Shapes of Molecules and Intermolecular Forces

Electron Pair Repulsion Theory (VSEPR Theory)

• Proposed by Sidgwick and Powell in 1940.
• Explains the shapes of molecules based on electron pair repulsion.

Key Points:

• Electron pairs repel each other and position themselves as far apart as possible.
• Two types of electron pairs:
  • Bond pairs (involved in bonding).
  • Lone pairs (not involved in bonding).
• Order of repulsion strength:
  • Lone pair: lone pair > Lone pair: bond pair > Bond pair: bond pair.
• The shape of the molecule is determined by:
  1. The number of electron pairs around the central atom.
  2. The type of electron pairs (lone or bond pairs).

Molecular Shapes

Linear

• Two bond pairs of electrons.
• Bond angle: 180°.

Triangular Planar

• Three bond pairs of electrons.
• Bond angle: 120°.

Tetrahedral

• Four bond pairs of electrons.
• Bond angle: 109.5°.

Pyramidal (Distorted Tetrahedral)

• Three bond pairs and one lone pair.
• Lone pair repulsion forces bond pairs closer.
• Bond angle: 107°.

V-shaped (Bent)

• Two bond pairs and two lone pairs.
• Lone pairs push bond pairs closer.
• Bond angle: 104°.

Intramolecular vs Intermolecular Forces

• Intramolecular forces: Forces within a molecule that hold atoms together (e.g. covalent bonds).
• Intermolecular forces: Forces between molecules.

Types of Intermolecular Forces

Van der Waals Forces

• Arise due to temporary dipoles.
• Weak forces, easily separated.

Permanent Dipole-Dipole Interactions

• Attraction between positive and negative poles of molecules.
• Stronger than Van der Waals forces.

Hydrogen Bonding

• Special dipole-dipole interaction involving hydrogen bonded to fluorine, oxygen, or nitrogen.
• Strong due to the significant difference in electronegativity.

Effect of intermolecular forces on boiling points

• Strength of intermolecular forces influences boiling points.
• Hydrogen bonding > Dipole-dipole interactions > Van der Waals forces.

Comparison of Boiling Points

Boiling Points Overview

The boiling points of substances are influenced by the strength of the intermolecular forces present. The stronger these forces, the more energy (heat) is required to separate molecules, resulting in a higher boiling point.

$H₂$ (Hydrogen) vs. $O₂$ (Oxygen)

• Hydrogen ($H₂$) has a boiling point of -253°C, while Oxygen ($O₂$) boils at -183°C.
• Both molecules are non-polar, meaning they only experience Van der Waals forces (temporary dipole-dipole attractions). However, oxygen molecules are larger and have more electrons than hydrogen, leading to stronger Van der Waals forces in $O₂$ and a higher boiling point compared to H₂. Conclusion: The difference in boiling points is due to the larger molecular size and stronger Van der Waals forces in $O₂$.

$C₂H₄$ (Ethene) vs. $HCHO$ (Methanal)

• Ethene ($C₂H₄$) has a boiling point of -104°C, while Methanal ($HCHO$) boils at -19°C.
• Ethene is a non-polar molecule, so it only experiences weak Van der Waals forces. On the other hand, methanal is a polar molecule and has dipole-dipole interactions, which are stronger than Van der Waals forces.
• Additionally, the molecular weight of $HCHO$ is slightly higher, further contributing to its higher boiling point. Methanal has a higher boiling point because of the presence of dipole-dipole interactions, whereas ethene only has weaker Van der Waals forces.

$H₂O$ (Water) vs. $H₂S$ (Hydrogen Sulfide)

• Water ($H₂O$) has a boiling point of 100°C, while Hydrogen Sulfide ($H₂S$) boils at -60°C.
• Both molecules are polar, but water experiences hydrogen bonding, a particularly strong form of dipole-dipole attraction that occurs when hydrogen is bonded to highly electronegative elements (such as oxygen).
• $H₂S$, however, lacks this type of bonding because sulfur is less electronegative than oxygen and cannot form strong hydrogen bonds.
• Hydrogen bonding in $H₂O$ results in a much higher boiling point compared to $H₂S$, which only experiences weaker dipole-dipole interactions. Water has a significantly higher boiling point due to hydrogen bonding, whereas $H₂S$ has weaker dipole-dipole attractions.