Chemical Bonding & Intermolecular Forces
1. Ionic Bonding
Ionic bond: the strong electrostatic force of attraction between a positive metal cation and a negative non-metal anion.
- Metal atom gives its electron(s) to the non-metal atom to reach a stable configuration → metal = cation, non-metal = anion.
- Ionic compounds always form a giant rigid lattice (regular arrangement of alternating + / − ions); the big lattice can be broken into pieces and it is still NaCl.
- High m.p. — ionic bonds are strong, lots of energy needed to break them (they are broken on melting/boiling).
- Solid: does not conduct; molten or aqueous: conducts — because ions are then free to move.
- Dissolve well in water — water can form new, strong intermolecular forces (ion–dipole) with the ions, favouring dissolving.
Why soluble: water is polar/has H-bonding, so it attracts the ions and new intermolecular forces are made — this favours dissolving. (Cl⁻ is surrounded with the δ+ hydrogens facing it.)
Dot-and-cross diagrams — ionic (outer shell electrons only)
2. Metallic Bonding
Metallic bond: strong force of electrostatic attraction between positive cations and the delocalised sea of electrons.
- Metal atoms lose valence electrons → positive cations; the electrons are delocalised and form a "sea" around them.
- Cations are packed in a regular lattice.
- High m.p. / b.p. — metallic bonds are very strong.
- Good conductors in any state — free-moving electrons (unlike ionic: no state restriction).
- Insoluble in water.
Q. Show metallic bonding in Na and in Mg
Ans: Na < Mg < Al — more charge, more delocalised e⁻, smaller radius ⇒ stronger metallic bond. Sir's note: if charge alone tells the story, stop — only add radius/electron count if the examiner asks for a reason.
3. Covalent Bonding
Covalent bond: the force of attraction between the nuclei and a shared pair of electrons.
- Atoms share electrons to achieve a stable configuration (each shared pair = one bond).
- Can form giant covalent or simple molecular structures.
- Usually insoluble in water.
- O only needs 2 more e⁻, so with C it shares 2 pairs (double bond) — an atom shares only as many pairs as it needs.
Dot-and-cross diagrams (outer shell only) — the course examples
Past-paper dot-and-cross questions
(i) Complete the dot-and-cross diagram for SCl₂ (outer electrons only).
S (group 16) has 6 outer e⁻ → uses 2 in two S–Cl bonds, keeps 2 lone pairs; each Cl keeps 3 lone pairs.
Answer: S × two shared pairs, 2 lone pairs on S, 3 lone pairs on each Cl.
Draw a dot-and-cross diagram of Cl₂O (outer shell only).
O is the central atom: 2 bonding pairs (one to each Cl) + 2 lone pairs on O; every Cl has 3 lone pairs.
Answer: O(centre) shares one pair with each Cl; 2 lone pairs on O, 3 on each Cl.
The peroxide ion has a single covalent bond between the two O atoms; each O carries a negative charge. Draw the dot-and-cross diagram.
Each O has 6 outer e⁻ + 1 extra (gained from the other species that makes the 2− charge) → 7. One e⁻ goes into the single bond, so each O keeps 3 lone pairs. Put square brackets and 2− outside.
Answer: [ :Ö–Ö: ]²⁻ with brackets and charge; extra electrons marked separately.
Carbon monoxide is a gas that contains a coordinate (dative) bond. Draw the dot-and-cross diagram showing outer electrons.
C (4 e⁻) and O (6 e⁻): two pairs formed by one × and one •, and the third pair has both electrons from O (its lone pair → dative bond O→C). Each atom keeps one lone pair: C = ××, O = ••. So C≡O has 3 bonds, 1 dative and 2 lone pairs in total.
Answer: 6 shared e⁻ (× • | × • | • •) + 1 lone pair on C (××) + 1 on O (••).
The dative bond is shown by an arrow pointing towards the atom receiving the lone pair. Which diagram of NH₄⁺ is correct? (A, B: arrow points into N, charge + / −; C, D: arrow from N to H, charge + / −)
N donates its lone pair to H⁺, so the arrow goes N → H, and the ion is +.
Answer: C
Which type of bonding is never found in elements? A covalent · B ionic · C metallic · D van der Waals
Elements: covalent (Cl₂, diamond), metallic (Na), van der Waals (I₂, noble gases). Ionic needs two different kinds of atom (metal + non-metal).
Answer: B (ionic)
4. Dative (Coordinate) Bonding
Dative bond: formed when an atom uses its lone pair and contributes both electrons in the shared pair of electrons.
Lone pair: a pair of valence electrons not bonded to any other species (unshared pair).
★ The big case study — Al₂Cl₆ (know this by heart)
Metal + non-metal, so AlCl₃ "should" be ionic — but it isn't (reason: Al³⁺ has a huge polarising power — see §5). In AlCl₃ Al has only 6 e⁻ instead of 8 → electron deficient!
- AlCl₃ is very unstable: ① exists as gas only at very high temperature ② Al is electron deficient.
- It dimerises → Al₂Cl₆: two Cl atoms use their lone pairs to bond to the 2nd Al (2 dative bonds) → Al is no longer electron deficient → more stable.
- Very high T breaks Al₂Cl₆ back into 2 AlCl₃. At 200 °C the gas has Mr = 267 = Al₂Cl₆.
- Al₂Cl₆ = 8 bonds: 6 covalent + 2 dative, and 16 lone pairs (4 terminal Cl × 3 + 2 bridging Cl × 2).
At 200 °C aluminium chloride is a gas with Mr = 267. How many covalent bonds, dative covalent bonds and lone pairs are in one molecule?
Answer: B
What favours dative bond formation?
In case of a base accepting H⁺ it is feasible because: ① it has a lone pair and ② an ionic bond can form (e.g. NH₄⁺Cl⁻) which is more stable than a covalent bond.
More examples (draw them as dot-and-cross, then as structural formula with →)
- Cross out the spectator ion (Cl⁻) to write the ionic equation — the H⁺ has an empty orbital, no electrons, so N/O uses its lone pair.
- H₃O⁺ and dative examples other than Al₂Cl₆/NH₄⁺ need not be memorised, but you must be able to work them out: find the lone pair (donor) and the electron-deficient / H⁺ (acceptor). The examiner often tells you it is a coordinate bond.
- In the dot-and-cross diagram of the product all 4 N–H pairs look identical — the dative bond is only visible in the structural formula (arrow).
- The examiner may draw the dot-and-cross differently to how you learned — don't get confused; just count electrons.
(ii) The product Cl₃Al←NH₃ forms when ammonia reacts with aluminium chloride. Name the bond between N and Al and explain how it is formed.
Dative (coordinate) bond. Al in AlCl₃ is electron deficient (6 e⁻); the N in NH₃ has a lone pair and donates both electrons of the shared pair to Al.
Answer: dative covalent bond — N uses its lone pair, contributing both electrons of the shared pair to the electron-deficient Al.
5. Polarisation of Ions (covalent character in ionic bonds)
In ionic bonding the cation can pull electron density from the anion, distorting its shape → some degree of covalent character develops.
Polarising power of a cation: the tendency of a cation to polarise an anion.
A cation has a greater polarising power if ① it has a greater positive charge ② it has a smaller
ionic radius (charge density ↑).
Polarisability of an anion: the ease with which an anion can be polarised.
An anion has greater polarisability if ① greater negative charge ② greater ionic radius.
F⁻ has the smaller radius, Cl⁻ the bigger radius; S²⁻ has the greatest charge (and a big radius) ⇒ greatest polarisability.
Answer: F⁻ < Cl⁻ < S²⁻
Sir: fluorine essentially never forms covalent compounds with a cation; chlorine does (AlCl₃).
Which has the least covalent character? A LiCl (Li⁺) · B MgCl₂ (Mg²⁺) · C KCl (K⁺) · D SiCl₄ (Si⁴⁺)
Least covalent character = least polarisation = cation with smallest charge and biggest radius → K⁺ (charge +1, large).
Answer: C (KCl)
Q. What factors favour ionic bonding?
Opposite of polarising power/polarisability:
Cation: smaller positive charge, greater cationic radius | Anion: smaller negative
charge, smaller anionic radius.
6. Polarity in Covalent Molecules
Electronegativity: the tendency of an atom to attract a shared pair of electrons towards itself.
- F most electronegative, O 2nd, N & Cl about the same (3rd/4th).
- C, H and I have similar electronegativity → C–H bond is (essentially) non-polar.
- Use FONCl to work out any trend: electronegativity increases across a period (→ F) and decreases down a group.
Trends across the periodic table
Across a period ↑ because:
- nuclear charge increases
- shielding remains constant
- atomic radius decreases
⇒ more attraction between shared pair and nucleus
Down a group ↓ because:
- nuclear charge increases
- shielding also increases (these two cancel each other out)
- atomic radius increases (more shells)
⇒ less attraction between shared pair and nucleus
| Pauling scale (no need to learn!) | H | C | N | O | F | Na | Mg | Al | Si | P | S | Cl |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| value | 2.1 | 2.5 | 3.0 | 3.5 | 4.0 | 0.9 | 1.2 | 1.5 | 1.8 | 2.2 | 2.5 | 3.0 |
Don't panic: the difference will never be exactly 1.8 — it will be clearly above or below.
Polarity in H–F (a polar bond)
Fluorine is much more electronegative than H and pulls most of the electron density towards itself. Electrons spend more time near F, so F becomes partially negative (δ−) and H partially positive (δ+). Draw as H→F with the arrow pointing from δ+ to δ−.
Careful: a symmetrical shape alone is not enough — the outer atoms must also be the same species (e.g. CH₃Cl is not symmetrical in polarity). C–H bond is non-polar, so CH₄ is non-polar.
Behaviour of polar molecules in an electric field
Which molecules have no overall dipole moment? 1 BF₃ ✓ 2 CH₄ ✓ 3 PF₅ ✓
BF₃ trigonal planar, CH₄ tetrahedral, PF₅ trigonal bipyramidal — all symmetrical with identical outer atoms ⇒ dipoles cancel. All three correct.
1 ammonia ✓ (pyramidal, N δ−) · 2 hydrogen sulfide ✓ (V-shaped) · 3 boron trifluoride ✗ (trigonal planar, symmetrical)
EN rises Na < Mg < Al < Si (Si highest). 1st IE rises Na < Al < Mg < Si (Al dips below Mg — group 2→13 dip). The correct plot has Si top-right, Na bottom-left, Al higher than Mg on EN but lower on IE. Diagram A wrongly shows Al IE > Mg IE.
Answer: B
EN increases steadily Na → Cl (no dips, unlike IE/m.p.) ⇒ smoothly rising line. B decreases (atomic radius), A is m.p. (peak at Si), C is IE (dips at Al and S).
Answer: D
CO has three bonds between C and O. Features: 1 two lone pairs ✓ (one on each atom) · 2 a dative bond from O to C ✓ · 3 two π bonds ✓ (triple bond = 1σ + 2π). All correct.
7. Intermolecular Forces
Intermolecular forces: forces of attraction between one molecule and a neighbouring molecule (these are broken in simple molecules on melting and boiling — not the covalent bonds).
Permanent dipole–dipole
Two polar molecules attract each other; the force is a permanent dipole–dipole interaction. Molecule must be polar. (Called "permanent" because of the δ+/δ− that is always there.) Somewhat strong — stronger than temporary dipoles.
Temporary (instantaneous) induced dipoles
Definition: the temporary force of attraction formed between electrons and nuclei due to electron density not being symmetrical at any instant. (Also called London forces / induced dipole — the examiner says "induced dipole".)
- Temporary dipoles exist in every substance (all molecules) — they are weaker than permanent dipoles.
- Collectively temporary + permanent dipoles = van der Waals' forces.
- If a molecule has both permanent and temporary dipoles, the strongest overall force is the permanent (+ temporary present too).
A C₂H₆ vs C₂H₅Cl: 2nd has permanent dipole ↑ (higher b.p.) ✗ · B CH₃OCH₃ vs C₂H₅OH: ethanol H-bonds ↑ ✗ · C Ne vs Ar: Ar more electrons ↑ ✗ · D CH₃NH₂ vs C₂H₆ ✓: CH₃NH₂ has permanent dipoles and hydrogen bonding; C₂H₆ only temporary ⇒ ethane is lower.
Answer: D
A C₂H₅OH(l)→(g) ✗ (H-bonding) · B H₂O(s)→(l) ✗ (H-bonding) · C O₂(s)→(l) ✓ (non-polar, only temporary dipoles) · D C₄H₁₀(l)→(s) ✗ (freezing: forces are formed, not overcome)
Answer: C
The greater number of electrons in iodine gives larger temporary induced dipole forces. (Not bond energy, EN or permanent dipoles — I₂ is non-polar.)
Answer: B
Millions of microscopic hairs make the pads extremely close to the glass ⇒ attraction between toe pads and glass = van der Waals forces (weak, but huge in number).
Answer: D
8. Hydrogen Bonding
Hydrogen bond: force of attraction between a hydrogen bonded to a very electronegative atom and a neighbouring lone pair.
Drawing H-bonds (2–4 marks!)
(i) Force between methanol and water molecules that makes them soluble in each other → hydrogen bonding. (ii) Diagram: H₂O:···H–O(CH₃) — lone pair on one O, dotted line to the H (δ+) of the other, δ−/δ+ on O and H, CH₃ on the second O.
Answer: hydrogen bond between lone pair on O (δ−) of one molecule and H (δ+) on O of the other; all lone pairs and partial charges shown.
Ammonia is very soluble in water because it forms hydrogen bonds with water. Draw the H-bond: N lone pair (:N, δ−) ··· H (δ+)–O (δ−, with 2 lone pairs) — hydrogen in the middle, N···H–O in a straight line, all lone pairs and partial charges shown.
Boiling points of hydrides — why the anomalies?
H₂O has hydrogen bonding, while H₂S can at best only have permanent dipoles.
The number of electrons increases so stronger temporary dipole forces are achieved.
Properties of water due to hydrogen bonding
- Higher m.p. and b.p. than similar isoelectronic molecules (same electron number, e.g. H₂S, HF vs H₂O) — H-bonds need extra energy to break.
- High viscosity — H-bonding does not allow molecules to slide over each other easily (thicker, flows slowly).
- High surface tension — surface molecules pulled inwards; a mosquito can stand on water.
- Good solvent — forms H-bonds to other polar molecules (ethanol) and stronger ion–dipole forces to ions (NaCl).
- Ice is less dense than water (anomalous expansion) — see below. If ice sank, lakes would freeze from below and life could not survive.
1 water b.p. higher than H₂S ✓ · 2 ice floats on water ✓ · 3 H–O–H angle ≈ 104° ✗ (that is lone-pair repulsion, not H-bonding)
Answer: 1 and 2 only
H₂O has 2 lone pairs and 2 H ⇒ up to 4 H-bonds per molecule; NH₃ has only 1 lone pair ⇒ fewer H-bonds. HCl (no H-bond), I₂ (temporary dipoles only).
Answer: A (H₂O)
Substance 1: strong covalent bonds + permanent dipoles → CH₂Cl₂ · Substance 2: only weak forces → CH₄ (non-polar) · Substance 3: H-bonding → CH₃OH (or H₂O).
Answer: C (CH₂Cl₂, CH₄, CH₃OH)
(a) CH₃OCH₃ has 2 lone pairs on O; C–O–C bond angle ≈ 104.5°.
(b) Strongest force (A =
permanent dipole, B = temporary, C = H-bond):
| ethanal CH₃CHO | ethanol CH₃CH₂OH | methoxymethane CH₃OCH₃ | 2-methylpropane |
|---|---|---|---|
| A | C | A | B |
Ethanal: C=O polar, no O–H. Ethanol: O–H ⇒ H-bond. Ether: polar C–O–C but no H on O. 2-methylpropane: non-polar alkane.
(a)(ii) H₂ non-polar; H₂O polar: O has a much higher electronegativity than H; in H₂ both H pull
equally.
(b)(ii) NH₃ bond angle 107° vs water 104.5°: N has only 1 lone pair, O has 2 — lone pair–lone pair
repulsion is larger in water.
(c)(i) 3-D shapes: NH₃ trigonal pyramidal (107°), AlCl₃ trigonal planar (120°).
9. Simple Molecular & Giant Structures
Simple molecular (small covalent molecules)
- e.g. H₂O, CO₂, Cl₂, I₂. Held together by weak intermolecular forces; on melting/boiling these are overcome ⇒ low m.p. and b.p.
- Do not conduct electricity; usually insoluble in water. Solid: regular lattice held by weak intermolecular forces.
Giant covalent structures (big covalent molecules)
- diamond, graphite, SiO₂, graphene, nanotubes. Many covalent bonds in a giant lattice; on melting/boiling covalent bonds are broken ⇒ very high m.p./b.p.
- May or may not conduct; insoluble in water.
| Graphite | Diamond | C₆₀ | |
|---|---|---|---|
| structure | giant covalent, hexagonal layers, 1 C–3 C | giant covalent, tetrahedral, 1 C–4 C | simple molecular |
| delocalised e⁻ | 1 per C, moves atom to atom ⇒ conducts | none ⇒ does not conduct | 1 per C but can't leave molecule ⇒ no |
| m.p./b.p. | high (covalent bonds) | high (covalent bonds) | low (temporary dipoles) |
| other | sheets slide (vdW) ⇒ lubricant; insoluble | hardest material known; insoluble | insoluble |
- Graphene = a single sheet of graphite — same properties as graphite.
- Nanotubes = graphene sheet rolled up — same properties as graphite & graphene; highest strength-to-weight ratio of any substance.
- All are allotropes of carbon.
Simple molecular: low-ish m.p. (113 °C), insoluble, always poor conductor ⇒ B. (A reacts with water and conducts when solid = metal; C soluble, good when solid… ionic/metal; D 1600 °C giant covalent.)
Answer: B
Ionic: high m.p. (680 °C), good conductor when molten (mobile ions) ⇒ C.
Answer: C
C₆₀ and iodine ✓ (graphene, graphite giant covalent).
Answer: B
1 both have delocalised e⁻ ✓ · 2 C₆₀ giant ✗ · 3 graphene tetrahedral ✗ (trigonal planar, sp²)
Answer: D (1 only)
Both solids have a (regular) lattice ✓ — CO₂ a molecular lattice, SiO₂ a giant lattice. B ✗ (SiO₂ giant) · C ✗ (CO₂ double bonds) · D ✗ (SiO₂ doesn't turn to gas)
Answer: A
1 all contain covalent bonds ✓ · 2 all giant ✗ (C₆₀ simple) · 3 all delocalised e⁻ ✗ (diamond none)
Answer: D (1 only)
10. Shapes of Molecules (VSEPR)
| e⁻ pairs | shape | angle | extra details |
|---|---|---|---|
| 2 bonding | linear | 180° | in one plane; also for single-bond-equivalents |
| 3 bonding | trigonal planar | 120° | in one plane (angles add to 360° — "tri-gonal") |
| 4 bonding | tetrahedral | 109.5° | 3-D, rotational symmetry |
| 5 bonding | trigonal bipyramidal | 120° & 90° | 3-D; 90° between equator and axis; 120° in equator plane |
| 6 bonding | octahedral | 90° | 3-D; all angles 90° |
| 3 bonding + 1 lone | trigonal pyramidal | 107° | 3-D |
| 2 bonding + 2 lone | V-shaped / bent | 104.5° | in one plane only |
Drawing structures of molecules — method
- Identify the central atom ② draw the dot-and-cross ③ deduce the structure ④ treat double/triple bonds as single bonds (each counts as one region of repulsion).
| molecule | pairs on central atom | shape | angle |
|---|---|---|---|
| BeCl₂ | 2 bp | linear | 180° |
| CO₂ (O=C=O) | 2 regions | linear | 180° |
| BF₃ / AlCl₃ | 3 bp | trigonal planar | 120° |
| CH₄ / SiCl₄ | 4 bp | tetrahedral | 109.5° |
| PCl₅ | 5 bp | trigonal bipyramidal | 120° & 90° |
| SF₆ | 6 bp | octahedral | 90° |
| NH₃ | 3 bp + 1 lp | trigonal pyramidal | 107° |
| H₂O | 2 bp + 2 lp | V-shaped / bent | 104.5° |
| NH₄⁺ | 4 bp | tetrahedral | 109.5° |
| H₃O⁺ | 3 bp + 1 lp | trigonal pyramidal | 107° |
x = H–C=C (trigonal planar) ≈ 120°; y = C=C–C (trigonal planar) ≈ 120°; z = C–S–C… (2 bp + 2 lp around S / or sp³ centre) ≈ 109°. Row C: 120°, 120°, 109°.
Answer: C
x = H–C–H in CH₃ (tetrahedral) 109.5° · y = C=N–… (trigonal planar) 120° · z = C–N–O / N–O–H with lone pair(s): O has 2 lone pairs ⇒ 104.5° (smallest). Order: z < x < y.
Answer: A (z, x, y)
Plane needs linear / trigonal planar / bent: C₂H₄ (planar ✓), CO₂ (linear ✓), H₂O (bent ✓). A has PH₃ (pyramidal), B has NH₃, C has C₃H₆ (has a CH₃ — tetrahedral).
Answer: D
1 BF₃ ✓ (trigonal planar) · 2 NCl₃ ✗ (pyramidal, 107°) · 3 H₃O⁺ ✗ (pyramidal)
Answer: 1 only
11. Orbital Overlap: σ and π Bonds
σ (sigma) bond — end-to-end overlap
- A single bond is made when 2 orbitals overlap end to end ⇒ a σ bond (1s+1s, 1s+2p, 2p+2p).
- e⁻ density lies along the internuclear axis; 1 region of overlap; stronger than π bonds.
π (pi) bond — sideways overlap of parallel p orbitals
- π bonds are always sideways overlap; e⁻ density above and below the axis (that is why π is weaker than σ); 2 regions of overlap.
- Only present in double and triple bonds — a σ bond has to be made before a π bond.
- π bonds prevent rotation about the bond ⇒ this gives rise to isomerism in alkenes.
N: 2p³ = [↑][↑][↑] → 3 unpaired ⇒ 1σ + 2π (2px+2px and 2py+2py).
O=C=O: each double bond = 1σ + 1π ⇒ 2σ and 2π.
Answer: B (2, 2)
12. Hybridisation
sp³ — methane (4 σ bonds)
s + 3 p fuse into 4 sp³ hybrid orbitals (1 part s, 3 part p), each one containing 1 e⁻. Now C forms 4 σ bonds with 1s of 4 H — tetrahedral 109.5°.
sp² — ethene (3 σ + 1 π per C)
Each C in ethene needs 3 σ bonds (3 orbitals for end-to-end overlap) and 1 π bond (1 p orbital left over for sideways overlap).
sp — ethyne, CO₂ (2 σ + 2 π)
| C atom bonded to… | σ bonds | hybridisation | examples |
|---|---|---|---|
| 4 C / atoms | 4 | sp³ | diamond, alkanes |
| 3 atoms (1 double bond) | 3 | sp² | graphite, graphene, nanotubes, C₆₀ (1 C to 3 C), alkenes |
| 2 atoms | 2 | sp | alkynes, CO₂ |
C₂H₄: each C has 3σ + 1π ⇒ sp²; whole molecule has C–H ×4 + C–C ×1 = 5σ + 1π.
Answer: B
13. Bond Lengths and Bond Energies
Bond length: the distance between the nuclei of two covalently bonded atoms. For a homonuclear diatomic (H₂, F₂, Cl₂): ½ × bond length = atomic radius.
Bond energy: the minimum amount of energy required to break 1 mole of a bond in 1 mole of gaseous molecules. e.g. H–H(g) → 2H(g).
| bond | bond length / nm | bond energy / kJ mol⁻¹ |
|---|---|---|
| H–F | 0.092* | +562 |
| H–Cl | 0.127 | +431 |
| H–Br | 0.141 | +366 |
| H–I | 0.161 | +299 |
*board table reads "0.009" — a slip for ≈0.092 nm (must be shorter than H–Cl 0.127 nm).
Down the group: bond length ↑ and bond energy ↓ (bond gets weaker) because the bond gets less polar and, with bigger nuclei, the orbital overlap becomes less efficient (H 1s with F 2p is more efficient than with I 5p).
Reactivity: H–I → H–F: reactivity decreases as it becomes more difficult to break the existing bond and make new ones. (Hence HI is the most easily decomposed hydrogen halide.)
★ Chapter Cheat-Sheet
- Ionic: strong electrostatic attraction between + metal cation and − non-metal anion.
- Metallic: strong electrostatic attraction between + cations and the delocalised sea of electrons.
- Covalent: attraction between nuclei and a shared pair of electrons.
- Dative: one atom uses its lone pair and contributes both electrons to the shared pair (→ donor to acceptor).
- Lone pair: pair of valence e⁻ not bonded to any other species.
- Electronegativity: tendency of an atom to attract a shared pair towards itself. Hydrogen bond: H bonded to F/O/N attracted to a neighbouring lone pair.
- Bond length: distance between nuclei; bond energy: minimum energy to break 1 mol of bond in 1 mol of gaseous molecules.
- Strength: temporary < permanent dipoles < H-bond < giant covalent < ionic/metallic. Melting/boiling overcomes: simple molecular → intermolecular forces; giant covalent → covalent bonds; ionic → ionic bonds; metals → metallic bonds.
- Temporary dipoles ↑ with more electrons; straight chain > branched. Every molecule has them.
- Metallic strength ↑: cation charge ↑, delocalised e⁻ ↑, radius ↓. Polarising power ↑: charge ↑, radius ↓. Polarisability ↑: charge ↑, radius ↑.
- Non-polar: same-atom molecules (H₂, Cl₂), symmetrical molecules (CO₂, CCl₄, BF₃, CH₄, PF₅). ΔEN > 1.8 ionic, < 1.8 covalent.
- Al₂Cl₆: 6 covalent + 2 dative, 16 lone pairs; AlCl₃, BF₃, BeCl₂ electron deficient. P and S can exceed the octet.
- Linear 180° · trigonal planar 120° · tetrahedral 109.5° · pyramidal 107° · bent 104.5° · trig. bipyramidal 120°/90° · octahedral 90°. Repulsion: bp–bp < bp–lp < lp–lp; single < double < triple.
- Single = σ; double = σ+π; triple = σ+2π. σ = end-on, along axis, stronger; π = sideways, above/below, weaker, stops rotation.
- Hybridisation hack: 4σ ⇒ sp³ (109.5°) · 3σ+1π ⇒ sp² (120°) · 2σ+2π ⇒ sp (180°). Graphite/graphene/nanotubes/C₆₀ sp²; diamond sp³.
- H–X: bond length ↑ and bond energy ↓ down group 17 (overlap less efficient) ⇒ HF strongest, HI weakest.