angles and explain any compression by lone pairs.

My quickest route is to identify the parent electron-region arrangement first: 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal and 6 octahedral. Then use the lone pairs to work out the molecular shape and the expected angle changes.

Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond. Keep the definition about a bonded atom, not a free ion or electron affinity.

When you compare bond polarities, a larger electronegativity difference usually means a more uneven bonding-pair distribution. Keep δ⁺/δ⁻ as partial charges; they are not full ionic charges.

The more electronegative atom attracts the shared pair more strongly, giving it δ− and leaving the other atom δ+. That is a polar covalent bond.

Show the polarity with δ symbols or a dipole arrow. These are partial charges caused by unequal sharing; the electron pair is still shared, so the bond remains covalent.

To decide whether a molecule has a permanent dipole, treat the bond dipoles as vectors. You only get a molecular dipole if those individual bond dipoles do not cancel because of the molecular shape.

First identify each bond dipole, then use the molecular geometry. Symmetric molecules such as CO₂ can contain polar bonds but have zero resultant dipole because those vectors cancel.

Put δ− on the more electronegative atom and δ+ on the less electronegative one. These are partial charges, not full ionic charges.

Use δ symbols, not full +/− charges. In H–Cl, for example, Cl is δ− and H is δ+ because chlorine is more electronegative.

If the bond dipoles point symmetrically in opposite directions, their vector sum can be zero even though each bond is polar. If the shape prevents cancellation, the molecule has a permanent dipole.

Compare CO₂ and H₂O: both contain polar bonds, but linear CO₂ cancels its two bond dipoles whereas bent H₂O does not. Shape decides the vector sum.

London forces act between all atoms and molecules. Permanent dipole–dipole attractions also act between polar molecules, and hydrogen bonding requires H bonded to N, O or F interacting with a lone pair on N, O or F.

When you list the attractions in a molecule, build them cumulatively: London forces are always present; polarity can add permanent dipole–dipole attraction; an H–N, H–O or H–F group can add hydrogen bonding.

Compare the intermolecular attractions that must be overcome. Stronger attractions need more energy, so they usually give higher boiling or melting temperatures; for London forces, size/polarisability often matters.

Use the structure to justify the comparison. An alcohol can boil much higher than a similar-sized alkane because hydrogen bonding adds a strong intermolecular attraction; within a non-polar homologous series, the London-force trend becomes more important.

Start with hydrogen bonding when you explain water’s unusual physical properties. It raises the boiling temperature because relatively strong intermolecular attractions must be overcome, and in ice it holds the molecules in an open arrangement.

That open hydrogen-bonded lattice occupies more volume for the same number of molecules, so ice is less dense than liquid water and floats.

Start by noting London forces are always present. Then decide whether the molecule has a permanent dipole and whether it has the N–H, O–H or F–H arrangement needed for hydrogen bonding.

Try the decision on familiar examples: CH₄ has London forces only; HCl has London plus permanent dipole–dipole attraction; NH₃ has London, permanent dipole–dipole attraction and hydrogen bonding. That checklist transfers to unfamiliar molecules.

For a boiling- or melting-point trend, identify what changes between the substances and then name the attraction affected. With a homologous series, increasing electron-cloud size usually strengthens London forces; hydrogen bonding needs its own structural justification.

State which attraction changes and why. “Bigger molecule, higher boiling point” earns its chemistry only when you connect size/electron cloud to stronger London forces that require more energy to overcome.