Bonding

Chemical bonding explains how atoms combine to form compounds. This topic covers ionic, covalent, And metallic bonding, molecular geometry, intermolecular forces, and electronegativity.

Types of Bonding

Ionic Bonding (OL/HL)

Ionic bonds form between metals and non-metals. Electrons are transferred from the metal to the Non-metal, producing oppositely charged ions that attract electrostatically.

Example (OL): Formation of sodium chloride.

\mathrm{Na \to \mathrm{Na^+ + e^- \quad (\mathrm{ionisation) \mathrm{Cl + e^- \to \mathrm{Cl^- \quad (\mathrm{electron gain) \mathrm{Na^+ + \mathrm{Cl^- \to \mathrm{NaCl

Properties of ionic compounds:

• High melting and boiling points.
• Conduct electricity when molten or in solution (not when solid).
• Brittle.
• Form crystalline lattices.

Worked Example 1 (OL): Explain why \mathrm{NaCl has a high melting point.

\mathrm{NaCl has a giant ionic lattice with strong electrostatic forces between \mathrm{Na^+ and \mathrm{Cl^- ions. A large amount of energy is required to overcome these forces, giving a high Melting point (801°C).

Covalent Bonding (OL/HL)

Covalent bonds form when non-metal atoms share electrons to achieve a stable electron configuration.

Single bond: one shared pair (e.g., H—H, Cl—Cl). Double bond: two shared pairs (e.g., O=O). Triple bond: three shared pairs (e.g., N$\equiv$N).

Bond length and bond strength:

Bond type	Bond length	Bond energy (kJ/mol)
C-C	154 pm	348
C=C	134 pm	612
C≡C	120 pm	839

Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single Bonds.

Coordinate (Dative) Covalent Bond (HL)

Both electrons in the bond come from one atom.

Example (HL): Formation of the ammonium ion.

\mathrm{NH_3 + \mathrm{H^+ \to \mathrm{NH_4^+

The lone pair on nitrogen forms a dative bond with the hydrogen ion.

Example (HL): Formation of the aluminium hexaaqua ion:

\mathrm{Al^{3+} + 6\mathrm{H_2\mathrm{O \to [\mathrm{Al(\mathrm{H_2\mathrm{O)_6]^{3+}

Each water molecule donates a lone pair from oxygen to form a dative bond with \mathrm{Al^{3+}.

Metallic Bonding (OL/HL)

Metals consist of a lattice of positive ions in a “sea” of delocalised electrons.

Properties of metals:

• Good conductors of heat and electricity (delocalised electrons).
• Malleable and ductile (ions can slide past each other).
• High melting points (strong electrostatic attraction).
• Shiny (delocalised electrons absorb and re-emit light).

Electronegativity and Bond Polarity (OL/HL)

Pauling Scale

Electronegativity is a measure of an atom’s ability to attract bonding electrons.

• \Delta\mathrm{EN < 0.4: non-polar covalent.
• 0.4 \leq \Delta\mathrm{EN < 1.7: polar covalent.
• \Delta\mathrm{EN \geq 1.7: ionic.

Example (OL): HCl: \Delta\mathrm{EN = 3.16 - 2.20 = 0.96 — polar covalent.

Dipole Moments (HL)

A polar bond has a dipole moment. The dipole moment vector points from the positive to the negative End.

In molecules with multiple polar bonds, the overall polarity depends on the molecular geometry (vector sum of individual dipoles).

• \mathrm{CO_2: linear — dipoles cancel — non-polar.
• \mathrm{H_2\mathrm{O: bent — dipoles do not cancel — polar.
• \mathrm{CCl_4: tetrahedral — dipoles cancel — non-polar.
• \mathrm{NH_3: trigonal pyramidal — dipoles do not cancel — polar.

Worked Example 2 (HL): Explain why \mathrm{CO_2 is non-polar but \mathrm{SO_2 is polar.

\mathrm{CO_2 is linear (\mathrm{O=\mathrm{C=\mathrm{O). The two C=O bond dipoles are equal in Magnitude and opposite in direction, so they cancel. \mathrm{SO_2 is bent (V-shaped) with a bond Angle of approximately $119^\circ$. The two S=O bond dipoles do not cancel, leaving a net dipole moment.

VSEPR Theory (OL/HL)

The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular shapes based on electron Pair repulsion.

Electron pairs	Shape	Bond angle
2	Linear	$180^\circ$
3	Trigonal planar	$120^\circ$
4	Tetrahedral	$109.5^\circ$
5	Trigonal bipyramidal	$120°, 90^\circ$
6	Octahedral	$90^\circ$

Effect of Lone Pairs

Lone pairs repel more strongly than bonding pairs, reducing bond angles.

Bonding pairs	Lone pairs	Shape	Example	Bond angle
3	1	Trigonal pyramidal	\mathrm{NH_3	$107^\circ$
2	2	Bent	\mathrm{H_2\mathrm{O	$104.5^\circ$
4	1	See-saw	\mathrm{SF_4	—
3	2	T-shaped	\mathrm{ClF_3	—
2	3	Linear	\mathrm{XeF_2	$180^\circ$

Worked Example 3 (OL): Use VSEPR theory to predict the shape of \mathrm{BF_3 and \mathrm{H_2\mathrm{O.

\mathrm{BF_3: Boron has 3 bonding pairs and 0 lone pairs. Electron geometry: trigonal planar. Molecular shape: trigonal planar. Bond angle: $120^\circ$.

\mathrm{H_2\mathrm{O: Oxygen has 2 bonding pairs and 2 lone pairs. Electron geometry: tetrahedral. Molecular shape: bent. Bond angle: $104.5^\circ$ (reduced from $109.5^\circ$ due to lone pair repulsion).

Worked Example 4 (HL): Describe the bonding in \mathrm{SF_4 using VSEPR theory. Include the Hybridisation of the central atom.

Sulfur has 6 valence electrons. Four are used in S-F bonds, leaving one lone pair. Total electron Domains = 5 (4 bonding + 1 lone pair). Electron geometry: trigonal bipyramidal. The lone pair Occupies an equatorial position. Molecular shape: see-saw. Hybridisation: $sp^3d$.

Hybridisation (HL)

sp Hybridisation

Two hybrid orbitals, linear geometry ($180^\circ$). Example: \mathrm{BeCl_2, \mathrm{C_2\mathrm{H_2.

sp2 Hybridisation

Three hybrid orbitals, trigonal planar ($120^\circ$). Example: \mathrm{BF_3, \mathrm{C_2\mathrm{H_4.

sp3 Hybridisation

Four hybrid orbitals, tetrahedral ($109.5^\circ$). Example: \mathrm{CH_4, \mathrm{NH_3 \mathrm{H_2\mathrm{O.

sp3d Hybridisation

Five hybrid orbitals, trigonal bipyramidal. Example: \mathrm{PCl_5, \mathrm{SF_4.

sp3d2 Hybridisation

Six hybrid orbitals, octahedral. Example: \mathrm{SF_6.

Example (HL): Describe the bonding in ethene (\mathrm{C_2\mathrm{H_4).

Each carbon is $sp^2$ hybridised. The three $sp^2$ orbitals form three sigma bonds (two C—H and one C—C). The remaining unhybridised $p$ orbital on each carbon overlaps to form a pi ($\pi$) bond Above and below the plane. The C=C double bond consists of one sigma and one pi bond.

Worked Example 5 (HL): Describe the bonding in ethyne (\mathrm{C_2\mathrm{H_2), including sigma And pi bonds and hybridisation.

Each carbon in ethyne is $sp$ hybridised. The two $sp$ orbitals form two sigma bonds (one C—H and One C—C). Each carbon has two remaining unhybridised $p$ orbitals ($p_y$ and $p_z$). These overlap Sideways to form two pi bonds perpendicular to each other. The C≡C triple bond consists of one sigma And two pi bonds. The molecule is linear ($180^\circ$).

Intermolecular Forces (OL/HL)

Van der Waals Forces (London Dispersion Forces)

Present in all molecules. Caused by temporary dipoles due to instantaneous electron movements.

Strength increases with molecular size (more electrons = larger temporary dipoles).

Derivation: As molecular size increases, the electron cloud becomes more diffuse and more Polarisable. Larger electron clouds are more distorted, creating larger instantaneous dipoles And hence stronger London forces. This explains why boiling points increase down Group 7: \mathrm{F_2 (85°\mathrm{K) < \mathrm{Cl_2 (239°\mathrm{K) < \mathrm{Br_2 (332°\mathrm{K) < \mathrm{I_2 (457°\mathrm{K).

Dipole-Dipole Forces

Present in polar molecules. The positive end of one molecule attracts the negative end of another.

Hydrogen Bonding

A special, strong dipole-dipole force between:

• H bonded to N, O, or F (highly electronegative).
• A lone pair on N, O, or F of another molecule.

Consequences of hydrogen bonding:

• Water has an unusually high boiling point for its molecular mass.
• Ice is less dense than liquid water (open lattice structure).
• DNA base pairing.

Example (OL): Explain why \mathrm{H_2\mathrm{O boils at 100^\circ\mathrm{C but \mathrm{H_2\mathrm{S boils At -60^\circ\mathrm{C despite having a similar molecular mass.

Water forms hydrogen bonds (O is highly electronegative), which require significant energy to Overcome. \mathrm{H_2\mathrm{S cannot form hydrogen bonds (S is not electronegative enough), so only Weak van der Waals forces act between molecules.

Worked Example 6 (HL): Explain why the boiling point of \mathrm{HF ($20°C$) is much higher than That of \mathrm{HCl ($-85°C$), despite \mathrm{HCl having a larger molecular mass.

\mathrm{HF can form hydrogen bonds because H is bonded to F (the most electronegative element). These strong intermolecular forces require significant energy to overcome. \mathrm{HCl cannot form Hydrogen bonds (Cl is not electronegative enough), so only weaker dipole-dipole and London forces Act. Despite \mathrm{HCl having stronger London forces (more electrons), hydrogen bonding in \mathrm{HF dominates.

Summary of Intermolecular Forces

Force	Strength	Present in	Dependence
London dispersion	Weakest	All molecules	Increases with size
Dipole-dipole	Moderate	Polar molecules	Depends on dipole moment
Hydrogen bonding	Strongest	H bonded to N, O, F	Depends on H-bond density

Metallic Bonding in Detail (HL)

Band Theory

In metals, atomic orbitals overlap to form energy bands. The valence band contains electrons That can conduct electricity. The conduction band overlaps with the valence band in metals, Allowing free movement of electrons.

In insulators, there is a large energy gap between the valence and conduction bands. In Semiconductors, the gap is small enough that thermal energy can promote some electrons.

Comparison: Conductors, Semiconductors, Insulators

Property	Conductor	Semiconductor	Insulator
Band gap	None (bands overlap)	Small (~1 eV)	Large (>5 eV)
Conductivity at 0 K	High	Zero	Zero
Conductivity with heating	Decreases slightly	Increases	Remains zero
Examples	Cu, Al	Si, Ge	Diamond, SiO$_2$

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Worked Examples

See the examples integrated throughout the sections above.

Common Pitfalls

1. Ionic vs covalent — ionic involves electron transfer; covalent involves electron sharing.

2. VSEPR — count lone pairs on the central atom, not total lone pairs.

3. Bond polarity vs molecular polarity — a molecule can have polar bonds but be non-polar overall (e.g., \mathrm{CO_2).

4. Hydrogen bonding — only occurs with H bonded to N, O, or F, not Cl.

5. Hybridisation — count the number of electron domains (bonding + lone pairs) to determine hybridisation.

6. Bond order — double bonds are NOT exactly twice as strong as single bonds. A double bond consists of one sigma and one pi bond; the pi bond is weaker than the sigma bond.

7. Sigma vs pi bonds — sigma bonds can rotate freely; pi bonds cannot (this is the basis of E/Z isomerism).

8. Electronegativity — fluorine is the most electronegative element, NOT oxygen.

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Practice Questions

Ordinary Level

1. Describe the difference between ionic and covalent bonding. Give one example of each.
2. Use VSEPR theory to predict the shape of \mathrm{BF_3 and \mathrm{H_2\mathrm{O.
3. Explain why \mathrm{NaCl has a high melting point.
4. What type of intermolecular force is responsible for the high boiling point of water?

Higher Level

1. Describe the bonding in \mathrm{SF_4 using VSEPR theory. Include the hybridisation of the central atom.
2. Explain why \mathrm{CO_2 is non-polar but \mathrm{SO_2 is polar.
3. Describe the bonding in ethyne (\mathrm{C_2\mathrm{H_2), including sigma and pi bonds and hybridisation.
4. Explain why the boiling point of \mathrm{HF ($20°C$) is much higher than that of \mathrm{HCl ($-85°C$), despite \mathrm{HCl having a larger molecular mass.

Extended Questions

5. Draw dot-and-cross diagrams for (a) \mathrm{MgO and (b) \mathrm{H_2\mathrm{S.
6. Explain why graphite conducts electricity but diamond does not, in terms of bonding and structure.
7. Use VSEPR theory to predict the shapes of \mathrm{XeF_4 and \mathrm{BrF_5.
8. Explain why the boiling point increases in the order \mathrm{CH_4 \lt \mathrm{SiH_4 \lt \mathrm{GeH_4 \lt \mathrm{SnH_4 but \mathrm{NH_3 \gt \mathrm{PH_3.
9. Compare and contrast the properties of ionic, covalent, and metallic bonding in a table.
10. Explain the term “hybridisation” and describe how it accounts for the shapes of \mathrm{CH_4 \mathrm{C_2\mathrm{H_4And \mathrm{C_2\mathrm{H_2.
11. Describe the structure and bonding in sodium chloride, including a description of the ionic lattice.
12. Explain why \mathrm{BeCl_2 is covalent while \mathrm{CaCl_2 is ionic, referring to the concept of polarising power.

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Polarisation and Fajans’ Rules (HL)

Polarising Power

Small, highly charged cations have high polarising power. Large, highly charged anions are Polarised. When a cation distorts the electron cloud of an anion, the bond acquires covalent Character.

Fajans’ Rules

1. A small cation has high polarising power (e.g., \mathrm{Li^+ vs. \mathrm{Cs^+).
2. A large anion is more polarised (e.g., \mathrm{I^- vs. \mathrm{F^-).
3. A high charge on either ion increases polarisation.

Consequences:

• \mathrm{AlCl_3 has significant covalent character despite the electronegativity difference.
• \mathrm{LiI is more covalent than \mathrm{LiF because \mathrm{I^- is more polarisable.
• \mathrm{SnCl_4 is a covalent liquid; \mathrm{SnCl_2 is an ionic solid.

Trend in Covalent Character of Group 1 Halides

Compound	Covalent character	Explanation
\mathrm{LiF	Very low	Small cation, small anion
\mathrm{LiI	Moderate	Small cation, large polarisable anion
\mathrm{CsF	Very low	Large cation, small anion
\mathrm{CsI	Moderate	Large cation, large anion

Resonance (HL)

Concept

Some molecules cannot be represented by a single Lewis structure. Resonance describes the Delocalisation of electrons over two or more equivalent structures.

Example: The carbonate ion, \mathrm{CO_3^{2-}:

Three equivalent resonance structures can be drawn, each with a C=O double bond in a different Position. The actual structure is a hybrid with three identical C-O bonds, each with bond order $1\frac{1}{3}$.

Evidence for resonance: All three C-O bonds in \mathrm{CO_3^{2-} have the same length (131 pm), Intermediate between a typical C-O single bond (143 pm) and C=O double bond (120 pm).

Benzene Resonance

Benzene (\mathrm{C_6\mathrm{H_6) can be represented by two Kekule structures with alternating single And double bonds. The actual structure has delocalised pi electrons above and below the ring, giving Six identical C-C bonds with bond order 1.5 (length 140 pm).

This delocalisation provides extra stability (resonance energy, approximately 150 kJ/mol).

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Molecular Orbital Theory (HL)

Basic Concept

Molecular orbital (MO) theory describes bonding as the combination of atomic orbitals to form Molecular orbitals that belong to the molecule as a whole.

Rules for Forming MOs

1. Atomic orbitals of similar energy combine.
2. The number of MOs formed equals the number of atomic orbitals combined.
3. Bonding MOs are lower in energy than the original atomic orbitals.
4. Antibonding MOs are higher in energy.
5. Electrons fill MOs following the Aufbau principle and Hund’s rule.

MO Diagram for \mathrm{O_2

For \mathrm{O_2 (and heavier diatomic molecules), the energy ordering is:

$\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$

\mathrm{O_2 has 12 valence electrons. The last two electrons go into the degenerate $\pi^*$ Orbitals with parallel spins (Hund’s rule), making \mathrm{O_2 paramagnetic (has unpaired Electrons).

Bond order:

\mathrm{Bond order = \frac{\mathrm{bonding electrons - \mathrm{antibonding electrons}{2}

For \mathrm{O_2: \mathrm{BO = (8 - 4)/2 = 2.

MO Diagram for \mathrm{N_2

For \mathrm{N_2 (and lighter diatomic molecules), the $\sigma_{2p}$ and $\pi_{2p}$ ordering is Reversed: $\pi_{2p} < \sigma_{2p}$.

\mathrm{N_2 has 10 valence electrons: \mathrm{BO = (8 - 2)/2 = 3 (triple bond), consistent with The Lewis structure \mathrm{N\equiv\mathrm{N.

Comparison of MO Theory with Lewis Theory

Feature	Lewis theory	MO theory
Bonds	Localised between two atoms	Delocalised over molecule
Paramagnetism of \mathrm{O_2	Cannot explain	Correctly predicts
Bond order	Single, double, triple	Can be fractional (e.g., 1.5)
Resonance	Requires multiple structures	Single structure with delocalisation

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Giant Covalent Structures (OL/HL)

Diamond

Each carbon is tetrahedrally bonded to four others ($sp^3$ hybridised). Very hard, high melting Point, does not conduct electricity (no free electrons).

Graphite

Each carbon is bonded to three others in layers ($sp^2$ hybridised). Soft, high melting point, Conducts electricity along layers (delocalised electrons in $p$ orbitals).

Silicon Dioxide (\mathrm{SiO_2)

Each silicon is bonded to four oxygens and each oxygen to two silicons, forming a giant covalent Network. Very high melting point, hard, insulator.

Comparison Table

Property	Diamond	Graphite	\mathrm{SiO_2
Bonding	4 covalent bonds per C	3 covalent bonds + delocalised	4 covalent bonds per Si
Hardness	Extremely hard	Soft (layers slide)	Very hard
Melting point	Very high	Very high	Very high
Electrical conductivity	None	Conducts along planes	None
Structure	3D tetrahedral	Layered hexagonal	3D network

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Ionic Structures in Detail (HL)

Sodium Chloride Structure

Face-centred cubic lattice. Each \mathrm{Na^+ is surrounded by 6 \mathrm{Cl^- ions (octahedral Coordination). Coordination number: 6:6.

Caesium Chloride Structure

Body-centred cubic. Each \mathrm{Cs^+ is surrounded by 8 \mathrm{Cl^- ions. Coordination number: 8:8.

Relationship Between Structure and Properties

Compound	Coordination number	Lattice energy (kJ/mol)	Melting point
NaCl	6:6	-787	801°C
CsCl	8:8	-657	645°C
MgO	6:6	-3791	2852°C

\mathrm{MgO has a much higher melting point than \mathrm{NaCl because the doubly charged ions Produce stronger electrostatic attraction.

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Bond Enthalpy (OL/HL)

Bond enthalpy is the energy required to break one mole of a particular bond in the gaseous State.

Average Bond Enthalpies

Bond	Enthalpy (kJ/mol)	Bond	Enthalpy (kJ/mol)
C-C	348	C-H	412
C=C	612	O-H	463
C≡C	839	C=O	743
C-O	358	H-H	436
C-Cl	328	O=O	496

Worked Example 7 (HL): Use bond enthalpies to estimate the enthalpy change for the combustion of Methane: \mathrm{CH_4 + 2\mathrm{O_2 \to \mathrm{CO_2 + 2\mathrm{H_2\mathrm{O.

Bonds broken: 4 \times \mathrm{C-H + 2 \times \mathrm{O=O = 4(412) + 2(496) = 1648 + 992 = 2640 \mathrm{ kJ/mol.

Bonds formed: 2 \times \mathrm{C=O + 4 \times \mathrm{O-H = 2(743) + 4(463) = 1486 + 1852 = 3338 \mathrm{ kJ/mol.

\Delta H = 2640 - 3338 = -698 \mathrm{ kJ/mol

The reaction is exothermic, consistent with combustion.

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Practice Questions (Extended)

13. Explain why the bond length of C=C is shorter than C-C, and why C≡C is shorter than C=C.
14. Draw all the resonance structures of the nitrate ion (\mathrm{NO_3^-) and explain the experimental evidence for resonance.
15. Use MO theory to explain why \mathrm{O_2 is paramagnetic but \mathrm{N_2 is diamagnetic.
16. Calculate the bond order of \mathrm{He_2 using MO theory and explain why \mathrm{He_2 does not exist as a stable molecule.
17. Explain why \mathrm{AlCl_3 sublimes at a relatively low temperature and forms dimer molecules \mathrm{Al_2\mathrm{Cl_6 in the gas phase.
18. Describe the structure and bonding in diamond and explain why diamond is used as an abrasive.
19. Explain why the melting point of \mathrm{NaF ($993°C$) is higher than that of \mathrm{NaCl ($801°C$), despite \mathrm{F^- being smaller than \mathrm{Cl^-.
20. Using bond enthalpy data, estimate $\Delta H$ for the reaction: \mathrm{N_2 + 3\mathrm{H_2 \to 2\mathrm{NH_3.

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Advanced VSEPR: Odd-Electron Molecules and Expanded Octets (HL)

Odd-Electron Molecules

Some molecules have an odd number of valence electrons, making the central electron count unusual.

Example: \mathrm{NO_2 has 17 valence electrons. Nitrogen has 5 valence electrons, each oxygen Contributes 6. Total: $5 + 2(6) = 17$. The unpaired electron counts as a “half” electron domain.

The molecule is bent (approximately $134^\circ$) with bond angle between that of \mathrm{BF_3 ($120^\circ$) And \mathrm{H_2\mathrm{O ($104.5^\circ$).

Expanded Octets

Elements in period 3 and below can have more than 8 electrons in their valence shell (expanded Octet) by using $d$ orbitals.

Example: \mathrm{SF_6 has sulfur with 12 electrons around it (6 bonding pairs). Electron Geometry: octahedral ($sp^3d^2$). Bond angle: $90^\circ$.

Example: \mathrm{PCl_5 has phosphorus with 10 electrons (5 bonding pairs). Electron geometry: Trigonal bipyramidal ($sp^3d$). Bond angles: $90^\circ$ and $120^\circ$.

Determining Hybridisation

The hybridisation of the central atom is determined by the number of electron domains (bonding Pairs + lone pairs):

Electron domains	Hybridisation	Geometry
2	$sp$	Linear
3	$sp^2$	Trigonal planar
4	$sp^3$	Tetrahedral
5	$sp^3d$	Trigonal bipyramidal
6	$sp^3d^2$	Octahedral

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Metallic Bonding: Properties Explained (OL/HL)

Why Metals Conduct Electricity

The delocalised electrons in the “sea” are free to move when a potential difference is applied. These mobile charge carriers explain why metals are good conductors. Conductivity decreases with Temperature because increased atomic vibrations impede electron flow.

Why Metals Are Malleable

When a force is applied, layers of metal ions can slide past each other. The delocalised electrons Adjust their positions to maintain the metallic bonds, preventing the lattice from shattering.

Why Alloys Are Harder

Alloying introduces atoms of different sizes into the metal lattice, distorting the regular Arrangement. This prevents layers from sliding , making the alloy harder than the pure metal.

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Intermolecular Forces and Physical Properties (OL/HL)

Boiling Points Across a Period

Consider the hydrides of Group 14-17:

Compound	Boiling point	Dominant IMF
\mathrm{CH_4	-162°C	London forces
\mathrm{NH_3	-33°C	Hydrogen bonding
\mathrm{H_2\mathrm{O	100°C	Hydrogen bonding
\mathrm{HF	20°C	Hydrogen bonding

Water has the highest boiling point because each molecule can form up to four hydrogen bonds (two as Donor, two as acceptor), creating an extensive three-dimensional network.

Solubility: “Like Dissolves Like”

• Polar solutes dissolve in polar solvents (e.g., \mathrm{NaCl in water).
• Non-polar solutes dissolve in non-polar solvents (e.g., grease in hexane).

This is because the intermolecular forces between solute and solvent must be comparable to those Within the solute and within the solvent for dissolution to be energetically favourable.

Vapour Pressure

Molecules with stronger intermolecular forces have lower vapour pressure because fewer molecules Have sufficient energy to escape from the liquid surface.

Order of vapour pressure (decreasing): London forces > dipole-dipole > hydrogen bonding.

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Comparison Table: All Bonding Types

Feature	Ionic	Covalent (simple)	Covalent (giant)	Metallic
Particles	Ions	Molecules	Atoms in network	Metal ions + delocalised e-
Forces	Electrostatic	Covalent bonds (intramolecular), IMF (intermolecular)	Covalent bonds	Metallic bonding
Melting point	High	Low	Very high	Moderate-high
Electrical conductivity	Molten/aqueous	None	Graphite only	Good
Solubility	Polar solvents	Non-polar solvents	Insoluble	Insoluble
Hardness	Hard, brittle	Soft	Very hard	Malleable
Examples	\mathrm{NaCl$$\mathrm{MgO	\mathrm{H_2\mathrm{O$$\mathrm{CO_2	Diamond, graphite	\mathrm{Fe$$\mathrm{Cu

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Lewis Structures (HL)

Rules for Drawing Lewis Structures

1. Count total valence electrons.
2. Draw the skeletal structure (least electronegative atom central).
3. Connect atoms with single bonds (2 electrons each).
4. Complete octets of outer atoms first.
5. Place remaining electrons on the central atom.
6. If the central atom lacks an octet, form double or triple bonds.

Formal Charge

\mathrm{Formal charge = \mathrm{valence electrons - \mathrm{non-bonding electrons - \frac{1}{2}\mathrm{bonding electrons

The best Lewis structure minimises formal charges and places negative formal charge on the most Electronegative atoms.

Worked Example 8 (HL): Draw the Lewis structure of \mathrm{SO_4^{2-} and calculate the formal Charge on each atom.

Total valence electrons: $6 + 4(6) + 2 = 32$.

Structure: S is central with four S=O bonds (each oxygen has 4 non-bonding electrons).

Formal charge on S: $6 - 0 - \frac{1}{2}(8) = +2$. Formal charge on each O: $6 - 4 - \frac{1}{2}(4) = 0$.

Total formal charge: $+2 + 4(0) = +2$But the ion has charge $-2$. This suggests the simple Structure with four double bonds is not ideal.

Alternative: Two S=O double bonds and two S-O single bonds (with the single-bonded oxygens carrying A negative formal charge):

Formal charge on S: $6 - 0 - \frac{1}{2}(12) = 0$. Formal charge on double-bonded O: $6 - 4 - \frac{1}{2}(4) = 0$. Formal charge on single-bonded O: $6 - 6 - \frac{1}{2}(2) = -1$.

Total: $0 + 2(0) + 2(-1) = -2$. This matches the charge on the ion.

In reality, all four S-O bonds are equivalent due to resonance.

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Practice Questions (Further Extended)

21. Draw Lewis structures for the following molecules/ions and assign formal charges: (a) \mathrm{CO_3^{2-}(b) \mathrm{SO_3(c) \mathrm{NO_2^-(d) \mathrm{XeO_4.

22. Explain the trend in boiling points for the noble gases and relate this to intermolecular forces.

23. Use MO theory to determine the bond order of \mathrm{F_2$$\mathrm{O_2And \mathrm{N_2. How does bond order relate to bond length and bond strength?

24. Describe how the properties of graphite make it useful as a lubricant and as an electrode material.

25. Explain why ice is less dense than liquid water, referring to hydrogen bonding and the structure of the ice lattice.

Summary

This topic covers the essential chemistry of bonding, including key reactions, underlying theories, and practical applications.

Key concepts include:

• ionic, covalent, and metallic bonding
• electronegativity and bond polarity
• intermolecular forces
• giant and simple molecular structures
• VSEPR theory

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.