While covalent and ionic bonds act inside molecules and lattices to hold atoms together, another set of weaker attractions operates between separate molecules. These attractions are known as intermolecular forces. They are responsible for determining physical properties such as boiling points, melting points, and solubility.
🔑 Key Principle
Intermolecular forces are attractions between molecules, whereas covalent bonds are attractions within molecules. When a simple molecular substance melts or boils, only the weak intermolecular forces are broken, the strong covalent bonds remain completely intact.
Types of Intermolecular Forces
You need to know the three main types of intermolecular forces. In order of increasing typical strength, they are:
- Induced dipole-dipole forces (often called London forces or van der Waals forces)
- Permanent dipole-dipole forces
- Hydrogen bonding (a special, highly strong type of permanent dipole-dipole force)
1. Induced Dipole-Dipole (Van der Waals) Forces
Induced dipole-dipole forces exist between all atoms and molecules, polar or non-polar. They are caused by the constant movement of electrons in an atom or molecule.
Temporary attractions that occur when the random movement of electrons in a molecule creates a temporary dipole, which induces an opposite dipole in a neighbouring molecule.
The mechanism of these forces is as follows:
- At any given instant, the electron density in a molecule may be slightly greater on one side than the other. This creates a temporary dipole.
- This temporary dipole repels or attracts the electrons in a neighbouring molecule, creating an induced dipole.
- The two dipoles attract each other weakly. These dipoles are constantly forming, shifting, and breaking.
Factors Affecting Van der Waals Forces
The strength of induced dipole-dipole forces increases with:
- More electrons: Larger molecules with more electrons have larger electron clouds that are more easily polarised (shifted), leading to stronger temporary and induced dipoles.
- Larger surface area: Long, straight-chain molecules have more points of contact with neighbours compared to branched, spherical isomers. This increases the total attraction.
Solution:
1. Identify the bonding and structure: Both compounds are isomers with the same molecular formula (\( \text{C}_4\text{H}_{10} \)) and the same number of electrons. They are simple molecular substances held together by van der Waals forces.
2. Describe molecular shape: Butane is a straight-chain alkane, whereas 2-methylpropane is branched and has a more spherical shape.
3. Compare surface area: The straight chain of butane allows the molecules to pack closer together and provides a larger surface area of contact between neighbouring molecules.
4. Relate to force strength: This results in stronger van der Waals forces between butane molecules compared to 2-methylpropane molecules.
5. Relate to boiling point: More thermal energy is needed to overcome these stronger forces, giving butane a higher boiling point.
2. Permanent Dipole-Dipole Forces
Permanent dipole-dipole forces only exist between polar molecules (molecules with a permanent, overall dipole due to differences in electronegativity and asymmetric shape).
Electrostatic attractions between the permanent positive end of one polar molecule and the permanent negative end of a neighbouring polar molecule.
For example, in hydrogen chloride (\( \text{HCl} \)), chlorine is much more electronegative than hydrogen, forming a permanent polar bond. The \( \delta^+ \) hydrogen of one \( \text{HCl} \) molecule is electrostatically attracted to the \( \delta^- \) chlorine of a neighbouring \( \text{HCl} \) molecule.
Remember that polar molecules experience both permanent dipole-dipole forces and van der Waals forces. In many small polar molecules, the van der Waals forces can actually contribute more to the overall boiling point than the dipole-dipole forces! Do not say that polar molecules only have dipole-dipole forces.
3. Hydrogen Bonding
Hydrogen bonding is the strongest type of intermolecular force. It is a special, extreme case of permanent dipole-dipole attraction.
A strong intermolecular attraction between a hydrogen atom covalently bonded to a highly electronegative atom (nitrogen, oxygen, or fluorine) and a lone pair on a neighbouring nitrogen, oxygen, or fluorine atom.
Hydrogen bonds only form when hydrogen is directly bonded to one of the three most electronegative elements: nitrogen (N), oxygen (O), or fluorine (F). This occurs because:
- The electronegativity difference is very large, creating a highly polar bond.
- The hydrogen atom is extremely small, meaning the positive charge density on the \( \delta^+ \) hydrogen is exceptionally high.
- The highly electronegative N, O, or F atom has at least one lone pair of electrons which strongly attracts the small, electron-deficient \( \delta^+ \) hydrogen.
Boiling Point Trends in Hydrides
The presence of hydrogen bonding explains the anomalous physical properties of water (\( \text{H}_2\text{O} \)), ammonia (\( \text{NH}_3 \)), and hydrogen fluoride (\( \text{HF} \)).
If we plot the boiling points of the hydrides of Groups 5, 6, and 7, we observe a clear trend:
Explanation of the graph:
- Anomalously high boiling points in Period 2: Water (\( \text{H}_2\text{O} \)), hydrogen fluoride (\( \text{HF} \)), and ammonia (\( \text{NH}_3 \)) have much higher boiling points than expected. This is because they form strong hydrogen bonds, which require significant thermal energy to break.
- The drop from Period 2 to Period 3: The hydrides of Period 3 (\( \text{H}_2\text{S} \), \( \text{HCl} \), \( \text{PH}_3 \)) cannot form hydrogen bonds. They are only held together by weaker permanent dipole-dipole and van der Waals forces, so their boiling points are much lower.
- The rise from Period 3 to Period 5: As we go down the groups from Period 3 to Period 5, the molecules gain more shells and thus more electrons. This increases the polarisation of the molecules, resulting in stronger van der Waals forces and higher boiling points.
When asked to explain this trend, make sure you compare the relative strengths of the forces. State clearly: "Hydrogen bonds in \( \text{H}_2\text{O} \) are stronger than the van der Waals and dipole-dipole forces in \( \text{H}_2\text{S} \), requiring more energy to break." You must explicitly state that the forces are stronger and require more energy to break to get full marks.
Why Water is Anomalous
Water has two major anomalous physical properties caused by its hydrogen bonding:
1. High Boiling Point
Water has a much higher boiling point than other hydrides of its group. Each water molecule can form up to four hydrogen bonds because it has two hydrogen atoms and two lone pairs on the oxygen atom, forming a highly cohesive network.
2. Ice is Less Dense than Water
When liquid water freezes, the molecules slow down and organise into a rigid, open tetrahedral lattice held together by hydrogen bonds. This positions the water molecules further apart than they are in the liquid state, creating more empty space. This is why ice floats.
Solution:
1. Analyze hydrogen bonding capacity: A water molecule (\( \text{H}_2\text{O} \)) has two hydrogen atoms and two lone pairs on the oxygen atom, allowing each molecule to participate in an average of four hydrogen bonds.
2. Analyze HF capacity: A hydrogen fluoride molecule (\( \text{HF} \)) has only one hydrogen atom (despite having three lone pairs on the fluorine atom). This limits HF to forming an average of only two hydrogen bonds per molecule.
3. Relate to total force strength: The extensive, three-dimensional network of hydrogen bonds in water results in a greater overall strength of attraction between molecules than in HF, despite HF forming stronger individual hydrogen bonds due to fluorine's higher electronegativity.
4. Relate to boiling point: More thermal energy is required to break the larger number of hydrogen bonds in water, leading to its higher boiling point.
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