Chapter 11: Problem 22
Which member in each pair has the stronger intermolecular dispersion forces? (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{CH}_{3} \mathrm{OH},\) (b) \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) or \(\mathrm{CCl}_{3} \mathrm{CCl}_{3}\) (c) \(\mathrm{C}\left(\mathrm{CH}_{3}\right)_{4}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\).
Short Answer
Step by step solution
Analyze (a) H2O vs CH3OH
Analyze (b) CBr3CBr3 vs CCl3CCl3
Analyze (c) C(CH3)4 vs CH3CH2CH2CH2CH3
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dispersion Forces
The strength of dispersion forces depends on several factors, including the total number of electrons in the molecule and its overall shape. Molecules with a larger number of electrons typically exhibit stronger dispersion forces. That's because a larger electron cloud is more easily distorted, leading to stronger charge fluctuations. Additionally, the molecular shape affects the surface area available for interactions.
Even small or light molecules can have significant dispersion forces if they have a high number of electrons or a shape that promotes surface stability. Understanding these forces is essential when predicting physical properties like boiling and melting points.
Molecular Weight
This is because larger molecules have more electrons, which increases the potential for temporary dipole formation. Take, for instance, the comparison between \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) and \(\mathrm{CCl}_{3} \mathrm{CCl}_{3}\). The bromine atoms in \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) are heavier than chlorine atoms. Therefore, \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) has a higher molecular weight, leading to stronger dispersion forces.
Knowledge of molecular weight allows chemists to estimate the strength of dispersion forces and predict how substances will behave under various conditions, such as how easily they might condense or evaporate.
Molar Mass
The molar mass can help predict and compare the impact of dispersion forces in different molecules. In example (a), comparing \(\mathrm{H}_{2} \mathrm{O}\) with \(\mathrm{CH}_{3} \mathrm{OH}\): Both have hydrogen bonding, but we focus on dispersion forces here. The molar mass of \(\mathrm{CH}_{3} \mathrm{OH}\) is slightly higher than \(\mathrm{H}_{2} \mathrm{O}\), leading to stronger dispersion forces.
Calculating and understanding molar mass is critical when determining the physical properties of substances, such as boiling points and solubility, which are influenced by the strength of intermolecular forces.
Molecular Shape
In scenario (c), comparing \(\mathrm{C}(\mathrm{CH}_{3})_{4}\) with \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\), we observe that the latter, also known as n-pentane, is linear, while the former, tetramethylmethane, is more spherical. Linear molecules like n-pentane can pack together more closely than branched molecules, leading to larger surface areas and stronger dispersion forces.
This understanding of molecular shape and its impact on dispersion forces helps chemists design and synthesize molecules with desired physical and chemical properties, such as optimizing pharmaceuticals for better absorption and interactions within the body.