Bonding

Ionic Bonding

Formation

Ionic bonds form between metals (which lose electrons to form cations) and non-metals (which gain electrons to form anions). The electrostatic attraction between oppositely charged ions is the ionic bond.

Energy Considerations

The formation of an ionic compound from its elements is governed by the balance between:

• Ionisation energy (endothermic): energy to remove electrons from the metal
• Electron affinity (exothermic): energy released when electrons are added to the non-metal
• Lattice energy (exothermic): energy released when gaseous ions form the ionic lattice

For ionic bonding to be favourable, the lattice energy must exceed the sum of ionisation energies minus electron affinities.

Giant Ionic Lattice

Ionic compounds form a regular 3D lattice where each ion is surrounded by ions of opposite charge. There are no discrete molecules.

Properties of Ionic Compounds

Property	Explanation
High melting/boiling points	Strong electrostatic forces throughout the lattice
Conduct when molten or dissolved	Ions are free to move and carry charge
Do not conduct when solid	Ions are fixed in position
Soluble in polar solvents	Polar solvent molecules attract and separate ions
Brittle	Shifting layers brings like charges together

Worked Example 1

Write the formula of aluminium oxide.

Solution

Aluminium (Group 13): $\mathrm{Al} \to \mathrm{Al}^{3+} + 3e^-$

Oxygen (Group 16): $\mathrm{O} + 2e^- \to \mathrm{O}^{2-}$

To balance charges: $2 \times (+3) = 3 \times (-2)$. Formula: $\mathrm{Al}_2\mathrm{O}_3$

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Covalent Bonding

Formation

A covalent bond is formed when two atoms share a pair of electrons. It occurs between non-metal atoms.

Types of Covalent Bonds

Bond Type	Description	Example
Single	One shared pair	H-H
Double	Two shared pairs	O=O
Triple	Three shared pairs	$\mathrm{N}\equiv\mathrm{N}$
Dative (coordinate)	Both electrons from one atom	$\mathrm{NH}_4^+$

Bond Polarity

When two atoms with different electronegativities form a covalent bond, the bonding electrons are displaced towards the more electronegative atom, creating a polar bond with a dipole.

VSEPR Theory

The Valence Shell Electron Pair Repulsion theory predicts molecular shapes:

Electron Pairs	Shape	Bond Angle	Example
2 bonding	Linear	$180^\circ$	$\mathrm{BeCl}_2$, $\mathrm{CO}_2$
3 bonding	Trigonal planar	$120^\circ$	$\mathrm{BF}_3$
2 bonding, 1 lone	Bent	$\lt 120^\circ$	$\mathrm{SO}_2$
4 bonding	Tetrahedral	$109.5^\circ$	$\mathrm{CH}_4$
3 bonding, 1 lone	Trigonal pyramidal	$\lt 109.5^\circ$	$\mathrm{NH}_3$
2 bonding, 2 lone	Bent	$\lt 109.5^\circ$	$\mathrm{H}_2\mathrm{O}$

Lone pairs exert greater repulsion than bonding pairs, reducing bond angles below the ideal values.

Worked Example 2

Predict the shape and bond angle of $\mathrm{H}_2\mathrm{O}$.

Solution

Oxygen has 6 valence electrons. Two are used in bonds with hydrogen, leaving 2 lone pairs.

Total electron pairs = 4 (2 bonding + 2 lone pairs). Electron pair geometry: tetrahedral.

Molecular shape: bent. Bond angle: approximately $104.5^\circ$ (reduced from $109.5^\circ$ by the two lone pairs).

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Metallic Bonding

The Sea of Electrons Model

In metallic bonding, metal atoms lose their valence electrons to form positive ions in a regular lattice. The valence electrons are delocalised, forming a "sea" of electrons that holds the lattice together through electrostatic attraction.

Properties of Metals

Property	Explanation
High melting/boiling points	Strong metallic bonding throughout the lattice
Good electrical conductivity	Delocalised electrons carry charge
Good thermal conductivity	Delocalised electrons transfer kinetic energy
Malleable and ductile	Layers of cations can slide without breaking bonds
Lustrous	Delocalised electrons absorb and re-emit light

Alloys

An alloy is a mixture of two or more elements, at least one of which is a metal. Different-sized atoms disrupt the regular lattice, preventing layers from sliding easily. This makes alloys harder and stronger than pure metals.

Examples: steel (Fe + C), brass (Cu + Zn), bronze (Cu + Sn).

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Intermolecular Forces

Intermolecular forces are weaker than intramolecular (covalent) bonds. They determine physical properties such as melting point, boiling point, and solubility.

van der Waals Forces (London Dispersion Forces)

• Present between all molecules (including non-polar ones)
• Caused by instantaneous dipoles from uneven electron distribution
• Strength increases with molecular size (more electrons) and surface area

Dipole-Dipole Interactions

• Occur between polar molecules
• The positive end of one molecule attracts the negative end of another
• Stronger than van der Waals forces but weaker than hydrogen bonding

Hydrogen Bonding

A special strong dipole-dipole interaction occurring when:

1. Hydrogen is covalently bonded to N, O, or F
2. The H atom interacts with a lone pair on another N, O, or F atom

Examples: $\mathrm{H}_2\mathrm{O}$, $\mathrm{NH}_3$, HF, DNA base pairing.

Relative Strengths

Hydrogen bonding $\gt$ dipole-dipole $\gt$ van der Waals forces $\lt\lt$ covalent/ionic bonds

Worked Example 3

Explain why $\mathrm{H}_2\mathrm{O}$ has a higher boiling point than $\mathrm{H}_2\mathrm{S}$.

Solution

$\mathrm{H}_2\mathrm{O}$ can form hydrogen bonds (H bonded to O), while $\mathrm{H}_2\mathrm{S}$ cannot (S is not electronegative enough). Although $\mathrm{H}_2\mathrm{S}$ has stronger van der Waals forces (larger molecule), hydrogen bonding in $\mathrm{H}_2\mathrm{O}$ is much stronger, resulting in a higher boiling point.

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Structures and Properties

Simple Molecular

Examples: $\mathrm{H}_2\mathrm{O}$, $\mathrm{CO}_2$, $\mathrm{I}_2$, $\mathrm{CH}_4$

• Weak intermolecular forces between molecules
• Low melting and boiling points
• Do not conduct electricity
• Usually gases or volatile liquids at room temperature

Giant Covalent (Network)

Substance	Structure	Bonding	Properties
Diamond	Tetrahedral 3D	Each C bonded to 4 others ($sp^3$)	Hardest natural substance, insulator, very high MP
Graphite	Layered sheets	Each C bonded to 3 others ($sp^2$)	Soft, conductor (delocalised electrons), lubricant
$\mathrm{SiO}_2$	3D network	Each Si bonded to 4 O, each O to 2 Si	Very high MP, insulator

Giant Ionic

Examples: $\mathrm{NaCl}$, $\mathrm{MgO}$, $\mathrm{CaF}_2$

• Strong electrostatic forces throughout the lattice
• High melting and boiling points
• Conduct when molten or dissolved
• Brittle

Worked Example 4

Explain why diamond is hard but graphite is soft.

Solution

In diamond, each carbon atom is covalently bonded to four others in a rigid 3D tetrahedral network. There are no weak planes. In graphite, carbon atoms are bonded in flat hexagonal layers. Strong covalent bonds hold atoms within each layer, but only weak van der Waals forces act between layers, allowing them to slide over each other.

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Common Pitfalls

• Confusing intramolecular forces (covalent bonds) with intermolecular forces. Covalent bonds hold atoms together within a molecule; intermolecular forces act between molecules.
• Stating that ionic compounds have weak bonds. The ionic bond is strong; it is the individual ion interactions that can be broken during melting/boiling.
• Forgetting that graphite conducts electricity because of delocalised electrons, not because it is a metal.
• Assuming all molecules with H bonded to N/O/F exhibit strong hydrogen bonding in bulk. The effect also depends on molecular geometry and the number of H-bonding sites.
• Confusing the terms "molecular" and "macromolecular" when describing giant covalent structures.

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Summary Table

Bond Type	Particles Involved	Strength	Typical MP/BP
Ionic	Cations and anions	Strong	High
Covalent	Shared electrons	Strong	Varies
Metallic	Cations and delocalised electrons	Strong	High
Hydrogen bonding	H bonded to N/O/F	Moderate	Moderate-High
Dipole-dipole	Polar molecules	Weak	Low-Moderate
van der Waals	All molecules	Very weak	Low

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Problem Set

Problem 1: Predict the shape and bond angle of $\mathrm{SF}_4$.

If you get this wrong, revise: VSEPR Theory

Solution

Sulphur has 6 valence electrons. Four are used in bonding with fluorine, leaving 1 lone pair. Total electron pairs = 5. Electron pair geometry: trigonal bipyramidal.

The lone pair occupies an equatorial position to minimise repulsion. Molecular shape: see-saw. Bond angles: approximately $120^\circ$ (equatorial) and $90^\circ$ (axial-equatorial), reduced by lone pair repulsion.

Problem 2: Explain why the boiling point of $\mathrm{HF}$ is higher than that of $\mathrm{HCl}$, even though $\mathrm{HCl}$ has a larger molecular mass.

If you get this wrong, revise: Hydrogen Bonding

Solution

$\mathrm{HF}$ forms hydrogen bonds between molecules (H bonded to F, which is highly electronegative). $\mathrm{HCl}$ has only dipole-dipole interactions and van der Waals forces. Hydrogen bonding is much stronger than these, so more energy is needed to overcome the intermolecular forces in $\mathrm{HF}$.

Problem 3: Why is $\mathrm{SiO}_2$ a solid with a very high melting point, while $\mathrm{CO}_2$ is a gas at room temperature?

If you get this wrong, revise: Structures and Properties — Giant Covalent vs Simple Molecular

Solution

$\mathrm{SiO}_2$ has a giant covalent structure: each Si atom is covalently bonded to four O atoms in a continuous 3D network. Breaking this requires breaking many strong covalent bonds.

$\mathrm{CO}_2$ consists of simple discrete molecules with strong covalent bonds within each molecule but only weak van der Waals forces between molecules. These weak intermolecular forces are easily overcome, so $\mathrm{CO}_2$ is a gas at room temperature.

Problem 4: Draw the dot-and-cross diagram for the ammonium ion $\mathrm{NH}_4^+$ and identify the dative bond.

If you get this wrong, revise: Covalent Bonding — Dative Bonds

Solution

Nitrogen has 5 valence electrons and forms three normal covalent bonds with three hydrogen atoms (3 shared pairs). The fourth hydrogen ion ($\mathrm{H}^+$, which has no electrons) forms a dative (coordinate) bond with the lone pair on nitrogen. The nitrogen now has four bonds and a positive charge. The dative bond is the $\mathrm{N\to H}$ bond where both electrons come from nitrogen.

Problem 5: Explain why steel is harder than pure iron.

If you get this wrong, revise: Metallic Bonding — Alloys

Solution

Pure iron has a regular lattice of iron atoms. The layers can slide over each other relatively easily, making pure iron soft. In steel, carbon atoms (which are smaller than iron atoms) occupy interstitial positions in the lattice. These carbon atoms disrupt the regular arrangement and prevent the iron atom layers from sliding, making steel harder and stronger than pure iron.

Problem 6: Predict the shape and bond angle of the $\mathrm{ClF}_3$ molecule.

If you get this wrong, revise: VSEPR Theory

Solution

Chlorine has 7 valence electrons. Three are used in bonding with fluorine, leaving 2 lone pairs. Total electron pairs = 5. Electron pair geometry: trigonal bipyramidal.

The two lone pairs occupy equatorial positions to minimise repulsion. Molecular shape: T-shaped. Bond angles: approximately $87.5^\circ$ (axial-equatorial), reduced from $90^\circ$ by lone pair repulsion.

Problem 7: Explain why the melting point of sodium chloride ($801^\circ\mathrm{C}$) is much higher than that of iodine ($114^\circ\mathrm{C}$).

If you get this wrong, revise: Ionic Bonding vs Simple Molecular Structures

Solution

$\mathrm{NaCl}$ has a giant ionic lattice with strong electrostatic forces between $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions throughout the entire structure. A large amount of energy is needed to overcome these forces.

Iodine ($\mathrm{I_2}$) has a simple molecular structure. Although the covalent bonds within each $\mathrm{I_2}$ molecule are strong, only weak van der Waals forces act between molecules. These are easily overcome, giving iodine a much lower melting point.

Problem 8: State and explain whether $\mathrm{BF}_3$ is a polar or non-polar molecule.

If you get this wrong, revise: Bond Polarity and Molecular Polarity

Solution

$\mathrm{BF}_3$ is a non-polar molecule. Each $\mathrm{B - F}$ bond is polar (F is more electronegative than B). However, the molecule has a trigonal planar geometry with bond angles of $120^\circ$. The three bond dipoles are arranged symmetrically and cancel each other out completely, resulting in a net dipole moment of zero.

Problem 9: Explain why graphite conducts electricity but diamond does not, even though both are forms of carbon.

If you get this wrong, revise: Giant Covalent Structures

Solution

In graphite, each carbon atom is bonded to only three others using three of its four valence electrons. The fourth electron from each carbon is delocalised and free to move between the layers. These mobile delocalised electrons carry charge, allowing graphite to conduct electricity.

In diamond, each carbon atom uses all four valence electrons to form four strong covalent bonds in a rigid 3D tetrahedral network. There are no delocalised electrons, so diamond cannot conduct electricity and is an insulator.

Problem 10: Explain why ethanol ($\mathrm{C_2H_5OH}$, b.p. $78^\circ\mathrm{C}$) has a higher boiling point than dimethyl ether ($\mathrm{CH_3OCH_3}$, b.p. $-24^\circ\mathrm{C}$), even though both have the same molecular formula $\mathrm{C_2H_6O}$.

If you get this wrong, revise: Intermolecular Forces — Hydrogen Bonding

Solution

Both molecules have the same molar mass ($46 \mathrm{ g/mol}$) and similar van der Waals forces. However, ethanol can form hydrogen bonds because it has an $-\mathrm{OH}$ group: the hydrogen of the $-\mathrm{OH}$ interacts with lone pairs on the oxygen of neighbouring ethanol molecules.

Dimethyl ether has no $-\mathrm{OH}$ group (the oxygen is bonded to two carbons), so it cannot form hydrogen bonds. It only has weaker dipole-dipole interactions and van der Waals forces. The hydrogen bonding in ethanol requires significantly more energy to overcome, resulting in a much higher boiling point.