Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Intermolecular forces (IMFs) are the forces of attraction between molecules, which determine physical properties such as boiling point, melting point, and solubility. They are much weaker than intramolecular forces like covalent or ionic bonds.
London (Dispersion) Forces: The weakest IMF, present in all molecules. They arise from temporary dipoles caused by the movement of electrons. The strength of LDFs increases with increasing molar mass () and larger surface area (e.g., has stronger LDFs than ).
Dipole-Dipole Forces: Forces of attraction between the permanent dipoles of polar molecules. A molecule is polar if it has polar bonds (difference in electronegativity ) and a non-symmetrical geometry (e.g., , ).
Hydrogen Bonding: A specific, strong type of dipole-dipole attraction. It occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (, , or ) and is attracted to the lone pair of an , , or atom in a neighboring molecule.
The relative strength of IMFs generally follows the order: . However, very large molecules with only LDFs can sometimes have higher boiling points than small molecules with hydrogen bonding.
Solubility: 'Like dissolves like.' Polar solutes (like ) dissolve in polar solvents (like ) due to favorable dipole-dipole or hydrogen bonding interactions. Non-polar solutes (like ) dissolve in non-polar solvents (like ) via LDFs.
📐Formulae
💡Examples
Problem 1:
Explain why ethanol () has a significantly higher boiling point () than dimethyl ether (, ), despite both having the same molecular formula .
Solution:
Ethanol contains an group, which allows it to form strong intermolecular hydrogen bonds. Dimethyl ether is a polar molecule but lacks bonded directly to , so it only exhibits dipole-dipole forces and London dispersion forces.
Explanation:
Hydrogen bonds are significantly stronger than dipole-dipole interactions, requiring more energy (higher temperature) to overcome during the phase change from liquid to gas.
Problem 2:
Predict the trend in boiling points for the following noble gases: , , , and .
Solution:
Explanation:
Noble gases are non-polar monatomic species that only experience London Dispersion Forces. As we move down Group 18, the number of electrons increases and the electron cloud becomes larger and more polarizable. This leads to stronger temporary dipoles and increased LDF strength, resulting in higher boiling points.
Problem 3:
Identify the dominant intermolecular force in , , and .
Solution:
: Hydrogen bonding; : Dipole-dipole; : London dispersion forces.
Explanation:
In , is bonded to , enabling hydrogen bonding. is a polar molecule (bent geometry) but is not electronegative enough for hydrogen bonding, so it uses dipole-dipole. is non-polar because its tetrahedral geometry causes the individual bond dipoles to cancel out, leaving only LDFs.