Intermolecular Forces — Drills · Grade 11 Science
Chapter: States of Matter & GasesWhat these drills covers
By the end of Intermolecular Forces you should be able to:
- I can identify the intermolecular forces present in a substance: London dispersion, dipole-dipole, and hydrogen bonding.
- I can use the relative strength of intermolecular forces to compare boiling points, melting points, and related properties.
6 questions, multiple choice and written answer, each with a worked explanation.
Vocabulary used: intermolecular forces (IMFs), London dispersion forces, dipole-dipole forces, hydrogen bonding, polarity, polarizability.
Problem 1
- between protons in a nucleus
- only inside metals
- between separate molecules
- within a molecule's covalent bonds
Answer and explanation
between separate molecules
Intermolecular forces act between molecules; the bonds within a molecule are much stronger intramolecular forces.
Problem 2
- HF
- H₂O
- NH₃
- CH₄
Answer and explanation
CH₄
CH₄ is nonpolar and cannot hydrogen-bond, so its only intermolecular force is London dispersion; the others can hydrogen-bond.
Problem 3
- London dispersion < dipole-dipole < hydrogen bonding
- hydrogen bonding < dipole-dipole < London dispersion
- dipole-dipole < London dispersion < hydrogen bonding
- they are all equal in strength
Answer and explanation
London dispersion < dipole-dipole < hydrogen bonding
For comparably sized molecules, London dispersion is weakest, then dipole-dipole, then hydrogen bonding.
Problem 4
- I₂ is larger with more electrons, giving stronger dispersion forces
- I₂ is polar while F₂ is not
- I₂ can form hydrogen bonds
- they actually melt at the same temperature
Answer and explanation
I₂ is larger with more electrons, giving stronger dispersion forces
I₂ has more electrons and a larger, more polarizable electron cloud, so its London dispersion forces are stronger, raising its melting point.
Problem 5
Answer and explanation
(1) A hydrogen atom must be bonded directly to a highly electronegative N, O, or F atom, and (2) that hydrogen must be attracted to a lone pair of electrons on an N, O, or F atom of a neighboring molecule.
Problem 6
- they all boil at the same temperature
- HCl < HBr < HI
- HI < HBr < HCl
- HBr < HCl < HI
Answer and explanation
HCl < HBr < HI
Since none can hydrogen-bond, London dispersion forces dominate and increase with size, so boiling point rises HCl < HBr < HI.