Topic 2 · Chemical Bonding · spread 1 of 2

Chemical Bonding

1

Types of Chemical Bonds

All bonds and intermolecular forces are electrostatic attractions between …

Strong bonds
metallicpositive metal cations and a sea of delocalised valence electrons
ionicoppositely charged cations and anions
covalentthe nuclei of each atom and the shared pair(s) of electrons
Weak intermolecular forces (IMFs)
id–idnon-polar molecules
pd–pdpolar molecules
H bondthe protonic H of one molecule and a lone pair on F, O or N of another

id–id: instantaneous dipole–induced dipole, pd–pd: permanent dipole–permanent dipole. Structures: giant metallic, giant ionic, giant molecular, simple molecular.

2

Metallic Bonding

2.1Structure and bonding

Metals have a giant metallic lattice structure, with strong electrostatic forces of attraction between a sea of delocalised valence electrons and a lattice of positive metal cations.

Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺

2.2Factors affecting strength

Strength of metallic bond ↑ when

  1. Charge of cation ↑ and ionic radius ↓ → charge density of cation ↑
  2. No. of valence e⁻ ↑ → more delocalised e⁻ in the sea
  3. → stronger attraction → more energy required to overcome it → m.p. ↑

Worked exampleExplain why the melting point of aluminium (660 °C) is higher than that of sodium (98 °C).

  1. Both Al and Na have a giant metallic lattice structure
  2. Al delocalises 3 valence e⁻ per atom, Na only 1. Al³⁺ has a higher charge and smaller ionic radius than Na⁺
  3. → more energy is needed to overcome the stronger electrostatic forces of attraction between Al³⁺ cations and the sea of delocalised e⁻ than between Na⁺ cations and its sea of delocalised e⁻

2.3Physical properties

  • Conduct electricity in all states: delocalised valence e⁻ become mobile and migrate freely when a potential difference is applied. More delocalised e⁻ → higher conductivity.
  • Good thermal conductors: delocalised e⁻ throughout the structure transfer kinetic energy quickly from one end to the other.
  • Malleable and ductile: metallic bonds are non-directional, so layers of cations slide over one another without overcoming the attraction between cations and the sea of e⁻.
14Topic 02 · Chemical Bonding
3

Ionic Bonding

3.1Structure and bonding

Ionic compounds have a giant ionic lattice structure with strong electrostatic forces of attraction between cations and anions.

DefinitionCoordination number: the number of oppositely charged ions surrounding an ion in a crystal lattice (6 for both Na⁺ and Cl⁻ in NaCl).
Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻Na⁺Cl⁻
Dot-and-cross diagram of MgO
  • Show valence e⁻ only, as dots and crosses
  • Square brackets with the charge outside
  • No. of each ion as a coefficient, e.g. Ca²⁺ 2[F]⁻
Mg2+O2−

3.2Factors affecting strength

Lattice energy ∝q+ × q−r+ + r−

Greater charge and smaller ionic radii → greater magnitude of lattice energy → more energy required to overcome the attraction.

Worked exampleExplain why the melting point of MgO (2852 °C) is higher than that of Na₂O (1132 °C).

  1. Both have a giant ionic lattice structure
  2. Mg²⁺ has a higher charge and smaller ionic radius than Na⁺ → greater magnitude of lattice energy
  3. → more energy is needed to overcome the stronger electrostatic forces of attraction between Mg²⁺ and O²⁻ ions than between Na⁺ and O²⁻ ions

3.3Physical properties

  • Conducts when molten or aqueous: ions are mobile and migrate to the electrodes when a p.d. is applied. In the solid, ions are held in fixed positions.
  • Hard: large amount of energy needed to overcome the strong attraction. Brittle: under force, layers slide, ions of like charges come together and repel → lattice shatters.
  • Soluble in polar solvents: energy released forming ion–dipole interactions compensates for overcoming the ionic attraction and the solvent–solvent forces. Insoluble in non-polar solvents.

3.4Ionic bonds with covalent character

  • A small, highly charged cation such as Al³⁺ polarises (distorts) the electron cloud of an anion such as Cl⁻ → electron density drawn between the nuclei → partial sharing of e⁻ → covalent character
  • Polarising power of cation ↑ with charge density ∝ q+ / r+
  • Polarisability of anion ↑ with anion size
−3+Al³⁺Cl⁻, its electronsdrawn towards Al³⁺

Worked exampleAre Al₂O₃ and AlBr₃ largely ionic or covalent? Compare with AlCl₃.

  1. Al³⁺ has a high charge density → high polarising power
  2. O²⁻ is smaller and less polarisable than Cl⁻ → electron density unshared with Al³⁺ → Al₂O₃ largely ionic
  3. Br⁻ is larger and more polarisable than Cl⁻ → electron density largely shared → AlBr₃ largely covalent
Topic 02 · Chemical Bonding15
4

Covalent Bonding

4.1Structure and bonding

DefinitionCovalent bond: the strong electrostatic forces of attraction between the nuclei of each atom and the shared pair(s) of electrons.
Definitionσ bond: a covalent bond formed by head-on overlap of orbitals. Electron density is concentrated between the nuclei.
Definitionπ bond: a covalent bond formed by sideways overlap of p orbitals, giving two regions of overlap above and below the inter-nuclear axis.
σ: s + sH₂
σ: s + pHCl
σ: p + p head-onCl₂
π: p + p sidewaysafter the σ bond
  1. π overlap lies above and below the inter-nuclear axis → less effective overlap → π bond weaker than σ
  2. A π bond only forms after a σ bond: the σ overlap brings the atoms close enough for the p orbitals to overlap sideways
Bondσπ
single10
double11
triple12
Co-ordinate (dative) bonds
DefinitionDative bond: a covalent bond formed when an atom donates a lone pair of electrons into an empty orbital of an electron-deficient atom for sharing.
  • Shown as an arrow from donor to acceptor, or as 2 e⁻ of the same type in a dot-and-cross diagram
  • Once formed, identical to a normal covalent bond
  • Lewis acid: electron pair acceptor. Lewis base: electron pair donor. E.g. BF₃ (acid) + NH₃ (base) → H₃N→BF₃
HHHHN+NH₄⁺: H⁺ accepts the lone pair on N
ClClClClClClAlAlAl₂Cl₆: 2 dative bonds, each Al gets an octet
16Topic 02 · Chemical Bonding
4

Covalent Bonding (continued)

4.2Dot-and-cross diagrams and the octet rule

Drawing steps
  1. Central atom: forms the most bonds, or is the least electronegative (not H)
  2. Anion: add e⁻ to the more electronegative atom(s). Cation: remove e⁻ from the less electronegative atom(s). Square brackets, charge outside
  3. Give each atom an octet with single, double or triple bonds
  4. Period 2 atoms: maximum 8 valence e⁻
Exceptions to the octet rule
TypeExamples (e⁻ around central atom)
Expanded octetPCl₅ (10), SF₆ (12), XeF₄ (12)
Electron-deficientBeCl₂ (4), BF₃ (6), AlCl₃ (6)
Odd no. of e⁻ (radicals)NO, NO₂, ClO₂ (11, 17, 19 valence e⁻ in all)
Key ideaPeriod 3 and higher atoms have vacant, energetically accessible d orbitals, so they can expand the octet. Period 2 atoms have no valence d orbitals, so NCl₅ does not exist.
NNN₂ a triple bond
OCOCO₂ two double bonds
CCHHHHC₂H₄ a C=C double bond
NHHHNH₃ one lone pair on N
BFFFBF₃ only 6 e⁻ around B
NHHHH+NH₄⁺ dative pair from N

× and • are electrons from different atoms.

4.3Electronegativity: covalent bonds with ionic character

DefinitionElectronegativity: the measure of an atom's ability to attract electrons in a bond.
  • Across a period ↑: nuclear charge ↑, shielding about constant → bonding e⁻ more strongly attracted
  • Down a group ↓: more shells → bonding e⁻ further away and more shielded
  • Different electronegativities → bonding e⁻ drawn to the more electronegative atom → permanent dipole: a polar bond
Topic 02 · Chemical Bonding17
4

Covalent Bonding (continued)

4.4Shapes of molecules (VSEPR)

VSEPR theory
  1. Electron pairs around the central atom are arranged as far apart as possible to minimise repulsion
  2. Repulsion: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair
  • A lone pair repels more strongly: it is non-bonding, so its electron density is closer to the nucleus
  • A double or triple bond counts as one region of electron density
2e⁻ pairs
linear180°CO₂ (2 bp, 0 lp)
3e⁻ pairs
trigonal planar120°BF₃ (3 bp, 0 lp)
bent<120°SO₂ (2 bp, 1 lp)
4e⁻ pairs
tetrahedral109.5°CH₄ (4 bp, 0 lp)
trigonal pyramidal107°NH₃ (3 bp, 1 lp)
bent104.5°H₂O (2 bp, 2 lp)
5e⁻ pairslone pairs equatorial
trigonal bipyramidal120°, 90°PCl₅ (5 bp, 0 lp)
see-saw<120°, <90°SF₄ (4 bp, 1 lp)
T-shaped<90°ClF₃ (3 bp, 2 lp)
linear180°XeF₂ (2 bp, 3 lp)
6e⁻ pairslone pairs axial
octahedral90°SF₆ (6 bp, 0 lp)
square pyramidal<90°BrF₅ (5 bp, 1 lp)
square planar90°XeF₄ (4 bp, 2 lp)
Answer structureShape of NH₃
  1. There are 3 bond pairs and 1 lone pair around N
  2. To minimise repulsion, the 4 electron pairs are arranged as far apart as possible
  3. → NH₃ is trigonal pyramidal, bond angle 107°
Electronegativity and bond angle
  1. NF₃ 102° < NH₃ 107°: F more electronegative → bonding e⁻ closer to F → less crowding around N → less repulsion → smaller angle
  2. H₂O 104.5° > H₂S 92°: O more electronegative → bonding e⁻ drawn towards O → more crowding → more repulsion → larger angle
18Topic 02 · Chemical Bonding
4

Covalent Bonding (continued)

4.5Finding the shape of an ion or an unfamiliar molecule

Counting the electron pairs
  1. Valence e⁻ of the central atom + one e⁻ per surrounding atom bonded to it
  2. Add 1 for each negative charge, subtract 1 for each positive charge
  3. Divide by 2 for the number of electron pairs. Surrounding atoms give the bond pairs, the rest are lone pairs

ICl₄⁻: 7 + 4 + 1 = 12, so 6 pairs: 4 bp, 2 lp, square planar. PH₄⁺: 5 + 4 − 1 = 8, so 4 pairs: 4 bp, tetrahedral.

Oxygen and substitution
  • Try =O first (CO₂, SO₂). In an oxoanion the charge gives the number of –O⁻ (CO₃²⁻, SO₄²⁻). Use →O when the central atom cannot take =O (NO₂). O joining two atoms has 2 single bonds (H₂O₂)
  • Swap an unfamiliar atom for one with the same covalency (number of bonds): CHCl₃ as CH₄, F₂O as H₂O, BH₄⁻ as CH₄, CH₃⁺ as BCl₃

4.6Polarity of molecules

Steps
  1. Identify any polar bonds (bond dipoles)
  2. Determine the shape of the molecule
  3. Resolve the dipoles: net dipole → polar
Answer structureX is a polar / non-polar molecule as the dipoles do not cancel / cancel each other out. Hence there is / isn't an overall dipole moment.
OCOdipoles cancelCO₂ non-polar
OHHnet dipoleH₂O polar
BFFFdipoles cancelBF₃ non-polar
NHHHnet dipoleNH₃ polar

4.7Factors affecting strength

DefinitionBond dissociation energy: the enthalpy change when one mole of a particular covalent bond between atoms in a gaseous molecule is broken. Cl₂(g) → 2Cl(g) +244 kJ mol⁻¹
DefinitionBond length: the distance between the nuclei of the atoms in a covalent bond. Shorter bond = stronger bond.
Factors affecting covalent bond strength
  1. Orbital overlap: larger bonding atoms → larger, more diffused valence orbitals → less effective overlap → weaker bond. Cl–Cl 244 > Br–Br 193 > I–I 151 kJ mol⁻¹
  2. Polarity: the more electronegative atom has a larger share of the electron density → dipole, ionic character, may ↑ or ↓ strength (deduce from data). N–H 390 < O–H 460 < F–H 562
  3. Multiplicity: triple > double > single. C–C 350, C=C 610, C≡C 840. BE(C=C) < 2 × BE(C–C) as the π bond is weaker than σ
Bond strength and reactivity
  • A weaker, longer bond breaks more easily → more reactive. N≡N 944 kJ mol⁻¹: N₂ is unreactive.
  • C–I 240 < C–Br 280 < C–Cl 340 < C–F 485 → R–I hydrolyses fastest: bond strength outweighs polarity.
  • A polar bond has a δ+ atom for nucleophiles to attack, e.g. C in C–Cl. Non-polar C–H has none, so alkanes are unreactive.
Topic 02 · Chemical Bonding19
5

Intermolecular Forces

5.1Instantaneous dipole–induced dipole (id–id)

Definitionid–id: weak electrostatic forces of attraction between non-polar molecules (present between all molecules).
How id–id forms
  1. Continuous movement of e⁻ → at an instant, one region has slightly higher electron density
  2. This instantaneous dipole induces a dipole in a nearby molecule
  3. The two molecules attract by weak electrostatic forces between opposite partial charges
Factors: strength and extent of id–id
  • No. of electrons ↑ (larger Mr) → electron cloud more polarisable → stronger id–id. F₂ −188, Cl₂ −34, Br₂ 59 °C
  • Surface area ↑ → more extensive id–id. pentane 36 °C > 2,2-dimethylpropane 10 °C

5.2Permanent dipole–permanent dipole (pd–pd)

Definitionpd–pd: weak electrostatic forces of attraction between polar molecules, in addition to id–id.

Strength of pd–pd: the larger the magnitude of the overall dipole moment, the stronger the pd–pd interactions.

Similar vs very different Mr
CompoundMrPolarityb.p. / °C
propene42non-polar−48
ethanal44polar20
CH₃Cl50.5more polar−24
CH₃Br94.9less polar4
  • Similar Mr: polar ethanal has pd–pd as well as id–id → higher b.p.
  • Very different Mr: CH₃Br has many more e⁻ → stronger id–id outweighs the pd–pd of CH₃Cl
20Topic 02 · Chemical Bonding
5

Intermolecular Forces (continued)

5.3Hydrogen bonding (H bond)

DefinitionHydrogen bond: weak electrostatic forces of attraction between the protonic hydrogen of one molecule and a lone pair on a highly electronegative atom of another molecule.
Conditions for hydrogen bonding
  1. At least one H atom directly bonded to F, O or N
  2. The neighbouring molecule has at least one lone pair on F, O or N
Why only F, O or N?They are highly electronegative, so H carries a large δ+, and their lone pair sits in a small 2p orbital with high electron density.
OHHOHHδ+δ−δ−hydrogen bond

Draw: the lone pair, both δ+ / δ− pairs, and label the hydrogen bond.

Strength of a hydrogen bondThe more polar the H–X bond, the stronger the hydrogen bond. Electronegativity F > O > N → H–F is the most polar → strength of one H bond: HF > H₂O > NH₃
Extent of hydrogen bonding
MoleculeH on F, O, NLone pairsH bonds / molecule
H₂O222
HF131
NH₃311

H bonds per molecule = the smaller number (HF: 1 H, NH₃: 1 lone pair).

Topic 02 · Chemical Bonding21
6

Physical Properties of Covalent Substances

6.1Simple molecular substances: boiling points

Low m.p. and b.p.: only the weak IMF between the discrete molecules are overcome, not the covalent bonds within them.

H₂O, HF and NH₃ boil far higher than the trend in id–id predicts, because of hydrogen bonding.

Boiling points of hydrides / °C

-150-100-50050100Period 2Period 3Period 4Period 5CH₄Group 14NH₃Group 15H₂OhydrogenbondsGroup 16HFGroup 17

Worked exampleExplain why the boiling points are H₂O (100 °C) > HF (20 °C) > NH₃ (−33 °C), using both the strength and the extent of hydrogen bonding.

  1. All three have hydrogen bonds between molecules, b.p. depends on both their strength and their extent
  2. Strength: H–F is the most polar bond, so one H bond between HF molecules is the strongest, then H₂O, then NH₃
  3. Extent: each H₂O can form 2 H bonds (2 H atoms, 2 lone pairs), but HF and NH₃ form only 1 on average
  4. H₂O: more energy is required to overcome the more extensive H bonds between H₂O molecules, which outweighs their slightly lower strength → highest b.p.
  5. HF vs NH₃: same extent, but the H bonds between HF molecules are stronger → b.p. HF > NH₃

6.2Giant molecular structures

Diamond
Graphiteweak forcesdelocalised electrons within each layer
PropertyDiamondGraphite
BondingEach C covalently bonded to 4 others, tetrahedrallyEach C bonded to 3 others in hexagonal layers. The 4th valence e⁻ is delocalised within each layer
m.p.Very high: a large amount of energy is needed to overcome the strong and extensive covalent bonds
HardnessHard: atoms are held rigidly and do not slideSoft: weak id–id between layers let them slide
Electrical conductivityNon-conductor: all 4 valence e⁻ are used in covalent bonding → no mobile electronsConducts parallel to the layers: delocalised e⁻ move when a p.d. is applied
SolubilityInsoluble in all solvents: energy released forming solute–solvent interactions cannot overcome the strong covalent bonds
Notem.p. diamond (3350 °C) > Si (1410 °C): C uses smaller, less diffused 2sp³ orbitals than Si (3sp³) → more effective overlap → stronger C–C bonds. Other giant molecular examples: Si, SiO₂, BN.
22Topic 02 · Chemical Bonding
6

Physical Properties of Covalent Substances (continued)

6.3Special effects of hydrogen bonding

Ice is less dense than water
  1. Ice adopts an open, crystalline structure with hexagonal holes
  2. Each H₂O forms H bonds with 4 other H₂O molecules
  3. Each O is surrounded tetrahedrally by 4 H atoms: 2 covalent bonds, 2 H bonds
  4. The regular arrangement spaces molecules more widely than in the liquid
  5. → water expands on freezing → ice is less dense than water
Intramolecular H bonds lower the boiling point

2-nitrophenol (b.p. 216 °C) < 4-nitrophenol (279 °C)

  1. The –OH and –NO₂ groups are in close proximity and correctly oriented for an intramolecular H bond
  2. → the H atom of –OH is less available for intermolecular H bonds
  3. Boiling overcomes intermolecular H bonds → less energy needed → lower b.p.

Ethanol, NH₃, HF and ethanoic acid are miscible with water: they form H bonds with water molecules. Ethanoic acid in benzene forms H-bonded dimers (apparent Mr 120, not 60).

7

Structure and Physical Properties

Structurem.p. / b.p.Electrical conductivitySolubilityHardness
Giant metallic
e.g. Cu		high: strong metallic bondsgood in all states: delocalised e⁻insoluble (some react with water)malleable, ductile
Giant ionic
e.g. NaCl, MgO		high: strong ionic bondsonly molten / aqueous: mobile ionssoluble in polar solvents (ion–dipole)hard, brittle
Giant molecular
e.g. diamond, graphite		very high: strong, extensive covalent bondsnon-conductor (graphite: along layers)insoluble in all solventsdiamond hard, graphite soft
Simple molecular
e.g. I₂		low: weak IMF between moleculesnon-conductor: discrete, uncharged moleculeslike dissolves like: polar in polar, non-polar in non-polarsoft: weak IMF easily overcome
Answer structureMelting pointState each structure → more energy is needed to overcome the stronger (type of attraction) between the (particles) of ZZ than the weaker … of YY. Same structure? Explain the factor that makes one attraction stronger, e.g. charge and ionic radius (ionic), no. of valence e⁻ and cation size (metallic), no. of e⁻ or surface area (id–id), extent of H bonding.
Answer structureSolubilityEnergy released forming solute–solvent interactions is / is not able to compensate for the energy needed to overcome the solute–solute and solvent–solvent interactions → soluble / insoluble.
Topic 02 · Chemical Bonding23